Patience, young “Padovan”

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.

edited yesterday

asked 2 days ago

Tau

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

|
show 3 more comments

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.

edited yesterday

asked 2 days ago

Tau

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

|
show 3 more comments

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.

edited yesterday

asked 2 days ago

Tau

946515

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

edited yesterday

asked 2 days ago

Tau

946515

edited yesterday

asked 2 days ago

Tau

946515

edited yesterday

asked 2 days ago

Tau

946515

asked 2 days ago

Tau

946515

asked 2 days ago

Tau

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

|
show 3 more comments

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

14 (0-indexed) is shown as outputting 28 while I believe it should yield 37

– Jonathan Allan
2 days ago

@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.

– Tau
2 days ago

@LuisMendo I believe so. I'll edit the post.

– Tau
yesterday

@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.

– Tau
23 hours ago

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

– Carmeister
23 hours ago

|
show 3 more comments

38 Answers
38

active

oldest

votes

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

answered 2 days ago

Lynn

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

|
show 3 more comments

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)

answered 2 days ago

Emigna

47.8k433145

add a comment |

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)

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

add a comment |

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!

answered 2 days ago

xnor

93.9k18192450

add a comment |

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.

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

add a comment |

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!

answered 2 days ago

J42161217

13.9k21353

add a comment |

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].

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

add a comment |

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.

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

add a comment |

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.

answered 2 days ago

MickyT

10.3k21637

add a comment |

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!

edited yesterday

answered 2 days ago

Jonah

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

add a comment |

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.

answered 2 days ago

Chas Brown

5,2191523

add a comment |

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

add a comment |

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.

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

add a comment |

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.

edited yesterday

answered 2 days ago

cyclaminist

1813

add a comment |

Ruby, 26 bytes

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

Try it online!

answered yesterday

G B

8,2561429

add a comment |

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

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.

edited yesterday

answered yesterday

Sherlock9

8,16411860

add a comment |

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

edited yesterday

answered yesterday

Galen Ivanov

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

|
show 2 more comments

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 :)

answered yesterday

anatolyg

7,2592166

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

add a comment |

Japt `-N`, 12 bytes

<3ªßUµ2 +ß´U

Try it

answered 2 days ago

Embodiment of Ignorance

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

add a comment |

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.

answered 2 days ago

RK.

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

add a comment |

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).

edited yesterday

answered yesterday

Mr. Xcoder

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

add a comment |

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

answered yesterday

Benjamin Urquhart

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

add a comment |

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.

edited yesterday

answered yesterday

rturnbull

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

add a comment |

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!

edited 15 hours ago

answered 2 days ago

Arnauld

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

add a comment |

C (clang), 41 bytes

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

Try it online!

answered 10 hours ago

peterzuger

213

New contributor

1

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

add a comment |

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

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

Try it online!

answered 2 days ago

Embodiment of Ignorance

2,886127

add a comment |

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:

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

answered 2 days ago

Tau

946515

add a comment |

Perl 5, 34 bytes

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

Try it online!

answered yesterday

Xcali

5,475520

add a comment |

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

answered 20 hours ago

attinat

4897

add a comment |

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 $frac{1+x}{1-x^2-x^3}$.

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

add a comment |

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38 Answers
38

active

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38 Answers
38

active

oldest

votes

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

Try it online!

(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

answered 2 days ago

Lynn

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

|
show 3 more comments

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

Try it online!

(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

answered 2 days ago

Lynn

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

|
show 3 more comments

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

Try it online!

(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

answered 2 days ago

Lynn

50.7k898233

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

Try it online!

(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

answered 2 days ago

Lynn

50.7k898233

answered 2 days ago

Lynn

50.7k898233

answered 2 days ago

Lynn

50.7k898233

answered 2 days ago

Lynn

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

|
show 3 more comments

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

this is clearly some kind of voodoo

– Pureferret
yesterday

This should be published.

– YSC
yesterday

@YSC It has already been published in A000931. I'd never have guess the primes trick:)

– flawr
yesterday

@flawr ho... I missed it

– YSC
yesterday

...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)

– Jonathan Allan
yesterday

|
show 3 more comments

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)

answered 2 days ago

Emigna

47.8k433145

add a comment |

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)

answered 2 days ago

Emigna

47.8k433145

add a comment |

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)

answered 2 days ago

Emigna

47.8k433145

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)

answered 2 days ago

Emigna

47.8k433145

answered 2 days ago

Emigna

47.8k433145

answered 2 days ago

Emigna

47.8k433145

answered 2 days ago

Emigna

47.8k433145

add a comment |

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)

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

add a comment |

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)

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

add a comment |

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)

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

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)

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

edited yesterday

answered 2 days ago

Jonathan Allan

54k536174

answered 2 days ago

Jonathan Allan

54k536174

answered 2 days ago

Jonathan Allan

54k536174

add a comment |

Haskell, 26 bytes

(l!!)

