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f(x+1)=(52x−1+1)(x+1)2f(x)\large f(x+1) = (5^{2^{x-1}} + 1)(x+1)^{2}f(x)f(x+1)=(52x−1+1)(x+1)2f(x)
Suppose we define a function of xxx as described above for xxx is a non-negative integer with f(0)=5−1f(0)=\sqrt{5} - 1f(0)=5−1.
If the value of f(2020)f(2020)f(2020) can be written in form (5p−1)(q!)2(5^{p} - 1)(q!)^{2}(5p−1)(q!)2, for positive integers ppp and qqq, find 2qp\frac{2^{q}}{p}p2q.
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