Patience, young “Padovan”
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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 $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(1-indexed, first $N$ terms)
Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16
Rules:
This is code-golf: the fewer bytes, the better!
Standard loopholes are forbidden.
code-golf number sequence
asked Apr 7 at 20:21
TauTau
1,001515
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|
show 3 more comments
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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 $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(1-indexed, first $N$ terms)
Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16
Rules:
This is code-golf: the fewer bytes, the better!
Standard loopholes are forbidden.
code-golf number sequence
asked Apr 7 at 20:21
TauTau
1,001515
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2
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14(0-indexed) is shown as outputting28while I believe it should yield37
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– Jonathan Allan
Apr 7 at 20:53
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@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.
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– Tau
Apr 7 at 20:55
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@LuisMendo I believe so. I'll edit the post.
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– Tau
2 days ago
1
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@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.
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– Tau
yesterday
1
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Note that the OEIS sequence you linked is slightly different, since it usesa_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because thena_5=a_6=a_7=1
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– Carmeister
yesterday
|
show 3 more comments
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(1-indexed, first $N$ terms)
Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16
Rules:
This is code-golf: the fewer bytes, the better!
Standard loopholes are forbidden.
code-golf number sequence
asked Apr 7 at 20:21
TauTau
1,001515
$endgroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(1-indexed, first $N$ terms)
Input | Output
--------------
1 | 1
3 | 1,1,1
4 | 1,1,1,2
7 | 1,1,1,2,2,3,4
10 | 1,1,1,2,2,3,4,5,7,9
12 | 1,1,1,2,2,3,4,5,7,9,12,16
Rules:
This is code-golf: the fewer bytes, the better!
Standard loopholes are forbidden.
code-golf number sequence
code-golf number sequence
asked Apr 7 at 20:21
TauTau
1,001515
asked Apr 7 at 20:21
TauTau
1,001515
edited 2 days ago
Tau
asked Apr 7 at 20:21
TauTau
1,001515
asked Apr 7 at 20:21
TauTau
1,001515
asked Apr 7 at 20:21
TauTau
1,001515
1,001515
2
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14(0-indexed) is shown as outputting28while I believe it should yield37
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– Jonathan Allan
Apr 7 at 20:53
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@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.
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– Tau
Apr 7 at 20:55
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@LuisMendo I believe so. I'll edit the post.
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– Tau
2 days ago
1
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@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.
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– Tau
yesterday
1
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Note that the OEIS sequence you linked is slightly different, since it usesa_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because thena_5=a_6=a_7=1
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– Carmeister
yesterday
|
show 3 more comments
2
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14(0-indexed) is shown as outputting28while I believe it should yield37
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– Jonathan Allan
Apr 7 at 20:53
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@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.
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– Tau
Apr 7 at 20:55
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@LuisMendo I believe so. I'll edit the post.
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– Tau
2 days ago
1
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@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.
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– Tau
yesterday
1
$begingroup$
Note that the OEIS sequence you linked is slightly different, since it usesa_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because thena_5=a_6=a_7=1
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– Carmeister
yesterday
2
2
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14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
Apr 7 at 20:53
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14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
Apr 7 at 20:53
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@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.
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– Tau
Apr 7 at 20:55
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@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.
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– Tau
Apr 7 at 20:55
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@LuisMendo I believe so. I'll edit the post.
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– Tau
2 days ago
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@LuisMendo I believe so. I'll edit the post.
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– Tau
2 days ago
1
1
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@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.
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– Tau
yesterday
$begingroup$
@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.
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– Tau
yesterday
1
1
$begingroup$
Note that the OEIS sequence you linked is slightly different, since it uses
a_0=1, a_1=0, a_2=0. It ends up being shifted by a bit because then a_5=a_6=a_7=1$endgroup$
– Carmeister
yesterday
$begingroup$
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
yesterday
|
show 3 more comments
39 Answers
39
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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 Apr 7 at 23:46
LynnLynn
50.7k898233
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31
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this is clearly some kind of voodoo
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– Pureferret
2 days ago
7
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This should be published.
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– YSC
2 days ago
6
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@YSC It has already been published in A000931. I'd never have guess the primes trick:)
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– flawr
2 days ago
1
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...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
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– Jonathan Allan
2 days ago
1
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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
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– Tasos Papastylianou
yesterday
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show 3 more comments
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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)
answered Apr 7 at 20:32
EmignaEmigna
47.9k434145
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add a comment |
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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 $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$
ŻṚm2Jc$S - Link: integer, n e.g. 20
Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
$ - last two links as a monad:
J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
S - sum 200
And here is a "twofer"
...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):
3ḊṗRẎ§ċ‘ - Link: n
3Ḋ - 3 dequeued = [2,3]
R - range = [1,2,3,...,n]
ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
§ - sums [ 2, 3, 4, 5, 5, 6, 6,...]
