# 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.

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<h2>Shortest Solution by Language</h2>
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• I am sort of waiting for a response like "f", 1 byte, in my math based golf language. Aug 11, 2020 at 11:57
• @ChrisJesterYoung can we use 1.0 are 1 only? May 11, 2022 at 2:45
• @NumberBasher 1.0 is fine. May 20, 2022 at 19:10
• What about 1.3? Aug 28, 2022 at 15:10
• Am I allowed to start the sequence with 0, 1? Oct 11, 2022 at 3:41

# PHP, 39 bytes

<?php for($b=1;;)echo$a=-$a+$b+=$a,' ';  Try it online! Explanation <?php  An infinite loop is started. The zero-th term in the series, initially $a, is 0, so needn't be assigned. $b is initially the second term and so is set to 1. for ($b = 1;;)


The part which does all the work is echo $a = -$a + $b +=$a, ' ';. Here it is expanded.

{


Calculate the new value for $b: the next term is the sum of the previous two. $b = $b +$a;


$a needs to be moved on one term as well. It is calculated by subtracting itself from the new value of $b.

    $a =$b - $a;  For byte-saving convenience, it is$a that is echoed each time—followed by a space!

    echo $a, ' '; }  • This can be 31 bytes since you don't need PHP's opening tag (you can run with php -r "code here" without opening tag) and you can use _ as separator instead of space: Try it online! Sep 25, 2019 at 9:42 # Klein, 2322 21 + 3 = 24 bytes (non-competing) Run with the 000 topology (1)\((@ ):?\1-+(:(+)$


## Explanation

When the program starts it executes (1) which will put a 1 under the input. It then deflects into the main loop.

The main loop is on the second line. It starts with the \ character. Unwrapped it looks like:

