Romanian Mathematical Olympiad

For $a=1$ or $a=2$ we get the constant sequence $1$ or $2$ respectively. We shall prove that these are the only values satisfying the problem. To see this, let $x_n=\frac{k}{l}$ and $x_{n+1}=\frac{p}{q}$, with $k,l$ coprime positive integers, $p,q$ coprime positive integers with $q\ge 2$. Then $$\frac{k}{l}=x_n=2^{x_{n+1}-1}=\sqrt[q]{2^{p-q}},$$ implying $l^q{2}^p=k^q{2}^q$. The last equality cannot be true, as the left hand side number has $2$ as factor at an exponent which is not divisible by $q$ ($p$ and $q$ are coprime), and the right hand side has the factor $2$ at an exponent divisible by $q$. So $q=1$, thus $l=1$, so, if all terms of the sequence would be rationals, they should be natural numbers. As $2^n>n+1$, for $n\ge 2,n\in \mathbb{N}$, we get $1+{\log }_{2}x<x$, for $x\in \mathbb{N},x\ge 3$. As a conclusion, if all terms are positive integers and $x>2$, an inductive argument shows that the sequence is decreasing, a contradiction. That is $a=x_1\in \{1,2\}$.