Patience, young “Padovan”












33












$begingroup$


Everyone knows the Fibonacci sequence:

You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.

The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:





But, what if we didn't want to use squares?



If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:





Task:



Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.



Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$



Input:




  • Any positive integer $Nge0$


  • Invalid input does not have to be taken into account



Output:




  • The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.


  • If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)


  • Can be either $0$-indexed or $1$-indexed



Test Cases:

(0-indexed, $N$th term)



Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739


(1-indexed, first $N$ terms)



Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16


Rules:




  • This is code-golf: the fewer bytes, the better!


  • Standard loopholes are forbidden.











share|improve this question











$endgroup$








  • 2




    $begingroup$
    14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
    $endgroup$
    – Jonathan Allan
    2 days ago










  • $begingroup$
    @JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
    $endgroup$
    – Tau
    2 days ago










  • $begingroup$
    @LuisMendo I believe so. I'll edit the post.
    $endgroup$
    – Tau
    yesterday






  • 1




    $begingroup$
    @sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
    $endgroup$
    – Tau
    23 hours ago






  • 1




    $begingroup$
    Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
    $endgroup$
    – Carmeister
    23 hours ago


















33












$begingroup$


Everyone knows the Fibonacci sequence:

You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.

The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:





But, what if we didn't want to use squares?



If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:





Task:



Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.



Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$



Input:




  • Any positive integer $Nge0$


  • Invalid input does not have to be taken into account



Output:




  • The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.


  • If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)


  • Can be either $0$-indexed or $1$-indexed



Test Cases:

(0-indexed, $N$th term)



Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739


(1-indexed, first $N$ terms)



Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16


Rules:




  • This is code-golf: the fewer bytes, the better!


  • Standard loopholes are forbidden.











share|improve this question











$endgroup$








  • 2




    $begingroup$
    14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
    $endgroup$
    – Jonathan Allan
    2 days ago










  • $begingroup$
    @JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
    $endgroup$
    – Tau
    2 days ago










  • $begingroup$
    @LuisMendo I believe so. I'll edit the post.
    $endgroup$
    – Tau
    yesterday






  • 1




    $begingroup$
    @sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
    $endgroup$
    – Tau
    23 hours ago






  • 1




    $begingroup$
    Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
    $endgroup$
    – Carmeister
    23 hours ago
















33












33








33


2



$begingroup$


Everyone knows the Fibonacci sequence:

You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.

The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:





But, what if we didn't want to use squares?



If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:





Task:



Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.



Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$



Input:




  • Any positive integer $Nge0$


  • Invalid input does not have to be taken into account



Output:




  • The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.


  • If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)


  • Can be either $0$-indexed or $1$-indexed



Test Cases:

(0-indexed, $N$th term)



Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739


(1-indexed, first $N$ terms)



Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16


Rules:




  • This is code-golf: the fewer bytes, the better!


  • Standard loopholes are forbidden.











share|improve this question











$endgroup$




Everyone knows the Fibonacci sequence:

You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.

The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:





But, what if we didn't want to use squares?



If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:





Task:



Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.



Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$



Input:




  • Any positive integer $Nge0$


  • Invalid input does not have to be taken into account



Output:




  • The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.


  • If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)


  • Can be either $0$-indexed or $1$-indexed



Test Cases:

(0-indexed, $N$th term)



Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739


(1-indexed, first $N$ terms)



Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16


Rules:




  • This is code-golf: the fewer bytes, the better!


  • Standard loopholes are forbidden.








code-golf number sequence






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







Tau

















asked 2 days ago









TauTau

946515




946515








  • 2




    $begingroup$
    14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
    $endgroup$
    – Jonathan Allan
    2 days ago










  • $begingroup$
    @JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
    $endgroup$
    – Tau
    2 days ago










  • $begingroup$
    @LuisMendo I believe so. I'll edit the post.
    $endgroup$
    – Tau
    yesterday






  • 1




    $begingroup$
    @sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
    $endgroup$
    – Tau
    23 hours ago






  • 1




    $begingroup$
    Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
    $endgroup$
    – Carmeister
    23 hours ago
















  • 2




    $begingroup$
    14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
    $endgroup$
    – Jonathan Allan
    2 days ago










  • $begingroup$
    @JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
    $endgroup$
    – Tau
    2 days ago










  • $begingroup$
    @LuisMendo I believe so. I'll edit the post.
    $endgroup$
    – Tau
    yesterday






  • 1




    $begingroup$
    @sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
    $endgroup$
    – Tau
    23 hours ago






  • 1




    $begingroup$
    Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
    $endgroup$
    – Carmeister
    23 hours ago










2




2




$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
$endgroup$
– Jonathan Allan
2 days ago




$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37
$endgroup$
– Jonathan Allan
2 days ago












$begingroup$
@JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
$endgroup$
– Tau
2 days ago




$begingroup$
@JonathanAllan yes, you are correct. I fixed the last two test cases for $N$th term but not that one. The post has been edited.
$endgroup$
– Tau
2 days ago












$begingroup$
@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
yesterday




$begingroup$
@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
yesterday




1




1




$begingroup$
@sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
$endgroup$
– Tau
23 hours ago




$begingroup$
@sharur this definition for the Fibonacci sequence is the visual definition. Each successive square added has a length of that term in the sequence. The sequence you describe is the numerical reasoning behind it. Both sequences work just as well as the other.
$endgroup$
– Tau
23 hours ago




1




1




$begingroup$
Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
$endgroup$
– Carmeister
23 hours ago






$begingroup$
Note that the OEIS sequence you linked is slightly different, since it uses a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1
$endgroup$
– Carmeister
23 hours ago












38 Answers
38






active

oldest

votes













1 2
next












45












$begingroup$


Jelly, 10 bytes



9s3’Ẓæ*³FṀ


Try it online!



1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$



(this is a total coincidence.)



9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum





share|improve this answer









$endgroup$









  • 25




    $begingroup$
    this is clearly some kind of voodoo
    $endgroup$
    – Pureferret
    yesterday






  • 7




    $begingroup$
    This should be published.
    $endgroup$
    – YSC
    yesterday






  • 5




    $begingroup$
    @YSC It has already been published in A000931. I'd never have guess the primes trick:)
    $endgroup$
    – flawr
    yesterday










  • $begingroup$
    @flawr ho... I missed it
    $endgroup$
    – YSC
    yesterday






  • 1




    $begingroup$
    ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
    $endgroup$
    – Jonathan Allan
    yesterday





















24












$begingroup$


Oasis, 5 bytes



nth term 0-indexed



cd+1V


Try it online!



Explanation



   1V   # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)





share|improve this answer









$endgroup$





















    21












    $begingroup$


    Jelly,  10 9  8 bytes



    ŻṚm2Jc$S


    A monadic Link accepting n (0-indexed) which yields P(n).



    Try it online!



    How?



    Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$



    ŻṚm2Jc$S - Link: integer, n       e.g. 20
    Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
    Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
    m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
    $ - last two links as a monad:
    J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
    c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
    S - sum 200




    And here is a "twofer"

    ...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):



    3ḊṗRẎ§ċ‘ - Link: n
    3Ḋ - 3 dequeued = [2,3]
    R - range = [1,2,3,...,n]
    ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
    Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
    § - sums [ 2, 3, 4, 5, 5, 6, 6,...]
    ‘ - increment n+1
    ċ - count occurrences P(n)





    share|improve this answer











    $endgroup$





















      17












      $begingroup$


      Haskell, 26 bytes





      (l!!)
      l=1:1:1:2:scanl(+)2l


      Try it online! Outputs the n'th term zero-indexed.



      I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.



      If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.



      27 bytes





      f n|n<3=1|1>0=f(n-2)+f(n-3)


      Try it online!






      share|improve this answer









      $endgroup$





















        9












        $begingroup$


        Python 2, 30 bytes





        f=lambda n:n<3or f(n-2)+f(n-3)


        Try it online!



        Returns the n'th term zero indexed. Outputs True for 1.






        share|improve this answer











        $endgroup$





















          6












          $begingroup$


          Wolfram Language (Mathematica), 33 bytes



          a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]   


          1-indexed, returns the nth term



          Try it online!






          share|improve this answer









          $endgroup$





















            6












            $begingroup$


            Octave / MATLAB, 35 33 bytes





            @(n)[1 filter(1,'cbaa'-98,2:n<5)]


            Outputs the first n terms.



            Try it online!



            How it works



            Anonymous function that implements a recursive filter.



            'cbaa'-98 is a shorter form to produce [1 0 -1 -1].



            2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).



            filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].






            share|improve this answer











            $endgroup$





















              5












              $begingroup$


              Retina, 47 42 bytes



              K`0¶1¶0
              "$+"+`.+¶(.+)¶.+$
              $&¶$.(*_$1*
              6,G`


              Try it online! Outputs the first n terms on separate lines. Explanation:



              K`0¶1¶0


              Replace the input with the terms for -2, -1 and 0.



              "$+"+`.+¶(.+)¶.+$
              $&¶$.(*_$1*


              Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.



              6,G`


              Discard the first six characters, i.e. the first three lines.






              share|improve this answer











              $endgroup$





















                5












                $begingroup$


                Cubix, 20 bytes



                This is 0 indexed and outputs the Nth term



                ;@UOI010+p?/sqq;W.(


                Try it online!



                Wraps onto a cube with side length 2



                    ; @
                U O
                I 0 1 0 + p ? /
                s q q ; W . (
                . .
                . .


                Watch it run





                • I010 - Initiates the stack


                • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests


                • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt


                • (sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.






                share|improve this answer









                $endgroup$





















                  5












                  $begingroup$


                  J, 23 bytes



                  -1 byte thanks to ngn and Galen



                  closed form, 26 bytes



                  0.5<.@+1.04535%~1.32472^<:


                  Try it online!



                  iterative, 23 bytes



                  (],1#._2 _3{ ::1:])^:[#


                  Try it online!






                  share|improve this answer











                  $endgroup$









                  • 1




                    $begingroup$
                    Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                    $endgroup$
                    – Galen Ivanov
                    yesterday












                  • $begingroup$
                    23 bytes thanks to ngn 1: -> # : Try it online!
                    $endgroup$
                    – Galen Ivanov
                    yesterday












                  • $begingroup$
                    @GalenIvanov tyvm, that's a great trick.
                    $endgroup$
                    – Jonah
                    yesterday



















                  4












                  $begingroup$


                  Python 2, 56 48 bytes





                  f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c


                  Try it online!



                  Returns nth value, 0-indexed.






                  share|improve this answer









                  $endgroup$





















                    3












                    $begingroup$


                    Jelly, 11 bytes



                    5B+Ɲ2ị;Ʋ⁸¡Ḣ


                    Try it online!



                    0-indexed.






                    share|improve this answer











                    $endgroup$





















                      3












                      $begingroup$


                      Perl 6, 24 bytes



                      {(1,1,1,*+*+!*...*)[$_]}


                      Try it online!



                      A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        Why is this community wiki?
                        $endgroup$
                        – Jo King
                        2 days ago










                      • $begingroup$
                        @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                        $endgroup$
                        – Sean
                        yesterday



















                      3












                      $begingroup$


                      Lua 5.3, 49 48 bytes





                      function f(n)return n<4 and 1or f(n-2)+f(n-3)end


                      Try it online!



                      Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.






                      share|improve this answer











                      $endgroup$





















                        3












                        $begingroup$


                        Ruby, 26 bytes





                        f=->n{n<3?1:f[n-2]+f[n-3]}


                        Try it online!






                        share|improve this answer









                        $endgroup$





















                          3












                          $begingroup$


                          APL (Dyalog Unicode), 20 18 17 bytesSBCS



                          This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.



                          Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn



                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3


                          Try it online!



                          Explanation



                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

                          ⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
                          inside the brackets (...) to its operands (here marked ⍵ and ⍺).
                          2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
                          where we save our results as the function is repeatedly applied
                          and our ⍺, 2, is our right argument and is immediately applied to +/,
                          so that we have 2+/ which will return the pairwise sums of our array.
                          2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
                          ⊢,⍨ And add it to the head of our array.
                          4⌷ When we've finished adding Padovan numbers to the end of our list,
                          the n-th Padovan number (1-indexed) is the 4th member of that list,
                          and so, we implicitly return that.





                          share|improve this answer











                          $endgroup$





















                            3












                            $begingroup$


                            K (ngn/k), 24 20 bytes



                            -4 bytes thanks to ngn!



                            {$[x<3;1;+/o'x-2 3]}


                            Try it online!



                            0-indexed, first N terms






                            share|improve this answer











                            $endgroup$









                            • 1




                              $begingroup$
                              f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                              $endgroup$
                              – ngn
                              yesterday










                            • $begingroup$
                              @ngn Thanks! I tried it (without success) in J; it's elegant here.
                              $endgroup$
                              – Galen Ivanov
                              yesterday










                            • $begingroup$
                              @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                              $endgroup$
                              – Galen Ivanov
                              yesterday










                            • $begingroup$
                              ah, right, base-1 decode is a train-friendly way to sum :)
                              $endgroup$
                              – ngn
                              yesterday






                            • 2




                              $begingroup$
                              1: -> # in the j solution
                              $endgroup$
                              – ngn
                              yesterday





















                            3












                            $begingroup$

                            x86 32-bit machine code, 17 bytes



                            53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3


                            Disassembly:



                            00CE1250 53                   push        ebx  
                            00CE1251 33 DB xor ebx,ebx
                            00CE1253 F7 E3 mul eax,ebx
                            00CE1255 43 inc ebx
                            00CE1256 83 C1 04 add ecx,4
                            00CE1259 03 D8 add ebx,eax
                            00CE125B 93 xchg eax,ebx
                            00CE125C 92 xchg eax,edx
                            00CE125D E2 FA loop myloop (0CE1259h)
                            00CE125F 5B pop ebx
                            00CE1260 C3 ret


                            It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.



