Patience, young “Padovan”
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(0-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
asked 5 hours ago
TauTau
801313
$endgroup$
add a comment |
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(0-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
asked 5 hours ago
TauTau
801313
$endgroup$
$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37
$endgroup$
– Jonathan Allan
5 hours 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
5 hours ago
add a comment |
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(0-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
asked 5 hours ago
TauTau
801313
$endgroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(0-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
code-golf number sequence
asked 5 hours ago
TauTau
801313
asked 5 hours ago
TauTau
801313
edited 5 hours ago
Tau
asked 5 hours ago
TauTau
801313
asked 5 hours ago
TauTau
801313
asked 5 hours ago
TauTau
801313
801313
$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37
$endgroup$
– Jonathan Allan
5 hours 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
5 hours ago
add a comment |
$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37
$endgroup$
– Jonathan Allan
5 hours 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
5 hours ago
$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
1
1
$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
5 hours 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
5 hours 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
5 hours ago
add a comment |
19 Answers
19
active
oldest
votes
$begingroup$
Haskell, 26 bytes
(l!!)
l=1:1:1:2:scanl(+)2l
Try it online! Outputs the n'th term zero-indexed.
I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.
If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.
27 bytes
f n|n<3=1|1>0=f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
xnorxnor
93.4k18190448
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add a comment |
$begingroup$
Oasis, 5 bytes
nth term 0-indexed
cd+1V
Try it online!
Explanation
1V # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)
answered 5 hours ago
EmignaEmigna
47.4k433144
$endgroup$
add a comment |
$begingroup$
Python 2, 30 bytes
f=lambda n:n<3or f(n-2)+f(n-3)
Try it online!
Returns the n'th term zero indexed. Outputs True for 1.
answered 5 hours ago
xnorxnor
93.4k18190448
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 33 bytes
a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]
1-indexed, returns the nth term
Try it online!
answered 5 hours ago
J42161217J42161217
13.8k21253
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add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
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$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
add a comment |
$begingroup$
Jelly, 10 bytes
‘HRcḤạ¥¥‘S
A monadic Link accepting n (1-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=1}^{lfloorfrac{n+1}2rfloor}binom{2i}{n+1-2i}$
‘HRcḤạ¥¥‘S - Link: integer, n e.g. 21
‘ - increment n 22
H - halve 11
R - range [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
- (note: floors input, e.g. n=6 -> ‘=7 -> H=3.5 -> R=[1,2,3] )
‘ - increment n 21
¥ - last two links as a dyad:
¥ - last two links as a dyad:
Ḥ - double [ 2, 4, 6, 8,10,12,14,16,18,20,22]
ạ - absolute difference [20,18,16,14,12,10, 8, 6, 4, 2, 0]
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0,28,126,45,1]
S - sum 200
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
$endgroup$
add a comment |
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered 2 hours ago
LynnLynn
50.4k897232
$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
Chas BrownChas Brown
5,1991523
$endgroup$
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
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$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered 4 hours ago
NeilNeil
82.6k745179
$endgroup$
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
J, 24 bytes
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 24 bytes
(],1#._2 _3{ ::1:])^:[1:
Try it online!
answered 5 hours ago
JonahJonah
2,5911017
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
ArnauldArnauld
80.5k797333
$endgroup$
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
(Explanation to follow when my exhaustion wears off!)
answered 3 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Lua 5.3, 49 bytes
function f(n)return n<=3 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.
answered 3 hours ago
cyclaministcyclaminist
1613
$endgroup$
$begingroup$
<=3can be<4
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
MickyTMickyT
10.3k21637
$endgroup$
add a comment |
$begingroup$
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
answered 3 hours ago
TauTau
801313
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
RK.RK.
397111
$endgroup$
add a comment |
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19 Answers
19
active
oldest
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19 Answers
19
active
oldest
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active
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active
oldest
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$begingroup$
Haskell, 26 bytes
(l!!)
l=1:1:1:2:scanl(+)2l
Try it online! Outputs the n'th term zero-indexed.
I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.
If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.
27 bytes
f n|n<3=1|1>0=f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
xnorxnor
93.4k18190448
$endgroup$
add a comment |
$begingroup$
Haskell, 26 bytes
(l!!)
l=1:1:1:2:scanl(+)2l
Try it online! Outputs the n'th term zero-indexed.