\1-+(:(+)$):?  This will redirect our pointer if the counter is zero or perform one iteration of the fibonacci sequence otherwise. Once the counter reaches zero we are deflected to the code ((@, this will hide the top two values (the counter and one of the fibonacci numbers) and terminate the program. # Chip-8, 36 bytes 6301 'LD v3,1 6D05 'LD vD,5 6E0A 'LD vE,A 8344 'ADD v3,v4 A200 'LD I,200 F333 'LDD [I],v3 8343 'XOR v3,v4 8433 'XOR v4,v3 8343 'XOR v3,v4 F265 'LD v2,[I] F029 'LDF I,v0 00E0 'CLS DFF5 'DRW vF,vF,5 F129 'LDF I,v1 DDF5 'DRW vD,vF,5 F229 'LDF I,v2 DEF5 'DRW vE,vF,5 1206 'JMP 206  Displays Fibonacci numbers (up to 233) in decimal. (It might be shorter to use hexadecimal, but I think that's cheating) This one writes the numbers into memory: 6001 A300 8014 F055 8013 8103 8013 1204  ... but it's actually longer than valid numbers it writes: 01 01 02 03 05 08 0D 15 22 37 59 90 E9 79 (overflow)  # J-uby, 8 6 bytes :++2.*  In J-uby, + on a proc (or a symbol in this case, as symbols can be used as procs in J-uby), defines a recurrence relation. It takes a starter array, and then produces a function that takes n, and then applies itself to the starter array n times, pushing the result to the end and removing the first element. Naturally :+ + [0,1] is a recurrence relation that starts with elements 0, 1 and adds them together n times. 2.* is shorthand for [0,1] # VBA, 28 Bytes Anonymous VBE immediate window function that takes no input and infinitely outputs the n-th term of the Fibonacci Sequence while iterating n i=1:Do:k=i+j:i=j:j=k:?j:Loop  # C, 64 bytes a,x,y,z=1;main(){for(;;){a=y;y=z;z=a;x+=y;y=x;printf("%d ",x);}}  Try it online! Uses the same method as my Implicit answer. • Aug 19, 2018 at 4:01 # Pyt, 1 byte Get the nth Fibonacci number: Ḟ  Explanation:  Implicit input Ḟ Return (input)-th Fibonacci number  # Pyt, 7 bytes Get an infinite list of Fibonacci numbers: 0ĐḞƤ⁺ł  Explanation: 0 Push 0 [this is the counter]  ł While the counter is not zero (checked at 'ł') Đ Duplicate the counter ḞƤ Print the (counter)-th Fibonacci number ⁺ Increment the counter  Try it online! # Japt, 3 bytes Just to add to the collection. 0-indexed, using 0 as the first number in the sequence. MgU  Try it # R16K1S60 Assembly, 36 bytes mov bx, ip mov ax, ip mov sp, data jmp inner prg: mov cx, [sp+ax] mov [sp+bx], cx inner: mov ex, [sp] mov dx, [sp+bx] mov [sp], dx add ex, dx mov [sp+ax], ex send ax, ex jmp prg data: dw 0x0000 dw 0x0001  Pretty simple. Abuses 7 registers, including the instruction pointer (for some predefines) To note why I used the IP instead of a constant, it's because the R16K1S60 has to use an extra word (two bytes) to encode a constant into an instruction. Alongside that, I used ax and bx instead of ex and dx for the offset because ex and dx cannot be referenced in only 3 bits (the size of the offset section of instructions that support it) Outputs the number as a word on port 2 # Haskell, 30 bytes (was 33) f=0:1:[f!!n+f!!(n+1)|n<-[0..]]  Try it online! • You should add a TIO link and some sample cases Aug 9, 2018 at 19:32 • @MuhammadSalman They do not have to add a link. I think "should" is a bit too forceful. Aug 9, 2018 at 19:52 • Ok I added the TIO link, this is my first time posting Aug 9, 2018 at 22:04 • @mrFoobles For demonstration purposes, I think main=print f would be more impressive as it shows the magnitude of infinite lists. Aug 9, 2018 at 22:19 • @JonathanFrech yeah, should have been you could add and not should add Aug 10, 2018 at 10:28 ## Prolog, 3635 29 bytes X+Y:-writeln(X),Z is X+Y,Y+Z.  Run with 1+1. (I don't think having to call the base case is cheating, but let me know.) Prints the first parameter and a newline, sets Z to X+Y, then does a recursive call. Edit 1: Can use writeln(X) instead of write(X),nl, saving one character. Edit 2: Can use X+Y as a predicate instead of f(X,Y), saving 6 characters. Also the initial call is shorter. • Welcome to PPCG! The usual consensus is to include the function invocation if it needs to be called with special arguments. Nov 4, 2018 at 10:46 • Should I include the call in the character count? Nov 4, 2018 at 16:54 • 26 bytes: X+Y+O:-O=X;Z is X+Y,Y+Z+O. Try it online! Feb 5, 2020 at 19:18 ## Burlesque, 8 bytes Update: With current WIP one can use 1J2q?+C~. Shortest way to produce [fib(0)..fib(n)] without trashing the stack (14B): {0 1q?+#RC!}RS  Explanation There's the concept of "Continuation" in Burlesque which basically means that you run a function on a stack without destroying the stack. Fibonacci is the perfect example use-case for what these continuations are good for. If you have a program like 1 1 add then this results in a stack of 2 because add destroys the data. If add were not to destroy the data the stack would look like 1 1 2 and if we just do 1 1 add add it would look like 1 1 2 3. So all we need to do to generate a Fibonacci sequence is to call add n-times without popping the arguments. A continuation takes a snapshot of the stack, runs the function, pops the result from the stack, reverts the stack to the snapshot and pushes the result of the function to it. C! is the Burlesque built-in for "run this continuation n-times". However, doing so would trash our stack (which is no problem if you just want to print out Fibonacci numbers). Otherwise we need to use the RS built-in which runs a function in a different stack environment. RS takes a value as an argument, creates an empty stack, pushes that value to it and then runs the given function on that stack and after the function has run it will collect that stack into a list and push that list to the main stack. #R rotates the stack because the stack layout will look like N 0 1 but we need that N because it's the argument for C! so we rotate the stack. q?