                            At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.



                            I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.



                            I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)






                            share|improve this answer









                            $endgroup$













                            • $begingroup$
                              I love to see machine code answers on PP&CG! +1
                              $endgroup$
                              – Tau
                              23 hours ago



















                            2












                            $begingroup$


                            Japt -N, 12 bytes



                            <3ªßUµ2 +ß´U


                            Try it






                            share|improve this answer









                            $endgroup$













                            • $begingroup$
                              Looks like 12 is the best we can do :
                              $endgroup$
                              – Shaggy
                              2 days ago










                            • $begingroup$
                              I stand corrected!
                              $endgroup$
                              – Shaggy
                              14 hours ago



















                            2












                            $begingroup$

                            Pyth, 16 bytes



                            L?<b3!b+y-b2y-b3


                            This defines the function y. Try it here!



                            Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.



                            +l{sa.pMf.Am&>d2%d2T./QY!


                            This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.






                            share|improve this answer









                            $endgroup$













                            • $begingroup$
                              y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                              $endgroup$
                              – Ven
                              yesterday












                            • $begingroup$
                              @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                              $endgroup$
                              – RK.
                              yesterday










                            • $begingroup$
                              Well done on hB2. Too bad it's the same byte count.
                              $endgroup$
                              – Ven
                              yesterday










                            • $begingroup$
                              Yeah, though I wonder if there's some way to get rid of the d in the map.
                              $endgroup$
                              – RK.
                              yesterday



















                            2












                            $begingroup$


                            05AB1E, 8 bytes



                            1Ð)λ£₂₃+


                            Try it online!



                            Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.



                            How?



                            1Ð)λ£₂₃+ | Full program.
                            1Ð) | Initialize the stack with [1, 1, 1].
                            λ | Begin the recursive generation of a list: Starting from some base case,
                            | this command generates an infinite list with the pattern function given.
                            £ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
                            ₂₃+ | Add a(n-2) and a(n-3).





                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                              $endgroup$
                              – Kevin Cruijssen
                              yesterday



















                            2












                            $begingroup$

                            Java, 41 bytes



                            Can't use a lambda (runtime error). Port of this Javascript answer



                            int f(int n){return n<3?1:f(n-2)+f(n-3);}


                            TIO






                            share|improve this answer









                            $endgroup$













                            • $begingroup$
                              I think you are missing some requirements: Have a look at my modification here.
                              $endgroup$
                              – Shaq
                              15 hours ago










                            • $begingroup$
                              Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                              $endgroup$
                              – Olivier Grégoire
                              11 hours ago










                            • $begingroup$
                              Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                              $endgroup$
                              – Benjamin Urquhart
                              6 hours ago





















                            2












                            $begingroup$


                            R + pryr, 38 36 bytes



                            Zero-indexed recursive function.





                            f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))


                            Try it online!



                            Thanks to @Giuseppe for pointing out two obviously needless bytes.






                            share|improve this answer











                            $endgroup$









                            • 2




                              $begingroup$
                              If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                              $endgroup$
                              – Giuseppe
                              yesterday












                            • $begingroup$
                              @Giuseppe thanks! Updated now.
                              $endgroup$
                              – rturnbull
                              yesterday



















                            2












                            $begingroup$

                            JavaScript (ES6), 23 bytes



                            Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.



                            Returns the $N$th term, 0-indexed.





                            f=n=>n<3||f(n-2)+f(n-3)


                            Try it online!






                            share|improve this answer











                            $endgroup$













                            • $begingroup$
                              I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                              $endgroup$
                              – Nit
                              yesterday






                            • 1




                              $begingroup$
                              @Nit Relevant meta post.
                              $endgroup$
                              – Arnauld
                              yesterday










                            • $begingroup$
                              I think you are missing some requirements: Have a look at my modification (version in Java) here.
                              $endgroup$
                              – Shaq
                              15 hours ago










                            • $begingroup$
                              @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                              $endgroup$
                              – Arnauld
                              15 hours ago










                            • $begingroup$
                              @Arnauld yep I can see it now. Sorry!
                              $endgroup$
                              – Shaq
                              14 hours ago



















                            2












                            $begingroup$


                            C (clang), 41 bytes





                            int a(int i){return i<3?1:a(i-2)+a(i-3);}


                            Try it online!






                            share|improve this answer








                            New contributor




                            peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$









                            • 1




                              $begingroup$
                              Welcome to PPCG :)
                              $endgroup$
                              – Shaggy
                              10 hours ago



















                            1












                            $begingroup$


                            C# (Visual C# Interactive Compiler), 34 bytes





                            int f(int g)=>g<3?1:f(g-2)+f(g-3);


                            Try it online!






                            share|improve this answer









                            $endgroup$





















                              1












                              $begingroup$

                              TI-BASIC (TI-84), 34 bytes



                              [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1


                              0-indexed $N$th term of the sequence.



                              Input is in Ans.

                              Output is in Ans and is automatically printed out.



                              I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.



                              Example:



                              0
                              0
                              prgmCDGFD
                              1
                              9
                              9
                              prgmCDGFD
                              9
                              16
                              16
                              prgmCDGFD
                              65


                              Explanation:



                              [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1      ;full program (example input: 6)

                              [[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
                              ; [0 1 0]
                              ; [0 0 1]
                              ; [1 1 0]
                              ^(Ans+5 ;then raise it to the power of: input + 5
                              ; [4 7 5]
                              ; [5 9 7]
                              ; [7 12 9]
                              Ans(1,1 ;get the top-left index and leave it in "Ans"
                              ;implicitly print Ans





                              share|improve this answer









                              $endgroup$





















                                1












                                $begingroup$


                                Perl 5, 34 bytes





                                sub f{"@_"<3||f("@_"-2)+f("@_"-3)}


                                Try it online!






                                share|improve this answer









                                $endgroup$





















                                  1












                                  $begingroup$


                                  Wolfram Language (Mathematica), 26 bytes



                                  If[#<3,1,#0[#-2]+#0[#-3]]&


                                  Try it online!






                                  share|improve this answer









                                  $endgroup$





















                                    1












                                    $begingroup$


                                    Pari/GP, 28 bytes



                                    0-indexed.



                                    f(n)=if(n<3,1,f(n-2)+f(n-3))


                                    Try it online!






                                    Pari/GP, 35 bytes



                                    1-indexed.



                                    n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]


                                    Try it online!



                                    The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.






                                    share|improve this answer











                                    $endgroup$

















                                      1 2
                                      next




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                                      38 Answers
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                                      38 Answers
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                                      1 2
                                      next










                                      45












                                      $begingroup$


                                      Jelly, 10 bytes



                                      9s3’Ẓæ*³FṀ


                                      Try it online!



                                      1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
                                      where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$



                                      (this is a total coincidence.)



                                      9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
                                      ’ [[0,1,2],[3,4,5],[6,7,8]] decrease
                                      Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
                                      æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
                                      FṀ flatten, maximum





                                      share|improve this answer









                                      $endgroup$









                                      • 25




                                        $begingroup$
                                        this is clearly some kind of voodoo
                                        $endgroup$
                                        – Pureferret
                                        yesterday






                                      • 7




                                        $begingroup$
                                        This should be published.
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 5




                                        $begingroup$
                                        @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                        $endgroup$
                                        – flawr
                                        yesterday










                                      • $begingroup$
                                        @flawr ho... I missed it
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 1




                                        $begingroup$
                                        ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                        $endgroup$
                                        – Jonathan Allan
                                        yesterday


















                                      45












                                      $begingroup$


                                      Jelly, 10 bytes



                                      9s3’Ẓæ*³FṀ


                                      Try it online!



                                      1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
                                      where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$



                                      (this is a total coincidence.)



                                      9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
                                      ’ [[0,1,2],[3,4,5],[6,7,8]] decrease
                                      Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
                                      æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
                                      FṀ flatten, maximum





                                      share|improve this answer









                                      $endgroup$









                                      • 25




                                        $begingroup$
                                        this is clearly some kind of voodoo
                                        $endgroup$
                                        – Pureferret
                                        yesterday






                                      • 7




                                        $begingroup$
                                        This should be published.
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 5




                                        $begingroup$
                                        @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                        $endgroup$
                                        – flawr
                                        yesterday










                                      • $begingroup$
                                        @flawr ho... I missed it
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 1




                                        $begingroup$
                                        ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                        $endgroup$
                                        – Jonathan Allan
                                        yesterday
















                                      45












                                      45








                                      45





                                      $begingroup$


                                      Jelly, 10 bytes



                                      9s3’Ẓæ*³FṀ


                                      Try it online!



                                      1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
                                      where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$



                                      (this is a total coincidence.)



                                      9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
                                      ’ [[0,1,2],[3,4,5],[6,7,8]] decrease
                                      Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
                                      æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
                                      FṀ flatten, maximum





                                      share|improve this answer









                                      $endgroup$




                                      Jelly, 10 bytes



                                      9s3’Ẓæ*³FṀ


                                      Try it online!



                                      1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
                                      where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$



                                      (this is a total coincidence.)



                                      9s3         [[1,2,3],[4,5,6],[7,8,9]]    9 split 3
                                      ’ [[0,1,2],[3,4,5],[6,7,8]] decrease
                                      Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
                                      æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
                                      FṀ flatten, maximum






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 2 days ago









                                      LynnLynn

                                      50.7k898233




                                      50.7k898233








                                      • 25




                                        $begingroup$
                                        this is clearly some kind of voodoo
                                        $endgroup$
                                        – Pureferret
                                        yesterday






                                      • 7




                                        $begingroup$
                                        This should be published.
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 5




                                        $begingroup$
                                        @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                        $endgroup$
                                        – flawr
                                        yesterday










                                      • $begingroup$
                                        @flawr ho... I missed it
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 1




                                        $begingroup$
                                        ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                        $endgroup$
                                        – Jonathan Allan
                                        yesterday
















                                      • 25




                                        $begingroup$
                                        this is clearly some kind of voodoo
                                        $endgroup$
                                        – Pureferret
                                        yesterday






                                      • 7




                                        $begingroup$
                                        This should be published.
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 5




                                        $begingroup$
                                        @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                        $endgroup$
                                        – flawr
                                        yesterday










                                      • $begingroup$
                                        @flawr ho... I missed it
                                        $endgroup$
                                        – YSC
                                        yesterday






                                      • 1




                                        $begingroup$
                                        ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                        $endgroup$
                                        – Jonathan Allan
                                        yesterday










                                      25




                                      25




                                      $begingroup$
                                      this is clearly some kind of voodoo
                                      $endgroup$
                                      – Pureferret
                                      yesterday




                                      $begingroup$
                                      this is clearly some kind of voodoo
                                      $endgroup$
                                      – Pureferret
                                      yesterday




                                      7




                                      7




                                      $begingroup$
                                      This should be published.
                                      $endgroup$
                                      – YSC
                                      yesterday




                                      $begingroup$
                                      This should be published.
                                      $endgroup$
                                      – YSC
                                      yesterday




                                      5




                                      5




                                      $begingroup$
                                      @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                      $endgroup$
                                      – flawr
                                      yesterday




                                      $begingroup$
                                      @YSC It has already been published in A000931. I'd never have guess the primes trick:)
                                      $endgroup$
                                      – flawr
                                      yesterday












                                      $begingroup$
                                      @flawr ho... I missed it
                                      $endgroup$
                                      – YSC
                                      yesterday




                                      $begingroup$
                                      @flawr ho... I missed it
                                      $endgroup$
                                      – YSC
                                      yesterday




                                      1




                                      1




                                      $begingroup$
                                      ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                      $endgroup$
                                      – Jonathan Allan
                                      yesterday






                                      $begingroup$
                                      ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
                                      $endgroup$
                                      – Jonathan Allan
                                      yesterday













                                      24












                                      $begingroup$


                                      Oasis, 5 bytes



                                      nth term 0-indexed



                                      cd+1V


                                      Try it online!



                                      Explanation



                                         1V   # a(0) = 1
                                      # a(1) = 1
                                      # a(2) = 1
                                      # a(n) =
                                      c # a(n-2)
                                      + # +
                                      d # a(n-3)





                                      share|improve this answer









                                      $endgroup$


















                                        24












                                        $begingroup$


                                        Oasis, 5 bytes



                                        nth term 0-indexed



                                        cd+1V


                                        Try it online!



                                        Explanation



                                           1V   # a(0) = 1
                                        # a(1) = 1
                                        # a(2) = 1
                                        # a(n) =
                                        c # a(n-2)
                                        + # +
                                        d # a(n-3)





                                        share|improve this answer









                                        $endgroup$
















                                          24












                                          24








                                          24





                                          $begingroup$


                                          Oasis, 5 bytes



                                          nth term 0-indexed



                                          cd+1V


                                          Try it online!



                                          Explanation



                                             1V   # a(0) = 1
                                          # a(1) = 1
                                          # a(2) = 1
                                          # a(n) =
                                          c # a(n-2)
                                          + # +
                                          d # a(n-3)





                                          share|improve this answer









                                          $endgroup$




                                          Oasis, 5 bytes



                                          nth term 0-indexed



                                          cd+1V


                                          Try it online!



