Algorithm Implementation/Mathematics/Fibonacci Number Program

C

Recursive version

 unsigned int fib(unsigned int n){
   if (n < 2)
     return n;
   else
     return fib(n - 1) + fib(n - 2);
 }

Ternary Recursive version

 unsigned int fib(unsigned int n){
     return (n < 2) ? n : fib(n - 1) + fib(n - 2);
 }

Tail recursive version

 // C++ default arguments used. If implemented in C, call with fib(n, 0, 1) instead.
 unsigned int fib(unsigned int n, unsigned int a = 0, unsigned int b = 1) {
     if(n < 1) return a;
     else return fib(n - 1, a + b, a);
}

Lucas form

#include <math.h>
 
long fib(unsigned int n) {
    double fi = (1 + sqrt(5))/2;
    return (pow(fi, n) - pow(-fi, -n)) / sqrt(5);
}

Iterative version

 unsigned int fib(unsigned int n)
 {
   unsigned int i = 1, j = 0, k, t;
   for (k = 1; k <= n; k++)
   {
     t = i + j;
     i = j;
     j = t;
   }
   return j;
 }

Exponentiation by squaring

 unsigned int fib(unsigned int n){
   unsigned int i = n - 1, a = 1, b = 0, c = 0, d = 1, t;
   if (n <= 0)
     return 0;
   while (i > 0){
     if (i % 2 == 1){
       t = d*(b + a) + c*b;
       a = d*b + c*a;
       b = t;
     }
     t = d*(2*c + d);
     c = c*c + d*d;
     d = t;
     i = i / 2;
   }
   return a + b;
 }

Alternate exponentiation by squaring

 unsigned int fib(unsigned int n){
   unsigned int i = n - 1, a = 1, b = 0, c = 0, d = 1, t;
   if (n <= 0)
     return 0;
   while (i > 0){
     while (i % 2 == 0){
       t = d*(2*c + d);
       c = c*c + d*d;
       d = t;
       i = i / 2;
     }
     t = d*(b + a) + c*b;
     a = d*b + c*a;
     b = t;
     i--;
   }
   return a + b;
 }

C#

Iterative version

 static int fib(int n)
 {
   int fib0 = 0, fib1 = 1;
   for (int i = 2; i <= n; i++)
   {
      int tmp = fib0;
      fib0 = fib1;
      fib1 = tmp + fib1;
   }
   return (n > 0 ? fib1 : 0);
 }

Binet's formula

 static int fibBINET(int n)
 {
    double sqrt5 = Math.Sqrt(5.0);
    double phi = (1 + sqrt5 ) / 2;
    return (int)((Math.Pow(phi, n+1) - Math.Pow(1-phi, n+1)) / sqrt5);
 }

Using long numbers

    static Num FibonacciNumber(int n)
    {
        Num n1 = new Num(0);
        Num n2 = new Num(1);
        Num n3 = new Num(1);
        for (int i = 2; i <= n; i++)
        {
            n3 = n2 + n1;
            n1 = n2;
            n2 = n3;
        }
        return n3;
    }
    struct Num
    {
        const int digit_base = 0x40000000; // 2^30
        List<int> digits;
        public int Length { get { return digits.Count; } }
        public int this[int index] { get { return digits[index]; } private set { digits[index] = value; } }
        public Num(int i)
        {
            digits = new List<int>();
            while (i > 0)
            {
                digits.Add(i % digit_base);
                i /= digit_base;
            }
        }
        public static Num operator +(Num a, Num b)
        {
            Num n = new Num();
            n.digits = new List<int>();
            int l = Math.Min(a.Length,b.Length);
            int remainder = 0;
            for (int i = 0; i < l; i++)
            {
                n.digits.Add((a[i] + b[i] + remainder) % digit_base);
                remainder = (a[i] + b[i] + remainder) / digit_base;
            }
            Num longer = a.Length > b.Length ? a : b;
            for (; l < longer.Length; l++)
            {
                n.digits.Add((longer[l] + remainder) % digit_base);
                remainder = (longer[l] + remainder) / digit_base;
            }
            if (remainder > 0) n.digits.Add(remainder);
            return n;
        }
        public override string ToString()
        {
            StringBuilder sb = new StringBuilder();
            for (int i = Length - 1; i >= 0; i--)
            {
                sb.AppendFormat("{0:D" + (digit_base.ToString().Length-1) + "}", this[i]);
            }
            return sb.ToString();
        }
    }

