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For each x in [0,1], define

 

{f(x)=2x,if0x12;f(x)=22x,if12<x1.


Let f[2](x)=f(f(x)), and f[n+1](x)=f[n](f(x)) for each integer n2. 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.

 Feb 19, 2020
 #1
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the answer is 2007(if that help)

 Oct 4, 2020

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