                                          Explanation



                                             1V   # a(0) = 1
                                          # a(1) = 1
                                          # a(2) = 1
                                          # a(n) =
                                          c # a(n-2)
                                          + # +
                                          d # a(n-3)






                                          share|improve this answer












                                          share|improve this answer



                                          share|improve this answer










                                          answered 2 days ago









                                          EmignaEmigna

                                          47.8k433145




                                          47.8k433145























                                              21












                                              $begingroup$


                                              Jelly,  10 9  8 bytes



                                              ŻṚm2Jc$S


                                              A monadic Link accepting n (0-indexed) which yields P(n).



                                              Try it online!



                                              How?



                                              Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$



                                              ŻṚm2Jc$S - Link: integer, n       e.g. 20
                                              Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
                                              Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
                                              m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
                                              $ - last two links as a monad:
                                              J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
                                              c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
                                              S - sum 200




                                              And here is a "twofer"

                                              ...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):



                                              3ḊṗRẎ§ċ‘ - Link: n
                                              3Ḋ - 3 dequeued = [2,3]
                                              R - range = [1,2,3,...,n]
                                              ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
                                              Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
                                              § - sums [ 2, 3, 4, 5, 5, 6, 6,...]
                                              ‘ - increment n+1
                                              ċ - count occurrences P(n)





                                              share|improve this answer











                                              $endgroup$


















                                                21












                                                $begingroup$


                                                Jelly,  10 9  8 bytes



                                                ŻṚm2Jc$S


                                                A monadic Link accepting n (0-indexed) which yields P(n).



                                                Try it online!



                                                How?



                                                Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$



                                                ŻṚm2Jc$S - Link: integer, n       e.g. 20
                                                Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
                                                Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
                                                m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
                                                $ - last two links as a monad:
                                                J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
                                                c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
                                                S - sum 200




                                                And here is a "twofer"

                                                ...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):



                                                3ḊṗRẎ§ċ‘ - Link: n
                                                3Ḋ - 3 dequeued = [2,3]
                                                R - range = [1,2,3,...,n]
                                                ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
                                                Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
                                                § - sums [ 2, 3, 4, 5, 5, 6, 6,...]
                                                ‘ - increment n+1
                                                ċ - count occurrences P(n)





                                                share|improve this answer











                                                $endgroup$
















                                                  21












                                                  21








                                                  21





                                                  $begingroup$


                                                  Jelly,  10 9  8 bytes



                                                  ŻṚm2Jc$S


                                                  A monadic Link accepting n (0-indexed) which yields P(n).



                                                  Try it online!



                                                  How?



                                                  Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$



                                                  ŻṚm2Jc$S - Link: integer, n       e.g. 20
                                                  Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
                                                  Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
                                                  m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
                                                  $ - last two links as a monad:
                                                  J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
                                                  c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
                                                  S - sum 200




                                                  And here is a "twofer"

                                                  ...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):



                                                  3ḊṗRẎ§ċ‘ - Link: n
                                                  3Ḋ - 3 dequeued = [2,3]
                                                  R - range = [1,2,3,...,n]
                                                  ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
                                                  Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
                                                  § - sums [ 2, 3, 4, 5, 5, 6, 6,...]
                                                  ‘ - increment n+1
                                                  ċ - count occurrences P(n)





                                                  share|improve this answer











                                                  $endgroup$




                                                  Jelly,  10 9  8 bytes



                                                  ŻṚm2Jc$S


                                                  A monadic Link accepting n (0-indexed) which yields P(n).



                                                  Try it online!



                                                  How?



                                                  Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$



                                                  ŻṚm2Jc$S - Link: integer, n       e.g. 20
                                                  Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
                                                  Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
                                                  m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
                                                  $ - last two links as a monad:
                                                  J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
                                                  c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
                                                  S - sum 200




                                                  And here is a "twofer"

                                                  ...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):



                                                  3ḊṗRẎ§ċ‘ - Link: n
                                                  3Ḋ - 3 dequeued = [2,3]
                                                  R - range = [1,2,3,...,n]
                                                  ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
                                                  Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
                                                  § - sums [ 2, 3, 4, 5, 5, 6, 6,...]
                                                  ‘ - increment n+1
                                                  ċ - count occurrences P(n)






                                                  share|improve this answer














                                                  share|improve this answer



                                                  share|improve this answer








                                                  edited yesterday

























                                                  answered 2 days ago









                                                  Jonathan AllanJonathan Allan

                                                  54k536174




                                                  54k536174























                                                      17












                                                      $begingroup$


                                                      Haskell, 26 bytes





                                                      (l!!)
                                                      l=1:1:1:2:scanl(+)2l


                                                      Try it online! Outputs the n'th term zero-indexed.



                                                      I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.



                                                      If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.



                                                      27 bytes





                                                      f n|n<3=1|1>0=f(n-2)+f(n-3)


                                                      Try it online!






                                                      share|improve this answer









                                                      $endgroup$


















                                                        17












                                                        $begingroup$


                                                        Haskell, 26 bytes





                                                        (l!!)
                                                        l=1:1:1:2:scanl(+)2l


                                                        Try it online! Outputs the n'th term zero-indexed.



                                                        I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.



                                                        If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.



                                                        27 bytes





                                                        f n|n<3=1|1>0=f(n-2)+f(n-3)


                                                        Try it online!






                                                        share|improve this answer









                                                        $endgroup$
















                                                          17












                                                          17








                                                          17





                                                          $begingroup$


                                                          Haskell, 26 bytes





                                                          (l!!)
                                                          l=1:1:1:2:scanl(+)2l


                                                          Try it online! Outputs the n'th term zero-indexed.



                                                          I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.



                                                          If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.



                                                          27 bytes





                                                          f n|n<3=1|1>0=f(n-2)+f(n-3)


                                                          Try it online!






                                                          share|improve this answer









                                                          $endgroup$




                                                          Haskell, 26 bytes





                                                          (l!!)
                                                          l=1:1:1:2:scanl(+)2l


                                                          Try it online! Outputs the n'th term zero-indexed.



                                                          I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.



                                                          If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.



                                                          27 bytes





                                                          f n|n<3=1|1>0=f(n-2)+f(n-3)


                                                          Try it online!







                                                          share|improve this answer












                                                          share|improve this answer



                                                          share|improve this answer










                                                          answered 2 days ago









                                                          xnorxnor

                                                          93.9k18192450




                                                          93.9k18192450























                                                              9












                                                              $begingroup$


                                                              Python 2, 30 bytes





                                                              f=lambda n:n<3or f(n-2)+f(n-3)


                                                              Try it online!



                                                              Returns the n'th term zero indexed. Outputs True for 1.






                                                              share|improve this answer











                                                              $endgroup$


















                                                                9












                                                                $begingroup$


                                                                Python 2, 30 bytes





                                                                f=lambda n:n<3or f(n-2)+f(n-3)


                                                                Try it online!



                                                                Returns the n'th term zero indexed. Outputs True for 1.






                                                                share|improve this answer











                                                                $endgroup$
















                                                                  9












                                                                  9








                                                                  9





                                                                  $begingroup$


                                                                  Python 2, 30 bytes





                                                                  f=lambda n:n<3or f(n-2)+f(n-3)


                                                                  Try it online!



                                                                  Returns the n'th term zero indexed. Outputs True for 1.






                                                                  share|improve this answer











                                                                  $endgroup$




                                                                  Python 2, 30 bytes





                                                                  f=lambda n:n<3or f(n-2)+f(n-3)


                                                                  Try it online!



                                                                  Returns the n'th term zero indexed. Outputs True for 1.







                                                                  share|improve this answer














                                                                  share|improve this answer



                                                                  share|improve this answer








                                                                  edited 2 days ago

























                                                                  answered 2 days ago









                                                                  xnorxnor

                                                                  93.9k18192450




                                                                  93.9k18192450























                                                                      6












                                                                      $begingroup$


                                                                      Wolfram Language (Mathematica), 33 bytes



                                                                      a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]   


                                                                      1-indexed, returns the nth term



                                                                      Try it online!






                                                                      share|improve this answer









                                                                      $endgroup$


















                                                                        6












                                                                        $begingroup$


                                                                        Wolfram Language (Mathematica), 33 bytes



                                                                        a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]   


                                                                        1-indexed, returns the nth term



                                                                        Try it online!






                                                                        share|improve this answer









                                                                        $endgroup$
















                                                                          6












                                                                          6








                                                                          6





                                                                          $begingroup$


                                                                          Wolfram Language (Mathematica), 33 bytes



                                                                          a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]   


                                                                          1-indexed, returns the nth term



                                                                          Try it online!






                                                                          share|improve this answer









                                                                          $endgroup$




                                                                          Wolfram Language (Mathematica), 33 bytes



                                                                          a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]   


                                                                          1-indexed, returns the nth term



                                                                          Try it online!







                                                                          share|improve this answer












                                                                          share|improve this answer



                                                                          share|improve this answer










                                                                          answered 2 days ago









                                                                          J42161217J42161217

                                                                          13.9k21353




                                                                          13.9k21353























                                                                              6












                                                                              $begingroup$


                                                                              Octave / MATLAB, 35 33 bytes





                                                                              @(n)[1 filter(1,'cbaa'-98,2:n<5)]


                                                                              Outputs the first n terms.



                                                                              Try it online!



                                                                              How it works



                                                                              Anonymous function that implements a recursive filter.



                                                                              'cbaa'-98 is a shorter form to produce [1 0 -1 -1].



                                                                              2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).



                                                                              filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].






                                                                              share|improve this answer











                                                                              $endgroup$


















                                                                                6












                                                                                $begingroup$


                                                                                Octave / MATLAB, 35 33 bytes





                                                                                @(n)[1 filter(1,'cbaa'-98,2:n<5)]


                                                                                Outputs the first n terms.



                                                                                Try it online!



                                                                                How it works



                                                                                Anonymous function that implements a recursive filter.



                                                                                'cbaa'-98 is a shorter form to produce [1 0 -1 -1].



                                                                                2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).



                                                                                filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].






                                                                                share|improve this answer











                                                                                $endgroup$
















                                                                                  6












                                                                                  6








                                                                                  6





                                                                                  $begingroup$


                                                                                  Octave / MATLAB, 35 33 bytes





                                                                                  @(n)[1 filter(1,'cbaa'-98,2:n<5)]


                                                                                  Outputs the first n terms.



                                                                                  Try it online!



                                                                                  How it works



                                                                                  Anonymous function that implements a recursive filter.



                                                                                  'cbaa'-98 is a shorter form to produce [1 0 -1 -1].



                                                                                  2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).



                                                                                  filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].






                                                                                  share|improve this answer











                                                                                  $endgroup$




                                                                                  Octave / MATLAB, 35 33 bytes





                                                                                  @(n)[1 filter(1,'cbaa'-98,2:n<5)]


                                                                                  Outputs the first n terms.



                                                                                  Try it online!



                                                                                  How it works



                                                                                  Anonymous function that implements a recursive filter.



                                                                                  'cbaa'-98 is a shorter form to produce [1 0 -1 -1].



                                                                                  2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).



                                                                                  filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].







                                                                                  share|improve this answer














                                                                                  share|improve this answer



                                                                                  share|improve this answer








                                                                                  edited yesterday

























                                                                                  answered yesterday









                                                                                  Luis MendoLuis Mendo

                                                                                  75.3k889292




                                                                                  75.3k889292























                                                                                      5












                                                                                      $begingroup$


                                                                                      Retina, 47 42 bytes



                                                                                      K`0¶1¶0
                                                                                      "$+"+`.+¶(.+)¶.+$
                                                                                      $&¶$.(*_$1*
                                                                                      6,G`


                                                                                      Try it online! Outputs the first n terms on separate lines. Explanation:



                                                                                      K`0¶1¶0


                                                                                      Replace the input with the terms for -2, -1 and 0.



                                                                                      "$+"+`.+¶(.+)¶.+$
                                                                                      $&¶$.(*_$1*


                                                                                      Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.



                                                                                      6,G`


                                                                                      Discard the first six characters, i.e. the first three lines.






                                                                                      share|improve this answer











                                                                                      $endgroup$


















                                                                                        5












                                                                                        $begingroup$


                                                                                        Retina, 47 42 bytes



                                                                                        K`0¶1¶0
                                                                                        "$+"+`.+¶(.+)¶.+$
                                                                                        $&¶$.(*_$1*
                                                                                        6,G`


                                                                                        Try it online! Outputs the first n terms on separate lines. Explanation:



                                                                                        K`0¶1¶0


                                                                                        Replace the input with the terms for -2, -1 and 0.



                                                                                        "$+"+`.+¶(.+)¶.+$
                                                                                        $&¶$.(*_$1*


                                                                                        Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.



                                                                                        6,G`


                                                                                        Discard the first six characters, i.e. the first three lines.






                                                                                        share|improve this answer











                                                                                        $endgroup$
















                                                                                          5












                                                                                          5








                                                                                          5





                                                                                          $begingroup$


                                                                                          Retina, 47 42 bytes



                                                                                          K`0¶1¶0
                                                                                          "$+"+`.+¶(.+)¶.+$
                                                                                          $&¶$.(*_$1*
                                                                                          6,G`


                                                                                          Try it online! Outputs the first n terms on separate lines. Explanation:



                                                                                          K`0¶1¶0


                                                                                          Replace the input with the terms for -2, -1 and 0.



