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, in accordance to the standard rules:

• Generates the Fibonacci sequence without end.

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

• Given n calculates the first n terms of the sequence

You may use standard forms of input and output.

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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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) {
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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],
});
else console.log(body);
});

valid.sort(function (a, b) {
var aB = a.size,
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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))
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langs.sort(function (a, b) {
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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>
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<table style="display: none">
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• I am sort of waiting for a response like "f", 1 byte, in my math based golf language. Commented Aug 11, 2020 at 11:57
• @ChrisJesterYoung can we use 1.0 are 1 only? Commented May 11, 2022 at 2:45
• @NumberBasher 1.0 is fine. Commented May 20, 2022 at 19:10
• What about 1.3? Commented Aug 28, 2022 at 15:10
• Am I allowed to start the sequence with 0, 1? Commented Oct 11, 2022 at 3:41

Ruby, 27

p a=b=1;loop{a,b=(p b),a+b}


Prints each number, starting correctly with the first two 1s, to STDOUT ad infinitum (VERY QUICKLY from irb in my environment - you've been warned). I've been learning Ruby lately, so I figured I'd contribute this. If it can be shortened in any way, let me know.

Julia - 20 Characters

f=n->([1 1;1 0]^n)[]


I used the same basic algorithm as the Octave answer. This starts with f(0)->1, f(1)->1, to avoid needing an explicit array index. This is 4 characters shorter than the naive recursive algorithm.

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


Python 3, 39 38 bytes

a=1
b=1
while 1:c=a+b;print(c);a=c;b=c


Ungolfed:

a = 1
b = 1
while 1:
c = a + b
print(c)
a = c
b = c


Is there some way of getting rid of the b=c statement?

• Welcome to PPCG! Does a=b=c work in Python? (Same for a=b=1.) Also, do you really need the space after :? Commented Oct 12, 2015 at 13:26
• @MartinBüttner It did print the Fibonnaci sequence, and it also showed a weird wavy "animation" :P
– m654
Commented Oct 12, 2015 at 14:07
• @MartinBüttner Assignments can be chained, just like comparisons, but they don't return a value so you can't do a=1+(b=c). Commented Oct 12, 2015 at 14:13
• This doesn't print the right sequence. It's not hard to fix though. Commented Jan 31, 2019 at 23:03

Rust, 44 bytes

fn f(n:u8)->u8{if n<2{n}else{f(n-1)+f(n-2)}}


dc, 29 chars

1ddppsa[+sblalbsalbplxx]sxlxx

• Verified. how the $^*# does that work?? Commented Aug 28, 2019 at 9:04 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. Nth Fibonacci (link): 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 Commented Oct 30, 2015 at 20:26 • .... ¯\_(ツ)_/¯ Commented Oct 30, 2015 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  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. ++++++>++++++++++>+>>>>>>>>>+<<<<<<<<<<<[->>[>>+>+<<<-]>>>[<<<+> >>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]+++++ +++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[ .[-]<]>>>>>>>>[->+<<<<<<<<<<+>>>>>>>>>]>[-<+>]<<<<<<<<<<<.>>>>>> >>>>[>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[- <+>]>+>>]<<<<<]>[-]++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[- >>++++++++[<++++++>-]]<[.[-]<]<<<<<<<<<<[->+>>>>>>>>+<<<<<<<<<]> [-<+>]<<.<]  (With added newlines for readability) 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: Commented Feb 2, 2016 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. Commented Mar 2, 2016 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 :) Commented Mar 2, 2016 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. Commented Mar 2, 2016 at 16:29 DUP, 10 bytes 1$[^^+2!]!


Try it here.

An infinite stream that leaves results on stack. Use the Step button to avoid setting off the infinite loop.

1${start w/ 2 1's} [ ]! {execute lambda} ^^ {take top 2 items on stack} + {add them} 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.

Uses the Alpax encoding.

• Alpax doesn't exist anymore.
– user85052
Commented Dec 24, 2019 at 12:43
• The Alpax repo is gone. Umm, is this language converted into Oasis?
– user92069
Commented May 11, 2020 at 8:32
• @Λ̸̸ Alpax is/was a different language and a public repository is not available anymore. Oasis was another attempt at a golfing language specialized in sequences, which became a bit more successful and 'took over' the place of Alpax. Commented May 11, 2020 at 16:17

PlatyPar, 7 bytes

0A1wAC+


Explanation:

0A1       ## push first two Fibonacci numbers to stack and print them
w     ## while last item != 0 (always true)
A      ## print the most recently calculated Fibonacci number
C+    ## push the sum of the last two items of the stack


This one is a sequence.

C#: 836968665853 51

I used a nasty trinary and recursive lambda expression to achieve this one.

