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 yesterday
TauTau
911515
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|
show 2 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 yesterday
TauTau
911515
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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
yesterday
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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
yesterday
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@LuisMendo I believe so. I'll edit the post.
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– Tau
16 hours ago
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Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
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– sharur
6 hours 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
3 hours ago
|
show 2 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 yesterday
TauTau
911515
$endgroup$
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 yesterday
TauTau
911515
asked yesterday
TauTau
911515
edited 16 hours ago
Tau
asked yesterday
TauTau
911515
asked yesterday
TauTau
911515
asked yesterday
TauTau
911515
911515
2
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14(0-indexed) is shown as outputting28while I believe it should yield37
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– Jonathan Allan
yesterday
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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
yesterday
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@LuisMendo I believe so. I'll edit the post.
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– Tau
16 hours ago
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Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
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– sharur
6 hours 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
3 hours ago
|
show 2 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
yesterday
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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
yesterday
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@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
16 hours ago
$begingroup$
Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
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– sharur
6 hours 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.
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– Tau
3 hours ago
2
2
$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
yesterday
$begingroup$
14 (0-indexed) is shown as outputting 28 while I believe it should yield 37$endgroup$
– Jonathan Allan
yesterday
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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.
$endgroup$
– Tau
yesterday
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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.
$endgroup$
– Tau
yesterday
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@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
16 hours ago
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@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
16 hours ago
$begingroup$
Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
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– sharur
6 hours ago
$begingroup$
Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
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– sharur
6 hours ago
1
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.
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– Tau
3 hours ago
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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
3 hours ago
|
show 2 more comments
32 Answers
32
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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 yesterday
LynnLynn
50.6k898233
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17
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this is clearly some kind of voodoo
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– Pureferret
19 hours ago
6
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This should be published.
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– YSC
18 hours ago
2
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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
14 hours ago
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@flawr ho... I missed it
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– YSC
14 hours 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
13 hours ago
|
show 1 more comment
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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 yesterday
EmignaEmigna
47.6k433145
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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 yesterday
xnorxnor
93.6k18190450
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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 yesterday
Jonathan AllanJonathan Allan
54k536174
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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 yesterday
xnorxnor
93.6k18190450
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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 yesterday
J42161217J42161217
13.9k21353
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add a comment |
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J, 23 bytes
-1 byte thanks to ngn and Galen
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 23 bytes
(],1#._2 _3{ ::1:])^:[#
Try it online!
answered yesterday
JonahJonah
2,6611017
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1
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Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
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– Galen Ivanov
10 hours ago
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23 bytes thanks to ngn1:->#: Try it online!
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– Galen Ivanov
9 hours ago
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@GalenIvanov tyvm, that's a great trick.
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– Jonah
8 hours ago
add a comment |
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Python 2, 56 48 bytes
f=lambda n,a=1,b=1,c=1:n>2and f(n-1,b,c,a+b)or c
Try it online!
Returns nth value, 0-indexed.
answered yesterday
Chas BrownChas Brown
5,2191523
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add a comment |
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Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$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 yesterday
NeilNeil
82.6k745179
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add a comment |
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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 yesterday
MickyTMickyT
10.3k21637
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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 19 hours ago
Luis MendoLuis Mendo
75.3k889292
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add a comment |
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Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
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Can you specify whether this answer is 0-indexed or 1-indexed?
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– Tau
yesterday
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@Tau It's 0-indexed. I've edited it in.
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– Erik the Outgolfer
yesterday
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
yesterday
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@JoKing Beats me. If I did it somehow, it wasn't on purpose.
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– Sean
12 hours ago
add a comment |
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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 yesterday
cyclaministcyclaminist
1813
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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 19 hours ago
Sherlock9Sherlock9
8,16411860
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add a comment |
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K (ngn/k), 24 20 bytes
-4 bytes thanks to ngn!
{$[x<3;1;+/o'x-2 3]}
Try it online!
0-indexed, first N terms
answered 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
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1
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f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
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– ngn
11 hours ago
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@ngn Thanks! I tried it (without success) in J; it's elegant here.