                                                                                          "$+"+`.+¶(.+)¶.+$
                                                                                          $&¶$.(*_$1*


                                                                                          Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.



                                                                                          6,G`


                                                                                          Discard the first six characters, i.e. the first three lines.






                                                                                          share|improve this answer











                                                                                          $endgroup$




                                                                                          Retina, 47 42 bytes



                                                                                          K`0¶1¶0
                                                                                          "$+"+`.+¶(.+)¶.+$
                                                                                          $&¶$.(*_$1*
                                                                                          6,G`


                                                                                          Try it online! Outputs the first n terms on separate lines. Explanation:



                                                                                          K`0¶1¶0


                                                                                          Replace the input with the terms for -2, -1 and 0.



                                                                                          "$+"+`.+¶(.+)¶.+$
                                                                                          $&¶$.(*_$1*


                                                                                          Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.



                                                                                          6,G`


                                                                                          Discard the first six characters, i.e. the first three lines.







                                                                                          share|improve this answer














                                                                                          share|improve this answer



                                                                                          share|improve this answer








                                                                                          edited 2 days ago

























                                                                                          answered 2 days ago









                                                                                          NeilNeil

                                                                                          82.6k745179




                                                                                          82.6k745179























                                                                                              5












                                                                                              $begingroup$


                                                                                              Cubix, 20 bytes



                                                                                              This is 0 indexed and outputs the Nth term



                                                                                              ;@UOI010+p?/sqq;W.(


                                                                                              Try it online!



                                                                                              Wraps onto a cube with side length 2



                                                                                                  ; @
                                                                                              U O
                                                                                              I 0 1 0 + p ? /
                                                                                              s q q ; W . (
                                                                                              . .
                                                                                              . .


                                                                                              Watch it run





                                                                                              • I010 - Initiates the stack


                                                                                              • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests


                                                                                              • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt


                                                                                              • (sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.






                                                                                              share|improve this answer









                                                                                              $endgroup$


















                                                                                                5












                                                                                                $begingroup$


                                                                                                Cubix, 20 bytes



                                                                                                This is 0 indexed and outputs the Nth term



                                                                                                ;@UOI010+p?/sqq;W.(


                                                                                                Try it online!



                                                                                                Wraps onto a cube with side length 2



                                                                                                    ; @
                                                                                                U O
                                                                                                I 0 1 0 + p ? /
                                                                                                s q q ; W . (
                                                                                                . .
                                                                                                . .


                                                                                                Watch it run





                                                                                                • I010 - Initiates the stack


                                                                                                • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests


                                                                                                • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt


                                                                                                • (sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.






                                                                                                share|improve this answer









                                                                                                $endgroup$
















                                                                                                  5












                                                                                                  5








                                                                                                  5





                                                                                                  $begingroup$


                                                                                                  Cubix, 20 bytes



                                                                                                  This is 0 indexed and outputs the Nth term



                                                                                                  ;@UOI010+p?/sqq;W.(


                                                                                                  Try it online!



                                                                                                  Wraps onto a cube with side length 2



                                                                                                      ; @
                                                                                                  U O
                                                                                                  I 0 1 0 + p ? /
                                                                                                  s q q ; W . (
                                                                                                  . .
                                                                                                  . .


                                                                                                  Watch it run





                                                                                                  • I010 - Initiates the stack


                                                                                                  • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests


                                                                                                  • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt


                                                                                                  • (sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.






                                                                                                  share|improve this answer









                                                                                                  $endgroup$




                                                                                                  Cubix, 20 bytes



                                                                                                  This is 0 indexed and outputs the Nth term



                                                                                                  ;@UOI010+p?/sqq;W.(


                                                                                                  Try it online!



                                                                                                  Wraps onto a cube with side length 2



                                                                                                      ; @
                                                                                                  U O
                                                                                                  I 0 1 0 + p ? /
                                                                                                  s q q ; W . (
                                                                                                  . .
                                                                                                  . .


                                                                                                  Watch it run





                                                                                                  • I010 - Initiates the stack


                                                                                                  • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests


                                                                                                  • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt


                                                                                                  • (sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.







                                                                                                  share|improve this answer












                                                                                                  share|improve this answer



                                                                                                  share|improve this answer










                                                                                                  answered 2 days ago









                                                                                                  MickyTMickyT

                                                                                                  10.3k21637




                                                                                                  10.3k21637























                                                                                                      5












                                                                                                      $begingroup$


                                                                                                      J, 23 bytes



                                                                                                      -1 byte thanks to ngn and Galen



                                                                                                      closed form, 26 bytes



                                                                                                      0.5<.@+1.04535%~1.32472^<:


                                                                                                      Try it online!



                                                                                                      iterative, 23 bytes



                                                                                                      (],1#._2 _3{ ::1:])^:[#


                                                                                                      Try it online!






                                                                                                      share|improve this answer











                                                                                                      $endgroup$









                                                                                                      • 1




                                                                                                        $begingroup$
                                                                                                        Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        @GalenIvanov tyvm, that's a great trick.
                                                                                                        $endgroup$
                                                                                                        – Jonah
                                                                                                        yesterday
















                                                                                                      5












                                                                                                      $begingroup$


                                                                                                      J, 23 bytes



                                                                                                      -1 byte thanks to ngn and Galen



                                                                                                      closed form, 26 bytes



                                                                                                      0.5<.@+1.04535%~1.32472^<:


                                                                                                      Try it online!



                                                                                                      iterative, 23 bytes



                                                                                                      (],1#._2 _3{ ::1:])^:[#


                                                                                                      Try it online!






                                                                                                      share|improve this answer











                                                                                                      $endgroup$









                                                                                                      • 1




                                                                                                        $begingroup$
                                                                                                        Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        @GalenIvanov tyvm, that's a great trick.
                                                                                                        $endgroup$
                                                                                                        – Jonah
                                                                                                        yesterday














                                                                                                      5












                                                                                                      5








                                                                                                      5





                                                                                                      $begingroup$


                                                                                                      J, 23 bytes



                                                                                                      -1 byte thanks to ngn and Galen



                                                                                                      closed form, 26 bytes



                                                                                                      0.5<.@+1.04535%~1.32472^<:


                                                                                                      Try it online!



                                                                                                      iterative, 23 bytes



                                                                                                      (],1#._2 _3{ ::1:])^:[#


                                                                                                      Try it online!






                                                                                                      share|improve this answer











                                                                                                      $endgroup$




                                                                                                      J, 23 bytes



                                                                                                      -1 byte thanks to ngn and Galen



                                                                                                      closed form, 26 bytes



                                                                                                      0.5<.@+1.04535%~1.32472^<:


                                                                                                      Try it online!



                                                                                                      iterative, 23 bytes



                                                                                                      (],1#._2 _3{ ::1:])^:[#


                                                                                                      Try it online!







                                                                                                      share|improve this answer














                                                                                                      share|improve this answer



                                                                                                      share|improve this answer








                                                                                                      edited yesterday

























                                                                                                      answered 2 days ago









                                                                                                      JonahJonah

                                                                                                      2,6611017




                                                                                                      2,6611017








                                                                                                      • 1




                                                                                                        $begingroup$
                                                                                                        Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        @GalenIvanov tyvm, that's a great trick.
                                                                                                        $endgroup$
                                                                                                        – Jonah
                                                                                                        yesterday














                                                                                                      • 1




                                                                                                        $begingroup$
                                                                                                        Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                        $endgroup$
                                                                                                        – Galen Ivanov
                                                                                                        yesterday












                                                                                                      • $begingroup$
                                                                                                        @GalenIvanov tyvm, that's a great trick.
                                                                                                        $endgroup$
                                                                                                        – Jonah
                                                                                                        yesterday








                                                                                                      1




                                                                                                      1




                                                                                                      $begingroup$
                                                                                                      Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                      $endgroup$
                                                                                                      – Galen Ivanov
                                                                                                      yesterday






                                                                                                      $begingroup$
                                                                                                      Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
                                                                                                      $endgroup$
                                                                                                      – Galen Ivanov
                                                                                                      yesterday














                                                                                                      $begingroup$
                                                                                                      23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                      $endgroup$
                                                                                                      – Galen Ivanov
                                                                                                      yesterday






                                                                                                      $begingroup$
                                                                                                      23 bytes thanks to ngn 1: -> # : Try it online!
                                                                                                      $endgroup$
                                                                                                      – Galen Ivanov
                                                                                                      yesterday














                                                                                                      $begingroup$
                                                                                                      @GalenIvanov tyvm, that's a great trick.
                                                                                                      $endgroup$
                                                                                                      – Jonah
                                                                                                      yesterday




                                                                                                      $begingroup$
                                                                                                      @GalenIvanov tyvm, that's a great trick.
                                                                                                      $endgroup$
                                                                                                      – Jonah
                                                                                                      yesterday











                                                                                                      4












                                                                                                      $begingroup$


                                                                                                      Python 2, 56 48 bytes





                                                                                                      f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c


                                                                                                      Try it online!



                                                                                                      Returns nth value, 0-indexed.






                                                                                                      share|improve this answer









                                                                                                      $endgroup$


















                                                                                                        4












                                                                                                        $begingroup$


                                                                                                        Python 2, 56 48 bytes





                                                                                                        f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c


                                                                                                        Try it online!



                                                                                                        Returns nth value, 0-indexed.






                                                                                                        share|improve this answer









                                                                                                        $endgroup$
















                                                                                                          4












                                                                                                          4








                                                                                                          4





                                                                                                          $begingroup$


                                                                                                          Python 2, 56 48 bytes





                                                                                                          f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c


                                                                                                          Try it online!



                                                                                                          Returns nth value, 0-indexed.






                                                                                                          share|improve this answer









                                                                                                          $endgroup$




                                                                                                          Python 2, 56 48 bytes





                                                                                                          f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c


                                                                                                          Try it online!



                                                                                                          Returns nth value, 0-indexed.







                                                                                                          share|improve this answer












                                                                                                          share|improve this answer



                                                                                                          share|improve this answer










                                                                                                          answered 2 days ago









                                                                                                          Chas BrownChas Brown

                                                                                                          5,2191523




                                                                                                          5,2191523























                                                                                                              3












                                                                                                              $begingroup$


                                                                                                              Jelly, 11 bytes



                                                                                                              5B+Ɲ2ị;Ʋ⁸¡Ḣ


                                                                                                              Try it online!



                                                                                                              0-indexed.






                                                                                                              share|improve this answer











                                                                                                              $endgroup$


















                                                                                                                3












                                                                                                                $begingroup$


                                                                                                                Jelly, 11 bytes



                                                                                                                5B+Ɲ2ị;Ʋ⁸¡Ḣ


                                                                                                                Try it online!



                                                                                                                0-indexed.






                                                                                                                share|improve this answer











                                                                                                                $endgroup$
















                                                                                                                  3












                                                                                                                  3








                                                                                                                  3





                                                                                                                  $begingroup$


                                                                                                                  Jelly, 11 bytes



                                                                                                                  5B+Ɲ2ị;Ʋ⁸¡Ḣ


                                                                                                                  Try it online!



                                                                                                                  0-indexed.






                                                                                                                  share|improve this answer











                                                                                                                  $endgroup$




                                                                                                                  Jelly, 11 bytes



                                                                                                                  5B+Ɲ2ị;Ʋ⁸¡Ḣ


                                                                                                                  Try it online!



                                                                                                                  0-indexed.







                                                                                                                  share|improve this answer














                                                                                                                  share|improve this answer



                                                                                                                  share|improve this answer








                                                                                                                  edited 2 days ago

























                                                                                                                  answered 2 days ago









                                                                                                                  Erik the OutgolferErik the Outgolfer

                                                                                                                  33k429106




                                                                                                                  33k429106























                                                                                                                      3












                                                                                                                      $begingroup$


                                                                                                                      Perl 6, 24 bytes



                                                                                                                      {(1,1,1,*+*+!*...*)[$_]}


                                                                                                                      Try it online!



                                                                                                                      A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.






                                                                                                                      share|improve this answer











                                                                                                                      $endgroup$













                                                                                                                      • $begingroup$
                                                                                                                        Why is this community wiki?
                                                                                                                        $endgroup$
                                                                                                                        – Jo King
                                                                                                                        2 days ago










                                                                                                                      • $begingroup$
                                                                                                                        @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                        $endgroup$
                                                                                                                        – Sean
                                                                                                                        yesterday
















                                                                                                                      3












                                                                                                                      $begingroup$


                                                                                                                      Perl 6, 24 bytes



                                                                                                                      {(1,1,1,*+*+!*...*)[$_]}


                                                                                                                      Try it online!



                                                                                                                      A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.






                                                                                                                      share|improve this answer











                                                                                                                      $endgroup$













                                                                                                                      • $begingroup$
                                                                                                                        Why is this community wiki?
                                                                                                                        $endgroup$
                                                                                                                        – Jo King
                                                                                                                        2 days ago










                                                                                                                      • $begingroup$
                                                                                                                        @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                        $endgroup$
                                                                                                                        – Sean
                                                                                                                        yesterday














                                                                                                                      3












                                                                                                                      3








                                                                                                                      3





                                                                                                                      $begingroup$


                                                                                                                      Perl 6, 24 bytes



                                                                                                                      {(1,1,1,*+*+!*...*)[$_]}


                                                                                                                      Try it online!