D

Basic Recursive Version

ulong fib(uint n){
	return (n < 2) ? n : fib(n - 1) + fib(n - 2);
}

Iterative Version

ulong fib(uint n) {
	ulong fib0 = 0;
	ulong fib1 = 1;
	for (auto i = 2; i <= n; i++) {
		auto tmp = fib0;
		fib0 = fib1;
		fib1 = tmp + fib1;
	}
	return (n > 0 ? fib1 : 0);
}

Memoized Recursive Version

ulong fib(uint n){
	static ulong[] memo;
	if (n < 0)
		return n;
	if (n < memo.length)
		return memo[n];
	auto result = (n < 2) ? n : fib(n - 1) + fib(n - 2);
	memo.length = n + 1;
	memo[n] = result;
	return result;
}

Erlang

 fib(0) -> 0;
 fib(1) -> 1;
 fib(N) -> fib(N-1) + fib(N-2).

Arithmetic version

 fib(N) ->
     S = math:sqrt(5),
     round(math:pow(((1 + S) / 2), N) / S).

algorithm taken from the Pascal "more efficient" version, below

F#

Simple recursive

 let rec fib x = if x < 2I then x else fib(x - 1I) + fib(x - 2I)

This version uses F# System.Numerics.BigInteger type

Memoized recursive

open System.Collections.Generic
 
let rec fib n =
    let memo = Dictionary<_, _>()
    let rec fibInner = function
        | n when n = 0I -> 0I
        | n when n = 1I -> 1I
        | n -> fib(n - 1I) + fib(n - 2I)
    if memo.ContainsKey(n) then memo.[n]
    else
       let res = fibInner n
       memo.[n] <- res
       res

Tail recursive

let fib n = 
    let rec fibInner (n, a, b) =
       if (n = 0I) then a
       else fibInner ((n - 1I), b, (a + b))
    fibInner (n, 0I, 1I)

Iterative

let fib n = 
    if n < 2I then
        n
    else
        let mutable fib1 = 0I
        let mutable fib2 = 1I
        let mutable i = 2I
        let mutable tmp = 0I
        while (i <= n) do
            i <- i + 1I
            tmp <- fib1
            fib1 <- fib2
            fib2 <- tmp + fib2
        fib2

Infinite Sequence Generator

let fibSeq =
    Seq.unfold (fun (a, b) -> Some(a, (b, a + b))) (0I, 1I)
 
let fib n =
    fibSeq |> (Seq.skip n) |> Seq.head

Forth

: fib ( n -- fib )
  0 1 rot 0 ?do over + swap loop drop ;

Go

Recursive Solution

func fib(n int) int {
    if n < 2 {
        return n;
    }
    return fib(n-1) + fib(n-2);
}

Iterative Solution

func fib(n int) int {
    if (n == 0) {
        return 0
    }
 
    a, b := 0, 1
 
    for i := 1; i < n; i++ {
        a, b = b, a+b
    }
    return b
}

Haskell

List version

 fib n = fibs 0 1 !! n
     where
       fibs a b = a : fibs b (a + b)

Tail-recursive version

 fib n | n < 0     = undefined
       | otherwise = fib' n 0 1
     where
       fib' 0 a _ = a
       fib' n a b = fib' (n - 1) b (a + b)

Simple recursive version

 fib 0 = 0
 fib 1 = 1
 fib n = fib (n-1) + fib (n-2)

Awesome recursive version

fibs = 0 : 1 : [ a + b | (a, b) <- zip fibs (tail fibs)]

Or :

fibs = 0 : 1 : zipWith (+) fibs (tail fibs)

Or :

fibs = map fst $ iterate (\(a,b)->(b,a+b)) (0,1)

Closed-form version

Defines arithmetic operations on a custom data type, and then uses it to run the explicit formula without going via floating point - no rounding or truncation. Calculates the ten millionth fibonacci number in a few seconds (it has roughly two million digits).

module Fib where
 
-- A type for representing a + b * sqrt n
-- The n is encoded in the type.
data PlusRoot n a = a :+/ a
 deriving (Eq, Read, Show)
infix 6 :+/
 