+ is just shorthand for {?+} (q wraps the next token into a block). If you don't care about trashing the stack you just drop the RS: blsq ) 10 0 1q?+#R!C 0 1 1 2 3 5 8 13 21 34 55 89  Shortest way to produce fib(n) as a reusable non stack-trashing piece of code I can think of is (17B): 0 1{Jx/?+}#RE!jvv  # Older Stuff There's dozens of ways to do that. These push the fibonacci numbers to the stack: blsq ) 0 1{#s2.+++}10E! blsq ) 0 1q?+10C!  However, the snippets above will also trash your stack. Alternatives for that are either: blsq ) 0 1{Jx/?+}10E!jvv  which just computes the 10th fibonacci number. Also by still using continuations you can let the whole thing run in a seperate stack environment like uhm so: blsq ) {10}{0 1q?+#RC!}rs {89 55 34 21 13 8 5 3 2 1 1 0} blsq ) 10{0 1q?+#RC!}RS {89 55 34 21 13 8 5 3 2 1 1 0}  Really depends on your needs. • This is code-golf, so please post the shortest solution you can find with its byte count. Oct 22, 2015 at 17:20 # Alchemist, 68 bytes y+_->y+a+b y+0_->z z+a->z+_ z+0a->x x+b->x+a x+0b->y+Out_a+Out_" "!y  Try it online! Outputs the 1-based sequence infinitely, If you want 0-based (i.e. 0 1 1 2 3 5...), you can change the trailing y to either x or z. ### Explanation: !y # Initialise the program with the y flag alongside the default _ y+_->y+a+b # Convert all _ atoms to a and b atoms y+0_->z # Once we're out of _ atoms, change to the z flag z+a->z+_ # Convert the a atoms back to _ atoms z+0a->x # Switch to the x flag x+b->x+a # Convert all b atoms to a atoms x+0b->y # Once we're out, change to y flag +Out_a # Print the number of a atoms +Out_" " # And a separator  If it makes you feel better, here's a more pseudo-codey version: _=1 while true: a=a+_ b=_ _=a a=b print a  # BitCycle, 21 bytes  1+ ~! CB0CA~ ^ 1 <  Outputs an unending sequence. Use the -u flag to get output in decimal. Try it online! Note: the current BitCycle interpreter doesn't play very well with infinite output. You have to halt the program (Ctrl-C) before it displays anything. On TIO, letting the program run until the 60-second timeout shows no output, either--you have to click the Run button (or hit Ctrl-Enter) again to halt it. ### Explanation This explanation assumes you're familiar with BitCycle. Conceptually, we store two numbers at a time, the smaller and the larger. At each step, we output the larger, set the new larger to be the larger plus the smaller, and set the new smaller to be the larger. We store and output the numbers in unary (using 1 bits), but we also need a separator (0 bit) after each number output. Our approach is to store the separator at the end of each number. When adding two numbers, we discard the separator from the first number added, and keep the separator from the second number added. In the code, the leftmost C collector holds the smaller number, while the rightmost C collector holds the larger. We're actually going to store everything negated, so the numbers are made of 0 bits and the separators are 1 bits. Thus, the leftmost C initially gets a single 1 (unary zero plus a separator bit) and the rightmost C gets 01 (unary one plus a separator bit). The C collectors open and dump their contents straight into the B and A collectors. Next, the A collector opens, holding the larger number. It goes through a couple of dupneg devices, with the following results: • A copy goes into the leftmost C collector, becoming the new smaller number. • A negated copy goes into the sink ! and is output. • A doubly-negated copy goes into the rightmost C collector, but the + ensures that it's only the 0 bits, not the trailing 1 separator. Finally, the B collector opens and dumps its contents into the rightmost C, adding the former smaller number to the former larger number to create the new larger number. The cycle repeats forever. ### Other versions Here's a modified version (still 21 bytes) that strips the separator off the smaller number (instead of the larger) before adding: 10>v ~! BA+BA~ ^ <  And here's an 18-byte version that starts at 0 instead of 1. (Thanks to Jo King for golfing it down from 21 bytes.) Here, we start with the "smaller" number at 1 and the "larger" number at 0, generating the extended Fibonacci sequence 1,0,1,1,2,3,... (Since the "larger" number is what we output, we don't see the first 1.)  1+ ~! CBCA~ ^10 <  # Python 3, 37 bytes lambda p,a=5**.5:round((.5+a/2)**p/a)  Try it online! # Explanation Fib(n) = Fib(n-1) + Fib(n-2) with Fib(0) = Fib(1) = 1 Using some simplification, this becomes: For n > 0, this becomes Where round is a function that rounds to the nearest integer. lambda p, # defines the anonymous function a=5**.5: # sets a to \sqrt{5} round((.5+a/2)**p/a) # Runs the function and returns the result  • This loses precision at only n=71, outputting 308061521170130 instead of 308061521170129 – Jo King Jul 26, 2019 at 4:58 # Jasmin, 120 bytes Defines a class F with a static method f that calculates the nth Fibonacci number. My implementation is essentially an iterative solution that stores partially computed Fibonacci numbers on the stack. .class F .super java/io/File .method static f(I)I ldc 0 ldc 1 dup_x1 iadd iinc 0 -1 iload_0 ifgt$-9
ireturn
.end method