I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.
If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.
27 bytes
f n|n<3=1|1>0=f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
xnorxnor
93.4k18190448
$endgroup$
add a comment |
$begingroup$
Haskell, 26 bytes
(l!!)
l=1:1:1:2:scanl(+)2l
Try it online! Outputs the n'th term zero-indexed.
I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.
If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.
27 bytes
f n|n<3=1|1>0=f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
xnorxnor
93.4k18190448
$endgroup$
Haskell, 26 bytes
(l!!)
l=1:1:1:2:scanl(+)2l
Try it online! Outputs the n'th term zero-indexed.
I thought that the "obvious" recursive solution below would be unbeatable, but then I found this. It's similar to the classic golfy expression l=1:scanl(+)1l for the infinite Fibonacci list, but here the difference between adjacent elements is the term 4 positions back. We can more directly write l=1:1:zipWith(+)l(0:l), but that's longer.
If this challenge allowed infinite list output, we could cut the first line and have 20 bytes.
27 bytes
f n|n<3=1|1>0=f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
xnorxnor
93.4k18190448
answered 4 hours ago
xnorxnor
93.4k18190448
answered 4 hours ago
xnorxnor
93.4k18190448
answered 4 hours ago
xnorxnor
93.4k18190448
93.4k18190448
add a comment |
add a comment |
$begingroup$
Oasis, 5 bytes
nth term 0-indexed
cd+1V
Try it online!
Explanation
1V # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)
answered 5 hours ago
EmignaEmigna
47.4k433144
$endgroup$
add a comment |
$begingroup$
Oasis, 5 bytes
nth term 0-indexed
cd+1V
Try it online!
Explanation
1V # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)
answered 5 hours ago
EmignaEmigna
47.4k433144
$endgroup$
add a comment |
$begingroup$
Oasis, 5 bytes
nth term 0-indexed
cd+1V
Try it online!
Explanation
1V # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)
answered 5 hours ago
EmignaEmigna
47.4k433144
$endgroup$
Oasis, 5 bytes
nth term 0-indexed
cd+1V
Try it online!
Explanation
1V # a(0) = 1
# a(1) = 1
# a(2) = 1
# a(n) =
c # a(n-2)
+ # +
d # a(n-3)
answered 5 hours ago
EmignaEmigna
47.4k433144
answered 5 hours ago
EmignaEmigna
47.4k433144
answered 5 hours ago
EmignaEmigna
47.4k433144
answered 5 hours ago
EmignaEmigna
47.4k433144
47.4k433144
add a comment |
add a comment |
$begingroup$
Python 2, 30 bytes
f=lambda n:n<3or f(n-2)+f(n-3)
Try it online!
Returns the n'th term zero indexed. Outputs True for 1.
answered 5 hours ago
xnorxnor
93.4k18190448
$endgroup$
add a comment |
$begingroup$
Python 2, 30 bytes
f=lambda n:n<3or f(n-2)+f(n-3)
Try it online!
Returns the n'th term zero indexed. Outputs True for 1.
answered 5 hours ago
xnorxnor
93.4k18190448
$endgroup$
add a comment |
$begingroup$
Python 2, 30 bytes
f=lambda n:n<3or f(n-2)+f(n-3)
Try it online!
Returns the n'th term zero indexed. Outputs True for 1.
answered 5 hours ago
xnorxnor
93.4k18190448
$endgroup$
Python 2, 30 bytes
f=lambda n:n<3or f(n-2)+f(n-3)
Try it online!