‘ - increment n+1
ċ - count occurrences P(n)
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
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add a comment |
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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!
answered Apr 7 at 21:36
xnorxnor
94k18192451
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add a comment |
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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.
answered Apr 7 at 20:47
xnorxnor
94k18192451
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add a comment |
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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!
answered Apr 7 at 20:34
J42161217J42161217
13.9k21353
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add a comment |
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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].
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
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add a comment |
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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!
answered Apr 7 at 20:43
JonahJonah
2,6711017
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1
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Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
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– Galen Ivanov
2 days ago
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23 bytes thanks to ngn1:->#: Try it online!
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– Galen Ivanov
2 days ago
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@GalenIvanov tyvm, that's a great trick.
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– Jonah
2 days ago
1
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1:->1. "adverse" works with a noun on the right, apparently
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– ngn
9 hours ago
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@ngn TIL... ty again!
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– Jonah
9 hours ago
add a comment |
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Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$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 $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered Apr 7 at 21:27
NeilNeil
82.7k745179
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Cubix, 20 bytes
This is 0 indexed and outputs the Nth term
;@UOI010+p?/sqq;W.(
Try it online!
Wraps onto a cube with side length 2
; @
U O
I 0 1 0 + p ? /
s q q ; W . (
. .
. .
Watch it run
I010- Initiates the stack
+p?- Adds the top of stack, pulls the counter from the bottom of stack and tests
/;UO@- If counter is 0, reflect onto top face, remove TOS, u-turn, output and halt
(sqq;W- If counter is positive, reflect, decrement counter, swap TOS, push top to bottom twice, remove TOS and shift lane back into the main loop.
answered Apr 7 at 22:32
MickyTMickyT
10.3k21637
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add a comment |
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Python 2, 56 48 bytes
f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c
Try it online!
Returns nth value, 0-indexed.
answered Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
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add a comment |
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Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
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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Why is this community wiki?
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– Jo King
Apr 7 at 23:14
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@JoKing Beats me. If I did it somehow, it wasn't on purpose.
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– Sean
2 days ago
add a comment |
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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).
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
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I don't think1Ð)can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
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– Kevin Cruijssen
2 days ago
add a comment |
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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.
answered 2 days ago
Sherlock9Sherlock9
8,17411860
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add a comment |
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K (ngn/k), 24 20 bytes
-4 bytes thanks to ngn!
{$[x<3;1;+/o'x-2 3]}
Try it online!
0-indexed, first N terms
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
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1
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f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
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– ngn
2 days ago
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@ngn Thanks! I tried it (without success) in J; it's elegant here.
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– Galen Ivanov
2 days ago
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@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
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– Galen Ivanov
2 days ago
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ah, right, base-1 decode is a train-friendly way to sum :)
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– ngn
2 days ago
2
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1:->#in the j solution
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– ngn
2 days ago
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show 2 more comments
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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 :)
answered 2 days ago
anatolyganatolyg
7,2692166
$endgroup$
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
$endgroup$
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 2 days ago
G BG B
8,2661429
$endgroup$
add a comment |
$begingroup$
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!
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
$endgroup$
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
2 days ago
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
|
show 1 more comment
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
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 Apr 7 at 23:01
TauTau
1,001515
$endgroup$
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 Apr 8 at 0:15
RK.RK.
407211
$endgroup$
$begingroup$
y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
$endgroup$
– Ven
2 days ago
$begingroup$
@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
$endgroup$
– RK.
2 days ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
2 days ago
add a comment |
$begingroup$
Java, 41 bytes
Can't use a lambda (runtime error). Port of this Javascript answer
int f(int n){return n<3?1:f(n-2)+f(n-3);}
TIO
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
$endgroup$
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
add a comment |
$begingroup$
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.
answered 2 days ago
rturnbullrturnbull
3,519925
$endgroup$
2
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
add a comment |
$begingroup$
C (clang), 41 33 bytes
a(i){return i<3?1:a(i-2)+a(i-3);}
Try it online!
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
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 Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 2 days ago
XcaliXcali
5,505520
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered yesterday
attinatattinat
4897
$endgroup$
add a comment |
$begingroup$
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 $frac{1+x}{1-x^2-x^3}$.
answered yesterday
alephalphaalephalpha
21.8k33094
$endgroup$
add a comment |
1 2
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$begingroup$
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 Apr 7 at 23:46
LynnLynn
50.7k898233
$endgroup$
31
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
7
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
6
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
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
yesterday
|
show 3 more comments
$begingroup$
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 Apr 7 at 23:46
LynnLynn
50.7k898233
$endgroup$
31
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
7
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
6
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
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
yesterday
|
show 3 more comments
$begingroup$
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 Apr 7 at 23:46
LynnLynn
50.7k898233
$endgroup$
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 Apr 7 at 23:46
LynnLynn
50.7k898233
answered Apr 7 at 23:46
LynnLynn
50.7k898233
answered Apr 7 at 23:46
LynnLynn
50.7k898233
answered Apr 7 at 23:46
LynnLynn
50.7k898233
50.7k898233
31
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
7
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
6
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
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
yesterday
|
show 3 more comments
31
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
7
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
6
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
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
yesterday
31
31
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
2 days ago
7
7
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
$begingroup$
This should be published.
