MathDB
2006 SMT Team Round #6 - Find p(2006)-q(2006)

Source:

August 22, 2011
functionratiogeometric series

Problem Statement

The expression 16n+4n+116^n+4^n+1 is equiavalent to the expression (2p(n)−1)/(2q(n)−1)(2^{p(n)}-1)/(2^{q(n)}-1) for all positive integers n>1n>1 where p(n)p(n) and q(n)q(n) are functions and p(n)q(n)\tfrac{p(n)}{q(n)} is constant. Find p(2006)−q(2006)p(2006)-q(2006).
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