Returns the n'th term zero indexed. Outputs True for 1.
answered 5 hours ago
xnorxnor
93.4k18190448
edited 5 hours ago
answered 5 hours ago
xnorxnor
93.4k18190448
answered 5 hours ago
xnorxnor
93.4k18190448
answered 5 hours ago
xnorxnor
93.4k18190448
93.4k18190448
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 33 bytes
a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]
1-indexed, returns the nth term
Try it online!
answered 5 hours ago
J42161217J42161217
13.8k21253
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 33 bytes
a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]
1-indexed, returns the nth term
Try it online!
answered 5 hours ago
J42161217J42161217
13.8k21253
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 33 bytes
a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]
1-indexed, returns the nth term
Try it online!
answered 5 hours ago
J42161217J42161217
13.8k21253
$endgroup$
Wolfram Language (Mathematica), 33 bytes
a@0=a@1=a@2=1;a@n_:=a[n-2]+a[n-3]
1-indexed, returns the nth term
Try it online!
answered 5 hours ago
J42161217J42161217
13.8k21253
answered 5 hours ago
J42161217J42161217
13.8k21253
answered 5 hours ago
J42161217J42161217
13.8k21253
answered 5 hours ago
J42161217J42161217
13.8k21253
13.8k21253
add a comment |
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
edited 5 hours ago
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 5 hours ago
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
add a comment |
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
5 hours ago
add a comment |
$begingroup$
Jelly, 10 bytes
‘HRcḤạ¥¥‘S
A monadic Link accepting n (1-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=1}^{lfloorfrac{n+1}2rfloor}binom{2i}{n+1-2i}$
‘HRcḤạ¥¥‘S - Link: integer, n e.g. 21
‘ - increment n 22
H - halve 11
R - range [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
- (note: floors input, e.g. n=6 -> ‘=7 -> H=3.5 -> R=[1,2,3] )
‘ - increment n 21
¥ - last two links as a dyad:
¥ - last two links as a dyad:
Ḥ - double [ 2, 4, 6, 8,10,12,14,16,18,20,22]
ạ - absolute difference [20,18,16,14,12,10, 8, 6, 4, 2, 0]
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0,28,126,45,1]
S - sum 200
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
$endgroup$
add a comment |
$begingroup$
Jelly, 10 bytes
‘HRcḤạ¥¥‘S
A monadic Link accepting n (1-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=1}^{lfloorfrac{n+1}2rfloor}binom{2i}{n+1-2i}$
‘HRcḤạ¥¥‘S - Link: integer, n e.g. 21
‘ - increment n 22
H - halve 11
R - range [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
- (note: floors input, e.g. n=6 -> ‘=7 -> H=3.5 -> R=[1,2,3] )
‘ - increment n 21
¥ - last two links as a dyad:
¥ - last two links as a dyad:
Ḥ - double [ 2, 4, 6, 8,10,12,14,16,18,20,22]
ạ - absolute difference [20,18,16,14,12,10, 8, 6, 4, 2, 0]
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0,28,126,45,1]
S - sum 200
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
$endgroup$
add a comment |
$begingroup$
Jelly, 10 bytes
‘HRcḤạ¥¥‘S
A monadic Link accepting n (1-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=1}^{lfloorfrac{n+1}2rfloor}binom{2i}{n+1-2i}$
‘HRcḤạ¥¥‘S - Link: integer, n e.g. 21
‘ - increment n 22
H - halve 11
R - range [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
- (note: floors input, e.g. n=6 -> ‘=7 -> H=3.5 -> R=[1,2,3] )
‘ - increment n 21
¥ - last two links as a dyad:
¥ - last two links as a dyad:
Ḥ - double [ 2, 4, 6, 8,10,12,14,16,18,20,22]
ạ - absolute difference [20,18,16,14,12,10, 8, 6, 4, 2, 0]
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0,28,126,45,1]
S - sum 200
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
$endgroup$
Jelly, 10 bytes
‘HRcḤạ¥¥‘S
A monadic Link accepting n (1-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=1}^{lfloorfrac{n+1}2rfloor}binom{2i}{n+1-2i}$
‘HRcḤạ¥¥‘S - Link: integer, n e.g. 21
‘ - increment n 22
H - halve 11
R - range [1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
- (note: floors input, e.g. n=6 -> ‘=7 -> H=3.5 -> R=[1,2,3] )
‘ - increment n 21
¥ - last two links as a dyad:
¥ - last two links as a dyad:
Ḥ - double [ 2, 4, 6, 8,10,12,14,16,18,20,22]
ạ - absolute difference [20,18,16,14,12,10, 8, 6, 4, 2, 0]
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0,28,126,45,1]
S - sum 200
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
edited 2 hours ago
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
answered 5 hours ago
Jonathan AllanJonathan Allan
53.8k535173
53.8k535173
add a comment |
add a comment |
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered 2 hours ago
LynnLynn
50.4k897232
$endgroup$
add a comment |
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered 2 hours ago
LynnLynn
50.4k897232
$endgroup$
add a comment |
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered 2 hours ago
LynnLynn
50.4k897232
$endgroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered 2 hours ago
LynnLynn
50.4k897232
answered 2 hours ago
LynnLynn
50.4k897232
answered 2 hours ago
LynnLynn
50.4k897232
answered 2 hours ago
LynnLynn
50.4k897232
50.4k897232
add a comment |
add a comment |
$begingroup$
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.