                                                                                                                      A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.






                                                                                                                      share|improve this answer











                                                                                                                      $endgroup$




                                                                                                                      Perl 6, 24 bytes



                                                                                                                      {(1,1,1,*+*+!*...*)[$_]}


                                                                                                                      Try it online!



                                                                                                                      A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.







                                                                                                                      share|improve this answer














                                                                                                                      share|improve this answer



                                                                                                                      share|improve this answer








                                                                                                                      answered 2 days ago


























                                                                                                                      community wiki





                                                                                                                      Sean













                                                                                                                      • $begingroup$
                                                                                                                        Why is this community wiki?
                                                                                                                        $endgroup$
                                                                                                                        – Jo King
                                                                                                                        2 days ago










                                                                                                                      • $begingroup$
                                                                                                                        @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                        $endgroup$
                                                                                                                        – Sean
                                                                                                                        yesterday


















                                                                                                                      • $begingroup$
                                                                                                                        Why is this community wiki?
                                                                                                                        $endgroup$
                                                                                                                        – Jo King
                                                                                                                        2 days ago










                                                                                                                      • $begingroup$
                                                                                                                        @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                        $endgroup$
                                                                                                                        – Sean
                                                                                                                        yesterday
















                                                                                                                      $begingroup$
                                                                                                                      Why is this community wiki?
                                                                                                                      $endgroup$
                                                                                                                      – Jo King
                                                                                                                      2 days ago




                                                                                                                      $begingroup$
                                                                                                                      Why is this community wiki?
                                                                                                                      $endgroup$
                                                                                                                      – Jo King
                                                                                                                      2 days ago












                                                                                                                      $begingroup$
                                                                                                                      @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                      $endgroup$
                                                                                                                      – Sean
                                                                                                                      yesterday




                                                                                                                      $begingroup$
                                                                                                                      @JoKing Beats me. If I did it somehow, it wasn't on purpose.
                                                                                                                      $endgroup$
                                                                                                                      – Sean
                                                                                                                      yesterday











                                                                                                                      3












                                                                                                                      $begingroup$


                                                                                                                      Lua 5.3, 49 48 bytes





                                                                                                                      function f(n)return n<4 and 1or f(n-2)+f(n-3)end


                                                                                                                      Try it online!



                                                                                                                      Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.






                                                                                                                      share|improve this answer











                                                                                                                      $endgroup$


















                                                                                                                        3












                                                                                                                        $begingroup$


                                                                                                                        Lua 5.3, 49 48 bytes





                                                                                                                        function f(n)return n<4 and 1or f(n-2)+f(n-3)end


                                                                                                                        Try it online!



                                                                                                                        Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.






                                                                                                                        share|improve this answer











                                                                                                                        $endgroup$
















                                                                                                                          3












                                                                                                                          3








                                                                                                                          3





                                                                                                                          $begingroup$


                                                                                                                          Lua 5.3, 49 48 bytes





                                                                                                                          function f(n)return n<4 and 1or f(n-2)+f(n-3)end


                                                                                                                          Try it online!



                                                                                                                          Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.






                                                                                                                          share|improve this answer











                                                                                                                          $endgroup$




                                                                                                                          Lua 5.3, 49 48 bytes





                                                                                                                          function f(n)return n<4 and 1or f(n-2)+f(n-3)end


                                                                                                                          Try it online!



                                                                                                                          Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.







                                                                                                                          share|improve this answer














                                                                                                                          share|improve this answer



                                                                                                                          share|improve this answer








                                                                                                                          edited yesterday

























                                                                                                                          answered 2 days ago









                                                                                                                          cyclaministcyclaminist

                                                                                                                          1813




                                                                                                                          1813























                                                                                                                              3












                                                                                                                              $begingroup$


                                                                                                                              Ruby, 26 bytes





                                                                                                                              f=->n{n<3?1:f[n-2]+f[n-3]}


                                                                                                                              Try it online!






                                                                                                                              share|improve this answer









                                                                                                                              $endgroup$


















                                                                                                                                3












                                                                                                                                $begingroup$


                                                                                                                                Ruby, 26 bytes





                                                                                                                                f=->n{n<3?1:f[n-2]+f[n-3]}


                                                                                                                                Try it online!






                                                                                                                                share|improve this answer









                                                                                                                                $endgroup$
















                                                                                                                                  3












                                                                                                                                  3








                                                                                                                                  3





                                                                                                                                  $begingroup$


                                                                                                                                  Ruby, 26 bytes





                                                                                                                                  f=->n{n<3?1:f[n-2]+f[n-3]}


                                                                                                                                  Try it online!






                                                                                                                                  share|improve this answer









                                                                                                                                  $endgroup$




                                                                                                                                  Ruby, 26 bytes





                                                                                                                                  f=->n{n<3?1:f[n-2]+f[n-3]}


                                                                                                                                  Try it online!







                                                                                                                                  share|improve this answer












                                                                                                                                  share|improve this answer



                                                                                                                                  share|improve this answer










                                                                                                                                  answered yesterday









                                                                                                                                  G BG B

                                                                                                                                  8,2561429




                                                                                                                                  8,2561429























                                                                                                                                      3












                                                                                                                                      $begingroup$


                                                                                                                                      APL (Dyalog Unicode), 20 18 17 bytesSBCS



                                                                                                                                      This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.



                                                                                                                                      Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn



                                                                                                                                      4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3


                                                                                                                                      Try it online!



                                                                                                                                      Explanation



                                                                                                                                      4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

                                                                                                                                      ⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
                                                                                                                                      inside the brackets (...) to its operands (here marked ⍵ and ⍺).
                                                                                                                                      2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
                                                                                                                                      where we save our results as the function is repeatedly applied
                                                                                                                                      and our ⍺, 2, is our right argument and is immediately applied to +/,
                                                                                                                                      so that we have 2+/ which will return the pairwise sums of our array.
                                                                                                                                      2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
                                                                                                                                      ⊢,⍨ And add it to the head of our array.
                                                                                                                                      4⌷ When we've finished adding Padovan numbers to the end of our list,
                                                                                                                                      the n-th Padovan number (1-indexed) is the 4th member of that list,
                                                                                                                                      and so, we implicitly return that.





                                                                                                                                      share|improve this answer











                                                                                                                                      $endgroup$


















                                                                                                                                        3












                                                                                                                                        $begingroup$


                                                                                                                                        APL (Dyalog Unicode), 20 18 17 bytesSBCS



                                                                                                                                        This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.



                                                                                                                                        Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn



                                                                                                                                        4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3


                                                                                                                                        Try it online!



                                                                                                                                        Explanation



                                                                                                                                        4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

                                                                                                                                        ⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
                                                                                                                                        inside the brackets (...) to its operands (here marked ⍵ and ⍺).
                                                                                                                                        2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
                                                                                                                                        where we save our results as the function is repeatedly applied
                                                                                                                                        and our ⍺, 2, is our right argument and is immediately applied to +/,
                                                                                                                                        so that we have 2+/ which will return the pairwise sums of our array.
                                                                                                                                        2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
                                                                                                                                        ⊢,⍨ And add it to the head of our array.
                                                                                                                                        4⌷ When we've finished adding Padovan numbers to the end of our list,
                                                                                                                                        the n-th Padovan number (1-indexed) is the 4th member of that list,
                                                                                                                                        and so, we implicitly return that.





                                                                                                                                        share|improve this answer











                                                                                                                                        $endgroup$
















                                                                                                                                          3












                                                                                                                                          3








                                                                                                                                          3





                                                                                                                                          $begingroup$


                                                                                                                                          APL (Dyalog Unicode), 20 18 17 bytesSBCS



                                                                                                                                          This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.



                                                                                                                                          Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn



                                                                                                                                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3


                                                                                                                                          Try it online!



                                                                                                                                          Explanation



                                                                                                                                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

                                                                                                                                          ⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
                                                                                                                                          inside the brackets (...) to its operands (here marked ⍵ and ⍺).
                                                                                                                                          2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
                                                                                                                                          where we save our results as the function is repeatedly applied
                                                                                                                                          and our ⍺, 2, is our right argument and is immediately applied to +/,
                                                                                                                                          so that we have 2+/ which will return the pairwise sums of our array.
                                                                                                                                          2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
                                                                                                                                          ⊢,⍨ And add it to the head of our array.
                                                                                                                                          4⌷ When we've finished adding Padovan numbers to the end of our list,
                                                                                                                                          the n-th Padovan number (1-indexed) is the 4th member of that list,
                                                                                                                                          and so, we implicitly return that.





                                                                                                                                          share|improve this answer











                                                                                                                                          $endgroup$




                                                                                                                                          APL (Dyalog Unicode), 20 18 17 bytesSBCS



                                                                                                                                          This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.



                                                                                                                                          Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn



                                                                                                                                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3


                                                                                                                                          Try it online!



                                                                                                                                          Explanation



                                                                                                                                          4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

                                                                                                                                          ⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
                                                                                                                                          inside the brackets (...) to its operands (here marked ⍵ and ⍺).
                                                                                                                                          2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
                                                                                                                                          where we save our results as the function is repeatedly applied
                                                                                                                                          and our ⍺, 2, is our right argument and is immediately applied to +/,
                                                                                                                                          so that we have 2+/ which will return the pairwise sums of our array.
                                                                                                                                          2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
                                                                                                                                          ⊢,⍨ And add it to the head of our array.
                                                                                                                                          4⌷ When we've finished adding Padovan numbers to the end of our list,
                                                                                                                                          the n-th Padovan number (1-indexed) is the 4th member of that list,
                                                                                                                                          and so, we implicitly return that.






                                                                                                                                          share|improve this answer














                                                                                                                                          share|improve this answer



                                                                                                                                          share|improve this answer








                                                                                                                                          edited yesterday

























                                                                                                                                          answered yesterday









                                                                                                                                          Sherlock9Sherlock9

                                                                                                                                          8,16411860




                                                                                                                                          8,16411860























                                                                                                                                              3












                                                                                                                                              $begingroup$


                                                                                                                                              K (ngn/k), 24 20 bytes



                                                                                                                                              -4 bytes thanks to ngn!



                                                                                                                                              {$[x<3;1;+/o'x-2 3]}


                                                                                                                                              Try it online!



                                                                                                                                              0-indexed, first N terms






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$









                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday






                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                1: -> # in the j solution
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday


















                                                                                                                                              3












                                                                                                                                              $begingroup$


                                                                                                                                              K (ngn/k), 24 20 bytes



                                                                                                                                              -4 bytes thanks to ngn!



                                                                                                                                              {$[x<3;1;+/o'x-2 3]}


                                                                                                                                              Try it online!



                                                                                                                                              0-indexed, first N terms






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$









                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday






                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                1: -> # in the j solution
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday
















                                                                                                                                              3












                                                                                                                                              3








                                                                                                                                              3





                                                                                                                                              $begingroup$


                                                                                                                                              K (ngn/k), 24 20 bytes



                                                                                                                                              -4 bytes thanks to ngn!



                                                                                                                                              {$[x<3;1;+/o'x-2 3]}


                                                                                                                                              Try it online!



                                                                                                                                              0-indexed, first N terms






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$




                                                                                                                                              K (ngn/k), 24 20 bytes



                                                                                                                                              -4 bytes thanks to ngn!



                                                                                                                                              {$[x<3;1;+/o'x-2 3]}


                                                                                                                                              Try it online!



                                                                                                                                              0-indexed, first N terms







                                                                                                                                              share|improve this answer














                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer








                                                                                                                                              edited yesterday

























                                                                                                                                              answered yesterday









                                                                                                                                              Galen IvanovGalen Ivanov

                                                                                                                                              7,41211034




                                                                                                                                              7,41211034








                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday






                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                1: -> # in the j solution
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday
















                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                                $endgroup$
                                                                                                                                                – Galen Ivanov
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday






                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                1: -> # in the j solution
                                                                                                                                                $endgroup$
                                                                                                                                                – ngn
                                                                                                                                                yesterday










                                                                                                                                              1




                                                                                                                                              1




                                                                                                                                              $begingroup$
                                                                                                                                              f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Galen Ivanov
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              @ngn Thanks! I tried it (without success) in J; it's elegant here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Galen Ivanov
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                              $endgroup$
                                                                                                                                              – Galen Ivanov
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
                                                                                                                                              $endgroup$
                                                                                                                                              – Galen Ivanov
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              ah, right, base-1 decode is a train-friendly way to sum :)
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday




                                                                                                                                              2




                                                                                                                                              2




                                                                                                                                              $begingroup$
                                                                                                                                              1: -> # in the j solution
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday






                                                                                                                                              $begingroup$
                                                                                                                                              1: -> # in the j solution
                                                                                                                                              $endgroup$
                                                                                                                                              – ngn
                                                                                                                                              yesterday













                                                                                                                                              3












                                                                                                                                              $begingroup$

                                                                                                                                              x86 32-bit machine code, 17 bytes



                                                                                                                                              53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3


                                                                                                                                              Disassembly:



                                                                                                                                              00CE1250 53                   push        ebx  
                                                                                                                                              00CE1251 33 DB xor ebx,ebx
                                                                                                                                              00CE1253 F7 E3 mul eax,ebx
                                                                                                                                              00CE1255 43 inc ebx
                                                                                                                                              00CE1256 83 C1 04 add ecx,4
                                                                                                                                              00CE1259 03 D8 add ebx,eax
                                                                                                                                              00CE125B 93 xchg eax,ebx
                                                                                                                                              00CE125C 92 xchg eax,edx
                                                                                                                                              00CE125D E2 FA loop myloop (0CE1259h)
                                                                                                                                              00CE125F 5B pop ebx
                                                                                                                                              00CE1260 C3 ret


                                                                                                                                              It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.



