2025 CMO 第 3 题

Let $\alpha > 1$ be an irrational number and $L$ be a integer such that $L > \frac{\alpha^2}{\alpha - 1}$ . A sequence $x_1, x_2, \cdots$ satisfies that $x_1 > L$ and for all positive integers $n$ , $$x_{n+1} = \begin{cases} \left \lfloor \alpha x_n \right \rfloor & \textup{if} \; x_n \leq L \left \lfloor \frac{x_n}{\alpha} \right \rfloor & \textup{if} \; x_n > L \end{cases}.$$

Prove that

(i) $\left\{x_n\right\}$ is eventually periodic.

(ii) The eventual fundamental period of $\left\{x_n\right\}$ is an odd integer which doesn't depend on the choice of $x_1$ .

设 $\alpha > 1$ 为无理数，$L$ 为整数，使得 $L > \frac{\alpha^2}{\alpha - 1}$ 。序列 $x_1, x_2, \cdots$ 满足 $x_1 > L$ 并且对于所有正整数 $n$ ， x_{n+1} = \begin{cases} \left \lfloor \alpha x_n \right \rfloor & \textup{if} \; x_n \leq L \left \lfloor \frac{x_n}{\alpha} \right \rfloor & \textup{if} \; x_n > L \end{案例}。

证明

(i) $\left\{x_n\right\}$ 最终是周期性的。

(ii) $\left\{x_n\right\}$ 最终的基本周期是一个奇整数，不依赖于 $x_1$ 的选择。