Note that all of these break at some point due to the limited precision of sqrt 5, it's not that noticable on the languages that don't provide bignum

Haskell:
let phi = ((sqrt 5) + 1) / 2
let fib n = round ((phi ** n) / (sqrt 5))

scheme:
(define phi (/ (+ 1 (sqrt 5)) 2))
(define (fib n) (inexact->exact (round (/ (expt phi n) (sqrt 5)))))

common lisp:
(defvar phi (/ (+ 1 (sqrt 5)) 2))
(defun fib (n) (round (/ (expt phi n) (sqrt 5))))

ocaml:
(* note: fib 44 max *)
let round n = if (n -. (floor n) > 0.5) then ceil n else floor n;;
let phi = ((sqrt 5.0) +. 1.0) /. 2.0;;
let fib n = int_of_float (round ((phi ** float_of_int n) /. sqrt 5.0));;

clean:
// note fib 46 max
module test

import StdEnv
import StdReal

pow x n 
    | n==0.0 = 1.0
    = pow x (n-1.0) * x

round n
    | n - (toReal (toInt n)) > 0.5 = (toInt n) + 1
    = toInt n

phi = ((sqrt 5.0) + 1.0) / 2.0

fib n = round ((pow phi (toReal n)) / (sqrt 5.0))
    
Start = fib 30

python:
from math import sqrt
from math import pow

def fib(n):
    return int(round(pow(((sqrt(5) + 1) / 2), n) / (sqrt(5))))