                                                                                                                                              At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.



                                                                                                                                              I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.



                                                                                                                                              I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I love to see machine code answers on PP&CG! +1
                                                                                                                                                $endgroup$
                                                                                                                                                – Tau
                                                                                                                                                23 hours ago
















                                                                                                                                              3












                                                                                                                                              $begingroup$

                                                                                                                                              x86 32-bit machine code, 17 bytes



                                                                                                                                              53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3


                                                                                                                                              Disassembly:



                                                                                                                                              00CE1250 53                   push        ebx  
                                                                                                                                              00CE1251 33 DB xor ebx,ebx
                                                                                                                                              00CE1253 F7 E3 mul eax,ebx
                                                                                                                                              00CE1255 43 inc ebx
                                                                                                                                              00CE1256 83 C1 04 add ecx,4
                                                                                                                                              00CE1259 03 D8 add ebx,eax
                                                                                                                                              00CE125B 93 xchg eax,ebx
                                                                                                                                              00CE125C 92 xchg eax,edx
                                                                                                                                              00CE125D E2 FA loop myloop (0CE1259h)
                                                                                                                                              00CE125F 5B pop ebx
                                                                                                                                              00CE1260 C3 ret


                                                                                                                                              It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.



                                                                                                                                              At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.



                                                                                                                                              I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.



                                                                                                                                              I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I love to see machine code answers on PP&CG! +1
                                                                                                                                                $endgroup$
                                                                                                                                                – Tau
                                                                                                                                                23 hours ago














                                                                                                                                              3












                                                                                                                                              3








                                                                                                                                              3





                                                                                                                                              $begingroup$

                                                                                                                                              x86 32-bit machine code, 17 bytes



                                                                                                                                              53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3


                                                                                                                                              Disassembly:



                                                                                                                                              00CE1250 53                   push        ebx  
                                                                                                                                              00CE1251 33 DB xor ebx,ebx
                                                                                                                                              00CE1253 F7 E3 mul eax,ebx
                                                                                                                                              00CE1255 43 inc ebx
                                                                                                                                              00CE1256 83 C1 04 add ecx,4
                                                                                                                                              00CE1259 03 D8 add ebx,eax
                                                                                                                                              00CE125B 93 xchg eax,ebx
                                                                                                                                              00CE125C 92 xchg eax,edx
                                                                                                                                              00CE125D E2 FA loop myloop (0CE1259h)
                                                                                                                                              00CE125F 5B pop ebx
                                                                                                                                              00CE1260 C3 ret


                                                                                                                                              It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.



                                                                                                                                              At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.



                                                                                                                                              I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.



                                                                                                                                              I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$



                                                                                                                                              x86 32-bit machine code, 17 bytes



                                                                                                                                              53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3


                                                                                                                                              Disassembly:



                                                                                                                                              00CE1250 53                   push        ebx  
                                                                                                                                              00CE1251 33 DB xor ebx,ebx
                                                                                                                                              00CE1253 F7 E3 mul eax,ebx
                                                                                                                                              00CE1255 43 inc ebx
                                                                                                                                              00CE1256 83 C1 04 add ecx,4
                                                                                                                                              00CE1259 03 D8 add ebx,eax
                                                                                                                                              00CE125B 93 xchg eax,ebx
                                                                                                                                              00CE125C 92 xchg eax,edx
                                                                                                                                              00CE125D E2 FA loop myloop (0CE1259h)
                                                                                                                                              00CE125F 5B pop ebx
                                                                                                                                              00CE1260 C3 ret


                                                                                                                                              It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.



                                                                                                                                              At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.



                                                                                                                                              I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.



                                                                                                                                              I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)







                                                                                                                                              share|improve this answer












                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer










                                                                                                                                              answered yesterday









                                                                                                                                              anatolyganatolyg

                                                                                                                                              7,2592166




                                                                                                                                              7,2592166












                                                                                                                                              • $begingroup$
                                                                                                                                                I love to see machine code answers on PP&CG! +1
                                                                                                                                                $endgroup$
                                                                                                                                                – Tau
                                                                                                                                                23 hours ago


















                                                                                                                                              • $begingroup$
                                                                                                                                                I love to see machine code answers on PP&CG! +1
                                                                                                                                                $endgroup$
                                                                                                                                                – Tau
                                                                                                                                                23 hours ago
















                                                                                                                                              $begingroup$
                                                                                                                                              I love to see machine code answers on PP&CG! +1
                                                                                                                                              $endgroup$
                                                                                                                                              – Tau
                                                                                                                                              23 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              I love to see machine code answers on PP&CG! +1
                                                                                                                                              $endgroup$
                                                                                                                                              – Tau
                                                                                                                                              23 hours ago











                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              Japt -N, 12 bytes



                                                                                                                                              <3ªßUµ2 +ß´U


                                                                                                                                              Try it






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                Looks like 12 is the best we can do :
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                2 days ago










                                                                                                                                              • $begingroup$
                                                                                                                                                I stand corrected!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                14 hours ago
















                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              Japt -N, 12 bytes



                                                                                                                                              <3ªßUµ2 +ß´U


                                                                                                                                              Try it






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                Looks like 12 is the best we can do :
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                2 days ago










                                                                                                                                              • $begingroup$
                                                                                                                                                I stand corrected!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                14 hours ago














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$


                                                                                                                                              Japt -N, 12 bytes



                                                                                                                                              <3ªßUµ2 +ß´U


                                                                                                                                              Try it






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$




                                                                                                                                              Japt -N, 12 bytes



                                                                                                                                              <3ªßUµ2 +ß´U


                                                                                                                                              Try it







                                                                                                                                              share|improve this answer












                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer










                                                                                                                                              answered 2 days ago









                                                                                                                                              Embodiment of IgnoranceEmbodiment of Ignorance

                                                                                                                                              2,886127




                                                                                                                                              2,886127












                                                                                                                                              • $begingroup$
                                                                                                                                                Looks like 12 is the best we can do :
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                2 days ago










                                                                                                                                              • $begingroup$
                                                                                                                                                I stand corrected!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                14 hours ago


















                                                                                                                                              • $begingroup$
                                                                                                                                                Looks like 12 is the best we can do :
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                2 days ago










                                                                                                                                              • $begingroup$
                                                                                                                                                I stand corrected!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                14 hours ago
















                                                                                                                                              $begingroup$
                                                                                                                                              Looks like 12 is the best we can do :
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              2 days ago




                                                                                                                                              $begingroup$
                                                                                                                                              Looks like 12 is the best we can do :
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              2 days ago












                                                                                                                                              $begingroup$
                                                                                                                                              I stand corrected!
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              14 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              I stand corrected!
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              14 hours ago











                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              Pyth, 16 bytes



                                                                                                                                              L?<b3!b+y-b2y-b3


                                                                                                                                              This defines the function y. Try it here!



                                                                                                                                              Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.



                                                                                                                                              +l{sa.pMf.Am&>d2%d2T./QY!


                                                                                                                                              This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Well done on hB2. Too bad it's the same byte count.
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday
















                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              Pyth, 16 bytes



                                                                                                                                              L?<b3!b+y-b2y-b3


                                                                                                                                              This defines the function y. Try it here!



                                                                                                                                              Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.



                                                                                                                                              +l{sa.pMf.Am&>d2%d2T./QY!


                                                                                                                                              This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Well done on hB2. Too bad it's the same byte count.
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$

                                                                                                                                              Pyth, 16 bytes



                                                                                                                                              L?<b3!b+y-b2y-b3


                                                                                                                                              This defines the function y. Try it here!



                                                                                                                                              Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.



                                                                                                                                              +l{sa.pMf.Am&>d2%d2T./QY!


                                                                                                                                              This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$



                                                                                                                                              Pyth, 16 bytes



                                                                                                                                              L?<b3!b+y-b2y-b3


                                                                                                                                              This defines the function y. Try it here!



                                                                                                                                              Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.



                                                                                                                                              +l{sa.pMf.Am&>d2%d2T./QY!


                                                                                                                                              This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.







                                                                                                                                              share|improve this answer












                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer










                                                                                                                                              answered 2 days ago









                                                                                                                                              RK.RK.

                                                                                                                                              407211




                                                                                                                                              407211












                                                                                                                                              • $begingroup$
                                                                                                                                                y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Well done on hB2. Too bad it's the same byte count.
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday


















                                                                                                                                              • $begingroup$
                                                                                                                                                y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Well done on hB2. Too bad it's the same byte count.
                                                                                                                                                $endgroup$
                                                                                                                                                – Ven
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                                $endgroup$
                                                                                                                                                – RK.
                                                                                                                                                yesterday
















                                                                                                                                              $begingroup$
                                                                                                                                              y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                              $endgroup$
                                                                                                                                              – Ven
                                                                                                                                              yesterday






                                                                                                                                              $begingroup$
                                                                                                                                              y-b2y-b3 could maybe be refactored with either bifurcate or L? Though declaring an array of 2 elements is costly. yL-Lb2,3 is longer :(
                                                                                                                                              $endgroup$
                                                                                                                                              – Ven
                                                                                                                                              yesterday














                                                                                                                                              $begingroup$
                                                                                                                                              @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                              $endgroup$
                                                                                                                                              – RK.
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              @Ven I was able to replace +y-b2y-b3 with smy-bdhB2 which is the same amount of bytes; hB2 results in the array [2, 3]
                                                                                                                                              $endgroup$
                                                                                                                                              – RK.
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              Well done on hB2. Too bad it's the same byte count.
                                                                                                                                              $endgroup$
                                                                                                                                              – Ven
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              Well done on hB2. Too bad it's the same byte count.
                                                                                                                                              $endgroup$
                                                                                                                                              – Ven
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                              $endgroup$
                                                                                                                                              – RK.
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              Yeah, though I wonder if there's some way to get rid of the d in the map.
                                                                                                                                              $endgroup$
                                                                                                                                              – RK.
                                                                                                                                              yesterday











                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              05AB1E, 8 bytes



                                                                                                                                              1Ð)λ£₂₃+


                                                                                                                                              Try it online!



                                                                                                                                              Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.



                                                                                                                                              How?



                                                                                                                                              1Ð)λ£₂₃+ | Full program.
                                                                                                                                              1Ð) | Initialize the stack with [1, 1, 1].
                                                                                                                                              λ | Begin the recursive generation of a list: Starting from some base case,
                                                                                                                                              | this command generates an infinite list with the pattern function given.
                                                                                                                                              £ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
                                                                                                                                              ₂₃+ | Add a(n-2) and a(n-3).





                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                                $endgroup$
                                                                                                                                                – Kevin Cruijssen
                                                                                                                                                yesterday
















                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              05AB1E, 8 bytes



                                                                                                                                              1Ð)λ£₂₃+


                                                                                                                                              Try it online!



                                                                                                                                              Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.



                                                                                                                                              How?



                                                                                                                                              1Ð)λ£₂₃+ | Full program.
                                                                                                                                              1Ð) | Initialize the stack with [1, 1, 1].
                                                                                                                                              λ | Begin the recursive generation of a list: Starting from some base case,
                                                                                                                                              | this command generates an infinite list with the pattern function given.
                                                                                                                                              £ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
                                                                                                                                              ₂₃+ | Add a(n-2) and a(n-3).





                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                                $endgroup$
                                                                                                                                                – Kevin Cruijssen
                                                                                                                                                yesterday














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$


                                                                                                                                              05AB1E, 8 bytes



                                                                                                                                              1Ð)λ£₂₃+


                                                                                                                                              Try it online!



                                                                                                                                              Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.



                                                                                                                                              How?



                                                                                                                                              1Ð)λ£₂₃+ | Full program.
                                                                                                                                              1Ð) | Initialize the stack with [1, 1, 1].
                                                                                                                                              λ | Begin the recursive generation of a list: Starting from some base case,
                                                                                                                                              | this command generates an infinite list with the pattern function given.
                                                                                                                                              £ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
                                                                                                                                              ₂₃+ | Add a(n-2) and a(n-3).





                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$




                                                                                                                                              05AB1E, 8 bytes



                                                                                                                                              1Ð)λ£₂₃+


                                                                                                                                              Try it online!



                                                                                                                                              Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.



                                                                                                                                              How?



                                                                                                                                              1Ð)λ£₂₃+ | Full program.
                                                                                                                                              1Ð) | Initialize the stack with [1, 1, 1].
                                                                                                                                              λ | Begin the recursive generation of a list: Starting from some base case,
                                                                                                                                              | this command generates an infinite list with the pattern function given.
                                                                                                                                              £ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
                                                                                                                                              ₂₃+ | Add a(n-2) and a(n-3).






