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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 $$p + a.$$

Apr 10, 2019