# Fibonacci function or sequence

The Fibonacci sequence is a sequence of numbers, where every number in the sequence is the sum of the two numbers preceding it. The first two numbers in the sequence are both 1. Here are the first few terms:

1 1 2 3 5 8 13 21 34 55 89 ...


Write the shortest code that either:

• Generates the Fibonacci sequence without end.

• Given n calculates the nth term of the sequence. (Either 1 or zero indexed)

You may use standard forms of input and output.

(I gave both options in case one is easier to do in your chosen language than the other.)

For the function that takes an n, a reasonably large return value (the largest Fibonacci number that fits your computer's normal word size, at a minimum) has to be supported.

/* Configuration */

var QUESTION_ID = 85; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 3; // This should be the user ID of the challenge author.

/* App */

return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(a) {
});
comment_page = 1;
}
});
}

jQuery.ajax({
method: "get",
dataType: "jsonp",
crossDomain: true,
success: function (data) {
data.items.forEach(function(c) {
if (c.owner.user_id === OVERRIDE_USER)
});
else process();
}
});
}

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

function getAuthorName(a) {
return a.owner.display_name;
}

function process() {
var valid = [];

var body = a.body;
if(OVERRIDE_REG.test(c.body))
body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
});

var match = body.match(SCORE_REG);
if (match)
valid.push({
user: getAuthorName(a),
size: +match[2],
language: match[1],
});
else console.log(body);
});

valid.sort(function (a, b) {
var aB = a.size,
bB = b.size;
return aB - bB
});

var languages = {};
var place = 1;
var lastSize = null;
var lastPlace = 1;
valid.forEach(function (a) {
if (a.size != lastSize)
lastPlace = place;
lastSize = a.size;
++place;

.replace("{{NAME}}", a.user)
.replace("{{LANGUAGE}}", a.language)
.replace("{{SIZE}}", a.size)

var lang = a.language;
lang = jQuery('<a>'+lang+'</a>').text();

languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
});

var langs = [];
for (var lang in languages)
if (languages.hasOwnProperty(lang))
langs.push(languages[lang]);

langs.sort(function (a, b) {
if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
return 0;
});

for (var i = 0; i < langs.length; ++i)
{
var language = jQuery("#language-template").html();
var lang = langs[i];
language = language.replace("{{LANGUAGE}}", lang.lang)
.replace("{{NAME}}", lang.user)
.replace("{{SIZE}}", lang.size)
language = jQuery(language);
jQuery("#languages").append(language);
}

}
body {
text-align: left !important;
display: block !important;
}

width: 290px;
float: left;
}

#language-list {
width: 290px;
float: left;
}

font-weight: bold;
}

table td {
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="language-list">
<h2>Shortest Solution by Language</h2>
<table class="language-list">
<tr><td>Language</td><td>User</td><td>Score</td></tr>
<tbody id="languages">

</tbody>
</table>
</div>
<tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>

</tbody>
</table>
</div>
<table style="display: none">
</tbody>
</table>
<table style="display: none">
<tbody id="language-template">
</tbody>
</table>

• I am sort of waiting for a response like "f", 1 byte, in my math based golf language. Aug 11 '20 at 11:57

# x86 Machine Code (ELF format) - 484 bytes

This program will calculate fibonacci digits until there is no memory left, so you might want to process the output to get Nth one you are looking for.