                                                                                                                                              share|improve this answer














                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer








                                                                                                                                              edited yesterday

























                                                                                                                                              answered yesterday









                                                                                                                                              Mr. XcoderMr. Xcoder

                                                                                                                                              32.2k759200




                                                                                                                                              32.2k759200












                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                                $endgroup$
                                                                                                                                                – Kevin Cruijssen
                                                                                                                                                yesterday


















                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                                $endgroup$
                                                                                                                                                – Kevin Cruijssen
                                                                                                                                                yesterday
















                                                                                                                                              $begingroup$
                                                                                                                                              I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                              $endgroup$
                                                                                                                                              – Kevin Cruijssen
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
                                                                                                                                              $endgroup$
                                                                                                                                              – Kevin Cruijssen
                                                                                                                                              yesterday











                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              Java, 41 bytes



                                                                                                                                              Can't use a lambda (runtime error). Port of this Javascript answer



                                                                                                                                              int f(int n){return n<3?1:f(n-2)+f(n-3);}


                                                                                                                                              TIO






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                                $endgroup$
                                                                                                                                                – Olivier Grégoire
                                                                                                                                                11 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                                $endgroup$
                                                                                                                                                – Benjamin Urquhart
                                                                                                                                                6 hours ago


















                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              Java, 41 bytes



                                                                                                                                              Can't use a lambda (runtime error). Port of this Javascript answer



                                                                                                                                              int f(int n){return n<3?1:f(n-2)+f(n-3);}


                                                                                                                                              TIO






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                                $endgroup$
                                                                                                                                                – Olivier Grégoire
                                                                                                                                                11 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                                $endgroup$
                                                                                                                                                – Benjamin Urquhart
                                                                                                                                                6 hours ago
















                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$

                                                                                                                                              Java, 41 bytes



                                                                                                                                              Can't use a lambda (runtime error). Port of this Javascript answer



                                                                                                                                              int f(int n){return n<3?1:f(n-2)+f(n-3);}


                                                                                                                                              TIO






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$



                                                                                                                                              Java, 41 bytes



                                                                                                                                              Can't use a lambda (runtime error). Port of this Javascript answer



                                                                                                                                              int f(int n){return n<3?1:f(n-2)+f(n-3);}


                                                                                                                                              TIO







                                                                                                                                              share|improve this answer












                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer










                                                                                                                                              answered yesterday









                                                                                                                                              Benjamin UrquhartBenjamin Urquhart

                                                                                                                                              40017




                                                                                                                                              40017












                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                                $endgroup$
                                                                                                                                                – Olivier Grégoire
                                                                                                                                                11 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                                $endgroup$
                                                                                                                                                – Benjamin Urquhart
                                                                                                                                                6 hours ago




















                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                                $endgroup$
                                                                                                                                                – Olivier Grégoire
                                                                                                                                                11 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                                $endgroup$
                                                                                                                                                – Benjamin Urquhart
                                                                                                                                                6 hours ago


















                                                                                                                                              $begingroup$
                                                                                                                                              I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              15 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              I think you are missing some requirements: Have a look at my modification here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              15 hours ago












                                                                                                                                              $begingroup$
                                                                                                                                              Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                              $endgroup$
                                                                                                                                              – Olivier Grégoire
                                                                                                                                              11 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
                                                                                                                                              $endgroup$
                                                                                                                                              – Olivier Grégoire
                                                                                                                                              11 hours ago












                                                                                                                                              $begingroup$
                                                                                                                                              Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                              $endgroup$
                                                                                                                                              – Benjamin Urquhart
                                                                                                                                              6 hours ago






                                                                                                                                              $begingroup$
                                                                                                                                              Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
                                                                                                                                              $endgroup$
                                                                                                                                              – Benjamin Urquhart
                                                                                                                                              6 hours ago













                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              R + pryr, 38 36 bytes



                                                                                                                                              Zero-indexed recursive function.





                                                                                                                                              f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))


                                                                                                                                              Try it online!



                                                                                                                                              Thanks to @Giuseppe for pointing out two obviously needless bytes.






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$









                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                                $endgroup$
                                                                                                                                                – Giuseppe
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Giuseppe thanks! Updated now.
                                                                                                                                                $endgroup$
                                                                                                                                                – rturnbull
                                                                                                                                                yesterday
















                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              R + pryr, 38 36 bytes



                                                                                                                                              Zero-indexed recursive function.





                                                                                                                                              f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))


                                                                                                                                              Try it online!



                                                                                                                                              Thanks to @Giuseppe for pointing out two obviously needless bytes.






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$









                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                                $endgroup$
                                                                                                                                                – Giuseppe
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Giuseppe thanks! Updated now.
                                                                                                                                                $endgroup$
                                                                                                                                                – rturnbull
                                                                                                                                                yesterday














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$


                                                                                                                                              R + pryr, 38 36 bytes



                                                                                                                                              Zero-indexed recursive function.





                                                                                                                                              f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))


                                                                                                                                              Try it online!



                                                                                                                                              Thanks to @Giuseppe for pointing out two obviously needless bytes.






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$




                                                                                                                                              R + pryr, 38 36 bytes



                                                                                                                                              Zero-indexed recursive function.





                                                                                                                                              f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))


                                                                                                                                              Try it online!



                                                                                                                                              Thanks to @Giuseppe for pointing out two obviously needless bytes.







                                                                                                                                              share|improve this answer














                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer








                                                                                                                                              edited yesterday

























                                                                                                                                              answered yesterday









                                                                                                                                              rturnbullrturnbull

                                                                                                                                              3,519925




                                                                                                                                              3,519925








                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                                $endgroup$
                                                                                                                                                – Giuseppe
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Giuseppe thanks! Updated now.
                                                                                                                                                $endgroup$
                                                                                                                                                – rturnbull
                                                                                                                                                yesterday














                                                                                                                                              • 2




                                                                                                                                                $begingroup$
                                                                                                                                                If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                                $endgroup$
                                                                                                                                                – Giuseppe
                                                                                                                                                yesterday












                                                                                                                                              • $begingroup$
                                                                                                                                                @Giuseppe thanks! Updated now.
                                                                                                                                                $endgroup$
                                                                                                                                                – rturnbull
                                                                                                                                                yesterday








                                                                                                                                              2




                                                                                                                                              2




                                                                                                                                              $begingroup$
                                                                                                                                              If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                              $endgroup$
                                                                                                                                              – Giuseppe
                                                                                                                                              yesterday






                                                                                                                                              $begingroup$
                                                                                                                                              If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes
                                                                                                                                              $endgroup$
                                                                                                                                              – Giuseppe
                                                                                                                                              yesterday














                                                                                                                                              $begingroup$
                                                                                                                                              @Giuseppe thanks! Updated now.
                                                                                                                                              $endgroup$
                                                                                                                                              – rturnbull
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              @Giuseppe thanks! Updated now.
                                                                                                                                              $endgroup$
                                                                                                                                              – rturnbull
                                                                                                                                              yesterday











                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              JavaScript (ES6), 23 bytes



                                                                                                                                              Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.



                                                                                                                                              Returns the $N$th term, 0-indexed.





                                                                                                                                              f=n=>n<3||f(n-2)+f(n-3)


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                                $endgroup$
                                                                                                                                                – Nit
                                                                                                                                                yesterday






                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                @Nit Relevant meta post.
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Arnauld yep I can see it now. Sorry!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                14 hours ago
















                                                                                                                                              2












                                                                                                                                              $begingroup$

                                                                                                                                              JavaScript (ES6), 23 bytes



                                                                                                                                              Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.



                                                                                                                                              Returns the $N$th term, 0-indexed.





                                                                                                                                              f=n=>n<3||f(n-2)+f(n-3)


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$













                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                                $endgroup$
                                                                                                                                                – Nit
                                                                                                                                                yesterday






                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                @Nit Relevant meta post.
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Arnauld yep I can see it now. Sorry!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                14 hours ago














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$

                                                                                                                                              JavaScript (ES6), 23 bytes



                                                                                                                                              Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.



                                                                                                                                              Returns the $N$th term, 0-indexed.





                                                                                                                                              f=n=>n<3||f(n-2)+f(n-3)


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer











                                                                                                                                              $endgroup$



                                                                                                                                              JavaScript (ES6), 23 bytes



                                                                                                                                              Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.



                                                                                                                                              Returns the $N$th term, 0-indexed.





                                                                                                                                              f=n=>n<3||f(n-2)+f(n-3)


                                                                                                                                              Try it online!







                                                                                                                                              share|improve this answer














                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer








                                                                                                                                              edited 15 hours ago

























                                                                                                                                              answered 2 days ago









                                                                                                                                              ArnauldArnauld

                                                                                                                                              80.7k797334




                                                                                                                                              80.7k797334












                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                                $endgroup$
                                                                                                                                                – Nit
                                                                                                                                                yesterday






                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                @Nit Relevant meta post.
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Arnauld yep I can see it now. Sorry!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                14 hours ago


















                                                                                                                                              • $begingroup$
                                                                                                                                                I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                                $endgroup$
                                                                                                                                                – Nit
                                                                                                                                                yesterday






                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                @Nit Relevant meta post.
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                yesterday










                                                                                                                                              • $begingroup$
                                                                                                                                                I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                                $endgroup$
                                                                                                                                                – Arnauld
                                                                                                                                                15 hours ago










                                                                                                                                              • $begingroup$
                                                                                                                                                @Arnauld yep I can see it now. Sorry!
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaq
                                                                                                                                                14 hours ago
















                                                                                                                                              $begingroup$
                                                                                                                                              I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                              $endgroup$
                                                                                                                                              – Nit
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              I don't think it's reasonable to say that returning true is the same as returning 1 if the rest of the output is numbers.
                                                                                                                                              $endgroup$
                                                                                                                                              – Nit
                                                                                                                                              yesterday




                                                                                                                                              1




                                                                                                                                              1




                                                                                                                                              $begingroup$
                                                                                                                                              @Nit Relevant meta post.
                                                                                                                                              $endgroup$
                                                                                                                                              – Arnauld
                                                                                                                                              yesterday




                                                                                                                                              $begingroup$
                                                                                                                                              @Nit Relevant meta post.
                                                                                                                                              $endgroup$
                                                                                                                                              – Arnauld
                                                                                                                                              yesterday












                                                                                                                                              $begingroup$
                                                                                                                                              I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              15 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              I think you are missing some requirements: Have a look at my modification (version in Java) here.
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              15 hours ago












                                                                                                                                              $begingroup$
                                                                                                                                              @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                              $endgroup$
                                                                                                                                              – Arnauld
                                                                                                                                              15 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              @Shaq The challenge clearly specifies that the first three terms of the sequence are all 1. So, it's not the sequence defined in A000931 (but the formula is the same).
                                                                                                                                              $endgroup$
                                                                                                                                              – Arnauld
                                                                                                                                              15 hours ago












                                                                                                                                              $begingroup$
                                                                                                                                              @Arnauld yep I can see it now. Sorry!
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              14 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              @Arnauld yep I can see it now. Sorry!
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaq
                                                                                                                                              14 hours ago











                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              C (clang), 41 bytes





                                                                                                                                              int a(int i){return i<3?1:a(i-2)+a(i-3);}


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer








                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.






                                                                                                                                              $endgroup$









                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                Welcome to PPCG :)
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                10 hours ago
















                                                                                                                                              2












                                                                                                                                              $begingroup$


                                                                                                                                              C (clang), 41 bytes





                                                                                                                                              int a(int i){return i<3?1:a(i-2)+a(i-3);}


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer








                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.






                                                                                                                                              $endgroup$









                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                Welcome to PPCG :)
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                10 hours ago














                                                                                                                                              2












                                                                                                                                              2








                                                                                                                                              2





                                                                                                                                              $begingroup$


                                                                                                                                              C (clang), 41 bytes





                                                                                                                                              int a(int i){return i<3?1:a(i-2)+a(i-3);}


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer








                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.






                                                                                                                                              $endgroup$




                                                                                                                                              C (clang), 41 bytes





                                                                                                                                              int a(int i){return i<3?1:a(i-2)+a(i-3);}


                                                                                                                                              Try it online!







                                                                                                                                              share|improve this answer








                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.









                                                                                                                                              share|improve this answer



                                                                                                                                              share|improve this answer






                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.









                                                                                                                                              answered 10 hours ago









                                                                                                                                              peterzugerpeterzuger

                                                                                                                                              213




                                                                                                                                              213




                                                                                                                                              New contributor




                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.





                                                                                                                                              New contributor





                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.






                                                                                                                                              peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                                                                                                              Check out our Code of Conduct.








                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                Welcome to PPCG :)
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                10 hours ago














                                                                                                                                              • 1




                                                                                                                                                $begingroup$
                                                                                                                                                Welcome to PPCG :)
                                                                                                                                                $endgroup$
                                                                                                                                                – Shaggy
                                                                                                                                                10 hours ago








                                                                                                                                              1




                                                                                                                                              1




                                                                                                                                              $begingroup$
                                                                                                                                              Welcome to PPCG :)
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              10 hours ago




                                                                                                                                              $begingroup$
                                                                                                                                              Welcome to PPCG :)
                                                                                                                                              $endgroup$
                                                                                                                                              – Shaggy
                                                                                                                                              10 hours ago











                                                                                                                                              1












                                                                                                                                              $begingroup$


                                                                                                                                              C# (Visual C# Interactive Compiler), 34 bytes





                                                                                                                                              int f(int g)=>g<3?1:f(g-2)+f(g-3);


                                                                                                                                              Try it online!






                                                                                                                                              share|improve this answer









                                                                                                                                              $endgroup$


















                                                                                                                                                1












                                                                                                                                                $begingroup$


                                                                                                                                                C# (Visual C# Interactive Compiler), 34 bytes





                                                                                                                                                int f(int g)=>g<3?1:f(g-2)+f(g-3);


                                                                                                                                                Try it online!