answered 5 hours ago
Chas BrownChas Brown
5,1991523
$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
Chas BrownChas Brown
5,1991523
$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
Chas BrownChas Brown
5,1991523
$endgroup$
Python 2, 56 48 bytes
f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c
Try it online!
Returns nth value, 0-indexed.
answered 5 hours ago
Chas BrownChas Brown
5,1991523
answered 5 hours ago
Chas BrownChas Brown
5,1991523
answered 5 hours ago
Chas BrownChas Brown
5,1991523
answered 5 hours ago
Chas BrownChas Brown
5,1991523
5,1991523
add a comment |
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
2,828127
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
add a comment |
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
3 hours ago
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered 4 hours ago
NeilNeil
82.6k745179
$endgroup$
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered 4 hours ago
NeilNeil
82.6k745179
$endgroup$
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered 4 hours ago
NeilNeil
82.6k745179
$endgroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered 4 hours ago
NeilNeil
82.6k745179
edited 4 hours ago
answered 4 hours ago
NeilNeil
82.6k745179
answered 4 hours ago
NeilNeil
82.6k745179
answered 4 hours ago
NeilNeil
82.6k745179
82.6k745179
add a comment |
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
answered 3 hours ago
community wiki
Sean
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
J, 24 bytes
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 24 bytes
(],1#._2 _3{ ::1:])^:[1:
Try it online!
answered 5 hours ago
JonahJonah
2,5911017
$endgroup$
add a comment |
$begingroup$
J, 24 bytes
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 24 bytes
(],1#._2 _3{ ::1:])^:[1:
Try it online!
answered 5 hours ago
JonahJonah
2,5911017
$endgroup$
add a comment |
$begingroup$
J, 24 bytes
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 24 bytes
(],1#._2 _3{ ::1:])^:[1:
Try it online!
answered 5 hours ago
JonahJonah
2,5911017
$endgroup$
J, 24 bytes
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 24 bytes
(],1#._2 _3{ ::1:])^:[1:
Try it online!
answered 5 hours ago
JonahJonah
2,5911017
edited 2 hours ago
answered 5 hours ago
JonahJonah
2,5911017
answered 5 hours ago
JonahJonah
2,5911017
answered 5 hours ago
JonahJonah
2,5911017
2,5911017
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
$endgroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
2,828127
2,828127
add a comment |
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
ArnauldArnauld
80.5k797333
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
ArnauldArnauld
80.5k797333
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
ArnauldArnauld
80.5k797333
$endgroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered 4 hours ago
ArnauldArnauld
80.5k797333
edited 4 hours ago
answered 4 hours ago
ArnauldArnauld
80.5k797333
answered 4 hours ago
ArnauldArnauld
80.5k797333
answered 4 hours ago
ArnauldArnauld
80.5k797333
80.5k797333
add a comment |
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
(Explanation to follow when my exhaustion wears off!)
answered 3 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
(Explanation to follow when my exhaustion wears off!)
answered 3 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
(Explanation to follow when my exhaustion wears off!)
answered 3 hours ago
ShaggyShaggy
18.9k21768
$endgroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
(Explanation to follow when my exhaustion wears off!)
answered 3 hours ago
ShaggyShaggy
18.9k21768
edited 3 hours ago
answered 3 hours ago
ShaggyShaggy
18.9k21768
answered 3 hours ago
ShaggyShaggy
18.9k21768
answered 3 hours ago
ShaggyShaggy
18.9k21768
18.9k21768
add a comment |
add a comment |
$begingroup$
Lua 5.3, 49 bytes
function f(n)return n<=3 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.
