Problem 1227

The condition is equivalent to the fact that starting from some $n$, the number $a_{n}$ is not divisible by 5. Let's prove this.

We will show that there will be two consecutive terms of the sequence that are not multiples of 5. Suppose the opposite. Then for any $n$, either $a_{n+1}$ is obtained from $a_{n}$ by dividing by 5, or $a_{n+2}$ is obtained from $a_{n+1}$ by dividing by 5. Note that $a_{k+1} \leq \sqrt{5} a_{k}$, so $a_{n+2} \leq 1 / 5 \sqrt{5} a_{n} < a_{n}$. This means that the sequence of natural numbers $a_{1}, a_{3}, a_{5}, \ldots$ is strictly decreasing. Contradiction.

Thus, there will be $a_{k}$ and $a_{k+1}$ that are not divisible by 5. We will prove that $a_{k+2}$ is also not a multiple of 5. Similarly, we will sequentially obtain that $a_{k+3}, a_{k+4}, \ldots$ are not divisible by 5, which is what we need.

$a_{k+1} = \left[\sqrt{5} a_{k}\right], a_{k+2} = \left[\sqrt{5} a_{k+1}\right]$. Let $a_{k} = m$, then $a_{k+1} = \sqrt{5} m - \alpha$, where $0 < \alpha < 1$. $a_{k+2} = [\sqrt{5}(\sqrt{5} m - \alpha)] = 5 m + [-\sqrt{5} \alpha]$. Since

$0 < \sqrt{5} \alpha < 3$, then $m - 3 \leq a_{k+2} < 5 m$, so $a_{k+2}$ is not divisible by 5.