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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 \(p+a.\)

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

 Oct 4, 2020

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