Is it possible for a recurrence relation to not have a solution?

Question

because I am trying to find the answer in wolfarmalpa but it show " no solution exist" !

$$a_{n}=a^{3}_{n-1}a^{2}_{n-2} ,\ a_{0}=2 , a_{1}=2a$$

I wrote it like $a(n)=a(n-1)^3*a(n-2)^2,a(0)=2,a(1)=2$

and $a(n)=(a(n-1))^3(a(n-2))^2,a(0)=2,a(1)=2 $

Answer 1

By induction it is easy to see that for each non-negative integer $n$ we have $a_n=2^{b_n}a^{c_n}$, where $(b_n)$ and $(c_n)$ are sequences of integer numbers such that $b_0=b_1=1$, $c_0=0$, $c_1=1$ and $b_n=3b_{n-1}+2b_{n-2}$, $c_n=3c_{n-1}+2c_{n-2}$ for each $n\ge 2$. These are linear recurrence with constant coefficients. Their characteristic polynomial $\lambda^2=3\lambda+2$ has distinct roots $\lambda_{1,2}=\tfrac 12\left(9\pm\sqrt{17}\right)$, so it has a general solution $C_1\lambda_1^n+ C_2\lambda_2^n$ for some constants $C_1$ and $C_2$. These coefficients can be easily found from the respective system of linear equations. Namely, for the sequence $(b_n)$ we have $C_1+ C_2=b_0=1$ and $C_1\lambda_1+ C_2\lambda_2=b_1=1$. For the sequence $(b_n)$ we have $C_1+ C_2=c_0=0$ and $C_1\lambda_1+ C_2\lambda_2=c_1=1$.