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Patience, young “Padovan”
Fibonacci function or sequenceGenerate a Padovan SpiralGenerate an ASCII Padovan SpiralGolf a Custom Fibonacci SequenceDivinacci SequenceGenerate unseen numbersImplement the Fibonacci sequence… Shifted to the rightDizzy integer enumerationModulus SummationTerms of the EKG sequence
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(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
$endgroup$
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
12 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
1 hour ago
add a comment |
$begingroup$
Everyone knows the Fibonacci sequence:
You take a square, attach an equal square to it, then repeatedly attach a square whose side length is equal to the largest side length of the resulting rectangle.
The result is a beautiful spiral of squares whose sequence of numbers is the Fibonacci sequence:
But, what if we didn't want to use squares?
If we use equilateral triangles—instead of squares—in a similar fashion, we get an equally beautiful spiral of triangles and a new sequence: the Padovan sequence, aka A000931:
Task:
Given a positive integer, $N$, output $a_N$, the $N$th term in the Padovan sequence OR the first $N$ terms.
Assume that the first three terms of the sequence are all $1$. Thus, the sequence will start as follows:
$$
1,1,1,2,2,3,...
$$
Input:
Any positive integer $Nge0$
Invalid input does not have to be taken into account
Output:
The $N$th term in the Padovan sequence OR the first $N$ terms of the Padovan sequence.
If the first $N$ terms are printed out, the output can be whatever is convenient (list/array, multi-line string, etc.)
Can be either $0$-indexed or $1$-indexed
Test Cases:
(0-indexed, $N$th term)
Input | Output
--------------
0 | 1
1 | 1
2 | 1
4 | 2
6 | 4
14 | 37
20 | 200
33 | 7739
(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 12 hours ago
Tau
asked yesterday
TauTau
911515
asked yesterday
TauTau
911515
asked yesterday
TauTau
911515
911515
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
12 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
1 hour ago
add a comment |
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
12 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
1 hour 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
$begingroup$
@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
$begingroup$
@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
12 hours ago
$begingroup$
@LuisMendo I believe so. I'll edit the post.
$endgroup$
– Tau
12 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
1 hour 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
1 hour ago
add a comment |
30 Answers
30
active
oldest
votes
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$beginbmatrix0&0&1 \ 1&0&1 \ 0&1&0endbmatrix^n$$
where the binary matrix is conveniently computed as: $$beginbmatrixmathsfisprime(0)&mathsfisprime(1)&mathsfisprime(2) \ mathsfisprime(3)&mathsfisprime(4)&mathsfisprime(5) \ mathsfisprime(6)&mathsfisprime(7)&mathsfisprime(8)endbmatrix$$
(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
50.5k898233
$endgroup$
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 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 |
$begingroup$
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 |
$begingroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_i=0^lfloorfracn2rfloorbinomi+1n-2i$
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
54k535174
$endgroup$
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 |
$begingroup$
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 |
$begingroup$
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
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f
Try it online!
answered 8 hours ago
XcaliXcali
5,465520
$endgroup$
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 6 hours ago
Neil A.Neil A.
1,348120
$endgroup$
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 5 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
$endgroup$
add a comment |
$begingroup$
Gaia, 16 14 bytes
7b@((⟨ṇ;(++⟩ₓ)
Try it online!
7b | push [1 1 1]
@(( | push input, decrement twice
⟨ ⟩ₓ | do the following that many times (0 times if 0 or less)
ṇ | pop the first element and leave the rest below
; | copy from below
( | take the first element
+ | add the two together
+ | and concatenate to the list. End loop.
