2024 CMO

Proof. First, since $\alpha$ is an irrational number, the integer part operation in the definition of $x_{n+1}$ always makes the corresponding number strictly smaller. For any integer $u$ satisfying $\frac{1}{\alpha -1}<u\le L$, we have $\lfloor \alpha u\rfloor >\alpha u-1>u$ and $\lfloor \alpha u\rfloor \le \alpha L$; while for any integer $v\ge L+1$, we have $\lfloor \frac{v}{\alpha }\rfloor >\frac{L}{\alpha }-1>\frac{1}{\alpha -1}$ and $\lfloor \frac{v}{\alpha }\rfloor <v$. Therefore, it is easy to verify by mathematical induction that for any positive integer $n$, $$\frac{1}{\alpha -1}<x_n\le \max \{x,\alpha L\},$$ which implies that the sequence $\{x_n\}$ is bounded. Since $\{x_n\}$ is a recursive sequence, it must eventually become periodic. Let $T$ denote the smallest positive period of $\{x_n\}$. By definition, there exists a positive integer $N$ such that for any integer $n\ge N$, we have $x_{n+N}=x_n$. Since $\lfloor \alpha L\rfloor >\alpha L-1>L$, it follows that $\lfloor \alpha L\rfloor \ge L+1$. Thus, for any integer $n\ge N$, we have $x_n\le \lfloor \alpha L\rfloor$. Consequently, for any integer $n\ge N$, if $x_n\ge L+1$, then $$x_{n+1}=\lfloor \frac{x_n}{\alpha }\rfloor <\frac{1}{\alpha }\cdot \alpha L=L,$$ and $$x_{n+2}=\lfloor \alpha x_{n+1}\rfloor .$$ Clearly, there exists an integer $n\ge N$ such that $x_n>L$. Let $m\ge N+1$ be the smallest integer such that $$x_{m-1}=\min \{x_n\mid n\ge N\text{ and }{x}_n>L\}.$$ From the previous analysis, we know $$x_m=\lfloor \frac{x_{m-1}}{\alpha }\rfloor <L\text{and}{x}_{m+1}=\lfloor \alpha x_m\rfloor >\alpha x_m-1>x_m.$$ Note that $x_{m+1}<\alpha x_m<\alpha \cdot \frac{x_{m-1}}{\alpha }=x_{m-1}$. By the minimality of $x_{m-1}$, we know $x_{m+1}\le L$. Since $\alpha x_m<x_{m+1}+1\le L+1$, it follows that $$x_m<\frac{L+1}{\alpha },$$

and hence $x_m\le \lfloor \frac{L+1}{\alpha }\rfloor$. Moreover, since $x_{m-1}\ge L+1$, we have $$x_m=\lfloor \frac{x_{m-1}}{\alpha }\rfloor \ge \lfloor \frac{L+1}{\alpha }\rfloor .$$ Therefore, $x_m=\lfloor \frac{L+1}{\alpha }\rfloor$. By the minimality of $x_{m-1}$, we conclude that for any integer $n\ge n_0$, $x_n\ge x_m$. For any integer $n\ge N$, if $x_m+1\le x_n\le L$, then $$x_{n+1}=\lfloor \alpha x_n\rfloor >\alpha (x_m+1)-1>\alpha \cdot \frac{L+1}{\alpha }-1=L,$$ which implies $x_{n+1}\ge L+1$. Moreover, $$x_{n+2}=\lfloor \frac{x_{n+1}}{\alpha }\rfloor <\frac{x_{n+1}}{\alpha }<\frac{1}{\alpha }\cdot \alpha x_n=x_n.$$ Combining the above analysis, we have $$x_m<x_{m+1}\le L<x_{m+2}.$$ Furthermore, for each $i=1,2,\dots ,T-1$, when $i$ is odd, $x_{m+i}\le L$; when $i$ is even, $x_{m+i}>L$. Note also that $x_{m+1}>x_{m+3}>\cdots >x_{m+T}=x_m$. Therefore, $T$ is an odd number. Finally, note that $x_m=\lfloor \frac{L+1}{\alpha }\rfloor$ is independent of $x$. Thus, the minimal positive period $T$ is also independent of $x$.