# Some interesting golfing tricks used

1. Extending java/io/File is shorter than extending java/lang/Object (The super line cannot be omitted). I've check and File is tied for the shortest fully qualified class name.
2. Making this an instance method would let me remove static from the method header but, then I would have to explicitly implement an empty constructor to make the function callable (costing quite a few bytes).
3. Juggling the Fibonacci values on the stack turned out to be shorter than storing them in local variables.
4. On the other hand it's worth storing the index in a local variable. This makes stack management easier (i.e. shorter) without too much extra length since there is an instruction for adding or subtracting from locals variables.
5. Although the JVM technically requires that you declare the maximum stack size before hand with .limit stack 5, this can be omitted if the class file is executed with the -noverify flag. I'm pretty sure this is some sort of undefined behavior but, it works in this this case.

# Test setup

To test the code you, need a main method to invoke the static method.

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


You then need to use jasmin.jar (obtained from the source forge linked in the title) to build F.class before building and executing the test file. Since the stack size was omitted, you need to execute the class with -noverify. The makefile below handles this.

test: FibTest.class
java -noverify FibTest

FibTest.class: FibTest.java F.class
javac FibTest.java

F.class: F.j
java -jar jasmin.jar F.j


# pure bash, 43 chars

for((r=l=i=1;i++<40;l+=r+=l)){ echo $r$l;}


Not really golfed, but I like it anyway.

Or

r=1 l=0;echo {,,}{,,}{,,}\ $[r+=l]\$[l+=r]


## 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.

# Zsh, 31 bytes

try it online

for ((a=1;b+=a;a+=b))echo $a$b


32 bytes, based on James Brown's awk solution:
for ((y=1;z=x+y;y=z))echo $[x=y] 42 bytes, to halt before int overflow: (for ((a=1;b+=a;a+=b))echo$a $b)|head -46 NB: For a properly "endless" solution I need logic for long long (..) long integers, per this post ## 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 # Cascade, 28 25 bytes ?01 ^/ |.# !9] -0 !0] +1  Try it online! Outputs the Fibonacci numbers separated by tabs starting from 1. This shows off the behaviour of variables in Cascade, in that the variables 1 and 0 aren't static in this program. Unfolded, this looks something like:  @ ^ ^ \ / . | # 9 | ] | 0 - | ] 0 | 1 + / 1 0 / |  Try it online! This initially branches twice, with the leftmost going down the tree until it sets ([) the variable 1 to the sum (+) of 1 and 0. Then it sets 0 to that value to the result of that minus 0. This has the effect of advancing one element in the Fibonacci sequence.For example, the values of repeated executions are: 0 1 1 1 1 2 2 3 3 5 5 8 8 13 ...  Finally it prints the total result of that, which is the new value of 0. The next branch prints the tab separator (.9), and the final branch loops back around to the top of the program. # Taktentus, 87 bytes a:=1 @wy_n:=a @wy:=32 b:=1 n:=44 @>=n _:=@stop @wy_n:=a @wy:=32 c:=a a+=b b:=c n-- _-=8  n are variable how many times we count (n-1 because first are writing directly) # 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 # International Phonetic Esoteric Language, 28 bytes <f>/b1ɨʌʟ|e|1zb1z<f>d<fib>s|e|\  A function that expects a number $$\n \ge 0\$$ to be on the stack, and leaves the $$\n\$$-th Fibonacci number. Explanation: <f>/ (n1 -- n2) (where n2 is the n1-th fibonacci number) (check for case n=1) b (dup) 1 (push 1) ɨ (n>1?) ʌ (skip if n>1) ʟ⟨e⟩ (if n<=1, jump to end label) 1 (push 1) z (subtract) b (dup) 1 (push 1) z (subtract) <f> (recurse) d (swap) <f> (recurse) s (add) ⟨e⟩ (end label) \  # R, 33 32 bytes CAUTION: This attempts to print the whole Fibonacci sequence. It does not stop. a=b=1;repeat print(a<-(b=b+a)-a)  Pretty simple. Initialize a and b. Then a repeat loop which adds them to find the next number and print it. This loop will not stop, though eventually the overflow means it just prints NaN repeatedly. Edit: saved 1 byte by switching to a=b=1 which required a different loop control mechanism to print the first few values, and then a different assignment location, etc. • 25 bytes by using T and F as variables. May 21, 2022 at 15:39 • 23 bytes by using T and F as variables, and additional inspiration from Giuseppe... May 26, 2022 at 21:22 • As far as I'm concerned, those are sufficient improvements you should submit them independently. Really like them by the way Jun 2, 2022 at 1:29 # Flurry, 46 bytes {}{<{}{<[<>{}]{{}[<><<>()>]}>}>}{{}{{}}{{}}}()  How to run: $ target/Flurry -nin -c "{}{<{}{<[<>{}]{{}[<><<>()>]}>}>}{{}{{}}{{}}}()" 8
34