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

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

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!

answered 2 days ago

xnor

93.9k18192450

add a comment |

Haskell, 26 bytes

(l!!)

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

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

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!

answered 2 days ago

xnor

93.9k18192450

add a comment |

Haskell, 26 bytes

(l!!)

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

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

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!

answered 2 days ago

xnor

93.9k18192450

Haskell, 26 bytes

(l!!)

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

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

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!

answered 2 days ago

xnor

93.9k18192450

answered 2 days ago

xnor

93.9k18192450

answered 2 days ago

xnor

93.9k18192450

answered 2 days ago

xnor

93.9k18192450

add a comment |

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.

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

add a comment |

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.

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

add a comment |

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.

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

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.

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

edited 2 days ago

answered 2 days ago

xnor

93.9k18192450

answered 2 days ago

xnor

93.9k18192450

answered 2 days ago

xnor

93.9k18192450

add a comment |

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!

answered 2 days ago

J42161217

13.9k21353

add a comment |

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!

answered 2 days ago

J42161217

13.9k21353

add a comment |

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!

answered 2 days ago

J42161217

13.9k21353

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!

answered 2 days ago

J42161217

13.9k21353

answered 2 days ago

J42161217

13.9k21353

answered 2 days ago

J42161217

13.9k21353

answered 2 days ago

J42161217

13.9k21353

add a comment |

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).

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

add a comment |

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).

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

add a comment |

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).

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

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).

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

edited yesterday

answered yesterday

Luis Mendo

75.3k889292

answered yesterday

Luis Mendo

75.3k889292

answered yesterday

Luis Mendo

75.3k889292

add a comment |

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*

6,G`

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

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

add a comment |

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*

6,G`

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

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

add a comment |

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*

6,G`

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

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

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*

6,G`

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

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

edited 2 days ago

answered 2 days ago

Neil

82.6k745179

answered 2 days ago

Neil

82.6k745179

answered 2 days ago

Neil

82.6k745179

add a comment |

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.

answered 2 days ago

MickyT

10.3k21637

add a comment |

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.

answered 2 days ago

MickyT

10.3k21637

add a comment |

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.

answered 2 days ago

MickyT

10.3k21637

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.

answered 2 days ago

MickyT

10.3k21637

answered 2 days ago

MickyT

10.3k21637

answered 2 days ago

MickyT

10.3k21637

answered 2 days ago

MickyT

10.3k21637

add a comment |

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!

edited yesterday

answered 2 days ago

Jonah

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

add a comment |

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!

edited yesterday

answered 2 days ago

Jonah

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

add a comment |

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!

edited yesterday

answered 2 days ago

Jonah

2,6611017

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!

edited yesterday

answered 2 days ago

Jonah

2,6611017

edited yesterday

answered 2 days ago

Jonah

2,6611017

answered 2 days ago

Jonah

2,6611017

answered 2 days ago

Jonah

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

add a comment |

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

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

– Galen Ivanov
yesterday

23 bytes thanks to ngn 1: -> # : Try it online!

– Galen Ivanov
yesterday

@GalenIvanov tyvm, that's a great trick.

– Jonah
yesterday

add a comment |

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.

answered 2 days ago

Chas Brown

5,2191523

add a comment |

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.

answered 2 days ago

Chas Brown

5,2191523

add a comment |

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.

answered 2 days ago

Chas Brown

5,2191523

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.

answered 2 days ago

Chas Brown

5,2191523

answered 2 days ago

Chas Brown

5,2191523

answered 2 days ago

Chas Brown

5,2191523

answered 2 days ago

Chas Brown

5,2191523

add a comment |

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

add a comment |

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

add a comment |

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

edited 2 days ago

answered 2 days ago

Erik the Outgolfer

33k429106

answered 2 days ago

Erik the Outgolfer

33k429106

answered 2 days ago

Erik the Outgolfer

33k429106

add a comment |

Perl 6, 24 bytes

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

Try it online!

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

add a comment |

Perl 6, 24 bytes

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

Try it online!

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

add a comment |

Perl 6, 24 bytes

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

Try it online!

answered 2 days ago

community wiki

Sean

Perl 6, 24 bytes

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

Try it online!

answered 2 days ago

community wiki

Sean

answered 2 days ago

community wiki

Sean

community wiki

Sean

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

add a comment |

$begingroup$
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

Why is this community wiki?

