Patience, young "Padovan"

Question

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:

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

• Standard loopholes are forbidden.

Answer 1

Jelly, 10 bytes

9s3’Ẓæ*³FṀ

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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

Answer 2

Oasis, 5 bytes

nth term 0-indexed

cd+1V

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Explanation

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

Answer 3

Jelly,  10 9  8 bytes

ŻṚm2Jc$S

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

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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)

Answer 4

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)

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Answer 5

Python 2, 30 bytes

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

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

Answer 6

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

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Answer 7

J, 22 bytes

-2 bytes thanks to ngn and Galen

closed form, 26 bytes

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

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iterative, 22 bytes

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

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Answer 8

Octave / MATLAB, 35 33 bytes

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

Outputs the first n terms.

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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].

Answer 9

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.

Answer 10

Cubix, 20 bytes

This is 0 indexed and outputs the Nth term

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

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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.

Answer 11

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

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Returns nth value, 0-indexed.

Answer 12

Perl 6, 24 bytes

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

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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.

Answer 13

05AB1E, 8 bytes

1Ð)λ£₂₃+

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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).

Answer 14

K (ngn/k), 24 20 bytes

-4 bytes thanks to ngn!

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

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0-indexed, first N terms

Answer 15

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

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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.

Answer 16

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 :)

Answer 17

Jelly, 11 bytes

5B+Ɲ2ị;Ʋ⁸¡Ḣ

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0-indexed.

Answer 18

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)

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Answer 19

Lua 5.3, 49 48 bytes

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

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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.

Answer 20

Ruby, 26 bytes

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

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Answer 21

Japt -N, 12 bytes

<3ªßUµ2 +ß´U

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Answer 22

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

Answer 23

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.

Answer 24

R + pryr, 38 36 bytes

Zero-indexed recursive function.

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

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

Answer 25

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

Answer 26

C (clang), 41 33 bytes

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

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Answer 27

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

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

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Answer 28

Japt, 12 9 bytes

Returns the nth term, 1-indexed.

@1gZÔä+}g

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@1gZÔä+}g     :Implicit input of integer U
        g     :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@             :  Pass the array through the following function as Z
 1g           :    Get the element at 0-based index 1, with wrapping, from the following
   ZÔ         :    Reverse Z
     ä+       :    Get the sums of each consecutive pair of elements
       }      :  End function
              :Implicit output of the last element in the array

Answer 29

Perl 5, 34 bytes

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

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Answer 30

C++ (gcc), 81 75 74 bytes

-6 bytes to small golfing

-1 byte thanks to @ceilingcat

int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;c=d)d=a+b,a=b,b=c;return c;}

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Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.

Answer 31

Wolfram Language (Mathematica), 26 bytes

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

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Answer 32

Pari/GP, 28 bytes

0-indexed.

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

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---

Pari/GP, 35 bytes

1-indexed.

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

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

Answer 33

05AB1E, 8 bytes

This implementation of the binomial formula: a(n) = Sum_{k=0..floor(n/2)} binomial(k+1, n-2k) is interestingly the same length as the recursive solution.

;Ý·-āscO

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Explanation

;Ý        # push [0 ... floor(input/2)]
  ·       # double each
   -      # subtract each from input
    ā     # push range [1 ... len(list)]
     s    # swap
      c   # choose
       O  # sum

Answer 34

Brain-Flak, 96 bytes

(()()()){({}[()]<{({}[()])<>(())(<>)}{}>)}{}{({}[()]<<>({}<(({}(({}))<>)<>[{}])>)<>({}<>)<>>)}<>

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Answer 35

Clojure, 46 bytes

(defn f[n](if(< n 3)1(+(f(- n 2))(f(- n 3)))))

Just for completeness sake :)

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Answer 36

Gaia, 16 14 12 bytes

7b@⟨ṇ;(++⟩ₓ(

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0-based index. Only holds the last 3 values.

