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For each x in [0,1], define \(\begin{cases} f(x) = 2x, \qquad\qquad \mathrm{if} \quad 0 \leq x \leq \frac{1}{2};\\ f(x) = 2-2x, \qquad \mathrm{if} \quad \frac{1}{2} < x \leq 1. \end{cases}\)
Let \(f^{[2]}(x) = f(f(x))\), and \(f^{[n + 1]}(x) = f^{[n]}(f(x))\) for each integer \(n \geq 2\). Then the number of values of x in [0,1] for which \(f^{[2005]}(x) = \frac {1}{2}\) can be expressed in the form \(p^a\), where p is a prime and a is a positive integer. Find p+a. 

 Jan 20, 2019
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p + a = 1002 + 4 = 1006.

 Dec 1, 2019

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