– Jo King
2 days ago

@JoKing Beats me. If I did it somehow, it wasn't on purpose.

– Sean
yesterday

add a comment |

Lua 5.3, 49 48 bytes

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

Try it online!

edited yesterday

answered 2 days ago

cyclaminist

1813

add a comment |

Lua 5.3, 49 48 bytes

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

Try it online!

edited yesterday

answered 2 days ago

cyclaminist

1813

add a comment |

Lua 5.3, 49 48 bytes

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

Try it online!

edited yesterday

answered 2 days ago

cyclaminist

1813

Lua 5.3, 49 48 bytes

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

Try it online!

edited yesterday

answered 2 days ago

cyclaminist

1813

edited yesterday

answered 2 days ago

cyclaminist

1813

answered 2 days ago

cyclaminist

1813

answered 2 days ago

cyclaminist

1813

add a comment |

Ruby, 26 bytes

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

Try it online!

answered yesterday

G B

8,2561429

add a comment |

Ruby, 26 bytes

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

Try it online!

answered yesterday

G B

8,2561429

add a comment |

Ruby, 26 bytes

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

Try it online!

answered yesterday

G B

8,2561429

Ruby, 26 bytes

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

Try it online!

answered yesterday

G B

8,2561429

answered yesterday

G B

8,2561429

answered yesterday

G B

8,2561429

answered yesterday

G B

8,2561429

add a comment |

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

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.

edited yesterday

answered yesterday

Sherlock9

8,16411860

add a comment |

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

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.

edited yesterday

answered yesterday

Sherlock9

8,16411860

add a comment |

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

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.

edited yesterday

answered yesterday

Sherlock9

8,16411860

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

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.

edited yesterday

answered yesterday

Sherlock9

8,16411860

edited yesterday

answered yesterday

Sherlock9

8,16411860

answered yesterday

Sherlock9

8,16411860

answered yesterday

Sherlock9

8,16411860

add a comment |

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

edited yesterday

answered yesterday

Galen Ivanov

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

|
show 2 more comments

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

edited yesterday

answered yesterday

Galen Ivanov

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

|
show 2 more comments

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

edited yesterday

answered yesterday

Galen Ivanov

7,41211034

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

edited yesterday

answered yesterday

Galen Ivanov

7,41211034

edited yesterday

answered yesterday

Galen Ivanov

7,41211034

answered yesterday

Galen Ivanov

7,41211034

answered yesterday

Galen Ivanov

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

|
show 2 more comments

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

f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")

– ngn
yesterday

@ngn Thanks! I tried it (without success) in J; it's elegant here.

– Galen Ivanov
yesterday

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

– Galen Ivanov
yesterday

ah, right, base-1 decode is a train-friendly way to sum :)

– ngn
yesterday

1: -> # in the j solution

– ngn
yesterday

|
show 2 more comments

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 :)

answered yesterday

anatolyg

7,2592166

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

add a comment |

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 :)

answered yesterday

anatolyg

7,2592166

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

add a comment |

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 :)

answered yesterday

anatolyg

7,2592166

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 :)

answered yesterday

anatolyg

7,2592166

answered yesterday

anatolyg

7,2592166

answered yesterday

anatolyg

7,2592166

answered yesterday

anatolyg

7,2592166

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

add a comment |

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

I love to see machine code answers on PP&CG! +1

– Tau
23 hours ago

add a comment |

Japt `-N`, 12 bytes

<3ªßUµ2 +ß´U

Try it

answered 2 days ago

Embodiment of Ignorance

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

add a comment |

Japt `-N`, 12 bytes

<3ªßUµ2 +ß´U

Try it

answered 2 days ago

Embodiment of Ignorance

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

add a comment |

Japt `-N`, 12 bytes

<3ªßUµ2 +ß´U

Try it

answered 2 days ago

Embodiment of Ignorance

2,886127

Japt `-N`, 12 bytes

<3ªßUµ2 +ß´U

Try it

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

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

add a comment |

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

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

Looks like 12 is the best we can do :

– Shaggy
2 days ago

I stand corrected!

– Shaggy
14 hours ago

add a comment |

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!

answered 2 days ago

RK.

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

add a comment |

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!

answered 2 days ago

RK.

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

add a comment |

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!

answered 2 days ago

RK.

407211

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!

answered 2 days ago

RK.

407211

answered 2 days ago

RK.

407211

answered 2 days ago

RK.

407211

answered 2 days ago

RK.

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

add a comment |

$begingroup$
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

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 :(

– Ven
yesterday

@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]

– RK.
yesterday

Well done on hB2. Too bad it's the same byte count.