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– Galen Ivanov
10 hours ago
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@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
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– Galen Ivanov
10 hours ago
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ah, right, base-1 decode is a train-friendly way to sum :)
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– ngn
10 hours ago
2
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1:->#in the j solution
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– ngn
10 hours ago
|
show 2 more comments
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JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
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I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
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– Nit
17 hours ago
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@Nit Relevant meta post.
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– Arnauld
17 hours ago
add a comment |
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Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
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Looks like 12 is the best we can do :
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– Shaggy
yesterday
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 yesterday
RK.RK.
407211
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y-b2y-b3could maybe be refactored with either bifurcate orL? Though declaring an array of 2 elements is costly.yL-Lb2,3is longer :(
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– Ven
22 hours ago
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@Ven I was able to replace+y-b2y-b3withsmy-bdhB2which is the same amount of bytes;hB2results in the array[2, 3]
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– RK.
14 hours ago
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Well done onhB2. Too bad it's the same byte count.
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– Ven
14 hours ago
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Yeah, though I wonder if there's some way to get rid of thedin the map.
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– RK.
14 hours 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 yesterday
Mr. XcoderMr. Xcoder
32.2k759200
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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
19 hours 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 11 hours ago
rturnbullrturnbull
3,499925
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1
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If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
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– Giuseppe
10 hours ago
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@Giuseppe thanks! Updated now.
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– rturnbull
9 hours ago
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 yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
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 yesterday
TauTau
911515
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add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 22 hours ago
G BG B
8,2161429
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add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
ÈnZs3n)x}hBì :Implicit input of integer U
B :11
ì :Convert to digit array
h :Repeat the following until the length of the array is U, pushing the result to the array each time
È : Take the last element X from the array Z and pass it through the following function
n : Subtract X from
Zs : Slice Z
3n : -3, giving the last 3 elements in the array
) : End slice
x : Reduce by addition
} : End function
answered yesterday
ShaggyShaggy
18.9k21768
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add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 13 hours ago
XcaliXcali
5,465520
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add a comment |
$begingroup$
C++ (gcc), 81 75 bytes
-6 bytes to small golfing
int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;)d=a+b,a=b,b=c,c=d;return c;}
Try it online!
Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.
answered 11 hours ago
Neil A.Neil A.
1,348120
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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 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
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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 8 hours ago
anatolyganatolyg
7,2392166
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I love to see machine code answers on PP&CG! +1
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– Tau
3 hours ago
add a comment |
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Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered 1 hour ago
attinatattinat
4897
$endgroup$
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 yesterday
LynnLynn
50.6k898233
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17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
13 hours ago
|
show 1 more comment
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered yesterday
LynnLynn
50.6k898233
$endgroup$
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
13 hours ago
|
show 1 more comment
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$begin{bmatrix}0&0&1 \ 1&0&1 \ 0&1&0end{bmatrix}^n$$
where the binary matrix is conveniently computed as: $$begin{bmatrix}mathsf{isprime}(0)&mathsf{isprime}(1)&mathsf{isprime}(2) \ mathsf{isprime}(3)&mathsf{isprime}(4)&mathsf{isprime}(5) \ mathsf{isprime}(6)&mathsf{isprime}(7)&mathsf{isprime}(8)end{bmatrix}$$
(this is a total coincidence.)
9s3 [[1,2,3],[4,5,6],[7,8,9]] 9 split 3
’ [[0,1,2],[3,4,5],[6,7,8]] decrease
Ẓ [[0,0,1],[1,0,1],[0,1,0]] isprime
æ*³ [[0,0,1],[1,0,1],[0,1,0]]^n matrix power by input
FṀ flatten, maximum
answered yesterday
LynnLynn
50.6k898233
$endgroup$
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 yesterday
LynnLynn
50.6k898233
answered yesterday
LynnLynn
50.6k898233
answered yesterday
LynnLynn
50.6k898233
answered yesterday
LynnLynn
50.6k898233
50.6k898233
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
13 hours ago
|
show 1 more comment
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours ago
1
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
13 hours ago
17
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
19 hours ago
6
6
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
$begingroup$
This should be published.
