two series recurrence relation

Given the recurrence

$$\begin{cases} F_n = 2F_{n-1}^2 H_{n-1} \\H_n = 2F_{n-1} H_{n-1}^2 \end{cases} \text{ for }n\geq 3$$

and $$F_2 = 1$$, $$H_2 = 3$$.

How can I find an explicit expression for $$F_n$$?

My approach so far was applying the logarithm, writing it in matrix/vector form an use linear algebra (via eigenvalue/eigenvectors). Then I got the result $$F_n = 3^{\frac{3^{n-2}-1}{2}} + 2^{\frac{3^{n-2}-1}{2}}.$$

Are there easier methods for solving this recurrence relation? I think one should be able to use the symmetry of $$F_n$$ and $$H_n$$.

• You could try multiplying and dividing both equations to see what you can get out of the symmetry present. Commented Dec 21, 2015 at 14:47

Note that $$\frac{F_n}{H_n}=\frac{2F_{n-1}^2 H_{n-1}}{2 F_{n-1}H_{n-1}^2}=\frac{F_{n-1}}{H_{n-1}};$$ i.e., the ratio is constant. So $H_n = 3 F_n$ always. Then the first recurrence becomes $$F_{n}=2 F_{n-1}^2 H_{n-1} = 6 F_{n-1}^3,$$ and you can proceed with the logarithm / linear recursion approach.