Just out of curiosity, how can you tell if a number x is a power of two (x = 2^n) without using recursion.
One way is to use bitwise AND. If a number
$x is a power of two (e.g., 8=1000), it will have no bits in common with its predecessor (7=0111). So you can write:
($x & ($x - 1)) == 0
Note: This will give a false positive for $x == 0.