Offset(h) 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0F

00000000  7F 45 4C 46 01 01 01 00 00 00 00 00 00 00 00 00  .ELF............
00000010  02 00 03 00 01 00 00 00 C0 80 04 08 34 00 00 00  ........Ŕ€..4...
00000020  00 00 00 00 00 00 00 00 34 00 20 00 02 00 28 00  ........4. ...(.
00000030  00 00 00 00 01 00 00 00 00 00 00 00 00 80 04 08  .............€..
00000040  00 00 00 00 E4 01 00 00 00 10 00 00 05 00 00 00  ....ä...........
00000050  00 10 00 00 01 00 00 00 00 00 00 00 00 90 04 08  ................
00000060  00 00 00 00 00 00 00 00 00 00 10 00 06 00 00 00  ................
00000070  00 10 00 00 00 00 00 00 00 00 00 00 00 00 00 00  ................
00000080  51 B9 00 90 04 08 88 01 31 C0 BA 01 00 00 00 EB  Qą......1Ŕş....ë
00000090  03 51 89 C1 31 C0 89 C3 43 B0 04 CD 80 31 C0 99  .Q‰Á1Ŕ‰ĂC°.Í€1Ŕ™
000000A0  42 59 C3 00 00 00 00 00 00 00 00 00 00 00 00 00  BYĂ.............
000000B0  00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00  ................
000000C0  31 C0 99 42 B9 03 90 04 08 C6 41 01 0A C6 41 02  1Ŕ™Bą....ĆA..ĆA.
000000D0  01 C6 41 03 01 3A 71 03 0F 84 FF 00 00 00 3A 71  .ĆA..:q..„˙...:q
000000E0  03 74 26 80 41 03 05 0F B6 41 03 6B C0 08 00 41  .t&€A...¶A.kŔ..A
000000F0  04 8A 41 04 E8 87 FF FF FF 80 69 04 30 C6 41 03  .ŠA.č‡˙˙˙€i.0ĆA.
00000100  01 83 E9 03 3A 71 03 75 DA 8A 41 04 E8 6F FF FF  ..é.:q.uÚŠA.čo˙˙
00000110  FF 3A 71 06 0F 84 BA 00 00 00 0F B6 41 05 88 41  ˙:q..„ş....¶A..A
00000120  06 0F B6 41 07 88 41 05 0F B6 41 07 00 41 06 C6  ..¶A..A..¶A..A.Ć
00000130  41 07 00 3A 71 06 0F 84 88 00 00 00 C6 41 07 01  A..:q..„....ĆA..
00000140  FE 49 06 3A 71 06 0F 84 78 00 00 00 C6 41 07 02  ţI.:q..„x...ĆA..
00000150  FE 49 06 3A 71 06 0F 84 68 00 00 00 C6 41 07 03  ţI.:q..„h...ĆA..
00000160  FE 49 06 3A 71 06 0F 84 58 00 00 00 C6 41 07 04  ţI.:q..„X...ĆA..
00000170  FE 49 06 3A 71 06 74 4C C6 41 07 05 FE 49 06 3A  ţI.:q.tLĆA..ţI.:
00000180  71 06 74 40 C6 41 07 06 FE 49 06 3A 71 06 74 34  q.t@ĆA..ţI.:q.t4
00000190  C6 41 07 07 FE 49 06 3A 71 06 74 28 C6 41 07 08  ĆA..ţI.:q.t(ĆA..
000001A0  FE 49 06 3A 71 06 74 1C C6 41 07 09 FE 49 06 3A  ţI.:q.t.ĆA..ţI.:
000001B0  71 06 74 10 FE 41 08 FE 41 09 FE 49 06 0F B6 41  q.t.ţA.ţA.ţI..¶A
000001C0  06 88 41 07 C6 41 06 01 83 C1 03 3A 71 06 0F 85  ..A.ĆA...Á.:q..…
000001D0  46 FF FF FF 3A 71 03 0F 85 01 FF FF FF B3 00 31  F˙˙˙:q..….˙˙˙ł.1
000001E0  C0 40 CD 80                                      Ŕ@Í€


## 8086/8088 machine code, 8 bytes

The machine code is on the left; the middle column is disassembly and the rightmost column is an explanation.

40      ; inc ax          set ax to 1
41      ; inc cx          set cx to 1
ef      ; out [dx], ax    output ax to port [dx] (that is, port 0)
03 c1   ; add ax, cx      set ax to ax + cx
91      ; xchg ax, cx     exchange ax with cx


Assumptions:

• The registers AX, CX and DX are initially 0.
• A number may be output by writing it as a word to port 0.

This runs forever, outputting the Fibonacci numbers modulo $$\2^{16}\$$ = 65,536, starting with 1, 1.

# Brain-Flak, 3634 32 bytes

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


Outputs the nth number of the zero indexed Fibonacci sequence (F(0) = 1, F(1) = 1, etc.)

Try it online!

## Oasis, 2 bytes

Answer to the open exercise on the Oasis repo.

+T


## Explanation

Expanded program:

bc+10


When + requires 1 parameter, it tries to calculate a(n-1). For the other parameter, it tries to calculate a(n-2). (Hence the expansion.)

In addition, the T instruction expands to 10 in the program, which are the base test cases (a(0) is 0. a(1) is 1. Since base test cases are popped from the end before the Oasis program is executed in reverse.)

TIO

# Gol><>, 7 bytes

12K+:N!


Byte knocked off courtesy of JoKing

Try it online!

9 bytes

01T2K+:Nt


Outputs forever with newlines

Try it online!