answered 3 hours ago
cyclaministcyclaminist
1613
$endgroup$
$begingroup$
<=3can be<4
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Lua 5.3, 49 bytes
function f(n)return n<=3 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.
answered 3 hours ago
cyclaministcyclaminist
1613
$endgroup$
$begingroup$
<=3can be<4
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Lua 5.3, 49 bytes
function f(n)return n<=3 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.
answered 3 hours ago
cyclaministcyclaminist
1613
$endgroup$
Lua 5.3, 49 bytes
function f(n)return n<=3 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.
answered 3 hours ago
cyclaministcyclaminist
1613
edited 3 hours ago
answered 3 hours ago
cyclaministcyclaminist
1613
answered 3 hours ago
cyclaministcyclaminist
1613
answered 3 hours ago
cyclaministcyclaminist
1613
1613
$begingroup$
<=3can be<4
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
<=3can be<4
$endgroup$
– Jo King
2 hours ago
$begingroup$
<=3 can be <4$endgroup$
– Jo King
2 hours ago
$begingroup$
<=3 can be <4$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
MickyTMickyT
10.3k21637
$endgroup$
add a comment |
$begingroup$
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.
answered 3 hours ago
MickyTMickyT
10.3k21637
$endgroup$
add a comment |
$begingroup$
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.
answered 3 hours ago
MickyTMickyT
10.3k21637
$endgroup$
Cubix, 20 bytes
This is 0 indexed and outputs the Nth term
;@UOI010+p?/sqq;W.(
Try it online!
Wraps onto a cube with side length 2
; @
U O
I 0 1 0 + p ? /
s q q ; W . (
. .
. .
Watch it run
I010- Initiates the stack
+p?- Adds the top of stack, pulls the counter from the bottom of stack and tests
/;UO@- If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt
(sqq;W- If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.
answered 3 hours ago
MickyTMickyT
10.3k21637
answered 3 hours ago
MickyTMickyT
10.3k21637
answered 3 hours ago
MickyTMickyT
10.3k21637
answered 3 hours ago
MickyTMickyT
10.3k21637
10.3k21637
add a comment |
add a comment |
$begingroup$
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
answered 3 hours ago
TauTau
801313
$endgroup$
add a comment |
$begingroup$
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
answered 3 hours ago
TauTau
801313
$endgroup$
add a comment |
$begingroup$
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
answered 3 hours ago
TauTau
801313
$endgroup$
TI-BASIC (TI-84), 34 bytes
[[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1
0-indexed $N$th term of the sequence.
Input is in Ans.
Output is in Ans and is automatically printed out.
I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.
Example:
0
0
prgmCDGFD
1
9
9
prgmCDGFD
9
16
16
prgmCDGFD
65
Explanation:
[[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1 ;full program (example input: 6)
[[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
; [0 1 0]
; [0 0 1]
; [1 1 0]
^(Ans+5 ;then raise it to the power of: input + 5
; [4 7 5]
; [5 9 7]
; [7 12 9]
Ans(1,1 ;get the top-left index and leave it in "Ans"
;implicitly print Ans
answered 3 hours ago
TauTau
801313
answered 3 hours ago
TauTau
801313
answered 3 hours ago
TauTau
801313
answered 3 hours ago
TauTau
801313
801313
add a comment |
add a comment |
$begingroup$
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.
answered 1 hour ago
RK.RK.
397111
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
RK.RK.
397111
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
RK.RK.
397111
$endgroup$
Pyth, 16 bytes
L?<b3!b+y-b2y-b3
This defines the function y. Try it here!
Here's a more fun solution, though it's 9 bytes longer; bytes could be shaved though.
+l{sa.pMf.Am&>d2%d2T./QY!
This uses the definition given by David Callan on the OEIS page: "a(n) = number of compositions of n into parts that are odd and >= 3." Try it here! It takes input directly instead of defining a function.
answered 1 hour ago
RK.RK.
397111
answered 1 hour ago
RK.RK.
397111
answered 1 hour ago
RK.RK.
397111
answered 1 hour ago
RK.RK.
397111
397111
add a comment |
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).
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$begingroup$
Sandbox post can be found here.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37$endgroup$
– Jonathan Allan
5 hours 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
5 hours ago