$endgroup$
– YSC
2 days ago
6
6
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
2 days ago
1
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
2 days ago
1
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
yesterday
$begingroup$
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
yesterday
|
show 3 more comments
$begingroup$
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 Apr 7 at 20:32
EmignaEmigna
47.9k434145
$endgroup$
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 Apr 7 at 20:32
EmignaEmigna
47.9k434145
$endgroup$
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 Apr 7 at 20:32
EmignaEmigna
47.9k434145
$endgroup$
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 Apr 7 at 20:32
EmignaEmigna
47.9k434145
answered Apr 7 at 20:32
EmignaEmigna
47.9k434145
answered Apr 7 at 20:32
EmignaEmigna
47.9k434145
answered Apr 7 at 20:32
EmignaEmigna
47.9k434145
47.9k434145
add a comment |
add a comment |
$begingroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$
ŻṚm2Jc$S - Link: integer, n e.g. 20
Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
$ - last two links as a monad:
J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
S - sum 200
And here is a "twofer"
...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):
3ḊṗRẎ§ċ‘ - Link: n
3Ḋ - 3 dequeued = [2,3]
R - range = [1,2,3,...,n]
ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
§ - sums [ 2, 3, 4, 5, 5, 6, 6,...]
‘ - increment n+1
ċ - count occurrences P(n)
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
add a comment |
$begingroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$
ŻṚm2Jc$S - Link: integer, n e.g. 20
Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
$ - last two links as a monad:
J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
S - sum 200
And here is a "twofer"
...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):
3ḊṗRẎ§ċ‘ - Link: n
3Ḋ - 3 dequeued = [2,3]
R - range = [1,2,3,...,n]
ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
§ - sums [ 2, 3, 4, 5, 5, 6, 6,...]
‘ - increment n+1
ċ - count occurrences P(n)
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
add a comment |
$begingroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$
ŻṚm2Jc$S - Link: integer, n e.g. 20
Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
$ - last two links as a monad:
J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
S - sum 200
And here is a "twofer"
...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):
3ḊṗRẎ§ċ‘ - Link: n
3Ḋ - 3 dequeued = [2,3]
R - range = [1,2,3,...,n]
ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
§ - sums [ 2, 3, 4, 5, 5, 6, 6,...]
‘ - increment n+1
ċ - count occurrences P(n)
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_{i=0}^{lfloorfrac{n}2rfloor}binom{i+1}{n-2i}$
ŻṚm2Jc$S - Link: integer, n e.g. 20
Ż - zero range [0, 1, 2, 3, 4, ..., 19, 20]
Ṛ - reverse [20, 19, ..., 4, 3, 2, 1, 0]
m2 - modulo-slice with 2 [20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0] <- n-2i
$ - last two links as a monad:
J - range of length [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] <- i+1
c - left-choose-right [ 0, 0, 0, 0, 0, 0, 0, 28,126, 45, 1]
S - sum 200
And here is a "twofer"
...a totally different method also for 8 bytes (this one is 1-indexed, but much slower):
3ḊṗRẎ§ċ‘ - Link: n
3Ḋ - 3 dequeued = [2,3]
R - range = [1,2,3,...,n]
ṗ - Cartesian power [[[2],[3]],[[2,2],[2,3],[3,2],[3,3]],[[2,2,2],...],...]
Ẏ - tighten [[2],[3],[2,2],[2,3],[3,2],[3,3],[2,2,2],...]
§ - sums [ 2, 3, 4, 5, 5, 6, 6,...]
‘ - increment n+1
ċ - count occurrences P(n)
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
edited 2 days ago
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
answered Apr 7 at 21:04
Jonathan AllanJonathan Allan
54.1k537174
54.1k537174
add a comment |
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 Apr 7 at 21:36
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 21:36
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 21:36
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 21:36
xnorxnor
94k18192451
answered Apr 7 at 21:36
xnorxnor
94k18192451
answered Apr 7 at 21:36
xnorxnor
94k18192451
answered Apr 7 at 21:36
xnorxnor
94k18192451
94k18192451
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 Apr 7 at 20:47
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 20:47
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 20:47
xnorxnor
94k18192451
$endgroup$
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 Apr 7 at 20:47
xnorxnor
94k18192451
edited Apr 7 at 21:02
answered Apr 7 at 20:47
xnorxnor
94k18192451
answered Apr 7 at 20:47
xnorxnor
94k18192451
answered Apr 7 at 20:47
xnorxnor
94k18192451
94k18192451
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 Apr 7 at 20:34
J42161217J42161217
13.9k21353
$endgroup$
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 Apr 7 at 20:34
J42161217J42161217
13.9k21353
$endgroup$
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 Apr 7 at 20:34
J42161217J42161217
13.9k21353
$endgroup$
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 Apr 7 at 20:34
J42161217J42161217
13.9k21353
answered Apr 7 at 20:34
J42161217J42161217
13.9k21353
answered Apr 7 at 20:34
J42161217J42161217
13.9k21353
answered Apr 7 at 20:34
J42161217J42161217
13.9k21353
13.9k21353
add a comment |
add a comment |
$begingroup$
Octave / MATLAB, 35 33 bytes
@(n)[1 filter(1,'cbaa'-98,2:n<5)]
Outputs the first n terms.
