44
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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 \$N\ge0\$

  • 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:

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  • 2
    \$\begingroup\$ 14 (0-indexed) is shown as outputting 28 while I believe it should yield 37 \$\endgroup\$ – Jonathan Allan Apr 7 at 20:53
  • \$\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 Apr 7 at 20:55
  • \$\begingroup\$ @LuisMendo I believe so. I'll edit the post. \$\endgroup\$ – Tau Apr 8 at 13:06
  • 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 Apr 9 at 2:12
  • 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 Apr 9 at 2:13

41 Answers 41

59
\$\begingroup\$

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

Try it online!

1-indexed. Computes the largest element of: $$\begin{bmatrix}0&0&1 \\ 1&0&1 \\ 0&1&0\end{bmatrix}^n$$ 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
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  • 33
    \$\begingroup\$ this is clearly some kind of voodoo \$\endgroup\$ – Pureferret Apr 8 at 10:51
  • 7
    \$\begingroup\$ This should be published. \$\endgroup\$ – YSC Apr 8 at 11:31
  • 6
    \$\begingroup\$ @YSC It has already been published in A000931. I'd never have guess the primes trick:) \$\endgroup\$ – flawr Apr 8 at 15:28
  • 1
    \$\begingroup\$ ...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter) \$\endgroup\$ – Jonathan Allan Apr 8 at 16:39
  • 1
    \$\begingroup\$ I'm so used to seeing absurdly small answers here, that I thought the comma after 'Jelly' was in fact the code for this problem \$\endgroup\$ – Tasos Papastylianou Apr 9 at 15:33
28
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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)
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26
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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}^{\lfloor\frac{n}2\rfloor}\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)
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18
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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!

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11
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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.

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7
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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!

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6
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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].

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6
\$\begingroup\$

J, 22 bytes

-2 bytes thanks to ngn and Galen

closed form, 26 bytes

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

Try it online!

iterative, 22 bytes

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

Try it online!

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  • 1
    \$\begingroup\$ Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online! \$\endgroup\$ – Galen Ivanov Apr 8 at 19:29
  • \$\begingroup\$ 23 bytes thanks to ngn 1: -> # : Try it online! \$\endgroup\$ – Galen Ivanov Apr 8 at 20:06
  • \$\begingroup\$ @GalenIvanov tyvm, that's a great trick. \$\endgroup\$ – Jonah Apr 8 at 20:56
  • 2
    \$\begingroup\$ 1: -> 1. "adverse" works with a noun on the right, apparently \$\endgroup\$ – ngn Apr 10 at 15:24
  • \$\begingroup\$ @ngn TIL... ty again! \$\endgroup\$ – Jonah Apr 10 at 15:45
5
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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.

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5
\$\begingroup\$

Cubix, 20 bytes

This is 0 indexed and outputs the Nth term

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

Try it online!

Wraps onto a cube with side length 2

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

Watch it run

  • I010 - Initiates the stack
  • +p? - Adds the top of stack, pulls the counter from the bottom of stack and tests
  • /;UO@ - If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt
  • \(sqq;W - If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.
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4
\$\begingroup\$

Python 2, 56 48 bytes

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

Try it online!

Returns nth value, 0-indexed.

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4
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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.

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  • \$\begingroup\$ Why is this community wiki? \$\endgroup\$ – Jo King Apr 7 at 23:14
  • \$\begingroup\$ @JoKing Beats me. If I did it somehow, it wasn't on purpose. \$\endgroup\$ – Sean Apr 8 at 16:58
4
\$\begingroup\$

05AB1E, 8 bytes

1Ð)λ£₂₃+

Try it online!

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

How?

1Ð)λ£₂₃+ | Full program.
1Ð)      | Initialize the stack with [1, 1, 1].
   λ     | Begin the recursive generation of a list: Starting from some base case,
         | this command generates an infinite list with the pattern function given.
    £    | Flag for λ. Instead of outputting an infinite stream, only print the first n.
     ₂₃+ | Add a(n-2) and a(n-3).
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4
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APL (Dyalog Unicode), 20 18 17 bytesSBCS

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

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

4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

Try it online!

Explanation

4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3

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

K (ngn/k), 24 20 bytes

-4 bytes thanks to ngn!

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

Try it online!