$endgroup$
– YSC
18 hours ago
2
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
14 hours 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
13 hours ago
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
13 hours ago
|
show 1 more 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 yesterday
EmignaEmigna
47.6k433145
$endgroup$
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 yesterday
EmignaEmigna
47.6k433145
$endgroup$
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 yesterday
EmignaEmigna
47.6k433145
$endgroup$
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 yesterday
EmignaEmigna
47.6k433145
answered yesterday
EmignaEmigna
47.6k433145
answered yesterday
EmignaEmigna
47.6k433145
answered yesterday
EmignaEmigna
47.6k433145
47.6k433145
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
answered yesterday
xnorxnor
93.6k18190450
answered yesterday
xnorxnor
93.6k18190450
answered yesterday
xnorxnor
93.6k18190450
93.6k18190450
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 yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
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 yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
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 yesterday
Jonathan AllanJonathan Allan
54k536174
$endgroup$
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 yesterday
Jonathan AllanJonathan Allan
54k536174
edited 5 hours ago
answered yesterday
Jonathan AllanJonathan Allan
54k536174
answered yesterday
Jonathan AllanJonathan Allan
54k536174
answered yesterday
Jonathan AllanJonathan Allan
54k536174
54k536174
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
$endgroup$
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 yesterday
xnorxnor
93.6k18190450
edited yesterday
answered yesterday
xnorxnor
93.6k18190450
answered yesterday
xnorxnor
93.6k18190450
answered yesterday
xnorxnor
93.6k18190450
93.6k18190450
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 yesterday
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 yesterday
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 yesterday
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 yesterday
J42161217J42161217
13.9k21353
answered yesterday
J42161217J42161217
13.9k21353
answered yesterday
J42161217J42161217
13.9k21353
answered yesterday
J42161217J42161217
13.9k21353
13.9k21353
add a comment |
add a comment |
$begingroup$
J, 23 bytes
-1 byte thanks to ngn and Galen
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 23 bytes
(],1#._2 _3{ ::1:])^:[#
Try it online!
answered yesterday
JonahJonah
2,6611017
$endgroup$
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
add a comment |
$begingroup$
J, 23 bytes
-1 byte thanks to ngn and Galen
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 23 bytes
(],1#._2 _3{ ::1:])^:[#
Try it online!
answered yesterday
JonahJonah
2,6611017
$endgroup$
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
add a comment |
$begingroup$
J, 23 bytes
-1 byte thanks to ngn and Galen
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 23 bytes
(],1#._2 _3{ ::1:])^:[#
Try it online!
answered yesterday
JonahJonah
2,6611017
$endgroup$
J, 23 bytes
-1 byte thanks to ngn and Galen
closed form, 26 bytes
0.5<.@+1.04535%~1.32472^<:
Try it online!
iterative, 23 bytes
(],1#._2 _3{ ::1:])^:[#
Try it online!
answered yesterday
JonahJonah
2,6611017
edited 8 hours ago
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
2,6611017
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
add a comment |
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
1
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
23 bytes thanks to ngn
1: -> # : Try it online!$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
23 bytes thanks to ngn
1: -> # : Try it online!$endgroup$
– Galen Ivanov
9 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
8 hours ago
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 yesterday
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 yesterday
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 yesterday
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 yesterday
Chas BrownChas Brown
5,2191523
answered yesterday
Chas BrownChas Brown
5,2191523
answered yesterday
Chas BrownChas Brown
5,2191523
answered yesterday
Chas BrownChas Brown
5,2191523
5,2191523
add a comment |
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 yesterday
NeilNeil
82.6k745179
$endgroup$
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered yesterday
NeilNeil
82.6k745179
$endgroup$
add a comment |
$begingroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered yesterday
NeilNeil
82.6k745179
$endgroup$
Retina, 47 42 bytes
K`0¶1¶0
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
6,G`
Try it online! Outputs the first n terms on separate lines. Explanation:
K`0¶1¶0
Replace the input with the terms for -2, -1 and 0.
"$+"+`.+¶(.+)¶.+$
$&¶$.(*_$1*
Generate the next n terms using the recurrence relation. *_ here is short for $&*_ which converts the (first) number in the match to unary, while $1* is short for $1*_ which converts the middle number to unary. The $.( returns the decimal sum of its unary arguments, i.e. the sum of the first and middle numbers.
