Problem with floor function and sequence $a_{n+1}= \lfloor a_n \rfloor \{ a_n \}$ I found this problem in a Romanian Magazine. Let $(a_n)_{n \ge 1}$ be a sequence such that $a_1 \lt 0$ and $a_{n+1}= \lfloor a_n \rfloor \{ a_n \}$. Show that there exists $n_0 \gt 1$ such that $a_{n+2}=a_n$, $ \forall n \ge n_0$.
 A: Note $a_n=p_n+x_n$, with $0\le x_n<1$, so that $\lfloor a_n\rfloor=p_n$ and $\{a_n\}=x_n$.
As $a_0<0$, all $a_n$ are non positive. 
If there is a $N$ such that $a_{N}$ is an integer, then for all $n>N$, $a_n=0$, and you can choose $n_0=N+1$.
If not, then the sequence $(p_n)$ is increasing, as
$$a_{n+1}=p_nx_n\ge p_n$$
So it has to be stationnary at a certain point. Let $p$ be the stationnary value. So there exists $N\in\mathbb N$ such that
$$(\forall n\ge N)\,a_n=p+x_n$$
But as $a_{n+1}=px_n=p+x_{n+1}$, when can see that :
$$x_{n+1}=px_n-p$$
The fixed point for this arithmetico-geometric relation is $\lambda=\frac{p}{p-1}$, and by posing $y_n=x_n-\lambda$, we find 
$$y_{n+1}=py_n$$
If $p$ is not $0$, this sequence would be divergent, and this would lead to $x_n$ not being able to stay in interval $[0,1[$, unless $y_N=0$, which means $x_N=\lambda$
Therefore, two cases can occur :
 1) for all $n\ge N+1$, $a_n=0$, and you can choose $n_0=N+1$,
 2) for all $n\ge N$, $a_n=p+\frac{p}{p-1}=\frac{p^2}{p-1}$, and you can choose $n_0=N$.