7b		| push [1 1 1]
  @		| push input
   ⟨     ⟩ₓ	| do the following that many times (0 times if 0 or less)
    ṇ		| pop the first element and leave the rest below
     ;		| copy from below
      (		| take the first element
       +	| add the two together
	+	| and concatenate to the list. End loop.
	   (	| finally, take the first element

Gaia, 14 bytes

ø@⟨18b+ₔ…Σ¦⟩ₓ<

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Returns the first n elements, 1-based index. < could also be E to get just the nth element.

Uses the identity from the OEIS page \$a(n+5)=\sum\limits_{i=1}^n a(i)\$.

This is quite inefficient, as it repeatedly copies the first few lists.

ø		| push empty list
 @⟨	   ⟩ₓ	| do (input) number of times:
   18b		| push [1 0 0 1 0] by converting 18 to bits
      +ₔ	| append cumulative list (initially empty) to end of bits
	…Σ¦	| and calculate the cumulative sums, replacing old cumulative list
	     <	| finally, take the first (input) elements of cumulative sums

Answer 37

Groovy, 22 bytes

a->a<3?1:f(a-2)+f(a-1)

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Answer 38

dc, 50 48 bytes

 2+sx3si1d:a[liddddd3-;ar2-;a+r:a1+silx>M]dsMx;ap

It's 1-indexed! Try it online! Previous version!

Golfed off two bytes by only populating the first element in the array.

Probably can be golfed a bit more, but it's tricky because you have to a: seed the initial values, and b: use arrays since dc doesn't let you pick from arbitrary stack positions.

2+sx3si adds two to the user input at top of stack into register x, and starts an increment counter register i at 3. We have to add two because the array is now populated in such a way that it's 3-indexed.

1d:a populates element 1 of array a with a 1. The 50 byte version populated elements 1 and 2 with 1d2:a1:a. The mechanism works with just the single element populated, however, it just takes a little while to get enough 1s in place such as to start the sequence. Even though going from 1d2:a1:a to 1d:a saves 4 bytes, we now have to 2+ our input. Still two bytes saved. Initially, I thought I had to seed the first three elements with 1dd3:a2:a1:a (+4 bytes), but... dc returns a zero for any unassigned array element, and if we start macro M (below) by creating the third element, adding 0 to 1... we're good to go.

Macro M:

liddddd loads register i and makes a bunch of copies of it. We subtract 3 from one of these, get the corresponding element from a and then swap top of stack. Subtract 2, get the corresponding element from a, add the two elements to get our new value, then swap the top of stack. At this point, :a puts our new value into a at position i. 1+si to increment i, and lx>M to keep running M until we hit our target (stored in x). dsMx runs M, and once M has finished running, the stack should be full of leftover i values, the topmost of which is the last one we made. Load the value from a with ;a and print it with p.

Answer 39

C# (.NET Core), 144 bytes, 0-indexed.

using C=System.Console;class A{static void Main(){int i=int.Parse(C.ReadLine());C.Write(p(i));}static int p(int i){return i<3?1:p(i-2)+p(i-3);}}

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Quite frankly, this is not my best work ever. But here it is.

Answer 40

CJam, 16 bytes

qi7Yb{((_j\(j+}j

0-indexed, returns nth item
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Explanation: It uses a recursive function to add a(n-2) to a(n-3) to find a(n).

qi7Yb{((_j\(j+}j
qi               - reads input as integer
  7Yb            - push an array of [1, 1, 1]. It does this by pushing 7, then 
                   turning it into binary.
     {        }j - define a recursive function to calculate a(n), where the 
                   predefined values for n = 0,1,2 are 1,1,1
      ((         - subtract 2 from n | Stack: n-2
        _        - duplicate | Stack: n-2, n-2
         j       - find a(n-2) | Stack: n-2, a(n-2)
          \      - swap the top two items in the stack | Stack: a(n-2), n-2
           (     - decrement | Stack: a(n-2), n-3
            j    - find a(n-3) | Stack: a(n-2), a(n-3)
             +   - add, finding a(n)
                 - implicit output

In pseudocode:

read input as integer
define padovan_sequence(n):
    return padovan_sequence(n-2) + padovan_sequence(n-3)
print(padovan_sequence(n))

Answer 41

Stax, 7 bytes

ÉKΦΘÄO¢

Run and debug it