Try it online!
How it works
Anonymous function that implements a recursive filter.
'cbaa'-98 is a shorter form to produce [1 0 -1 -1].
2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).
filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
$endgroup$
add a comment |
$begingroup$
Octave / MATLAB, 35 33 bytes
@(n)[1 filter(1,'cbaa'-98,2:n<5)]
Outputs the first n terms.
Try it online!
How it works
Anonymous function that implements a recursive filter.
'cbaa'-98 is a shorter form to produce [1 0 -1 -1].
2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).
filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
$endgroup$
add a comment |
$begingroup$
Octave / MATLAB, 35 33 bytes
@(n)[1 filter(1,'cbaa'-98,2:n<5)]
Outputs the first n terms.
Try it online!
How it works
Anonymous function that implements a recursive filter.
'cbaa'-98 is a shorter form to produce [1 0 -1 -1].
2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).
filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
$endgroup$
Octave / MATLAB, 35 33 bytes
@(n)[1 filter(1,'cbaa'-98,2:n<5)]
Outputs the first n terms.
Try it online!
How it works
Anonymous function that implements a recursive filter.
'cbaa'-98 is a shorter form to produce [1 0 -1 -1].
2:n<5 is a shorter form to produce [1 1 1 0 0 ··· 0] (n−1 terms).
filter(1,[1 0 -1 -1],[1 1 1 0 0 ··· 0]) passes the input [1 1 1 0 0 ··· 0] through a discrete-time filter defined by a transfer function with numerator coefficient 1 and denominator coefficients [1 0 -1 -1].
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
edited 2 days ago
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
answered 2 days ago
Luis MendoLuis Mendo
75.3k889292
75.3k889292
add a comment |
add a comment |
$begingroup$
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!
answered Apr 7 at 20:43
JonahJonah
2,6711017
$endgroup$
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
1
$begingroup$
1:->1. "adverse" works with a noun on the right, apparently
$endgroup$
– ngn
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 hours ago
add a comment |
$begingroup$
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!
answered Apr 7 at 20:43
JonahJonah
2,6711017
$endgroup$
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
1
$begingroup$
1:->1. "adverse" works with a noun on the right, apparently
$endgroup$
– ngn
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 hours ago
add a comment |
$begingroup$
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!
answered Apr 7 at 20:43
JonahJonah
2,6711017
$endgroup$
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!
answered Apr 7 at 20:43
JonahJonah
2,6711017
edited 9 hours ago
answered Apr 7 at 20:43
JonahJonah
2,6711017
answered Apr 7 at 20:43
JonahJonah
2,6711017
answered Apr 7 at 20:43
JonahJonah
2,6711017
2,6711017
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
1
$begingroup$
1:->1. "adverse" works with a noun on the right, apparently
$endgroup$
– ngn
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 hours ago
add a comment |
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
1
$begingroup$
1:->1. "adverse" works with a noun on the right, apparently
$endgroup$
– ngn
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 hours ago
1
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn
1: -> # : Try it online!$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
23 bytes thanks to ngn
1: -> # : Try it online!$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
2 days ago
1
1
$begingroup$
1: -> 1. "adverse" works with a noun on the right, apparently$endgroup$
– ngn
9 hours ago
$begingroup$
1: -> 1. "adverse" works with a noun on the right, apparently$endgroup$
– ngn
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 hours ago
$begingroup$
@ngn TIL... ty again!
$endgroup$
– Jonah
9 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 Apr 7 at 21:27
NeilNeil
82.7k745179
$endgroup$
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 Apr 7 at 21:27
NeilNeil
82.7k745179
$endgroup$
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 Apr 7 at 21:27
NeilNeil
82.7k745179
$endgroup$
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 Apr 7 at 21:27
NeilNeil
82.7k745179
edited Apr 7 at 21:37
answered Apr 7 at 21:27
NeilNeil
82.7k745179
answered Apr 7 at 21:27
NeilNeil
82.7k745179
answered Apr 7 at 21:27
NeilNeil
82.7k745179
82.7k745179
add a comment |
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 Apr 7 at 22:32
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 Apr 7 at 22:32
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 Apr 7 at 22:32
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 Apr 7 at 22:32
MickyTMickyT
10.3k21637
answered Apr 7 at 22:32
MickyTMickyT
10.3k21637
answered Apr 7 at 22:32
MickyTMickyT
10.3k21637
answered Apr 7 at 22:32
MickyTMickyT
10.3k21637
10.3k21637
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 Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
$endgroup$
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 Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
$endgroup$
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 Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
$endgroup$
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 Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
answered Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
answered Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
answered Apr 7 at 20:44
Chas BrownChas Brown
5,2191523
5,2191523
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
Apr 7 at 23:14
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
2 days 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
Apr 7 at 23:14
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
2 days 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 Apr 7 at 22:23
community wiki
Sean
$begingroup$
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
2 days ago
add a comment |
$begingroup$
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
2 days ago
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
Apr 7 at 23:14
$begingroup$
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
2 days ago
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
2 days ago
add a comment |
$begingroup$
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).