6,G`
Discard the first six characters, i.e. the first three lines.
answered yesterday
NeilNeil
82.6k745179
edited yesterday
answered yesterday
NeilNeil
82.6k745179
answered yesterday
NeilNeil
82.6k745179
answered yesterday
NeilNeil
82.6k745179
82.6k745179
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 yesterday
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 yesterday
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 yesterday
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 yesterday
MickyTMickyT
10.3k21637
answered yesterday
MickyTMickyT
10.3k21637
answered yesterday
MickyTMickyT
10.3k21637
answered yesterday
MickyTMickyT
10.3k21637
10.3k21637
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 19 hours 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 19 hours 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 19 hours 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 19 hours ago
Luis MendoLuis Mendo
75.3k889292
edited 19 hours ago
answered 19 hours ago
Luis MendoLuis Mendo
75.3k889292
answered 19 hours ago
Luis MendoLuis Mendo
75.3k889292
answered 19 hours ago
Luis MendoLuis Mendo
75.3k889292
75.3k889292
add a comment |
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
add a comment |
$begingroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Jelly, 11 bytes
5B+Ɲ2ị;Ʋ⁸¡Ḣ
Try it online!
0-indexed.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
edited yesterday
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
add a comment |
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
Can you specify whether this answer is 0-indexed or 1-indexed?
$endgroup$
– Tau
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
$begingroup$
@Tau It's 0-indexed. I've edited it in.
$endgroup$
– Erik the Outgolfer
yesterday
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
yesterday
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
yesterday
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
add a comment |
$begingroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
$endgroup$
Perl 6, 24 bytes
{(1,1,1,*+*+!*...*)[$_]}
Try it online!
A pretty standard generated sequence, with each new element generated by the expression * + * + !*. That adds the third-previous element, the second-previous element, and the logical negation of the previous element, which is always False, which is numerically zero.
answered yesterday
community wiki
Sean
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
yesterday
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
add a comment |
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
yesterday
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
yesterday
$begingroup$
Why is this community wiki?
$endgroup$
– Jo King
yesterday
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
$begingroup$
@JoKing Beats me. If I did it somehow, it wasn't on purpose.
$endgroup$
– Sean
12 hours ago
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 yesterday
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 yesterday
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 yesterday
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 yesterday
cyclaministcyclaminist
1813
edited 22 hours ago
answered yesterday
cyclaministcyclaminist
1813
answered yesterday
cyclaministcyclaminist
1813
answered yesterday
cyclaministcyclaminist
1813
1813
add a comment |
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 19 hours ago
Sherlock9Sherlock9
8,16411860
$endgroup$
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 19 hours ago
Sherlock9Sherlock9
8,16411860
$endgroup$
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 19 hours ago
Sherlock9Sherlock9
8,16411860
$endgroup$
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 19 hours ago
Sherlock9Sherlock9
8,16411860
edited 15 hours ago
answered 19 hours ago
Sherlock9Sherlock9
8,16411860
answered 19 hours ago
Sherlock9Sherlock9
8,16411860
answered 19 hours ago
Sherlock9Sherlock9
8,16411860
8,16411860
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 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
11 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
10 hours 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 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
11 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
10 hours 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 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
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 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
edited 10 hours ago
answered 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
answered 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
answered 22 hours ago
Galen IvanovGalen Ivanov
7,41211034
7,41211034
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
11 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
10 hours ago
|
show 2 more comments
1
$begingroup$
f[x-2]+f[x-3]->+/o'x-2 3(ois "recur")
$endgroup$
– ngn
11 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
2
$begingroup$
1:->#in the j solution
$endgroup$
– ngn
10 hours ago
1
1
$begingroup$
f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")$endgroup$
– ngn
11 hours ago
$begingroup$
f[x-2]+f[x-3] -> +/o'x-2 3 (o is "recur")$endgroup$
– ngn
11 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn Thanks! I tried it (without success) in J; it's elegant here.