• You don't need the leading 0, which means you can just strip out the teleporters and put a skip at the end instead. Try it online!. You can also do 1:@+:N! for the same amount of bytes if you prefer the stack not filling up
– Jo King
Oct 17 '19 at 0:54

# JVM Byte Code, 79 bytes

The byte count given is for a complete class file that defines a class Code containing single static method Code which implement Fibonacci. A hex dump of the file is given below.

0000000: cafe babe 0003 002d 0004 0100 0428 4929  .......-.....(I)
0000010: 4901 0004 436f 6465 0700 0200 2000 0300  I...Code.... ...
0000020: 0000 0000 0000 0100 0800 0200 0100 0100  ................
0000030: 0200 0000 1800 0400 0100 0000 0c03 045a  ...............Z
0000040: 6084 00ff 1a9d fffa ac00 0000 0000 00    ..............


A helper class is required to invoke the function.

class Test {
public static void main(String[] args){
for(int i = 0; i < 20; i++){
System.out.println(Code.Code(i));
}
}
}


Try it online!

The disassembly of the class file with javap -c shows actual byte code implementation of Fibonacci. This requires only 11 bytes, the rest of the 79 bytes in the class file are various header tables that can't be omitted.

class Code {
static int Code(int);
Code:
0: iconst_0
1: iconst_1
2: dup_x1
4: iinc          0, -1
8: ifgt          2
11: ireturn
}


# Pyth, 13 bytes

A(Z1)#HA(H+HG


Try it online!

### Explanation

A(Z1)#HA(H+HG

(Z1)           : The tuple (0, 1)
A               : Assign the first value of the tuple to G and the second to H
#          : Loop until error statement
H         : Print H
(H+HG   : The tuple (H, H + G)
A        : Assign the first value of the tuple to G and the second to H


# Ral, 27 bytes

11=,1*:0*+1=0=1/-:1:+:+?0*.


Try it online!

## Go 114113 106 chars

I'm sure this can be improved but I'd like to share my approach using the closed-form expression to calculate Fibonacci numbers.

package main
import(."fmt"
."math")
func main(){for i:=0.0;i<31;i++{Printf("%.0f\n",Pow(Phi,i)/Sqrt(5));}}


# dc, 21 17 bytes

0z[dp_3R+lmx]dsmx


Try it online!

This prints the Fibonacci sequence endlessly.

My previous (21-byte) version accepted an input $$\n\$$ on stdin, outputting the $$\n^\text{th}\$$ Fibonacci number on stdout (1-indexed):

9k5v1+2/?^5v/.5+0k1/p

# Actually, 16 bytes

"1,"◙01W;a+;◙',◙


Try it online!

# GAS x86-64 for Linux (Machine code), 76 68 bytes

### Loop-free, because I thought it would be a fun idea. GCC 10.1.

Note/warning: for this to work, you have to turn off DEP and PIE (ASLR). See compilation instructions below.

Look, ma! No loops! Byte count is for the assembled opcodes of func + e + f. Works on 32 bit input. Sure, you'll eventually run out of stack space, but the number will overflow before then. Obviously, a loop would have been much smaller (and easier) to write.

How it works:

1. Copy payload code to stack-relative space (No modification of %rsp)
3. Payload calculates the Fibonacci number in %rax and self-unwinds without ever looping

### Full testing program:

// 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
.global main

message: .asciz "%d"
main:
mov $N, %rdi call func mov %eax, %esi mov$message, %rdi
xor %eax, %eax
call printf
xor %eax, %eax
ret

func:
// We'll take %rdi, but we'll slice it to 32 bits
// %edi is arg, r8d is our counter.
mov %edi, %r8d
// Make 2 copies for arithmetic
mov %r8d, %eax
// Set up stack offset
// N * 32, need 64 bits to work with %rsp
imul $(eof - e), %eax neg %rax mov %rsp, %r9 lea (%rax, %r9), %r9 // Initialize for Fibonacci xor %ebx, %ebx xor %eax, %eax inc %eax // Copy initial payload: // Because PIE is turned off, we can use 32-bit addressing for local // Stack still seems to be in 64-land, however mov$e, %esi
mov %r9, %rdi
mov $(eof - e), %ecx rep movsb // Go to stack offset jmp *%r9 e: // The payload: dec %r8d jg f ret f: // Do actual Fibonacci stuff lea (%ebx, %eax), %edx mov %eax, %ebx mov %edx, %eax // Self-replicate mov %r9, %rsi mov$(eof - e), %cl
rep movsb
eof:


I decided on 32-bit input because the stack is not all valid for 64 bits, so we'd run out anyway. It overflows after a we get high enough input, but that's expected.