) | finally, take the last element
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
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 4 hours ago
anatolyganatolyg
7,2292166
$endgroup$
add a comment |
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30 Answers
30
active
oldest
votes
30 Answers
30
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$beginbmatrix0&0&1 \ 1&0&1 \ 0&1&0endbmatrix^n$$
where the binary matrix is conveniently computed as: $$beginbmatrixmathsfisprime(0)&mathsfisprime(1)&mathsfisprime(2) \ mathsfisprime(3)&mathsfisprime(4)&mathsfisprime(5) \ mathsfisprime(6)&mathsfisprime(7)&mathsfisprime(8)endbmatrix$$
(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
50.5k898233
$endgroup$
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 hours ago
|
show 1 more comment
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$beginbmatrix0&0&1 \ 1&0&1 \ 0&1&0endbmatrix^n$$
where the binary matrix is conveniently computed as: $$beginbmatrixmathsfisprime(0)&mathsfisprime(1)&mathsfisprime(2) \ mathsfisprime(3)&mathsfisprime(4)&mathsfisprime(5) \ mathsfisprime(6)&mathsfisprime(7)&mathsfisprime(8)endbmatrix$$
(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
50.5k898233
$endgroup$
17
$begingroup$
this is clearly some kind of voodoo
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– Pureferret
14 hours ago
6
$begingroup$
This should be published.
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– YSC
13 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
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– flawr
9 hours ago
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@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 hours ago
|
show 1 more comment
$begingroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$beginbmatrix0&0&1 \ 1&0&1 \ 0&1&0endbmatrix^n$$
where the binary matrix is conveniently computed as: $$beginbmatrixmathsfisprime(0)&mathsfisprime(1)&mathsfisprime(2) \ mathsfisprime(3)&mathsfisprime(4)&mathsfisprime(5) \ mathsfisprime(6)&mathsfisprime(7)&mathsfisprime(8)endbmatrix$$
(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
50.5k898233
$endgroup$
Jelly, 10 bytes
9s3’Ẓæ*³FṀ
Try it online!
1-indexed. Computes the largest element of: $$beginbmatrix0&0&1 \ 1&0&1 \ 0&1&0endbmatrix^n$$
where the binary matrix is conveniently computed as: $$beginbmatrixmathsfisprime(0)&mathsfisprime(1)&mathsfisprime(2) \ mathsfisprime(3)&mathsfisprime(4)&mathsfisprime(5) \ mathsfisprime(6)&mathsfisprime(7)&mathsfisprime(8)endbmatrix$$
(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
50.5k898233
answered yesterday
LynnLynn
50.5k898233
answered yesterday
LynnLynn
50.5k898233
answered yesterday
LynnLynn
50.5k898233
50.5k898233
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 hours ago
|
show 1 more comment
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
6
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 hours ago
17
17
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
$begingroup$
this is clearly some kind of voodoo
$endgroup$
– Pureferret
14 hours ago
6
6
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
$begingroup$
This should be published.
$endgroup$
– YSC
13 hours ago
2
2
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@YSC It has already been published in A000931. I'd never have guess the primes trick:)
$endgroup$
– flawr
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 hours ago
$begingroup$
@flawr ho... I missed it
$endgroup$
– YSC
9 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
8 hours ago
$begingroup$
...make that "unless someone can golf two bytes off this one" :) (now that I have a 9 byter)
$endgroup$
– Jonathan Allan
8 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 $P(n) = sum_i=0^lfloorfracn2rfloorbinomi+1n-2i$
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
54k535174
$endgroup$
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 $P(n) = sum_i=0^lfloorfracn2rfloorbinomi+1n-2i$
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
54k535174
$endgroup$
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 $P(n) = sum_i=0^lfloorfracn2rfloorbinomi+1n-2i$
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
54k535174
$endgroup$
Jelly, 10 9 8 bytes
ŻṚm2Jc$S
A monadic Link accepting n (0-indexed) which yields P(n).
Try it online!
How?
Implements $P(n) = sum_i=0^lfloorfracn2rfloorbinomi+1n-2i$
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
54k535174
edited 1 hour ago
answered yesterday
Jonathan AllanJonathan Allan
54k535174
answered yesterday
Jonathan AllanJonathan Allan
54k535174
answered yesterday
Jonathan AllanJonathan Allan
54k535174
54k535174
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 _3improve this answer
answered yesterday
JonahJonah
2,6611017
$endgroup$
1
$begingroup$
Another 24-byte solution (boring) : (1#.2 3$:@-~])`1:@.(3&>) Try it online!
