2025 Putnam A6

Let $b_0 = 0$ and, for $n \geq 0$, define $b_{n+1} = 2b_n^2 + b_n + 1$. For each $k \geq 1$,

show that $b_{2^{k+1}} - 2b_{2^k}$ is divisible by $2^{2k+2}$ but not by $2^{2k+3}$.

令$b_0 = 0$，并且对于$n \geq 0$，定义$b_{n+1} = 2b_n^2 + b_n + 1$。对于每个 $k \geq 1$，

证明 $b_{2^{k+1}} - 2b_{2^k}$ 能被 $2^{2k+2}$ 整除，但不能被 $2^{2k+3}$ 整除。