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a_{n+1} = a^2_n-5 for odd a_n - All-Russian MO 2000 Regional (R4) 10.6

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September 26, 2024
recurrence relationnumber theory

Problem Statement

Given a natural number a0a_0, we construct the sequence {an}\{a_n\} as follows an+1=an2−5a_{n+1} = a^2_n-5 if ana_n is odd, and an2\frac{a_n}{2} if ana_n is even. Prove that for any odd a0>5a_0 > 5 in the sequence {an}\{a_n\} arbitrarily large numbers will occur.
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