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
$endgroup$
$begingroup$
I don't think1Ð)can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
$endgroup$
– Kevin Cruijssen
2 days ago
add a comment |
$begingroup$
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).
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
$endgroup$
$begingroup$
I don't think1Ð)can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
$endgroup$
– Kevin Cruijssen
2 days ago
add a comment |
$begingroup$
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).
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
$endgroup$
05AB1E, 8 bytes
1Ð)λ£₂₃+
Try it online!
Bear with me, I haven't golfed in a while. I wonder if there's a shorter substitute for 1Ð) which works in this case (I've tried 1D), 3Å1 etc. but none of them save bytes). Outputs the first $n$ terms of the sequence. Or, without the £, it would output an infinite stream of the terms of the sequence.
How?
1Ð)λ£₂₃+ | Full program.
1Ð) | Initialize the stack with [1, 1, 1].
λ | Begin the recursive generation of a list: Starting from some base case,
| this command generates an infinite list with the pattern function given.
£ | Flag for λ. Instead of outputting an infinite stream, only print the first n.
₂₃+ | Add a(n-2) and a(n-3).
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
edited 2 days ago
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
answered 2 days ago
Mr. XcoderMr. Xcoder
32.3k759200
32.3k759200
$begingroup$
I don't think1Ð)can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
$endgroup$
– Kevin Cruijssen
2 days ago
add a comment |
$begingroup$
I don't think1Ð)can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
I don't think
1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
I don't think
1Ð) can be 2 bytes tbh. I can think of six different 3-bytes alternatives, but no 2-byters.$endgroup$
– Kevin Cruijssen
2 days ago
add a comment |
$begingroup$
APL (Dyalog Unicode), 20 18 17 bytesSBCS
This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.
Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
Try it online!
Explanation
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
inside the brackets (...) to its operands (here marked ⍵ and ⍺).
2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
where we save our results as the function is repeatedly applied
and our ⍺, 2, is our right argument and is immediately applied to +/,
so that we have 2+/ which will return the pairwise sums of our array.
2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
⊢,⍨ And add it to the head of our array.
4⌷ When we've finished adding Padovan numbers to the end of our list,
the n-th Padovan number (1-indexed) is the 4th member of that list,
and so, we implicitly return that.
answered 2 days ago
Sherlock9Sherlock9
8,17411860
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 20 18 17 bytesSBCS
This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.
Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
Try it online!
Explanation
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
inside the brackets (...) to its operands (here marked ⍵ and ⍺).
2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
where we save our results as the function is repeatedly applied
and our ⍺, 2, is our right argument and is immediately applied to +/,
so that we have 2+/ which will return the pairwise sums of our array.
2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
⊢,⍨ And add it to the head of our array.
4⌷ When we've finished adding Padovan numbers to the end of our list,
the n-th Padovan number (1-indexed) is the 4th member of that list,
and so, we implicitly return that.
answered 2 days ago
Sherlock9Sherlock9
8,17411860
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 20 18 17 bytesSBCS
This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.
Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
Try it online!
Explanation
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
inside the brackets (...) to its operands (here marked ⍵ and ⍺).
2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
where we save our results as the function is repeatedly applied
and our ⍺, 2, is our right argument and is immediately applied to +/,
so that we have 2+/ which will return the pairwise sums of our array.
2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
⊢,⍨ And add it to the head of our array.
4⌷ When we've finished adding Padovan numbers to the end of our list,
the n-th Padovan number (1-indexed) is the 4th member of that list,
and so, we implicitly return that.
answered 2 days ago
Sherlock9Sherlock9
8,17411860
$endgroup$
APL (Dyalog Unicode), 20 18 17 bytesSBCS
This code is 1-indexed. It's the same number of bytes to get n items of the Padovan sequence, as you have to drop the last few extra members. It's also the same number of bytes to get 0-indexing.
Edit: -2 bytes thanks to ngn. -1 byte thanks to ngn
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
Try it online!
Explanation
4⌷2(⊢,⍨2⌷+/)⍣⎕×⍳3
⍺(. . . .)⍣⎕⍵ This format simply takes the input ⎕ and applies the function
inside the brackets (...) to its operands (here marked ⍵ and ⍺).
2(. . .+/)⍣⎕×⍳3 In this case, our ⍵, the left argument, is the array 1 1 1,
where we save our results as the function is repeatedly applied
and our ⍺, 2, is our right argument and is immediately applied to +/,
so that we have 2+/ which will return the pairwise sums of our array.
2⌷ We take the second pairwise sum, f(n-2) + f(n-3)
⊢,⍨ And add it to the head of our array.
