The problem is: given the following recursive constraints, find the set of all sequences $\{X_k\}_{k = 1}^{\infty}$ and their corresponding $\alpha$, where $\alpha \neq 0$.
- $\alpha X_1 = 0$
- $\forall k \ge 1,\alpha X_{3k} = X_{6k}$
- $\forall k \ge 1, \alpha X_{3k + 1} = X_{6k + 2}$
- $\forall k \ge 0, \alpha X_{3k + 2} = X_{2k + 1} + X_{6k + 4}$
Other than the trivial solution, here's the progress I've made so far: $$X_{6k+4} = X_{3(2k + 1) + 1} = \frac{1}{\alpha}X_{12k + 8} \\ \therefore \alpha X_{3k + 2} = X_{2k + 1} + \frac{1}{\alpha} X_{12k + 8} \implies \alpha^2 X_{3k + 2} - \alpha X_{2k + 1} - X_{12k + 8} = 0 \\ \therefore \forall k \ge 0, \alpha = \frac{X_{2k + 1} \pm \sqrt{X_{2k + 1}^2 + 4(X_{3k + 2})(X_{12k + 8})}}{2X_{3k + 2}}$$
Plugging in $k = 0$ we get $\alpha = \pm \sqrt{\frac{X_8}{X_2}}$.
I'm not sure where to go from here. Am I even going in the right direction? I'm starting to wonder if the trivial solution is the only solution.
Any and all help would be greatly appreciated!
Edit: I've since been able to prove another statement, which is below in case it helps anybody solve this question, however it was a lengthy proof so I will leave that out.
$$X_{3k + 2} = \sum_{n = 0}^{\infty}\frac{X_{2^{2n + 1}k + \frac{2 \times 2 ^ {2n + 1} - 1}{3}}}{\alpha ^ {2n + 1}}$$