# 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;
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if (c.owner.user_id === OVERRIDE_USER)
});
else process();
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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;
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function process() {
var valid = [];

var body = a.body;
if(OVERRIDE_REG.test(c.body))
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var match = body.match(SCORE_REG);
if (match)
valid.push({
user: getAuthorName(a),
size: +match[2],
language: match[1],
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else console.log(body);
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valid.sort(function (a, b) {
var aB = a.size,
bB = b.size;
return aB - bB
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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;
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{
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language = language.replace("{{LANGUAGE}}", lang.lang)
.replace("{{NAME}}", lang.user)
.replace("{{SIZE}}", lang.size)
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jQuery("#languages").append(language);
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text-align: left !important;
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width: 290px;
float: left;
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font-weight: bold;
}

table td {
}
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<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>
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<tbody id="language-template">
</tbody>
</table>

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

# Red, 47 bytes

F: func[N][either N < 2[n][(F N - 2)+ F N - 1]]


Try it online!

# Vyxal, 2 bytes

ÞF


Try it Online!

Before you go saying that the online link doesn't match the submission here, that's because the extra , is needed to actually make the output appear online. If you use the offline version, then you will see that the above works just fine. Also, the 5 flag makes sure that the online interpreter times out after 5 seconds.

## Explained

ÞF  # Push every Fibonacci number


And now for the non-trivial version

## Vyxal5, 6 bytes

⁽+dk≈Ḟ


Try it Online!

Once again, discrepancies between online link and actual version are for the purposes of making it work online.

### Explained

⁽+dk≈Ḟ
⁽+d     # lambda x, y: x + y
k≈   # the list [0, 1]
Ḟ  # Create an infinite sequence based on the function and the initial list.


Fun fact: the infinite sequence function you see was inspired by the sequence blocks of the golfing language Arn by ZippyMagician.

• Fun fact: I was inspired by Raku when I added sequences – ZippyMagician Apr 11 at 2:36

## Bash 100

This is a very slow, but hey no performance penalty. First line needed.

#!/bin/bash
if [ $1 -lt 2 ]; then echo$1; exit; fi
expr $0 \expr$1 - 1\ + $0 \expr$1 - 2\

• More a question: You don't need a shebang, do you? See ruby, python and xy-script. – user unknown Apr 12 '11 at 1:39

# Vitsy, 11 Bytes

I'm certain there's a way to shorten these.

Print out all fibonacci (to Integer.MAX_VALUE)

01[D}+DNaO]
01          Push 0 and 1 to the stack.
[       ] Repeat infinitely.
D        Duplicate the top item of the stack.
}       Rotate the stack to right.
+      Add the top two items.
D     Duplicate the top item.
N    Print the top item out as a number.
aO  Print a return.


Print out to input fibonacci (13 bytes):

01}\[D}+DNaO]
01            Push 0 and 1 to the stack.
}\[       ] Get the input and repeat that many times.
D        Duplicate the top item of the stack.
}       Rotate the stack to right.
+      Add the top two items.
D     Duplicate the top item.
N    Print the top item out as a number.
aO  Print a return.


## Minkolang 0.10, 10 bytes

This language was created after this challenge but not for it.

Stream (link, do not click "Run"):

01d1R+dN2@


A mite clever, if I do think so. The 2@ at the end is a 2-trampoline that jumps over the 01 at the beginning, allowing the sequence to rise unabated.

01nd,7&[d1R+]rN.


Worse than I expected, 16 bytes. 01 sets it up, nd,7&...N. prints out 0 if the input is 0 and does the loop otherwise. [d1R+] builds up the sequence, then r reverses the stack and the correct number is outputted and the program ends with N..

• Grar! Again? You beat me by one again. grumble – Addison Crump Oct 30 '15 at 20:26
• .... ¯\_(ツ)_/¯ – El'endia Starman Oct 30 '15 at 20:38

# Turing machine code, 389

I wrote this the other day and decided to post it. Generates an infinite Fibonacci sequence in unary on the tape. See a commented version in action here.

0 _ 1 r 1
1 _ _ r 2
2 _ 0 r 3
3 _ _ r 4
4 _ 0 l 5
5 0 * l 5
5 _ * l 5
5 1 * r f
a 0 1 r b
b 0 * r b
b _ * r c
c 0 * r c
c _ * r d
d _ 0 l e
e 0 * l e
e _ * l e
e 1 * r f
f 0 1 r g
f _ * r k
g 0 * r g
g _ * r h
h 0 * r h
h _ * r i
i 0 * r i
i _ 0 l j
j 0 * l j
j _ * l j
j 1 * r f
k 0 1 r l
k _ * l R
l 0 * r l
l _ * r m
m 0 * r m
m _ 0 l n
n 0 * l n
n _ * l n
n 1 * r k
R _ * r a
R 1 0 l R


# ShapeScript, 16 14 bytes

_1@0@'@1?+'*!#


This reads an integer n (in unary) from STDIN and prints the nth Fibonacci number.

The submission is non-competing, since this challenge predates ShapeScript's creation by a few years.

Try it online!

### How it works

        Input: a string of n 1's
_       Get the length of the input to push n.
1@      Swap it with 1 (F[-1]).
0@      Swap it with 0 (F[0]).
STACK: F[-1]   F[0]   n
'       Push a string that, when evaluated for the i-th time,
does the following:
@       Swap F[i-2] on top of F[i-1].
1?      Push a copy of F[i-1].
+       Add the copy of F[i+1] to F[i].
'       STACK: F[i-1]   F[i]
*!      Repeat the string n times and evaluate it.
#       Discard F[n] from the stack.