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
@ngn In fact here's one possibillity how it looks in J: (1#.2 3$:@-~])`1:@.(3&>)
$endgroup$
– Galen Ivanov
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
$begingroup$
ah, right, base-1 decode is a train-friendly way to sum :)
$endgroup$
– ngn
10 hours ago
2
2
$begingroup$
1: -> # in the j solution$endgroup$
– ngn
10 hours ago
$begingroup$
1: -> # in the j solution$endgroup$
– ngn
10 hours ago
|
show 2 more comments
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
$endgroup$
JavaScript (ES6), 23 bytes
Implements the recursive definition of A000931. Returns the $N$th term, 0-indexed.
f=n=>n<3||f(n-2)+f(n-3)
Try it online!
answered yesterday
ArnauldArnauld
80.6k797334
edited yesterday
answered yesterday
ArnauldArnauld
80.6k797334
answered yesterday
ArnauldArnauld
80.6k797334
answered yesterday
ArnauldArnauld
80.6k797334
80.6k797334
$begingroup$
I don't think it's reasonable to say that returningtrueis the same as returning1if the rest of the output is numbers.
$endgroup$
– Nit
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours ago
add a 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
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours 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
17 hours 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
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours ago
$begingroup$
@Nit Relevant meta post.
$endgroup$
– Arnauld
17 hours ago
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
Japt -N, 12 bytes
<3ªßUµ2 +ß´U
Try it
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
2,866127
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
add a comment |
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
$begingroup$
Looks like 12 is the best we can do :
$endgroup$
– Shaggy
yesterday
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 yesterday
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
22 hours 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.
14 hours ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
14 hours 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 yesterday
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
22 hours 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.
14 hours ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
14 hours 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 yesterday
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 yesterday
RK.RK.
407211
answered yesterday
RK.RK.
407211
answered yesterday
RK.RK.
407211
answered yesterday
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
22 hours 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.
14 hours ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
14 hours 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
22 hours 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.
14 hours ago
$begingroup$
Well done onhB2. Too bad it's the same byte count.
$endgroup$
– Ven
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of thedin the map.
$endgroup$
– RK.
14 hours 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
22 hours 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
22 hours 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.
14 hours 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.
14 hours ago
$begingroup$
Well done on
hB2. Too bad it's the same byte count.$endgroup$
– Ven
14 hours ago
$begingroup$
Well done on
hB2. Too bad it's the same byte count.$endgroup$
– Ven
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of the
d in the map.$endgroup$
– RK.
14 hours ago
$begingroup$
Yeah, though I wonder if there's some way to get rid of the
d in the map.$endgroup$
– RK.
14 hours 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 yesterday
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
$begingroup$
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
19 hours 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 yesterday
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
$begingroup$
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
19 hours 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 yesterday
Mr. XcoderMr. Xcoder
32.2k759200
$endgroup$
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 yesterday
Mr. XcoderMr. Xcoder
32.2k759200
edited yesterday
answered yesterday
Mr. XcoderMr. Xcoder
32.2k759200
answered yesterday
Mr. XcoderMr. Xcoder
32.2k759200
answered yesterday
Mr. XcoderMr. Xcoder
32.2k759200
32.2k759200
$begingroup$
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
19 hours 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
19 hours 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
19 hours 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
19 hours 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 11 hours ago
rturnbullrturnbull
3,499925
$endgroup$
1
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
10 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours 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 11 hours ago
rturnbullrturnbull
3,499925
$endgroup$
1
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
10 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours 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 11 hours ago
rturnbullrturnbull
3,499925
$endgroup$
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 11 hours ago
rturnbullrturnbull
3,499925
edited 9 hours ago
answered 11 hours ago
rturnbullrturnbull
3,499925
answered 11 hours ago
rturnbullrturnbull
3,499925
answered 11 hours ago
rturnbullrturnbull
3,499925
3,499925
1
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
10 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours ago
add a comment |
1
$begingroup$
If you're going to be usingpryr, the language should beR + pryrand this can be 36 bytes
$endgroup$
– Giuseppe
10 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours ago
1
1
$begingroup$
If you're going to be using
pryr, the language should be R + pryr and this can be 36 bytes$endgroup$
– Giuseppe
10 hours ago
$begingroup$
If you're going to be using
pryr, the language should be R + pryr and this can be 36 bytes$endgroup$
– Giuseppe
10 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours ago
$begingroup$
@Giuseppe thanks! Updated now.
