1. $a[n] = \sum_{k=0}^\infty a[k] c[n-k], \; n\geq 2$

2. $a[n] > 0, \; n\geq 0$

3. $a[n] = 0, \; n < 0$

where $c[n] = 0.7\delta[n+1] + 0.3\delta[n-1]$. I felt that I needed to use the Wiener-Hopf technique to solve this problem. I do know that the solution is

$$

a[n] = \left(\frac{4}{7}\right) \left(\frac{3}{7}\right)^n

$$

But I'm trying to find it myself. Here is my preliminary solution:

1. Define $g^+[n]$ and $h^-[n]$ so that

$$

\sum_{k=0}^\infty a_k c_{n-k} =

\begin{cases}

g[n] = a[n], &\mbox{if $n\geq 2$,} \\

g^+[1], &\mbox{if $n=1$,} \\

g^+[0], &\mbox{if $n=0$,}\\

h^-[n], &\mbox{if $n\leq -1$,}

\end{cases}

$$

and $g^+[n] = 0$ for $n<0$ and $h^-[n] = 0$ for $n>0$.

2. Take the bilateral $z$-transform of the above equation:

\begin{align*}

A(z)C(z) &= G^+(z) + H^-(z) \\

&= \left(A(z) -a_0 -a_1 z^{-1} +g_0 +g_1 z^{-1} \right) +H^-(z) \\

&= A(z) + p(z) +H^-(z)

\end{align*}

3. $C(z)$ has a pole at $z=0$ and two zeros at $\pm i\sqrt{3/7}$. But I don't know what to do with this fact and I'm stuck here.