-- Fetch the n in the root.
class WithRoot n where
  getRoot :: Num b => PlusRoot n a -> b
 
instance (WithRoot n, Num a) => Num (PlusRoot n a) where
  (a :+/ b) + (c :+/ d) = (a + c) :+/ (b + d)
  x@(a :+/ b) * (c :+/ d) = (a * c + getRoot x * b * d) :+/ (a * d + b * c)
  negate (a :+/ b) = negate a :+/ negate b
  fromInteger = (:+/ 0) . fromInteger
 
  -- I could implement these with (Ord a) but then we can't use the type
  -- with e.g. complex numbers.
  abs _ = error "PlusRoot.abs: unimplemented"
  signum _ = error "PlusRoot.signum: unimplemented"
 
instance (WithRoot n, Fractional a) => Fractional (PlusRoot n a) where
  fromRational = (:+/ 0) . fromRational
  recip x@(a :+/ b) = (a / r) :+/ (negate b / r)
   where
    r = a*a - getRoot x * b*b
 
-- Type parameter to PlusRoot. It would be easy to declare similar
-- types for Two or whatever, and get all the above arithmetic for free.
newtype Five = Five Five
 
instance WithRoot Five where
  getRoot _ = 5
 
-- The formula is phi^n - xi^n / sqrt 5
-- but it's always an integer, i.e. phi^n - xi^n is always a multiple
-- of sqrt 5, so the division isn't strictly necessary - just grab the
-- relevant coefficient.
fib :: Integer -> Integer
fib n = case phi^n - xi^n of
  -- The 'round' here is to make the types match; as discussed previously
  -- n must be an integer so no actual rounding is done.
  _ :+/ n -> round n
 where
  phi :: PlusRoot Five Rational
  phi = (1 :+/ 1) / 2
  xi = (1 :+/ negate 1) / 2

For other versions, see Haskell/Overview.

Dyalog APL

Basic Tail-Recursive Version

fibonacci?{    
   ??0 1
   ?=0:???
   (1??,+/?)? ?-1
}

Array-Oriented Version

fibonacci?{+/{?!??}(??)-?IO}

Other Versions

See Fibonacci at the Dynamic Functions Database

Io

Generic method version

 fib := method(n,
   if(n < 4,
     (n + n % 2) / 2,
     (n % 2 * 2 - 1) * fib((n + n % 3) / 2 - 1) ** 2 + fib((n - n % 3) / 2 + 1) ** 3
   )
 )

Polymorphic method version

  Number fibonacci := method((self - 1) fibonacci + (self -2) fibonacci)
  1 fibonacci = 1
  0 fibonacci = 0

Java

Recursive version

    public void run(int n)
    {
        if (n <= 0)
        {
            return;
        }
 
        run(n,1,0);
 
    }
 
    private void run(int n, int eax, int ebx)
    {
        n--;
 
        if (n == 0)
        {
            System.out.println(eax+ebx);
            return;
        }
 
        run(n,ebx,eax+ebx);
 
    }

Variations on the recursive version

    /* Recursive versions. Horribly inefficient. Use iterative/memoized versions instead */
    public long fib(long n) {
        if (n<2) {
            return n;
        }
        return fib(n-1) + fib(n-2);
    }
 
    public long fib2(long n) {
        return (n < 2) ? n : getValue(n - 1) + getValue(n - 2);
    }

Iterative version

    /**
     * Source based on 
     * http://20bits.com/2007/05/08/introduction-to-dynamic-programming/
     * as at 9-May-2007
     */
    private long fibonacci(int n) 
    {
          long n2 = 0;
          long n1 = 1;
          long tmp;
          for (int i=n ; i>2 ; i--) {
              tmp = n2;
              n2 = n1;
              n1 = n1 + tmp;
          }
          return n2 + n1;
    }

Simpler Iterative Version

  /**
    * returns the Nth number in the Fibonacci sequence
    */
  public int fibonacci(int N) {
      int lo = 0;
      int hi = 1;
      for (int i = 0; i < N; i++) {
          hi = lo + hi;
          lo = hi - lo;
      }
      return lo;
  }