$endgroup$
– rturnbull
9 hours ago
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 yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
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 yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
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 yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
C# (Visual C# Interactive Compiler), 34 bytes
int f(int g)=>g<3?1:f(g-2)+f(g-3);
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
2,866127
add a comment |
add a comment |
$begingroup$
TI-BASIC (TI-84), 34 bytes
[[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1
0-indexed $N$th term of the sequence.
Input is in Ans.
Output is in Ans and is automatically printed out.
I figured that enough time had passed, plus multiple answers had been posted, of which there were many which out-golfed this answer.
Example:
0
0
prgmCDGFD
1
9
9
prgmCDGFD
9
16
16
prgmCDGFD
65
Explanation:
[[0,1,0][0,0,1][1,1,0]]^(Ans+5:Ans(1,1 ;full program (example input: 6)
[[0,1,0][0,0,1][1,1,0]] ;generate the following matrix:
; [0 1 0]
; [0 0 1]
; [1 1 0]
^(Ans+5 ;then raise it to the power of: input + 5
; [4 7 5]
; [5 9 7]
; [7 12 9]
Ans(1,1 ;get the top-left index and leave it in "Ans"
;implicitly print Ans
answered yesterday
TauTau
911515
$endgroup$
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 yesterday
TauTau
911515
$endgroup$
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 yesterday
TauTau
911515
$endgroup$
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 yesterday
TauTau
911515
answered yesterday
TauTau
911515
answered yesterday
TauTau
911515
answered yesterday
TauTau
911515
911515
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 22 hours ago
G BG B
8,2161429
$endgroup$
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 22 hours ago
G BG B
8,2161429
$endgroup$
add a comment |
$begingroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 22 hours ago
G BG B
8,2161429
$endgroup$
Ruby, 26 bytes
f=->n{n<3?1:f[n-2]+f[n-3]}
Try it online!
answered 22 hours ago
G BG B
8,2161429
answered 22 hours ago
G BG B
8,2161429
answered 22 hours ago
G BG B
8,2161429
answered 22 hours ago
G BG B
8,2161429
8,2161429
add a comment |
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
ÈnZs3n)x}hBì :Implicit input of integer U
B :11
ì :Convert to digit array
h :Repeat the following until the length of the array is U, pushing the result to the array each time
È : Take the last element X from the array Z and pass it through the following function
n : Subtract X from
Zs : Slice Z
3n : -3, giving the last 3 elements in the array
) : End slice
x : Reduce by addition
} : End function
answered yesterday
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
ÈnZs3n)x}hBì :Implicit input of integer U
B :11
ì :Convert to digit array
h :Repeat the following until the length of the array is U, pushing the result to the array each time
È : Take the last element X from the array Z and pass it through the following function
n : Subtract X from
Zs : Slice Z
3n : -3, giving the last 3 elements in the array
) : End slice
x : Reduce by addition
} : End function
answered yesterday
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
ÈnZs3n)x}hBì :Implicit input of integer U
B :11
ì :Convert to digit array
h :Repeat the following until the length of the array is U, pushing the result to the array each time
È : Take the last element X from the array Z and pass it through the following function
n : Subtract X from
Zs : Slice Z
3n : -3, giving the last 3 elements in the array
) : End slice
x : Reduce by addition
} : End function
answered yesterday
ShaggyShaggy
18.9k21768
$endgroup$
Japt, 12 bytes
Returns the first n terms, 0-indexed. Replace h with g to return the nth term, 1-indexed.
