olymmath

Putnam 2025 A6

  • Session A

Let b0=0b_0 = 0 and, for n≥0n \geq 0, define bn+1=2bn2+bn+1b_{n+1} = 2b_n^2 + b_n + 1. For each k≥1k \geq 1, show that b2k+1−2b2kb_{2^{k+1}} - 2b_{2^k} is divisible by 22k+22^{2k+2} but not by 22k+32^{2k+3}.

Curated text Kedlaya Putnam archive. Quoted with source.

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