$a=sqrt 5;print+(((1+$a)/2)**$_-((1-$a)/2)**$_)/$a,$/for 0..73

# For comparison, here is the result of the golfing
# that Tilly and I participated in earlier:
$a=1;print$a-=$b+=$a*=-1,$/for 0..73
[download]

In case you were wondering, 73 is an arbitrary stopping place. I picked it because Fib(74) is too big a number for Perl, and results in the scientific notation: 1.30496954492866e+015 which just isn't as much fun.

Update: Of course, what we really need is a benchmark! So here is the output:

Benchmark: timing 100000 iterations of Binet, Golf...
Binet:  1 wallclock secs ( 0.70 usr +  0.00 sys =  0.70 CPU) @ 142653.
+35/s (n=100000)
 Golf:  5 wallclock secs ( 4.75 usr +  0.00 sys =  4.75 CPU) @ 21065.9
+4/s (n=100000)
[download]

Recursive: 118 wallclock secs (116.37 usr +  0.00 sys = 116.37 CPU) @ 
+ 0.09/s (n=10)
[download]

#!perl -w
use strict;
use Benchmark;

# Fibonnaci Number to calculate. 
use vars ('$n'); # Make it global, but appease strict.
$n = 30;

my $iter = 100_000;
timethese( $iter, {
    Binet => '&FibBinet($n)',
    Golf => '&FibGolf($n)',
#    Recursive => '&FibRecursive($n)',
});


sub FibRecursive
{
  my $index = shift;
  return 0 if 0 == $index;
  return 1 if 1 == $index;
  return FibRecursive( $index - 1 ) + FibRecursive( $index - 2 );
}

sub FibGolf
{
  my $index = shift;
  my $acc = 1;
  my $tmp;

  while( $index-- )
  {
    $acc-=$tmp+=$acc*=-1;
  }

  return $acc;
}

sub FibBinet
{
  my $index = shift;
  my $root5 = sqrt 5;

  return ( ( ( 1 + $root5 ) / 2 ) ** $index 
         - ( ( 1 - $root5 ) / 2 ) ** $index ) / $root5;
}
[download]

Note that this is a classic example of a "useless formula". In real life you would never want to use it because square roots are slow to calculate. Instead you use the formula:

F(n+m) = F(n)F(m-1) + F(n-1)F(m+1)
[download]

This is where F is the Fibonacci function, F(0)=0, F(1)=F(2)=1. In particular you use the special cases:

F(2n) = F(n)F(n-1) + F(n-1)F(n+1)
      = F(n)F(n-1) + F(n-1)(F(n) + F(n-1))
      = 2F(n)F(n-1) + F(n-1)F(n-1)
[download]

and

F(2n-1) = F(n-1)F(n-1) + F(n-2)F(n)
        = F(n-1)F(n-1) + (F(n)-F(n-1))F(n)
        = F(n)F(n) - F(n-1)F(n) + F(n-1)F(n-1)
[download]

to calculate the powers of two and one off from the powers of two, then when you are done go to the general formula to calculate the number.

Now why would one do this? Well it turns out that certain numbers can be tested for primality by answering a divisibility question about some close relative of the Fibonacci numbers (eg the Lucas numbers), so most of the "record primes" that you hear about involved some very big Fibonacci numbers being calculated somewhere.

A good exercise for anyone who is interested, use the very slow Math::BigInt to come up with a function to calculate very large Fibonacci numbers.