0-indexed, first N terms

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  • 1
    \$\begingroup\$ f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur") \$\endgroup\$ – ngn Apr 8 at 18:04
  • \$\begingroup\$ @ngn Thanks! I tried it (without success) in J; it's elegant here. \$\endgroup\$ – Galen Ivanov Apr 8 at 18:54
  • \$\begingroup\$ @ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>) \$\endgroup\$ – Galen Ivanov Apr 8 at 19:32
  • \$\begingroup\$ ah, right, base-1 decode is a train-friendly way to sum :) \$\endgroup\$ – ngn Apr 8 at 19:35
  • 2
    \$\begingroup\$ 1: -> # in the j solution \$\endgroup\$ – ngn Apr 8 at 19:44
4
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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 :)

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  • 2
    \$\begingroup\$ I love to see machine code answers on PP&CG! +1 \$\endgroup\$ – Tau Apr 9 at 2:14
3
\$\begingroup\$

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

Try it online!

0-indexed.

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3
\$\begingroup\$

Lua 5.3, 49 48 bytes

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

Try it online!

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

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3
\$\begingroup\$

Ruby, 26 bytes

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

Try it online!

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3
\$\begingroup\$

JavaScript (ES6), 23 bytes

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

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

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

Try it online!

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  • \$\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 Apr 8 at 12:27
  • 2
    \$\begingroup\$ @Nit Relevant meta post. \$\endgroup\$ – Arnauld Apr 8 at 12:30
  • \$\begingroup\$ I think you are missing some requirements: Have a look at my modification (version in Java) here. \$\endgroup\$ – Shaq Apr 9 at 9:43
  • \$\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 Apr 9 at 9:49
  • \$\begingroup\$ @Arnauld yep I can see it now. Sorry! \$\endgroup\$ – Shaq Apr 9 at 11:22
2
\$\begingroup\$

Japt -N, 12 bytes

<3ªßUµ2 +ß´U

Try it

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  • \$\begingroup\$ Looks like 12 is the best we can do :\ \$\endgroup\$ – Shaggy Apr 7 at 22:26
  • \$\begingroup\$ I stand corrected! \$\endgroup\$ – Shaggy Apr 9 at 10:49
2
\$\begingroup\$

TI-BASIC (TI-84), 34 bytes

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

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

Input is in Ans.
Output is in Ans and is automatically printed out.

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

Example:

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

Explanation:

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

[[0,1,0][0,0,1][1,1,0]]                     ;generate the following matrix:
                                            ; [0 1 0]
                                            ; [0 0 1]
                                            ; [1 1 0]
                       ^(Ans+5              ;then raise it to the power of: input + 5
                                            ; [4  7 5]
                                            ; [5  9 7]
                                            ; [7 12 9]
                               Ans(1,1      ;get the top-left index and leave it in "Ans"
                                            ;implicitly print Ans
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2
\$\begingroup\$

Pyth, 16 bytes

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

This defines the function y. Try it here!

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

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

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

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  • \$\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 Apr 8 at 7:42
  • \$\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. Apr 8 at 15:44
  • \$\begingroup\$ Well done on hB2. Too bad it's the same byte count. \$\endgroup\$ – Ven Apr 8 at 15:44
  • \$\begingroup\$ Yeah, though I wonder if there's some way to get rid of the d in the map. \$\endgroup\$ – RK. Apr 8 at 15:52
2
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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

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  • \$\begingroup\$ I think you are missing some requirements: Have a look at my modification here. \$\endgroup\$ – Shaq Apr 9 at 9:43
  • \$\begingroup\$ Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12). \$\endgroup\$ – Olivier Grégoire Apr 9 at 13:59
  • \$\begingroup\$ Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer. \$\endgroup\$ – Benjamin Urquhart Apr 9 at 18:34
2
\$\begingroup\$

R + pryr, 38 36 bytes

Zero-indexed recursive function.

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

Try it online!

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

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  • 2
    \$\begingroup\$ If you're going to be using pryr, the language should be R + pryr and this can be 36 bytes \$\endgroup\$ – Giuseppe Apr 8 at 19:34
  • \$\begingroup\$ @Giuseppe thanks! Updated now. \$\endgroup\$ – rturnbull Apr 8 at 20:35
2
\$\begingroup\$

C (clang), 41 33 bytes

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

Try it online!

\$\endgroup\$
  • 2
    \$\begingroup\$ Welcome to PPCG :) \$\endgroup\$ – Shaggy Apr 9 at 15:17
1
\$\begingroup\$

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

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

Try it online!

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1
\$\begingroup\$

Perl 5, 34 bytes

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

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Wolfram Language (Mathematica), 26 bytes

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

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Pari/GP, 28 bytes

0-indexed.

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

Try it online!


Pari/GP, 35 bytes

1-indexed.

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

Try it online!

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

\$\endgroup\$

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