$endgroup$
– Galen Ivanov
5 hours ago
$begingroup$
23 bytes thanks to ngn1:->#: Try it online!
$endgroup$
– Galen Ivanov
5 hours ago
$begingroup$
@GalenIvanov tyvm, that's a great trick.
$endgroup$
– Jonah
4 hours ago
add a comment 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 16 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
Try it online!
answered 8 hours ago
XcaliXcali
5,465520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f
Try it online!
answered 8 hours ago
XcaliXcali
5,465520
$endgroup$
add a comment |
$begingroup$
Perl 5, 34 bytes
sub f
Try it online!
answered 8 hours ago
XcaliXcali
5,465520
$endgroup$
Perl 5, 34 bytes
sub f
Try it online!
answered 8 hours ago
XcaliXcali
5,465520
answered 8 hours ago
XcaliXcali
5,465520
answered 8 hours ago
XcaliXcali
5,465520
answered 8 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 6 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 6 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 6 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 6 hours ago
Neil A.Neil A.
1,348120
edited 6 hours ago
answered 6 hours ago
Neil A.Neil A.
1,348120
answered 6 hours ago
Neil A.Neil A.
1,348120
answered 6 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 5 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 5 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 5 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 5 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 5 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 5 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
answered 5 hours ago
Benjamin UrquhartBenjamin Urquhart
37017
37017
add a comment |
add a comment |
$begingroup$
Gaia, 16 14 bytes
7b@((⟨ṇ;(++⟩ₓ)
Try it online!
7b | push [1 1 1]
@(( | push input, decrement twice
⟨ ⟩ₓ | do the following that many times (0 times if 0 or less)
ṇ | pop the first element and leave the rest below
; | copy from below
( | take the first element
+ | add the two together
+ | and concatenate to the list. End loop.
) | finally, take the last element
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 16 14 bytes
7b@((⟨ṇ;(++⟩ₓ)
Try it online!
7b | push [1 1 1]
@(( | push input, decrement twice
⟨ ⟩ₓ | do the following that many times (0 times if 0 or less)
ṇ | pop the first element and leave the rest below
; | copy from below
( | take the first element
+ | add the two together
+ | and concatenate to the list. End loop.
) | finally, take the last element
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 16 14 bytes
7b@((⟨ṇ;(++⟩ₓ)
Try it online!
7b | push [1 1 1]
@(( | push input, decrement twice
⟨ ⟩ₓ | do the following that many times (0 times if 0 or less)
ṇ | pop the first element and leave the rest below
; | copy from below
( | take the first element
+ | add the two together
+ | and concatenate to the list. End loop.
) | finally, take the last element
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
$endgroup$
Gaia, 16 14 bytes
7b@((⟨ṇ;(++⟩ₓ)
Try it online!
7b | push [1 1 1]
@(( | push input, decrement twice
⟨ ⟩ₓ | do the following that many times (0 times if 0 or less)
ṇ | pop the first element and leave the rest below
; | copy from below
( | take the first element
+ | add the two together
+ | and concatenate to the list. End loop.
) | finally, take the last element
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
edited 5 hours ago
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
answered 5 hours ago
GiuseppeGiuseppe
17.6k31153
17.6k31153
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 4 hours ago
anatolyganatolyg
7,2292166
$endgroup$
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 4 hours ago
anatolyganatolyg
7,2292166
$endgroup$
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 4 hours ago
anatolyganatolyg
7,2292166
$endgroup$
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 4 hours ago
anatolyganatolyg
7,2292166
answered 4 hours ago
anatolyganatolyg
7,2292166
answered 4 hours ago
anatolyganatolyg
7,2292166
answered 4 hours ago
anatolyganatolyg
7,2292166
7,2292166
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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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
12 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
1 hour ago