Func<ulong,ulong> f=null;f=x=>x<2?x:f(x-2)+f(x-1);


Usage:

    public static void Main()
{
// Recursive lambda expression...
Func<ulong, ulong> f = null;
f = x => (x < 2) ? x : f(x - 2) + f(x - 1);

Console.WriteLine("Please enter a whole number to obtain the Fibonacci sequence number for:");

long numValue;
if(UInt64.TryParse(value, out numValue))
Console.WriteLine(f(numValue));

Console.WriteLine("Press any key to end the program.");
}

• You don't have no support negative indices. Also, the "ternary" conditional operator isn't nasty if you use it right. :-) Commented Apr 8, 2013 at 15:21
• That helps, thanks! I don't consider ternaries nasty usually, but in a case like the one I've posted, I would do everything in my power to avoid that getting into a codebase. It gets points for clever/short, but not readable. Commented Apr 8, 2013 at 15:28
• lol - I posted mine without ever seeing yours. Funny to see that they're almost identical. :) Commented Apr 17, 2013 at 17:54
• Yeah, but trying to calculate anything above f(45) will cause either a StackOverflow, or just take forever and some time to calculate. Commented Apr 17, 2013 at 18:09
• I don't think you need the parentheses around (n<2) Commented May 2, 2016 at 15:00

Detour, 20 bytes

This one is going for the "infinite sequence" option.

v1vq:+
p,p^
^ q


Try it online!

Branch 1 takes a number, prints it, adds it with the number from Branch 2, then puts the result in Branch 2
Branch 2 takes a number, feeds it to the addition with branch 1 then puts the original number (not the sum) in Branch 1.

For a better explanation click the link and you'll see it in action.

Detour, 267 bytes

:\$v  1v   q   # split into branches

+   # push sum of last 2 fibonacci numbers to branch 2
{

p , p   ^   # print branch 1, merge with branch 3

}

^   q       # push branch 2 into branch 1 for printing and recycling

# 1   2   3


Try it online!

Julia, 20 bytes

!n=n>1?!~-n+!~-~-n:n


Straightforward implementation of the recursive definition. No match for the matrix approach, but a lovely opportunity to abuse Julia's ability to redefine operators.

Try it online!

Maple, 27 bytes

ifelse(n<3,1,f(n-1)+f(n-2))


Usage:

> f := n -> ifelse(n<3,1,f(n-1)+f(n-2));
> f(2);
1
> f(3);
2


Sesos, 11 bytes (non-competing)

Not in-place, linear memory.

Hexdump:

0000000: ae8583 ef6bc7 045fe7 b907                         ....k.._...

Size   : 11 byte(s)


Try it online!

Assembler

set numin
set numout
jmp
fwd 1
rwd 1
sub 1
fwd 1
jnz
fwd 2
put


Java, 71 chars

Single number: (Binet formula, considering 1.62 as the golden ratio))

int f(int n){return(Math.pow(1.62,n)-(Math.pow(-1.62,-n))/Math.sqrt(5)}


I know this isn't surprisingly short, however Math is beautiful and this formula is even more!

Ruby

Ungolfed, 60 bytes

def fib(prev,nxt)
x = prev + nxt
puts x
fib(nxt,x)
end


Golfed, 33 bytes

def f(a,b)x=a+b;puts x;f(b,x)end


Pretty simple to call, use f(first, next).

• You can still golf it further. Try taking out the unnecessary whitespace. Also, there are some good tips here Commented Sep 12, 2016 at 5:14
• x=a+b;puts x can become puts x=a+b Commented Dec 11, 2016 at 3:26

Java 8 29 bytes

Using Java 8 lambdas. This is a valid statement if there exists a function interface with a method that returns an int and takes an int as a parameter. Also the variable that stores the lambda must be declared as a member (static or non static) of the class it is in so that it can be used recursively.

f=n->n<2?0:f.f(n-1)+f.f(n-2);


Ungolfed:

@FunctionalInterface interface F
{
int f(int n);
}

public class Main
{
static F f;

public static void main(String[] args)
{
f=n->n<2?0:f.f(n-1)+f.f(n-2);
}
}


Prismatic, 113 bytes (can be smaller)

right wideness wideness left forward up vertex longness backward right vertex tallness forward down vertex vertex


Inspired by Brainfuck, Cubix and Hexagony.

Python, 75 bytes

a=[1,1]
while True:
a.append(a[-1]+a[-2])
a.pop(0)
print(a[-2])


Yes, I know, way too big, but I don't know golfing languages that well.

• 1) while True can be replaced with while 1. 2) You only need one space of indentation. 3) The while loop contents can be replaced with print(a[0]);a=[a[1],sum(a)] (on the same line as the while statement). 4) The parentheses on the print call can be removed if you're using Python 2. With all these tricks applied: a=[1,1]\nwhile 1:print(a[0]);a=[a[1],sum(a)] (replace the '\n' with a literal newline).
– user45941
Commented Nov 17, 2016 at 13:41
• If you take Mego's approach you can replace a=[a[1],sum(a)] with a=a[1],sum(a) to save two bytes. This works because the comma creates an implicit tuple rather than a literal list. Commented Nov 17, 2016 at 13:56
• The same can also be done with a=[1,1]. It can be shortened to a=1,1 Commented Nov 17, 2016 at 14:00