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 $a = x_1 \in \{1, 2\}$.