– Ven
yesterday

Yeah, though I wonder if there's some way to get rid of the d in the map.

– RK.
yesterday

add a comment |

05AB1E, 8 bytes

1Ð)λ£₂₃+

Try it online!

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).

edited yesterday

answered yesterday

Mr. Xcoder

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

add a comment |

05AB1E, 8 bytes

1Ð)λ£₂₃+

Try it online!

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).

edited yesterday

answered yesterday

Mr. Xcoder

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

add a comment |

05AB1E, 8 bytes

1Ð)λ£₂₃+

Try it online!

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).

edited yesterday

answered yesterday

Mr. Xcoder

32.2k759200

05AB1E, 8 bytes

1Ð)λ£₂₃+

Try it online!

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).

edited yesterday

answered yesterday

Mr. Xcoder

32.2k759200

edited yesterday

answered yesterday

Mr. Xcoder

32.2k759200

answered yesterday

Mr. Xcoder

32.2k759200

answered yesterday

Mr. Xcoder

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

add a comment |

$begingroup$
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

I don't think 1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.

– Kevin Cruijssen
yesterday

add a comment |

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

answered yesterday

Benjamin Urquhart

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

add a comment |

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

answered yesterday

Benjamin Urquhart

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

add a comment |

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

answered yesterday

Benjamin Urquhart

40017

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

answered yesterday

Benjamin Urquhart

40017

answered yesterday

Benjamin Urquhart

40017

answered yesterday

Benjamin Urquhart

40017

answered yesterday

Benjamin Urquhart

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

add a comment |

$begingroup$
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

I think you are missing some requirements: Have a look at my modification here.

– Shaq
15 hours ago

Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).

– Olivier Grégoire
11 hours ago

Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.

– Benjamin Urquhart
6 hours ago

add a comment |

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.

edited yesterday

answered yesterday

rturnbull

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

add a comment |

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.

edited yesterday

answered yesterday

rturnbull

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

add a comment |

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.

edited yesterday

answered yesterday

rturnbull

3,519925

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.

edited yesterday

answered yesterday

rturnbull

3,519925

edited yesterday

answered yesterday

rturnbull

3,519925

answered yesterday

rturnbull

3,519925

answered yesterday

rturnbull

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

add a comment |

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

If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes

– Giuseppe
yesterday

@Giuseppe thanks! Updated now.

– rturnbull
yesterday

add a comment |

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!

edited 15 hours ago

answered 2 days ago

Arnauld

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

add a comment |

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!

edited 15 hours ago

answered 2 days ago

Arnauld

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

add a comment |

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!

edited 15 hours ago

answered 2 days ago

Arnauld

80.7k797334

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!

edited 15 hours ago

answered 2 days ago

Arnauld

80.7k797334

edited 15 hours ago

answered 2 days ago

Arnauld

80.7k797334

answered 2 days ago

Arnauld

80.7k797334

answered 2 days ago

Arnauld

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

add a comment |

$begingroup$
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

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.

– Nit
yesterday

@Nit Relevant meta post.

– Arnauld
yesterday

I think you are missing some requirements: Have a look at my modification (version in Java) here.

– Shaq
15 hours ago

@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).

– Arnauld
15 hours ago

@Arnauld yep I can see it now. Sorry!

– Shaq
14 hours ago

add a comment |

C (clang), 41 bytes

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

Try it online!

answered 10 hours ago

peterzuger

213

New contributor

1

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

add a comment |

C (clang), 41 bytes

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

Try it online!

answered 10 hours ago

peterzuger

213

New contributor

1

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

add a comment |

C (clang), 41 bytes

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

Try it online!

answered 10 hours ago

peterzuger

213

New contributor

C (clang), 41 bytes

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

Try it online!

answered 10 hours ago

peterzuger

213

New contributor

answered 10 hours ago

peterzuger

213

New contributor

answered 10 hours ago

peterzuger

213

answered 10 hours ago

peterzuger

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.

1

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

add a comment |

1

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

Welcome to PPCG :)

– Shaggy
10 hours ago

add a comment |

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

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

Try it online!

answered 2 days ago

Embodiment of Ignorance

2,886127

add a comment |

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

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

Try it online!

answered 2 days ago

Embodiment of Ignorance

2,886127

add a comment |

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

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

Try it online!

answered 2 days ago

Embodiment of Ignorance

2,886127

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

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

Try it online!