ÈnZs3n)x}hBì
Try it
ÈnZs3n)x}hBì :Implicit input of integer U
B :11
ì :Convert to digit array
h :Repeat the following until the length of the array is U, pushing the result to the array each time
È : Take the last element X from the array Z and pass it through the following function
n : Subtract X from
Zs : Slice Z
3n : -3, giving the last 3 elements in the array
) : End slice
x : Reduce by addition
} : End function
answered yesterday
ShaggyShaggy
18.9k21768
edited 20 hours ago
answered yesterday
ShaggyShaggy
18.9k21768
answered yesterday
ShaggyShaggy
18.9k21768
answered yesterday
ShaggyShaggy
18.9k21768
18.9k21768
add a comment |
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 13 hours ago
XcaliXcali
5,465520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 13 hours ago
XcaliXcali
5,465520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 13 hours ago
XcaliXcali
5,465520
$endgroup$
Perl 5, 34 bytes
sub f{"@_"<3||f("@_"-2)+f("@_"-3)}
Try it online!
answered 13 hours ago
XcaliXcali
5,465520
answered 13 hours ago
XcaliXcali
5,465520
answered 13 hours ago
XcaliXcali
5,465520
answered 13 hours ago
XcaliXcali
5,465520
5,465520
add a comment |
add a comment |
$begingroup$
C++ (gcc), 81 75 bytes
-6 bytes to small golfing
int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;)d=a+b,a=b,b=c,c=d;return c;}
Try it online!
Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.
answered 11 hours ago
Neil A.Neil A.
1,348120
$endgroup$
add a comment |
$begingroup$
C++ (gcc), 81 75 bytes
-6 bytes to small golfing
int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;)d=a+b,a=b,b=c,c=d;return c;}
Try it online!
Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.
answered 11 hours ago
Neil A.Neil A.
1,348120
$endgroup$
add a comment |
$begingroup$
C++ (gcc), 81 75 bytes
-6 bytes to small golfing
int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;)d=a+b,a=b,b=c,c=d;return c;}
Try it online!
Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.
answered 11 hours ago
Neil A.Neil A.
1,348120
$endgroup$
C++ (gcc), 81 75 bytes
-6 bytes to small golfing
int a(int n){int a=1,b=1,c=1,d,i=2;for(;i++<n;)d=a+b,a=b,b=c,c=d;return c;}
Try it online!
Simple function to compute the values iteratively. No loop occurs for n<3, so the first cases default to the initial 1.
answered 11 hours ago
Neil A.Neil A.
1,348120
edited 10 hours ago
answered 11 hours ago
Neil A.Neil A.
1,348120
answered 11 hours ago
Neil A.Neil A.
1,348120
answered 11 hours ago
Neil A.Neil A.
1,348120
1,348120
add a comment |
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 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
$endgroup$
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 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
$endgroup$
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 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
$endgroup$
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 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 9 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
37017
add a comment |
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 8 hours ago
anatolyganatolyg
7,2392166
$endgroup$
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
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 8 hours ago
anatolyganatolyg
7,2392166
$endgroup$
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
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 8 hours ago
anatolyganatolyg
7,2392166
$endgroup$
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 8 hours ago
anatolyganatolyg
7,2392166
answered 8 hours ago
anatolyganatolyg
7,2392166
answered 8 hours ago
anatolyganatolyg
7,2392166
answered 8 hours ago
anatolyganatolyg
7,2392166
7,2392166
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
add a comment |
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
$begingroup$
I love to see machine code answers on PP&CG! +1
$endgroup$
– Tau
3 hours ago
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered 1 hour ago
attinatattinat
4897
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered 1 hour ago
attinatattinat
4897
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered 1 hour ago
attinatattinat
4897
$endgroup$
Wolfram Language (Mathematica), 26 bytes
If[#<3,1,#0[#-2]+#0[#-3]]&
Try it online!
answered 1 hour ago
attinatattinat
4897
answered 1 hour ago
attinatattinat
4897
answered 1 hour ago
attinatattinat
4897
answered 1 hour ago
attinatattinat
4897
4897
add a comment |
add a comment |
1 2
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2
$begingroup$
14(0-indexed) is shown as outputting28while I believe it should yield37$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@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
yesterday
$begingroup$
@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
16 hours ago
$begingroup$
Not to detract from the question, but is this the actual definition of the Fibonacci sequence? I was taught it as a sequence of numbers, in which the first two numbers are 1, and the 3rd and subsequent numbers are the sum of the prior two numbers. Then again, I was taught this as an example of a problem to solve with recursion...
$endgroup$
– sharur
6 hours 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
3 hours ago