### Compilation:

# Prints 55
# Replace 10 with the input number (1-indexed):
gcc -no-pie -z execstack execstack_fibonacci.sx -DN=10 -g

• The -z execstack tells the linker to turn off DEP for this program. (Allow the stack to be executable)
• The -no-pie turns off ASLR.
• The -DN10 is just easy input control.
• The -g is for debugging control. Turn it off if you want.

Note: this will NOT work in MinGW because the DEP policy is managed by the OS on Windows. In fact, -z isn't even recognized by MinGW's ld.

### Hex dump:

funcs='func e f'
for el in $funcs; do echo 'Dump$el'
gdb -batch -ex "disas/r $el" ./a.out | sed '/^$/d'
done
gdb -batch -ex "print eof - e" ./a.out

• You may be able to save quite a few bytes here, as there are a lot of unnecessary REX prefixes, and there isn't nearly enough push/pop abuse. Jan 16 at 14:56

# Javascript, 24 bytes - recursive version, 33 bytes - usage of Binet's formula, 46 bytes - continuous approximation with the "phi" itself

Usage of "Binet's formula" and a bitwise "OR" operator to round the result. Originally in the "Binet's formula" you have to subtract the so-called "smaller phi to the n-th power". "Smaller phi" (-0.618...) is inside the (-1;1) interval, so it gets closer to 0 with each positive power - that's why we can leave it, and just round the meaningful part. Function itself is an anonymous one, declared with the arrow function declaration.

n=>(((5**.5/2+.5)**n)/5**.5)+.5|0


Try it online!

Recursive version - arrow function declaration. Check whether n is less than 3, if so return 1, else do it again (at least) 2 more times, but with an argument of value n-1 and n-2:

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


Try it online!

Continuous approximation. Ni * phi = Ni+1.

WARNING - RUNNING THIS CODE WILL END UP AS AN INFINITE AMOUNT OF ALERTS (next after clicking "ok")

f=n=>{l=n/2+5**.5/2*n+.5|0;alert(l);f(l)};f(1)

# V (vim), 32 bytes

i1
1<esc>qqkyjGp:s:\n:+
C<C-r>=<C-r>"
<esc>@qq@q


(don't)Try it online!

Prints the sequence forever.

Output is not visible on TIO, so here's the first 99 iterations: Try it online!

Last accurate value is $$\7540113804746346429\$$ after which it exceeds the integer limit.