Memoized version

	private int[]	fibs; // array for memoized fibonacci numbers
 
	public int fib(int n) {
 
		if (n < 2) {
			return n;
		}
 
		if (fibs == null) { // initialise array to first size asked for
			fibs = new int[n + 1];
		} else if (fibs.length < n) { // expand array
			int[] newfibs = new int[n + 1]; // inefficient if looping through values of n
			System.arraycopy(fibs, 0, newfibs, 0, fibs.length);
			fibs = newfibs;
		}
 
		if (fibs[n] == 0) {
			fibs[n] = fib(n - 1) + fib(n - 2);
		}
 
		return fibs[n];
	}

Iterative Memoized version

	public int fib(int n) {
 
		if (n < 2) {
			return n;
		}
                int[] f = new int[n+1];
                f[0] = 0;
	        f[1] = 1;
	        for(int i = 2;i<=n;i++) {
                        f[i] = f[i-1] + f[i-2];
	        }
		return f[n];
	}

Linotte

Fibonacci:
Principal :
Rôles :
        n :: nombre
Actions :
        "Entrez un nombre :" !
        n ?
        fibo(n) !

Fibo :
Rôles :
        * n :: nombre
Actions :
        si n < 2 alors retourne n
        retourne fibo(n-1) + fibo(n-2)

Lexico (in spanish)

clase Fib
publicos:
mensajes:
Fib nop
Fibonacci(deme n es una cantidad) es_funcion cantidad
        {
        los objetos uno, dos, tres, i, respuesta son cantidades
        copie 0 en uno
        copie 1 en dos
        variando i desde 1 hasta n haga:
                {
                copie uno en respuesta
                copie uno + dos en tres
                copie dos en uno
                copie tres en dos
                }
        retornar uno
        }
/**********************************/
tarea
{
el objeto f es un Fib
muestre "el 5: ", f.Fibonacci(doy 5)
}

Lua

 function fib(n)
   local a, b = 0, 1
   while n > 0 do
     a, b = b, a + b
     n = n - 1
   end
   return a
 end

Recursive version

 function fib(n)
   if n > 1 then n = fib(n - 1) + fib(n - 2) end
   return n
 end

Matlab

Recursive snippet

function F = fibonacci_recursive(n)
if n < 2 
    F = n;
else
    F = fibonacci_recursive(n-1) + fibonacci_recursive(n-2);
end

Iterative snippet

function F = fibonacci_iterative(n)
first  = 0;
second = 1;
third  = 0;
for q = 1:n,
    third = first + second;
    first = second;
    second = third;
end
F = first;

Maxima

Recursive version

fib(n):=
    if n < 2 then
        n
    else
        fib(n - 1) + fib(n - 2)
$

Lucas form

fib(n):=(%phi^n-(-%phi)^-n)/sqrt(5);

Iterative version

fib(n) := block(
    [i,j,k],
    i : 1,
    j : 0,
    for k from 1 thru n do
        [i,j] : [j,i + j],
    return(j)
)$

Exponentiation by squaring

fib(n) := block(
    [i,F,A],
    if n <= 0 then
        return(0),
    i : n - 1,
    F : matrix([1,0],[0,1]),
    A : matrix([0,1],[1,1]),
    while i > 0 do block(
        if oddp(i) then
            F : F.A,
        A : A^^2,
        i : quotient(i,2)
    ),
    return(F[2,2])
)$

O'Caml

 let fib n = 
   let rec fibonacci n = 
     match n with 
     | 0 -> (0, 0)
     | 1 -> (0, 1) 
     | m -> 
       let (a, b) = fibonacci (m-1) in 
         (b, a+b)
   in 
   let (_, k) = fibonacci n in 
     k;;

Pascal

  function F(n: integer): integer;
  begin
    case n of
    1,2: Result:=1
    else Result:=F(n-1)+F(n-2) end;
  end;

A bit more efficient

  function F(n: integer): integer;
  begin
    Result:=Round(Power((1+sqrt(5))/2, n)/sqrt(5));
  end;

Note that Power is usually defined in Math, which is not included by default.

For most compilers it's possible to improve performance by using the Math.IntPower instead of the Math.Power.

