Question 128
Fix an irrational number $\alpha \gt 1$ and a positive integer $L$ such that $L \gt \frac{\alpha^2}{\alpha - 1}$.
Given an integer $x_1 \gt L$, define a sequence $\{x_n\}$ as follows: for every integer $n \ge 1$,
$$
x_{n+1} = \begin{cases} \lfloor \alpha x_n \rfloor, & \text{if } x_n \le L, \\ \lfloor \frac{x_n}{\alpha} \rfloor, & \text{if } x_n \gt L. \end{cases}
$$
Here, $\lfloor u \rfloor$ denotes the largest integer that is less than or equal to $u$.
(1) Prove that the sequence $\{x_n\}$ is eventually periodic, i.e., there exist positive integers $T$ and $N$ such that for any integer $n \gt N$, we have $x_{n+T} = x_n$.
(2) Prove that the smallest integer $T$ satisfying (1) is an odd integer that is independent of $x_1$.
Given an integer $x_1 \gt L$, define a sequence $\{x_n\}$ as follows: for every integer $n \ge 1$,
$$
x_{n+1} = \begin{cases} \lfloor \alpha x_n \rfloor, & \text{if } x_n \le L, \\ \lfloor \frac{x_n}{\alpha} \rfloor, & \text{if } x_n \gt L. \end{cases}
$$
Here, $\lfloor u \rfloor$ denotes the largest integer that is less than or equal to $u$.
(1) Prove that the sequence $\{x_n\}$ is eventually periodic, i.e., there exist positive integers $T$ and $N$ such that for any integer $n \gt N$, we have $x_{n+T} = x_n$.
(2) Prove that the smallest integer $T$ satisfying (1) is an odd integer that is independent of $x_1$.