# Duocentehexaquinquagesimal, 5 bytes

±∊YO$ Try it online! Link is to a version with output; this one just writes the sequence to memory. Outputs codepoints of the entire sequence. Stops eventually because of memory limitations. # Branch, 18 bytes 1XY[/x#^\yX^+Y10.]  Try it on the online Branch interpreter! Outputs infinitely. Eventually starts producing garbage values because long long int overflows but that seems to be acceptable. ## Explanation 1 Set the node's value to 1 XY Set the X and Y registers to 1 [ ] While value is not 0 (this will always be true in this program) /x# Move to the left child, set to the X register, and output as number ^\ Move to the parent and then right child (go to right sibling) yX Set to the Y register, then set the X register to that value ^+Y Move to root, sum the children (X + Y), and set the Y register 10. Place 10 and output as character; this also keeps the loop going  # C (clang), 797047 46 bytes y;z;main(x){for(;printf("%i ",z=x+y);y=z)x=y;}  Try it online! Uses a for-loop instead of a while-loop just because I use it more and doing while() gives the same byte count. Thanks to ceilingcat for golfing 9 bytes. Thanks to Jo King for golfing 23 bytes. Thanks to ceilingcat for golfing another byte. # Knight, 20 19 bytes ;=x=y 1W!Ox=y+x=x y  Try it online! -1 byte: W1;Ox -> W!Ox The "print infinitely" variant. (It will eventually overflow). It is a simple add and swap loop. Ungolfed: # init x and y to 1 ; = x (= y 1) # loop forever # since OUTPUT evaluates to NULL, we # can just invert the condition : WHILE !(OUTPUT x) { # Add and swap : = y + x (= x y) }  # Nim-Lang, 44 bytes, thanks to @Jo King and @caird coinheringaahing: var a,b=1 while 1>0: a=b-a echo a b+=a  ## Arbitrary precision version, 106 bytes, thanks to @Jo King and @caird coinheringaahing: import bigints var a:BigInt=initBigInt(0) var b:BigInt=initBigInt(1) while 1>0: a=b-a echo a b+=a  • Can you golf true to 1 and temp to c? Jun 28 at 22:32 • You can also golf the declaration to var a,b=1 if you then change the update section to a=b-a;echo a;b+=a. You can also update the true part to be 1>0 like in your arbitrary precision code – Jo King Jun 29 at 4:33 • Right thanks, stupid of me to leave it like that. Jun 30 at 1:35 # Python 3.8 (pre-release), 126 bytes Neither short nor pretty, but possibly a new method! from decimal import* def f(n):l=n//4+1;p=10**l;getcontext().prec=n*l-l+1;return int(str(Decimal(p**2)/Decimal(p**2-p-1))[-l:])  Uses, for example: 1/(1000000-1000-1) = 0.000 001 001 002 003 005 008 013 021 034 055 089 ... l can be any upper bound for the number of digits of the Fibonacci number after the one we want. Try it online! # Quipu, 33 bytes 1&0&\n [][]/\ ^^/\0& --++?? 1& ++  Try it online! Saved 4 bytes thanks to Jo King. It prints the Fibonacci sequence separated by newlines. Equivalent pseudocode:  a = [0, 0, 0] // implicitly 0: a[0] = a[1] - a[0] + 1 1: print a[0] a[1] = a[0] + a[1] 2: print "\n" goto 0  # C / Objective-c, 62 c;f(a,b){printf("%d ",a+b);if(c++<40)f(a+b,a);}main(){f(0,1);}  This will print the first 40 fibonacci numbers. I assume the compiler will set c=0. If it is trash, than it will not work; This version is smaller, but it infite show all sequence number # C / Objective-c, 50 (infinite) f(a,b){printf("%d ",a+b);f(a+b,a);}main(){f(0,1);}  • Variables with static duration are zero-initialized if not explicitly initialized. This behavior is required by the standard. Jul 30 at 17:48 ## Python (56 chars) n=input() x=5**0.5 print ((1+x)**n-(1-x)**n)/((2**n)*x)  And for the sequence n=input() i=1 x=5**0.5 while i<=n: print ((1+x)**i-(1-x)**i)/((2**i)*x) i+=1  PHP, Finite - 46 chars <?for($b=1;$i++<$n;)echo$b-$a=($b+=$a)-$a," ";  where$n is the length of the sequence

PHP, Infinite - 39 chars

<?for($b=1;;)echo$b-$a=($b+=$a)-$a,"
";


# MATLAB/Octave, n first numbers, 41 39 chars

a=0:1;for(i=3:n);a(i)=a(i-2)+a(i-1);end


# Python 3 (53)

def f(n):
l,p=0,1
while n:n,l,p=n-1,p,l+p
return l

• Aug 12 '18 at 17:31

Clojure, 46

(defn f[x y z](if(= 0 z)x(recur y(+ y x)(- z 1))))


Although, technically 50 since Clojure requires the recur for pseudo tail call:

(defn f[x y z](if(= 0 z)x(recur y(+ y x)(- z 1))))


Non compressed:

(defn fib [left right iteration]
(if (= 0  iteration)
left
(fib right (+ right left)  (- iteration 1))))


It almost feels like cheating to use Haskell for something like this. It just prints Fibonacci numbers ad infinitum.

f=1:scanl(+)1f
main=print f


And if using GHCi only 21 characters, including two newlines, are necessary:

Prelude>let f=1:scanl(+)1f
Prelude>f
[1,1,2,3,5,8,13,21...

• this solution was already posted here. FYI. Nov 28 '17 at 10:49

JAVA - 108 characters:

int[]f={0,1};System.out.println(0);for(int i=0;i<9;i+=2)System.out.printf("%d\n%d\n",f[0]+=f[1],f[1]+=f[0]);

• If a space is required in the code, it should be included in the character count.
– Iszi
Nov 27 '13 at 21:53
• Alright, I will update it.
– user10766
Nov 27 '13 at 21:57
• Already fixed it for you - looks like someone approved my edit suggestion.
– Iszi
Nov 27 '13 at 22:05
• That was me - I went to fix it, and found you had already.
– user10766
Nov 27 '13 at 22:06
• Ah. All good, then. Welcome to Code Golf!
– Iszi
Nov 27 '13 at 22:10

# C 64 Characters

a;main(f,n){scanf("%d",&n);while(--n)f+=a=f-a;printf("%d",f-a);}


This will print the nth Fibonacci number.

a;
`