It doesn't use the stack at all (except for fetching the input number from the pre-populated stack). Instead, it uses pure functional construction developed by Anders Kaseorg in my SKI golf challenge.

one = \f. \x. f x
= I
= {{}}

one-pair = \f.f one one
= {{}{{}}{{}}}

succ = \n. \f. \x. f (n f x)
= \n. \f. S (\x. f) (n f)
= \n. S (\f. S (K f)) n
= S (S . K)
= <><<>()>

next-pair-helper = \f. \m. \n. f n (m succ n)
= \f. \m. S f (m succ)
= \f. S f ∘ (\m. m succ)
= {<[<>{}]{{}[<><<>()>]}>}

next-pair = \p. \f. p (next-pair-helper f)
= \p. p ∘ next-pair-helper
= {<{}{<[<>{}]{{}[<><<>()>]}>}>}

fib = {} next-pair one-pair K
= {}{<{}{<[<>{}]{{}[<><<>()>]}>}>}{{}{{}}{{}}}()


# Arn, 5 bytes

Since I finally implemented sequences in my interpreter this is now a valid submission :)

╔Tò”7


# Explained

Unpacked: [1 1{+

[ Sequence...
1 1 ...with 2 values initialized at 1...
{ ...the rest of which are determined by the block...
+ ...that adds the top two values
} Implied, can be removed
] Implied, can be removed


Since Arn supports infinite sequences and BigNums, this will continuously output fibonacci numbers infinitely (hypothetically)

# Barrel, 26 bytes

Disclaimer: The language is newer than the question, but I didn't even think of golfing this until after I'd created the language. I did update the language after I originally wrote the answer, and changed my answer, but that was because I was fixing the interpreter and made some changes to the spec to make the language work better. I wasn't cheating, I promise :)

+&1:0¤n &0:@1&1:a#@0+←1


Explanation:

+                          // set the accumulator to one by incrementing (initialization)
&1:0                      // set register 1 to value 0 (initialization)
¤               ←1   // define a target that can be jumped to; then, jump to the previously defined jump target
n                    // print the accumulator as a number and implicitly print the following space
&0:@1              // set register 0 to register 1
&1:a          // set register 1 to the value of the accumulator
#         // for as many times...
@0       //     ... as [value of register 0]...
+      //         ... increment the accumulator


I find it a bit hard to explain this further, so here's a rough chart of the accumulator and the two registers used during execution which will hopefully remove any confusion:

acc   reg[0]    reg[1] |
---------------------- |
1     <uninit>  0      | initialize; print acc("1")
1     0         1      | set reg[0] to reg[1]; set reg[1] to acc
1     0         1      | add reg[0] to acc; jump back and print acc ("1")
1     1         1      | set reg[0] to reg[1]; set reg[1] to acc
2     1         1      | add reg[0] to acc; jump back and print acc ("2")
2     1         2      | set reg[0] to reg[1]; set reg[1] to acc
3     1         2      | add reg[0] to acc; jump back and print acc ("3")
3     2         3      | set reg[0] to reg[1]; set reg[1] to acc
5     2         3      | add reg[0] to acc; jump back and print acc ("5")
5     3         5      | set reg[0] to reg[1]; set reg[1] to acc
8     3         5      | add reg[0] to acc; jump back and print acc ("8")


...and so forth and so on.

• Added the bounty! Also, you don't need the disclaimer, there used to be a rule banning languages newer than the challenge but it was removed a while back :) Apr 14, 2021 at 1:52

# Pinecone, 35 bytes

b:0;a:1;tru@(print:a;t:a;a:a+b;b:t)


• How did you get to know Pinecone? For me it was because I wanted to learn how to make my own language
– user100690
Apr 21, 2021 at 16:14
• @ophact I was researching on creating Lexers, and I found an article on it, that's how i reached there :) Apr 21, 2021 at 16:15

# Red, 47 bytes

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


Try it online!