# Solution to recursive sequence involving floor function?

I want to find an explicit formula for the following recursive sequence: $$E_n = \begin{cases} E_0 & n=0 \\ \lfloor k E_{n-1} \rfloor & n \ge 1 \end{cases}, \ n \in \mathbb{N}_0 \ , \ k \in \left( 0 , 1 \right)$$ Without the floor function, the solution looks like: $$E_n = E_0 k^n,$$ but I'm not sure how to approach this with the floor function involved...

• It generally helps questions get accepted and get attention when a bit more context and/or information is provided. What's this for, what have you tried, anything like that you can give? Jan 1 '21 at 20:01
• Why do you think an explicit formula exists? Jan 1 '21 at 22:31
• @Greg Martin I am not sure if an explicit formula exists :( Jan 1 '21 at 22:52

The exact behavior of the sequence $$\langle E_n:n\in\Bbb N\rangle$$ seems very hard to pin down, but the long-term behavior is fairly simple.
Suppose first that $$E_0\ge 0$$. Then $$E_{n+1}=\lfloor kE_n\rfloor\le kE_n for each $$n\in\Bbb N$$, so $$0\le E_n\le k^nE_0$$ for $$n\in\Bbb N$$, so $$\lim\limits_{n\to\infty}E_n=0$$, and since $$E_n$$ is an integer for each $$n\in\Bbb Z^+$$, there is an $$n_0\in\Bbb N$$ such that $$E_n=0$$ for all $$n\ge n_0$$.
If $$E_0<0$$, however, matters are somewhat different. For instance:
• If $$E_0=-1$$, for instance, $$E_n=-1$$ for all $$n\in\Bbb N$$.
• If $$E_0=-2$$, there are two cases. If $$\frac12, then $$E_n=-2$$ for all $$n\in\Bbb N$$, but if $$0, then $$E_n=-1$$ for all $$n\in\Bbb Z^+$$.
More generally, let $$m\in\Bbb Z^+$$. If $$E_0=-m$$ and $$\frac{m-1}m, then $$E_n=-m$$ for all $$n\in\Bbb N$$. If $$0, there are a positive integer $$m' and an $$n_0\in\Bbb Z^+$$ such that $$E_n=-m'$$ for all $$n\ge n_0$$. And with a little more thought it’s clear that if $$r\in\Bbb R^+$$, and $$E_0=-r$$, then there are a positive integer $$m$$ and an $$n_0\in\Bbb Z^+$$ such that $$m\le\lceil r\rceil$$ and $$E_n=-m$$ for all $$n\ge n_0$$.
In short, the sequence $$\langle E_n:n\in\Bbb N\rangle$$ is always eventually constant; its limit is $$0$$ if $$E_n\ge 0$$ and a negative integer greater than or equal to $$\lfloor E_0\rfloor$$ if $$E_0<0$$.