Iterative version also for negative arguments

function fib(n:integer):extended;
var i:integer;
    fib0,fib1:extended;
begin
fib0:=0;
fib1:=1;
for i:=1 to abs(n) do 
begin
fib0:=fib0+fib1;
fib1:=fib0-fib1;
end; 
if (n<0)and(not odd(n)) then fib0:=-fib0; 
fib:=fib0;
end:

Perl

 sub fib {
   my ($n, $a, $b) = (shift, 0, 1);
   ($a, $b) = ($b, $a + $b) while $n-- > 0;
   $a;
 }

Recursive versions

 sub fib {
   my $n = shift;
   return $n if $n < 2;
   return fib($n - 1) + fib($n - 2);
 }
 
 # returns F_n in a scalar context
 # returns all elements in the sequence up to F_n in a list context
 # only one recursive call
 sub fib {
    my ($n) = @_;
    return (0) if ($n == 0);
    return (0, 1) if ($n == 1);
    my @fib = fib($n - 1);
    return (@fib, $fib[-1] + $fib[-2]);
 }

Binary recursion, snippet

sub fibo;
sub fibo {$_ [0] < 2 ? $_ [0] : fibo ($_ [0] - 1) + fibo ($_ [0] - 2)}

Runs in Θ(F(n)) time, which is Ω(1.6ⁿ).

Binary recursion with special Perl "caching", snippet

use Memoize;
memoize 'fibo';
sub fibo;
sub fibo {$_ [0] < 2 ? $_ [0] : fibo ($_ [0] - 1) + fibo ($_ [0] - 2)}

Iterative, snippet

sub fibo
{
    my ($n, $a, $b) = (shift, 0, 1);
    ($a, $b) = ($b, $a + $b) while $n-- > 0;
    $a;
}

PHP

 function generate_fibonacci_sequence( $length ) {
   for( $l = array(0,1), $i = 2, $x = 0; $i < $length; $i++ )
        $l[] = $l[$x++] + $l[$x];
   return $l;
 }

Recursive version

 function fib( $n ){
   return ( $n < 2 )
          ? $n
          : fib( $n-1 )+fib( $n-2 );}

OOP version

 class fibonacci {
 
 	public $Begin	= 0;
 	public $Next;
 	public $Amount;
 	public $i;
 
 	public function __construct( $Begin, $Amount ) {
 
 		$this->Begin	= 0;
 		$this->Next		= 1;
 		$this->Amount	= $Amount;
 
 	}
 
 	public function _do() {
 
 		for( $this->i = 0; $this->i < $this->Amount; $this->i++ ) {
 
 			$Value = ( $this->Begin + $this->Next );
 
 			echo $this->Begin . ' + ' . $this->Next . ' = ' . $Value . '<br />';
 
 			$this->Begin	= $this->Next;
 			$this->Next		= $Value;
 
 		}
 
 	}
 
 }
 
 $Fib = new fibonacci( 0, 6 );
 echo $Fib->_do();

Alternate version

 function fib($n) {
   return round(pow(1.6180339887498948482, $n) / 2.2360679774998);
 }

Python

Recursive version

def fib(n):
    if n < 2: return n
    return fib(n - 1) + fib(n - 2)

Recursive with memorization

m = {0: 1, 1: 1}
def fib(n):
    #assert n >= 0
    if n not in m:
        m[n] = fib(n-1) + fib(n-2)
    return m[n]

Lucas form

def fib(n):
    fi = (1 + sqrt(5))/2
    return (fi**n - (-fi)**-n)/sqrt(5)

Iterative version

def fib(n):
    i,j = 1,0
    for k in range(1,n + 1):
        i,j = j, i + j
    return j

Exponentiation by squaring

def fib(n):
    if n <= 0:
        return 0
    i = n - 1
    a,b = 1,0
    c,d = 0,1
    while i > 0:
        if i % 2 == 1:
            a,b = d*b + c*a, d*(b + a) + c*b
        c,d = c**2 + d**2, d*(2*c + d)
        i = i / 2
    return a + b

Lucas sequence identities

def fib(n):
    if n <= 0:
        return 0
 
    # n = 2**r*s where s is odd
    s, r = n, 0
    while s & 1 == 0:
        r, s = r+1, s/2
 
    # calculate the bit reversal t of (odd) s
    # e.g. 19 (10011) <=> 25 (11001)
    t = 0
    while s > 0:
        if s & 1 == 1:
            t, s = t+1, s-1
        else:
            t, s = t*2, s/2
 
