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ChallengeSuperchallengeAdvanced Number Theory

Advanced Number Theory · Superchallenge problem

A sequence is defined by a1=3a_1=3 and an+1=11−ana_{n+1}=\dfrac{1}{1-a_n} for n≥1n\ge1. Then a2025=mna_{2025}=\dfrac{m}{n} in lowest terms. Find m+nm+n.