# R - 39

Shortest - recursive, until SO:

f=function(i,j){cat(i);f(j,i+j)};f(1,1)


Until n:

i=j=1;for(x in 1:n){print(i);k=i;i=i+j;j=k}


or (a bit vectorized):

a=c(1,1);for(x in 1:n)print((a=c(a[2],sum(a)))[1])


or (without any loop or recursion):

a=c(1,1);sapply(1:n,function(i)a<<-c(a[2],sum(a)))[1,]


# Brainf*ck, 489 466 characters

Granted, this is a bit overkill, not to mention that it could be optimised a lot. I will get to improving it tomorrow, since it's too late today.

EDIT: Improved by a few bytes by putting stuff closer together on the tape.

++++++>++++++++++>+>>>>>>>>>+<<<<<<<<<<<[->>[>>+>+<<<-]>>>[<<<+>
>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]+++++
+++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[
.[-]<]>>>>>>>>[->+<<<<<<<<<<+>>>>>>>>>]>[-<+>]<<<<<<<<<<<.>>>>>>
>>>>[>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-
<+>]>+>>]<<<<<]>[-]++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[-
>>++++++++[<++++++>-]]<[.[-]<]<<<<<<<<<<[->+>>>>>>>>+<<<<<<<<<]>
[-<+>]<<.<]


# Oration, 135 bytes

I believe that this is "optimal"... takes a deep breath here we go!

Inhale
Start a function f with n
If n<2
Return n
Backtracking
Inhale
Here
Literally, f(n-2)+f(n-1)
I'm done
Listen
Invoke f with number


The little ~> is input. This outputs the (input)th Fibonacci number. This transpiles to (in Python):

def f(n):
if n<2:
return  n
return f(n-2)+f(n-1)
print(f(eval(input("~>"))))

• Why is the transpiled Python code not golfed D: – Downgoat Feb 2 '16 at 4:01

# Oracle SQL 9.2, 80 bytes

SELECT ROUND(POWER((1+SQRT(5))/2,LEVEL-1)/SQRT(5))FROM DUAL CONNECT BY LEVEL<:1;


# Lua, 51 bytes

function f(n) return n<2 and n or f(n-1)+f(n-2)end

It creates a function called f(n), that takes an input (n). If n = 1, returns n. This function uses recursion.

# beeswax, 12 bytes (sequence), 42 bytes (n-th Fib.)

Beeswax is newer than the question, so no competition here.

Fibonacci sequence.

p{N<P{*
>~+d


No promotion to higher bit widths implemented in my solution, so 64-bit overflow starts at the 93rd or 92nd Fibonacci number, depending if you start counting your sequence at 0 or 1:

0
1
1
2
3
5
8
13
21
34
55
89
.
.
.
4660046610375530309
7540113804746346429
12200160415121876738   ← 93rd Fibonacci number
1293530146158671551    ← 1st. 64-bit overflow/wraparound
13493690561280548289


N-th Fibonacci number:

;{#'<>~P~L#MM@>+@'p@{;
_TNX~P~K#{; d~@M<


The same limit applies to this solution.

# CJam, noncompeting, 11 bytes

0X{_@+}q~*;

• F(0) = 0. You should eliminate the backslash. – Dennis Mar 2 '16 at 15:40
• Ah I was assuming that we were starting from 1,1..... so I guess this is a good convention since it saves a byte :) – A Simmons Mar 2 '16 at 16:18
• People start the Fibonacci sequence at different values, so F(0) = 0 may or may not be defined. However, when it comes to indexing, F(1), F(2) = 1, since a lot of the sequence's properties depend on that. – Dennis Mar 2 '16 at 16:29

1$[^^+2!]!  Try it here. An infinite stream that leaves results on stack. Use the Step button to avoid setting off the infinite loop. # Explanation 1$         {start w/ 2 1's}
[     ]! {execute lambda}
^^      {take top 2 items on stack}
2!   {self recurse!}


# Gogh, 10 bytes

¹Ƥ{Ƥ÷®+Ø}x


Executed from the command line like this:

\$ ./gogh "" "¹Ƥ{Ƥ÷®+Ø}x"


### Explanation

¹       “ Push two ones to the stack.                 ”
Ƥ       “ Print the TOS.                              ”
{       “ Open a code block.                          ”
Ƥ      “ Print the TOS.                              ”
÷      “ Duplicate the TOS.                          ”
®      “ Rotate the stack leftward.                  ”
+      “ Destructively add the TOS to the STOS.      ”
Ø      “ Loop all preceding code (within the block). ”
}       “ Close a code block.                         ”
x       “ Execute the TOS.                            ”


# Scratch, 106 characters

This isn't impressive at all but someone had to do it.

when gf clicked
forever
add((item[last v]of[f v])+(item((length of[f v])-(1))of[f v]))to[f v


scratchblocks2 render

Fairly bog-standard solution. "f" is a list which starts off empty. Runs as long as you let it.

Since it's not easy to define what is and isn't a "character" in Scratch I've used the forum plugin's formatting. This allows me to cheat off some additional characters (scratchblocks2 is very lenient with dropping closing parenthesis, "end"s, and shaving off whitespace here and there)

# Alpax, 5 bytes (non-competing)

Non-competing since the language postdates the challenge. Code:

⇇+
1¹


Yes, that's right mates. My newest invention, which is more mathematically based than 05AB1E. This language uses a lot of recursion, so be aware. This is a bit like a stack based language, but a little bit different. The elaborated version of the above code is:

a(n) = ⇇+
a(0) = 1, a(1) = 1


Explanation:

⇇ is short for pushing a(n - 1), a(n - 2)

It then implicitly prints the result of a(n), whereas n` is the input.