74th Romanian Mathematical Olympiad

The sequence $(x_n{)}_{n\ge 1}$ has positive terms (proof by induction). There is $k\in {\mathbb{N}}^{\ast }$ such that $x_k>p$. Indeed, if we assume, by reductio ad absurdum, that $x_n\le p,\forall n\in {\mathbb{N}}^{\ast }$, we obtain $x_{n+1}\ge x_n+1,\forall n\in {\mathbb{N}}^{\ast }$, so $x_n\ge a+n-1,\forall n\in {\mathbb{N}}^{\ast }$. In particular, we find $x_{p+1}\ge a+p>p$. Contradiction. Let us denote $k_0=\min \{k\in {\mathbb{N}}^{\ast }\mid x_k>p\}$. Since $\lfloor \frac{p}{x}\rfloor =0,\forall x>p$, we get $x_n=x_{k_0},\forall n\ge k_0$. Therefore $(x_n{)}_{n\ge 1}$ is convergent, with ${\lim }_{n\to \infty }{x}_n=x_{k_0}$. For the value of the limit, we analyze three cases.

Case 1. $a\in (p,\infty )$. Then ${\lim }_{n\to \infty }{x}_n=x_1=a$.

Case 2. $a\in (0,1)$. Then $x_2=a+\left[\frac{p}{a}\right]>\left[\frac{p}{a}\right]\ge p$, hence ${\lim }_{n\to \infty }{x}_n=x_2=a+\left[\frac{p}{a}\right]$.

Case 3. $a\in [1,p]$. The sequence has the terms $x_n=\{a\}+y_n$, where $\{a\}=a-[a]\in [0,1)$ is the fractional part of $a$, and $y_n\in {\mathbb{N}}^{\ast }$. We have $(x-1)(x-p)\le 0$, for all $x\in [1,p]$. It follows that $x+\left[\frac{p}{x}\right]\le x+\frac{p}{x}\le p+1$, for all $x\in [1,p]$. Therefore, if $x_n\in [1,p]$, then $x_{n+1}\le p+1$. Thus $x_{k_0}\in (p,p+1]\cap \{\{a\}+k\mid k\in {\mathbb{N}}^{\ast }\}$. We find $$\underset{n\to \infty }{\lim }{x}_n=x_{k_0}=\{\begin{array}{ll}p+\{a\},& a\in [1,p]\setminus \mathbb{N}\\ p+1,& a\in \{1,2,\dots ,p\}\end{array}$$