Vomchenkov S.G.

Find all infinite bounded sequences of natural numbers $a_{1}, a_{2}, a_{3}, \ldots$, for all members of which, starting from the third, the following holds

$$
a_{n}=\frac{a_{n-1}+a_{n-2}}{\left(a_{n-1}, a_{n-2}\right)}
$$
Let $\left(a_{k}, a_{k+1}\right)=1$. Then $\left(a_{k}, a_{k+1}\right)=\left(a_{k}+a_{k+1}, a_{k+1}\right)=\left(a_{k+2}, a_{k+1}\right)$, which means that the greatest common divisor (GCD) for all subsequent terms of the sequence will also be 1. At the same time, starting from the $k$-th term, the sequence turns into $a_{n}=a_{n-1}+a_{n-2}$, which increases without bound, which is impossible.

Therefore, $a_{k+2} \leq 1 / 2 \max \left\{a_{k+1}, a_{k}\right\}$ and $\max \left\{a_{k+1}, a_{k}\right\}$ does not increase. Consequently, at some point it stabilizes: $\max \left\{a_{k+1}, a_{k}\right\}=\max \left\{a_{k+2}, a_{k+1}\right\}=\max \left\{a_{k+3}, a_{k+2}\right\}=\ldots=d$.

If at this point $a_{k+1} \neq a_{k}$, then $a_{k+2}<d, a_{k+3}<d$, which is impossible. Therefore, $a_{k}=a_{k+1}=a_{k+2}=\ldots=d$, from which $d=$

2. Consequently, $\frac{a_{k-1}+2}{\left(a_{k-1}, 2\right)}=2$, from which $a_{k-1}=2$, and the sequence consists only of twos.

## Answer

$a_{1}=a_{2}=\ldots=2$