For each x in [0,1], define
{f(x)=2x,if0≤x≤12;f(x)=2−2x,if12<x≤1.
Let f[2](x)=f(f(x)), and f[n+1](x)=f[n](f(x)) for each integer n≥2. Then the number of values of x in [0,1] for which f[2005](x)=12 can be expressed in the form pa, where p is a prime and a is a positive integer. Find p+a.