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

2,886127

answered 2 days ago

Embodiment of Ignorance

2,886127

add a comment |

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:

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

answered 2 days ago

Tau

946515

add a comment |

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:

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

answered 2 days ago

Tau

946515

add a comment |

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:

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

answered 2 days ago

Tau

946515

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:

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

answered 2 days ago

Tau

946515

answered 2 days ago

Tau

946515

answered 2 days ago

Tau

946515

answered 2 days ago

Tau

946515

add a comment |

Perl 5, 34 bytes

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

Try it online!

answered yesterday

Xcali

5,475520

add a comment |

Perl 5, 34 bytes

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

Try it online!

answered yesterday

Xcali

5,475520

add a comment |

Perl 5, 34 bytes

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

Try it online!

answered yesterday

Xcali

5,475520

Perl 5, 34 bytes

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

Try it online!

answered yesterday

Xcali

5,475520

answered yesterday

Xcali

5,475520

answered yesterday

Xcali

5,475520

answered yesterday

Xcali

5,475520

add a comment |

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

answered 20 hours ago

attinat

4897

add a comment |

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

answered 20 hours ago

attinat

4897

add a comment |

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

answered 20 hours ago

attinat

4897

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

answered 20 hours ago

attinat

4897

answered 20 hours ago

attinat

4897

answered 20 hours ago

attinat

4897

answered 20 hours ago

attinat

4897

add a comment |

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 $frac{1+x}{1-x^2-x^3}$.

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

add a comment |

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 $frac{1+x}{1-x^2-x^3}$.

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

add a comment |

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 $frac{1+x}{1-x^2-x^3}$.

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

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 $frac{1+x}{1-x^2-x^3}$.

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

edited 19 hours ago

answered 20 hours ago

alephalpha

21.8k33094

answered 20 hours ago

alephalpha

21.8k33094

answered 20 hours ago

alephalpha

21.8k33094

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

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This page is only for reference, If you need detailed information, please check here

Patience, young “Padovan”

38 Answers 38

Jelly, 10 bytes

Oasis, 5 bytes

Jelly, 10 9 8 bytes

How?

Haskell, 26 bytes

Python 2, 30 bytes

Wolfram Language (Mathematica), 33 bytes

Octave / MATLAB, 35 33 bytes

How it works

Retina, 47 42 bytes

Cubix, 20 bytes

J, 23 bytes

closed form, 26 bytes

iterative, 23 bytes

Python 2, 56 48 bytes

Jelly, 11 bytes

Perl 6, 24 bytes

Lua 5.3, 49 48 bytes

Ruby, 26 bytes

APL (Dyalog Unicode), 20 18 17 bytesSBCS

K (ngn/k), 24 20 bytes

x86 32-bit machine code, 17 bytes

Japt -N, 12 bytes

Pyth, 16 bytes

05AB1E, 8 bytes

How?

Java, 41 bytes

R + pryr, 38 36 bytes

JavaScript (ES6), 23 bytes

C (clang), 41 bytes

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

TI-BASIC (TI-84), 34 bytes

Perl 5, 34 bytes

Wolfram Language (Mathematica), 26 bytes

Pari/GP, 28 bytes

Pari/GP, 35 bytes

Your Answer

Sign up or log in

Post as a guest

Post as a guest

38 Answers 38

38 Answers 38

Jelly, 10 bytes

Jelly, 10 bytes

Jelly, 10 bytes

Jelly, 10 bytes

Oasis, 5 bytes

Oasis, 5 bytes

Oasis, 5 bytes

Oasis, 5 bytes

Jelly, 10 9 8 bytes

How?

Jelly, 10 9 8 bytes

How?

Jelly, 10 9 8 bytes

How?

Jelly, 10 9 8 bytes

How?

Haskell, 26 bytes

Haskell, 26 bytes

Haskell, 26 bytes

Haskell, 26 bytes

Python 2, 30 bytes

Python 2, 30 bytes

Python 2, 30 bytes

Python 2, 30 bytes

Wolfram Language (Mathematica), 33 bytes

Wolfram Language (Mathematica), 33 bytes

Wolfram Language (Mathematica), 33 bytes

Wolfram Language (Mathematica), 33 bytes

Octave / MATLAB, 35 33 bytes

How it works

Octave / MATLAB, 35 33 bytes

How it works

Octave / MATLAB, 35 33 bytes

How it works

Octave / MATLAB, 35 33 bytes

How it works

38 Answers
38

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

Japt `-N`, 12 bytes

38 Answers
38

38 Answers
38

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

APL (Dyalog Unicode), 20 18 17 bytes^SBCS

Japt `-N`, 12 bytes

Japt `-N`, 12 bytes

Japt `-N`, 12 bytes

Japt `-N`, 12 bytes