4⌷ When we've finished adding Padovan numbers to the end of our list,
the n-th Padovan number (1-indexed) is the 4th member of that list,
and so, we implicitly return that.
answered 2 days ago
Sherlock9Sherlock9
8,17411860
edited 2 days ago
answered 2 days ago
Sherlock9Sherlock9
8,17411860
answered 2 days ago
Sherlock9Sherlock9
8,17411860
answered 2 days ago
Sherlock9Sherlock9
8,17411860
8,17411860
add a comment |
add a comment |
$begingroup$
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
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
$endgroup$
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
2 days ago
|
show 2 more comments
$begingroup$
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
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
$endgroup$
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
2 days ago
|
show 2 more comments
$begingroup$
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
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
$endgroup$
K (ngn/k), 24 20 bytes
-4 bytes thanks to ngn!
{$[x<3;1;+/o'x-2 3]}
Try it online!
0-indexed, first N terms
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
edited 2 days ago
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
answered 2 days ago
Galen IvanovGalen Ivanov
7,42211034
7,42211034
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
2 days ago
|
show 2 more comments
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
2 days ago
1
1
$begingroup$
f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")$endgroup$
– ngn
2 days ago
$begingroup$
f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")$endgroup$
– ngn
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
2 days ago
2
2
$begingroup$
1: -> # in the j solution$endgroup$
– ngn
2 days ago
$begingroup$
1: -> # in the j solution$endgroup$
– ngn
2 days ago
|
show 2 more comments
$begingroup$
x86 32-bit machine code, 17 bytes
53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3
Disassembly:
00CE1250 53 push ebx
00CE1251 33 DB xor ebx,ebx
00CE1253 F7 E3 mul eax,ebx
00CE1255 43 inc ebx
00CE1256 83 C1 04 add ecx,4
00CE1259 03 D8 add ebx,eax
00CE125B 93 xchg eax,ebx
00CE125C 92 xchg eax,edx
00CE125D E2 FA loop myloop (0CE1259h)
00CE125F 5B pop ebx
00CE1260 C3 ret
It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.
At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.
I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.
I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)
answered 2 days ago
anatolyganatolyg
7,2692166
$endgroup$
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
add a comment |
$begingroup$
x86 32-bit machine code, 17 bytes
53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3
Disassembly:
00CE1250 53 push ebx
00CE1251 33 DB xor ebx,ebx
00CE1253 F7 E3 mul eax,ebx
00CE1255 43 inc ebx
00CE1256 83 C1 04 add ecx,4
00CE1259 03 D8 add ebx,eax
00CE125B 93 xchg eax,ebx
00CE125C 92 xchg eax,edx
00CE125D E2 FA loop myloop (0CE1259h)
00CE125F 5B pop ebx
00CE1260 C3 ret
It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.
At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.
I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.
I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)
answered 2 days ago
anatolyganatolyg
7,2692166
$endgroup$
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
add a comment |
$begingroup$
x86 32-bit machine code, 17 bytes
53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3
Disassembly:
00CE1250 53 push ebx
00CE1251 33 DB xor ebx,ebx
00CE1253 F7 E3 mul eax,ebx
00CE1255 43 inc ebx
00CE1256 83 C1 04 add ecx,4
00CE1259 03 D8 add ebx,eax
00CE125B 93 xchg eax,ebx
00CE125C 92 xchg eax,edx
00CE125D E2 FA loop myloop (0CE1259h)
00CE125F 5B pop ebx
00CE1260 C3 ret
It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.
At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.
I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.
I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)
answered 2 days ago
anatolyganatolyg
7,2692166
$endgroup$
x86 32-bit machine code, 17 bytes
53 33 db f7 e3 43 83 c1 04 03 d8 93 92 e2 fa 5b c3
Disassembly:
00CE1250 53 push ebx
00CE1251 33 DB xor ebx,ebx
00CE1253 F7 E3 mul eax,ebx
00CE1255 43 inc ebx
00CE1256 83 C1 04 add ecx,4
00CE1259 03 D8 add ebx,eax
00CE125B 93 xchg eax,ebx
00CE125C 92 xchg eax,edx
00CE125D E2 FA loop myloop (0CE1259h)
00CE125F 5B pop ebx
00CE1260 C3 ret
It is 0-indexed. The initialization is conveniently achieved by calculating eax * 0. The 128-bit result is 0, and it goes in edx:eax.
At the beginning of each iteration, the order of the registers is ebx, eax, edx. I had to choose the right order to take advantage of the encoding for the xchg eax instruction - 1 byte.
I had to add 4 to the loop counter in order to let the output reach eax, which holds the function's return value in the fastcall convention.
I could use some other calling convention, which doesn't require saving and restoring ebx, but fastcall is fun anyway :)
answered 2 days ago
anatolyganatolyg
7,2692166
answered 2 days ago
anatolyganatolyg
7,2692166
answered 2 days ago
anatolyganatolyg
7,2692166
answered 2 days ago
anatolyganatolyg
7,2692166
7,2692166
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
add a comment |
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
1
1
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
yesterday
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
edited Apr 7 at 21:02
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
answered Apr 7 at 20:58
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
add a comment |
add a comment |
$begingroup$
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.
