Solve the Recurrence relation : $a_n= (a_{n-1})^3a_{n-2}$ , $a_0=1, a_1=3$

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$a_n= (a_{n-1})^3a_{n-2}$ , $a_0=1, a_1=3$

I'm ask to get an expression for $a_n$.

So i tried to solve with induction:

$a_n=(a_{n-1})^3a_{n-2}=(a_{n-1})^3(a_{n-3})^3a_{n-4}=(a_{n-1})^3(a_{n-3})^3(a_{n-5})^3 \ldots (a_1)^3a_0$

So $a_n= \prod _{k=1}^n a_{n-k}$ , (k is an odd number, so all odd numbers from 1 to n)

Is this the expression? i feel like this is still a reccurence relation, solving with the methods of linear-homog will not work cause this relation is non-linear.

any help would be appreciated!