                                                                                                                                                share|improve this answer









                                                                                                                                                $endgroup$
















                                                                                                                                                  1












                                                                                                                                                  1








                                                                                                                                                  1





                                                                                                                                                  $begingroup$


                                                                                                                                                  C# (Visual C# Interactive Compiler), 34 bytes





                                                                                                                                                  int f(int g)=>g<3?1:f(g-2)+f(g-3);


                                                                                                                                                  Try it online!






                                                                                                                                                  share|improve this answer









                                                                                                                                                  $endgroup$




                                                                                                                                                  C# (Visual C# Interactive Compiler), 34 bytes





                                                                                                                                                  int f(int g)=>g<3?1:f(g-2)+f(g-3);


                                                                                                                                                  Try it online!







                                                                                                                                                  share|improve this answer












                                                                                                                                                  share|improve this answer



                                                                                                                                                  share|improve this answer










                                                                                                                                                  answered 2 days ago









                                                                                                                                                  Embodiment of IgnoranceEmbodiment of Ignorance

                                                                                                                                                  2,886127




                                                                                                                                                  2,886127























                                                                                                                                                      1












                                                                                                                                                      $begingroup$

                                                                                                                                                      TI-BASIC (TI-84), 34 bytes



                                                                                                                                                      [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1


                                                                                                                                                      0-indexed $N$th term of the sequence.



                                                                                                                                                      Input is in Ans.

                                                                                                                                                      Output is in Ans and is automatically printed out.



                                                                                                                                                      I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.



                                                                                                                                                      Example:



                                                                                                                                                      0
                                                                                                                                                      0
                                                                                                                                                      prgmCDGFD
                                                                                                                                                      1
                                                                                                                                                      9
                                                                                                                                                      9
                                                                                                                                                      prgmCDGFD
                                                                                                                                                      9
                                                                                                                                                      16
                                                                                                                                                      16
                                                                                                                                                      prgmCDGFD
                                                                                                                                                      65


                                                                                                                                                      Explanation:



                                                                                                                                                      [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1      ;full program (example input: 6)

                                                                                                                                                      [[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
                                                                                                                                                      ; [0 1 0]
                                                                                                                                                      ; [0 0 1]
                                                                                                                                                      ; [1 1 0]
                                                                                                                                                      ^(Ans+5 ;then raise it to the power of: input + 5
                                                                                                                                                      ; [4 7 5]
                                                                                                                                                      ; [5 9 7]
                                                                                                                                                      ; [7 12 9]
                                                                                                                                                      Ans(1,1 ;get the top-left index and leave it in "Ans"
                                                                                                                                                      ;implicitly print Ans





                                                                                                                                                      share|improve this answer









                                                                                                                                                      $endgroup$


















                                                                                                                                                        1












                                                                                                                                                        $begingroup$

                                                                                                                                                        TI-BASIC (TI-84), 34 bytes



                                                                                                                                                        [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1


                                                                                                                                                        0-indexed $N$th term of the sequence.



                                                                                                                                                        Input is in Ans.

                                                                                                                                                        Output is in Ans and is automatically printed out.



                                                                                                                                                        I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.



                                                                                                                                                        Example:



                                                                                                                                                        0
                                                                                                                                                        0
                                                                                                                                                        prgmCDGFD
                                                                                                                                                        1
                                                                                                                                                        9
                                                                                                                                                        9
                                                                                                                                                        prgmCDGFD
                                                                                                                                                        9
                                                                                                                                                        16
                                                                                                                                                        16
                                                                                                                                                        prgmCDGFD
                                                                                                                                                        65


                                                                                                                                                        Explanation:



                                                                                                                                                        [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1      ;full program (example input: 6)

                                                                                                                                                        [[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
                                                                                                                                                        ; [0 1 0]
                                                                                                                                                        ; [0 0 1]
                                                                                                                                                        ; [1 1 0]
                                                                                                                                                        ^(Ans+5 ;then raise it to the power of: input + 5
                                                                                                                                                        ; [4 7 5]
                                                                                                                                                        ; [5 9 7]
                                                                                                                                                        ; [7 12 9]
                                                                                                                                                        Ans(1,1 ;get the top-left index and leave it in "Ans"
                                                                                                                                                        ;implicitly print Ans





                                                                                                                                                        share|improve this answer









                                                                                                                                                        $endgroup$
















                                                                                                                                                          1












                                                                                                                                                          1








                                                                                                                                                          1





                                                                                                                                                          $begingroup$

                                                                                                                                                          TI-BASIC (TI-84), 34 bytes



                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1


                                                                                                                                                          0-indexed $N$th term of the sequence.



                                                                                                                                                          Input is in Ans.

                                                                                                                                                          Output is in Ans and is automatically printed out.



                                                                                                                                                          I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.



                                                                                                                                                          Example:



                                                                                                                                                          0
                                                                                                                                                          0
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          1
                                                                                                                                                          9
                                                                                                                                                          9
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          9
                                                                                                                                                          16
                                                                                                                                                          16
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          65


                                                                                                                                                          Explanation:



                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1      ;full program (example input: 6)

                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
                                                                                                                                                          ; [0 1 0]
                                                                                                                                                          ; [0 0 1]
                                                                                                                                                          ; [1 1 0]
                                                                                                                                                          ^(Ans+5 ;then raise it to the power of: input + 5
                                                                                                                                                          ; [4 7 5]
                                                                                                                                                          ; [5 9 7]
                                                                                                                                                          ; [7 12 9]
                                                                                                                                                          Ans(1,1 ;get the top-left index and leave it in "Ans"
                                                                                                                                                          ;implicitly print Ans





                                                                                                                                                          share|improve this answer









                                                                                                                                                          $endgroup$



                                                                                                                                                          TI-BASIC (TI-84), 34 bytes



                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1


                                                                                                                                                          0-indexed $N$th term of the sequence.



                                                                                                                                                          Input is in Ans.

                                                                                                                                                          Output is in Ans and is automatically printed out.



                                                                                                                                                          I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.



                                                                                                                                                          Example:



                                                                                                                                                          0
                                                                                                                                                          0
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          1
                                                                                                                                                          9
                                                                                                                                                          9
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          9
                                                                                                                                                          16
                                                                                                                                                          16
                                                                                                                                                          prgmCDGFD
                                                                                                                                                          65


                                                                                                                                                          Explanation:



                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1      ;full program (example input: 6)

                                                                                                                                                          [[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
                                                                                                                                                          ; [0 1 0]
                                                                                                                                                          ; [0 0 1]
                                                                                                                                                          ; [1 1 0]
                                                                                                                                                          ^(Ans+5 ;then raise it to the power of: input + 5
                                                                                                                                                          ; [4 7 5]
                                                                                                                                                          ; [5 9 7]
                                                                                                                                                          ; [7 12 9]
                                                                                                                                                          Ans(1,1 ;get the top-left index and leave it in "Ans"
                                                                                                                                                          ;implicitly print Ans






                                                                                                                                                          share|improve this answer












                                                                                                                                                          share|improve this answer



                                                                                                                                                          share|improve this answer










                                                                                                                                                          answered 2 days ago









                                                                                                                                                          TauTau

                                                                                                                                                          946515




                                                                                                                                                          946515























                                                                                                                                                              1












                                                                                                                                                              $begingroup$


                                                                                                                                                              Perl 5, 34 bytes





                                                                                                                                                              sub f{"@_"<3||f("@_"-2)+f("@_"-3)}


                                                                                                                                                              Try it online!






                                                                                                                                                              share|improve this answer









                                                                                                                                                              $endgroup$


















                                                                                                                                                                1












                                                                                                                                                                $begingroup$


                                                                                                                                                                Perl 5, 34 bytes





                                                                                                                                                                sub f{"@_"<3||f("@_"-2)+f("@_"-3)}


                                                                                                                                                                Try it online!






                                                                                                                                                                share|improve this answer









                                                                                                                                                                $endgroup$
















                                                                                                                                                                  1












                                                                                                                                                                  1








                                                                                                                                                                  1





                                                                                                                                                                  $begingroup$


                                                                                                                                                                  Perl 5, 34 bytes





                                                                                                                                                                  sub f{"@_"<3||f("@_"-2)+f("@_"-3)}


                                                                                                                                                                  Try it online!






                                                                                                                                                                  share|improve this answer









                                                                                                                                                                  $endgroup$




                                                                                                                                                                  Perl 5, 34 bytes





                                                                                                                                                                  sub f{"@_"<3||f("@_"-2)+f("@_"-3)}


                                                                                                                                                                  Try it online!







                                                                                                                                                                  share|improve this answer












                                                                                                                                                                  share|improve this answer



                                                                                                                                                                  share|improve this answer










                                                                                                                                                                  answered yesterday









                                                                                                                                                                  XcaliXcali

                                                                                                                                                                  5,475520




                                                                                                                                                                  5,475520























                                                                                                                                                                      1












                                                                                                                                                                      $begingroup$


                                                                                                                                                                      Wolfram Language (Mathematica), 26 bytes



                                                                                                                                                                      If[#<3,1,#0[#-2]+#0[#-3]]&


                                                                                                                                                                      Try it online!






                                                                                                                                                                      share|improve this answer









                                                                                                                                                                      $endgroup$


















                                                                                                                                                                        1












                                                                                                                                                                        $begingroup$


                                                                                                                                                                        Wolfram Language (Mathematica), 26 bytes



                                                                                                                                                                        If[#<3,1,#0[#-2]+#0[#-3]]&


                                                                                                                                                                        Try it online!






                                                                                                                                                                        share|improve this answer









                                                                                                                                                                        $endgroup$
















                                                                                                                                                                          1












                                                                                                                                                                          1








                                                                                                                                                                          1





                                                                                                                                                                          $begingroup$


                                                                                                                                                                          Wolfram Language (Mathematica), 26 bytes



                                                                                                                                                                          If[#<3,1,#0[#-2]+#0[#-3]]&


                                                                                                                                                                          Try it online!






                                                                                                                                                                          share|improve this answer









                                                                                                                                                                          $endgroup$




                                                                                                                                                                          Wolfram Language (Mathematica), 26 bytes



                                                                                                                                                                          If[#<3,1,#0[#-2]+#0[#-3]]&


                                                                                                                                                                          Try it online!







                                                                                                                                                                          share|improve this answer












                                                                                                                                                                          share|improve this answer



                                                                                                                                                                          share|improve this answer










                                                                                                                                                                          answered 20 hours ago









                                                                                                                                                                          attinatattinat

                                                                                                                                                                          4897




                                                                                                                                                                          4897























                                                                                                                                                                              1












                                                                                                                                                                              $begingroup$


                                                                                                                                                                              Pari/GP, 28 bytes



                                                                                                                                                                              0-indexed.



                                                                                                                                                                              f(n)=if(n<3,1,f(n-2)+f(n-3))


                                                                                                                                                                              Try it online!






                                                                                                                                                                              Pari/GP, 35 bytes



                                                                                                                                                                              1-indexed.



                                                                                                                                                                              n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]


                                                                                                                                                                              Try it online!



                                                                                                                                                                              The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.






                                                                                                                                                                              share|improve this answer











                                                                                                                                                                              $endgroup$


















                                                                                                                                                                                1












                                                                                                                                                                                $begingroup$


                                                                                                                                                                                Pari/GP, 28 bytes



                                                                                                                                                                                0-indexed.



                                                                                                                                                                                f(n)=if(n<3,1,f(n-2)+f(n-3))


                                                                                                                                                                                Try it online!






                                                                                                                                                                                Pari/GP, 35 bytes



                                                                                                                                                                                1-indexed.



                                                                                                                                                                                n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]


                                                                                                                                                                                Try it online!



                                                                                                                                                                                The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.






                                                                                                                                                                                share|improve this answer











                                                                                                                                                                                $endgroup$
















                                                                                                                                                                                  1












                                                                                                                                                                                  1








                                                                                                                                                                                  1





                                                                                                                                                                                  $begingroup$


                                                                                                                                                                                  Pari/GP, 28 bytes



                                                                                                                                                                                  0-indexed.



                                                                                                                                                                                  f(n)=if(n<3,1,f(n-2)+f(n-3))


                                                                                                                                                                                  Try it online!






                                                                                                                                                                                  Pari/GP, 35 bytes



                                                                                                                                                                                  1-indexed.



                                                                                                                                                                                  n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]


                                                                                                                                                                                  Try it online!



                                                                                                                                                                                  The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.






                                                                                                                                                                                  share|improve this answer











                                                                                                                                                                                  $endgroup$




                                                                                                                                                                                  Pari/GP, 28 bytes



                                                                                                                                                                                  0-indexed.



                                                                                                                                                                                  f(n)=if(n<3,1,f(n-2)+f(n-3))


                                                                                                                                                                                  Try it online!






                                                                                                                                                                                  Pari/GP, 35 bytes



                                                                                                                                                                                  1-indexed.



                                                                                                                                                                                  n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]


                                                                                                                                                                                  Try it online!



                                                                                                                                                                                  The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.







                                                                                                                                                                                  share|improve this answer














                                                                                                                                                                                  share|improve this answer



                                                                                                                                                                                  share|improve this answer








                                                                                                                                                                                  edited 19 hours ago

























                                                                                                                                                                                  answered 20 hours ago









                                                                                                                                                                                  alephalphaalephalpha

                                                                                                                                                                                  21.8k33094




                                                                                                                                                                                  21.8k33094






















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