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
$endgroup$
Lua 5.3, 49 48 bytes
function f(n)return n<4 and 1or f(n-2)+f(n-3)end
Try it online!
Vanilla Lua doesn't have coercion of booleans to strings (even tonumber(true) returns nil), so you have to use a pseudo-ternary operator. This version is 1-indexed, like all of Lua. The 1or part has to be changed to 1 or in Lua 5.1, which has a different way of lexing numbers.
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
edited 2 days ago
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
answered Apr 7 at 22:12
cyclaministcyclaminist
1813
1813
add a comment |
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 2 days ago
G BG B
8,2661429
$endgroup$
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 2 days ago
G BG B
8,2661429
$endgroup$
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 2 days ago
G BG B
8,2661429
$endgroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 2 days ago
G BG B
8,2661429
answered 2 days ago
G BG B
8,2661429
answered 2 days ago
G BG B
8,2661429
answered 2 days ago
G BG B
8,2661429
8,2661429
add a comment |
add a comment |
$begingroup$
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!
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
$endgroup$
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
2 days ago
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
|
show 1 more comment
$begingroup$
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!
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
$endgroup$
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
2 days ago
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
|
show 1 more comment
$begingroup$
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!
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
$endgroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931, but with $a(0)=a(1)=a(2)=1$, as specified in the challenge.
Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
edited yesterday
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
answered Apr 7 at 21:18
ArnauldArnauld
80.7k797334
80.7k797334
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
2 days ago
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
|
show 1 more comment
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
2 days ago
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
$begingroup$
I don't think it's reasonable to say that returning
true is the same as returning 1 if the rest of the output is numbers.$endgroup$
– Nit
2 days ago
$begingroup$
I don't think it's reasonable to say that returning
true is the same as returning 1 if the rest of the output is numbers.$endgroup$
– Nit
2 days ago
2
2
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
2 days ago
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
I think you are missing some requirements: Have a look at my modification (version in Java) here.
$endgroup$
– Shaq
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
$begingroup$
@Arnauld yep I can see it now. Sorry!
$endgroup$
– Shaq
yesterday
|
show 1 more comment
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:26
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
2,916127
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
Apr 7 at 22:26
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
$begingroup$
I stand corrected!
$endgroup$
– Shaggy
yesterday
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 Apr 7 at 23:01
TauTau
1,001515
$endgroup$
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 Apr 7 at 23:01
TauTau
1,001515
$endgroup$
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 Apr 7 at 23:01
TauTau
1,001515
$endgroup$
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 Apr 7 at 23:01
TauTau
1,001515
answered Apr 7 at 23:01
TauTau
1,001515
answered Apr 7 at 23:01
TauTau
1,001515
answered Apr 7 at 23:01
TauTau
1,001515
1,001515
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 Apr 8 at 0:15
RK.RK.
407211
$endgroup$
$begingroup$
y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
$endgroup$
– Ven
2 days ago
$begingroup$
@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
$endgroup$
– RK.
2 days ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
2 days ago
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 Apr 8 at 0:15
RK.RK.
407211
$endgroup$
$begingroup$
y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
$endgroup$
– Ven
2 days ago
$begingroup$
@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
$endgroup$
– RK.
2 days ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
2 days ago
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 Apr 8 at 0:15
RK.RK.
407211
$endgroup$
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 Apr 8 at 0:15
RK.RK.
407211
answered Apr 8 at 0:15
RK.RK.
407211
answered Apr 8 at 0:15
RK.RK.
407211
answered Apr 8 at 0:15
RK.RK.
407211
407211
$begingroup$
y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
$endgroup$
– Ven
2 days ago
$begingroup$
@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
$endgroup$
– RK.
2 days ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
2 days ago
add a comment |
$begingroup$
y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
$endgroup$
– Ven
2 days ago
$begingroup$
@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
$endgroup$
– RK.
2 days ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
@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.
2 days ago
$begingroup$
@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.
2 days ago
$begingroup$
Well done on
hB2. Too bad it's the same byte count.$endgroup$
– Ven
2 days ago
$begingroup$
Well done on
hB2. Too bad it's the same byte count.$endgroup$
– Ven
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of the
d in the map.$endgroup$
– RK.
2 days ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of the
d in the map.$endgroup$
– RK.
2 days ago
add a comment |
$begingroup$
Java, 41 bytes
Can't use a lambda (runtime error). Port of this Javascript answer
int f(int n){return n<3?1:f(n-2)+f(n-3);}
TIO
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
$endgroup$
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
add a comment |
$begingroup$
Java, 41 bytes
Can't use a lambda (runtime error). Port of this Javascript answer
int f(int n){return n<3?1:f(n-2)+f(n-3);}
TIO
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
$endgroup$
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
add a comment |
$begingroup$
Java, 41 bytes
Can't use a lambda (runtime error). Port of this Javascript answer
int f(int n){return n<3?1:f(n-2)+f(n-3);}
TIO
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
$endgroup$
Java, 41 bytes
Can't use a lambda (runtime error). Port of this Javascript answer
int f(int n){return n<3?1:f(n-2)+f(n-3);}
TIO
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
answered 2 days ago
Benjamin UrquhartBenjamin Urquhart
44017
44017
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
add a comment |
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
I think you are missing some requirements: Have a look at my modification here.