    # use the same bit reversal process
    # to calculate the sth Fibonacci number
    # using Lucas sequence identities
    u, v, q = 0, 2, 2
    while t > 0:
        if t & 1 == 1:
            # u, v of x+1
            u, v = (u + v) / 2, (5*u + v) / 2
            q, t = -q, t-1
        else:
            # u, v of 2*x
            u, v = u * v, v * v - q
            q, t = 2, t/2
 
    # double s until we have
    # the 2**r*sth Fibonacci number
    while r > 0:
        u, v = u * v, v * v - q
        q, r = 2, r-1
 
    return u

Lucas sequence identities, recursion

As with the iterative version, this solution is also O(log n) with arbitrary precision.

def fib(n):
    def fib_inner(n):
        if n == 0:
            return 0, 2
        m = n >> 1
        # q = 2*(-1)**m
        q = -2 if (m & 1) == 1 else 2
        u, v = fib_inner(m)
        u, v = u * v, v * v - q
        if n & 1 == 1:
            # u, v of 2m+1
            u1 = (u + v) >> 1
            return u1, 2*u + u1
        else:
            # u, v of 2m
            return u, v
 
    if n <= 0:
        return 0
    # the outermost loop is unrolled
    # to avoid calculating an unnecessary v
    m = n >> 1
    u, v = fib_inner(m)
    if n & 1 == 1:
        # u of m+1
        u1 = (u + v) >> 1
        # u of 2m+1
        return u*u + u1*u1
    else:
        # u of 2m
        return u * v

REBOL

Recursive version

fib: func [n [integer!]] [
    either n < 2 [n] [(fib n - 1) + (fib n - 2)]
]

Ruby

 class Integer
     def fib
         @n = self.abs
         if @n < 2
             return @n
         else
             return (@n-1).fib + (@n-2).fib
         end
     end
 end

Alternate:

 class Integer
     def fib
         @n = self.abs
         (@n<2)?(return @n):(return (@n-1).fib+(@n-2).fib)
     end
 end
 
 # you run it like this
 puts 10.fib # output: 55
 puts 15.fib # output: 610

Recursive

def fib n
   return n if n < 2
   fib(n - 1) + fib(n - 2)
end

Generator

 class FibGenerator
   def initialize(n)
     @n = n		
   end	
 
   def each
     a, b = 1, 1
     @n.times do
       yield a
       a, b = b, a+b
     end
   end
 
   include Enumerable
 end
 
 def fibs(n)
   FibGenerator.new(n)
 end
 
 #use like this
 fibs(6).each do |x|
   puts x
 end

Arithmetic version

 def f(n)
   ((((1+Math.sqrt(5))/2)**n)/Math.sqrt(5)+0.5).floor
 end

Memoized Version

 fibmemo=Hash.new{|h,k| h[k-1]+h[k-2]}
 fibmemo[0]=1
 fibmemo[1]=1
 
 def fib n
   fibmemo[n]
 end

Scheme

Tree-recursive version

 (define (fib n)
   (if (<= n 1)
       n
       (+ (fib (- n 1)) (fib (- n 2)))))

Iterative (tail-recursive) version

 (define (fib n)
   (define (iter a b count)
     (if (<= count 0)
         a
         (iter b (+ a b) (- count 1))))
   (iter 0 1 n))

Named-let, Iterative version

 (define (fib n)
   (let loop ((a 0) (b 1)
              (count n))
     (if (<= count 0) a
         (loop b (+ a b) (- count 1))))))

Lucas form

 (define fib
   (let* ((sqrt5 (inexact->exact (sqrt 5)))
          (fi (/ (+ sqrt5 1) 2)))
     (lambda (n)
       (round (/ (- (expt fi n) (expt (- fi 1) n)) sqrt5)))))

Logarithmic-time Version

This version squares the Fibonacci transformation, allowing calculations in log₂(n) time:

(define (fib-log n)
  "Fibonacci, in logarithmic time."
  (define (fib-iter a b p q count)
    (cond ((= count 0) b)
	  ((even? count)
	   (fib-iter a b
		     (+ (* p p) (* q q))
		     (+ (* 2 p q) (* q q))
		     (/ count 2)))
	  (else (fib-iter (+ (* b q) (* a q) (* a p))
			  (+ (* b p) (* a q))
			  p q
			  (- count 1)))))
 
