XV OBM

We have $a_n=2^{b_n}{3}^{c_n}$, where $b_{n+2}=b_n+b_{n+1}$, and $c_{n+2}=c_n+c_{n+1}$.

$c_1$ and $c_2$ are even, so all $c_n$ are even.

$b_1$ and $b_2$ are odd, so $b_n$ is even for $n$ multiple of $3$.

We need both $b_n$ and $c_n$ even for $a_n$ to be a perfect square.

So the answer is all $n$ multiple of $3$.