$endgroup$
– Shaq
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Please disregard Shaq's comment: your answer is correct and is the shortest Java answer possible (as of Java 12).
$endgroup$
– Olivier Grégoire
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
$begingroup$
Ok then. I'm not sure what I "missed" but ok. Edit: nvm I read the JS answer.
$endgroup$
– Benjamin Urquhart
yesterday
add a comment |
$begingroup$
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.
answered 2 days ago
rturnbullrturnbull
3,519925
$endgroup$
2
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
add a comment |
$begingroup$
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.
answered 2 days ago
rturnbullrturnbull
3,519925
$endgroup$
2
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
add a comment |
$begingroup$
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.
answered 2 days ago
rturnbullrturnbull
3,519925
$endgroup$
R + pryr, 38 36 bytes
Zero-indexed recursive function.
f=pryr::f(`if`(n<3,1,f(n-2)+f(n-3)))
Try it online!
Thanks to @Giuseppe for pointing out two obviously needless bytes.
answered 2 days ago
rturnbullrturnbull
3,519925
edited 2 days ago
answered 2 days ago
rturnbullrturnbull
3,519925
answered 2 days ago
rturnbullrturnbull
3,519925
answered 2 days ago
rturnbullrturnbull
3,519925
3,519925
2
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
add a comment |
2
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
2
2
$begingroup$
If you're going to be using
pryr, the language should be R + pryr and this can be 36 bytes$endgroup$
– Giuseppe
2 days ago
$begingroup$
If you're going to be using
pryr, the language should be R + pryr and this can be 36 bytes$endgroup$
– Giuseppe
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
2 days ago
add a comment |
$begingroup$
C (clang), 41 33 bytes
a(i){return i<3?1:a(i-2)+a(i-3);}
Try it online!
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
C (clang), 41 33 bytes
a(i){return i<3?1:a(i-2)+a(i-3);}
Try it online!
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
C (clang), 41 33 bytes
a(i){return i<3?1:a(i-2)+a(i-3);}
Try it online!
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
C (clang), 41 33 bytes
a(i){return i<3?1:a(i-2)+a(i-3);}
Try it online!
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 hours ago
answered yesterday
peterzugerpeterzuger
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
peterzugerpeterzuger
235
answered yesterday
peterzugerpeterzuger
235
235
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
peterzuger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
add a comment |
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
2
2
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
$begingroup$
Welcome to PPCG :)
$endgroup$
– Shaggy
yesterday
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 Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
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 Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
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 Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
$endgroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
answered Apr 7 at 21:17
Embodiment of IgnoranceEmbodiment of Ignorance
2,916127
2,916127
add a comment |
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 2 days ago
XcaliXcali
5,505520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 2 days ago
XcaliXcali
5,505520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 2 days ago
XcaliXcali
5,505520
$endgroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 2 days ago
XcaliXcali
5,505520
answered 2 days ago
XcaliXcali
5,505520
answered 2 days ago
XcaliXcali
5,505520
answered 2 days ago
XcaliXcali
5,505520
5,505520
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered yesterday
attinatattinat
4897
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered yesterday
attinatattinat
4897
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered yesterday
attinatattinat
4897
$endgroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered yesterday
attinatattinat
4897
answered yesterday
attinatattinat
4897
answered yesterday
attinatattinat
4897
answered yesterday
attinatattinat
4897
4897
add a comment |
add a comment |
$begingroup$
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 $frac{1+x}{1-x^2-x^3}$.
answered yesterday
alephalphaalephalpha
21.8k33094
$endgroup$
add a comment |
$begingroup$
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 $frac{1+x}{1-x^2-x^3}$.
answered yesterday
alephalphaalephalpha
21.8k33094
$endgroup$
add a comment |
$begingroup$
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 $frac{1+x}{1-x^2-x^3}$.
answered yesterday
alephalphaalephalpha
21.8k33094
$endgroup$
Pari/GP, 28 bytes
0-indexed.
f(n)=if(n<3,1,f(n-2)+f(n-3))
Try it online!
Pari/GP, 35 bytes
1-indexed.
n->Vec((1+x+O(x^n))/(1-x^2-x^3))[n]
Try it online!
The generating function of the sequence is $frac{1+x}{1-x^2-x^3}$.
answered yesterday
alephalphaalephalpha
21.8k33094
edited yesterday
answered yesterday
alephalphaalephalpha
21.8k33094
answered yesterday
alephalphaalephalpha
21.8k33094
answered yesterday
alephalphaalephalpha
21.8k33094
21.8k33094
add a comment |
add a comment |
1 2
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$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37$endgroup$
– 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
2 days ago
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
yesterday
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 thena_5=a_6=a_7=1$endgroup$
– Carmeister
yesterday