The condition simply says that the set $\{a_n|n\in\mathbb N^*\}$ has at most one non-null value. It's clear that if this condition hiolds, then the sequence is bounded, so the other implication is the interesting one. We assume the sequence is bounded. There are two cases: Case 1: the maximum $M$ of $\{a_n\}$ is reached. We can see that if $a_n=M$, then $a_{n+1}$ is either $M$ or $0$, and the same holds for $a_{n-1}$. Moreover, $a_{n+1}=M\iff a_{n-1}=0$, and we can construct the sequence in two possible ways, in both directions, according to whether $a_{n+1}=M$ or $0$. In both these situations, the sequence contains only two values, in both directions, $0$ and $M$, so the conclusion is proven. Case 2: the maximum $M$ is never reached. Let $\varepsilon>0$ be small enough (small enough that $2(M-\varepsilon)>M$, for instance) so that we can find $a_n=M-\varepsilon$. It's easy to show that we can also find $n$ s.t. $a_n=M-\varepsilon$ and $a_{n+1}=t,\ 0<t<M-\varepsilon$. We thus have $a_{n+2}=M-\varepsilon+t$, and if $a_{n+3}=M-\varepsilon$, then $a_{n+4}=2(M-\varepsilon)+t>M$ , which is a contradiction. This shows that, in fact, $a_{n+3}=M-\varepsilon+2t, a_{n+4}=t$, and we can repeat the procedure with $\varepsilon$ replaced with $\varepsilon-2t$ (which must necessarily be $>0$). We find that $\varepsilon-2nt>0,\ \forall n\ge 0$, which is absurd. Hope it's Ok.

I hope my memory still works (I try to stay beyond Alzheimer's club of Pierre, but it becomes harder and harder), but I think this problem was already solved on the forum. I think Darij posted it, after a kind of TST of Germany. Probably I'm wrong, but it sounds too familiar to me and I do not have the ISL 2004.

Yes, I believe that problem asked whether such a sequence could be bounded .

yea it was with the conditions given that $x_0$ and $x_1$ and positive and not equal to each other. And the answer is no. Its in the ISL 2004, and came up in the British FST.

Only the main idea: write all $x_i-s$ in terms of the first 2 distinct $x_i-s$. By induction one can easily obtain that all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\in \mathbb{N}$ . There are finitely many $(m,n)-s$ with $m<\frac M{x_1}, n<\frac M{x_2}$, with $M$ being the bound so $M$ is attained . Let $x_{i+1}$ be the first term that achieves the maximum, so $0<x_i<x_{i+1}$, $x_{i+2}=x_{i+1}-x_i$, since it cannot be their sum (M is max) so $x_{i+3}=2x_{i+1}-x_i>x_{i+1}=M$, false. There are still some cases to be considered for the first few terms of the sequence and for the case $0$ occurs in the sequence.

ikap wrote: all the terms of the sequence are of the form $mx_1+nx_2$ with $m,n\in \mathbb{N}$ Why is that? ($m,n \in \mathbb{N}$) I can only prove it with $m,n \in \mathbb{Z}$ and that doesn't help very much...