  (fib-iter 1 0 0 1 n))

UCBLogo

to fib :n
  output (cascade :n [?1+?2] 1 [?1] 0)
end

Recursive version

to fib :n
  if :n<2 [output 1]
  output (fib :n-1)+(fib :n-2)
end

VB.NET

Array oriented version

 Dim i As Integer = 2
 Dim sequencelength As Integer = 50
 Dim fibonacci(sequencelength) As Integer
 fibonacci(0) = 0
 fibonacci(1) = 1
 While i <> sequencelength
      fibonacci(i) = fibonacci(i - 1) + fibonacci(i - 2)
      i += 1
 End While

Recursive Version

      Private Function fibonacci(ByVal i as integer) As Integer
            If i < 1 Then
                  Return -1
            ElseIf i < 2 Then
                  Return i
            Else
                  Return fibonacci(i-1) + fibonacci(i-2)
            End If
      End Function

JavaScript

Recursive version

function fib(n) {
  return n < 2 ? n : fib(n - 1) + fib(n - 2);
}

Alternative recursive version

function fib(n, prev, cur) {
  if (prev == null) prev = 0;
  if (cur == null) cur = 1;
  if (n < 2) return cur;
  return fib(n--, cur, cur + prev);
}

Prev and cur is optional arguments.

Iterative version

function fibonacci(n) {
  var i = 1, j = 0, k, t;
  for (k = 1; k <= Math.abs(n); k++) {
     t = i + j;
     i = j;
     j = t;
  }
  if (n < 0 && n % 2 === 0) j = -j;
  return j;
}

This example supports negative arguments.

Lucas form

function fibonacci(n) {
  var sqrt5 = Math.sqrt(5);
  var fi = (1 + sqrt5) / 2;
  return Math.round((Math.pow(fi, n) - Math.pow(-fi, -n)) / sqrt5);
}

Binets formula

function fibonacci(n) {
  var sqrt5 = Math.sqrt(5);
  var fi = (1 + sqrt5) / 2;
  return Math.round((Math.pow(fi, n + 1) - Math.pow(1 - fi, n + 1)) / sqrt5);
}

Algorithm from the Pascal "more efficient" version

function fibonacci(n) {
  var sqrt5 = Math.sqrt(5);
  return Math.round(Math.pow(((1 + sqrt5) / 2), n) / sqrt5);
}

Common Lisp

Lucas form

(defun fib (n)
  (cond
   ((= n 0) 0)
   ((or (= n 1) (= n 2)) 1)
   ((= 0 (mod n 2)) (-
                     (expt (fib (+ (truncate n 2) 1)) 2)
                     (expt (fib (- (truncate n 2) 1)) 2)))
   (t (+ (expt (fib (truncate n 2)) 2)
         (expt (fib (+ (truncate n 2) 1)) 2)))))
 
(fib (parse-integer (second *posix-argv*))) ;

Recursive version

(defun fib (x)
  (if (or (zerop x) (= x 1)) 1
    (+ (fib (- x 1)) (fib (- x 2)))))
 
(print (fib 10))

PostScript

Iterative

20 % how many Fibonacci numbers to print
                                                                               
1 dup
3 -1 roll
{
       dup
       3 -1 roll
       dup
       4 1 roll
       add
       3 -1 roll
       =
}
repeat

Stack recursion

This example uses recursion on the stack.

% the procedure
/fib
{
   dup dup 1 eq exch 0 eq or not
   {
      dup 1 sub fib
      exch 2 sub fib
      add
   } if
} def

% prints the first twenty fib numbers 
/ntimes 20 def
  
/i 0 def
ntimes {
   i fib =
   /i i 1 add def
} repeat

PL/SQL

Iterative snippet

CREATE OR REPLACE PROCEDURE fibonacci(lim NUMBER) AS
  fibupper NUMBER(38);
  fiblower NUMBER(38);
  fibnum NUMBER(38);
  i NUMBER(38);
  BEGIN
    fiblower := 0;
    fibupper := 1;
    fibnum := 1;
    FOR i IN 1 .. lim
    LOOP
      fibnum := fiblower + fibupper;
      fiblower := fibupper;
      fibupper := fibnum;
      DBMS_OUTPUT.PUT_LINE(fibnum);
    END LOOP;
  END;