I've been asking fragments of this question for the past week, here it is: I want to find a sequence/function $a[n]$ that satisfies:
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.