$a_n=$ $2^{n+1}$ for $n:3|n$
$a_n=$ $0$ for the rest
So theres what I tried to do:
$$\sum_{n=0}^{\infty}a_nz^{n}=2*z^0+0*z^1+0*z^2+2^4z^3+0*z^4+0*z^5+2^7z^6+...2^{k+1}z^k$$
$$2^{k+1}z^k=b^k$$
$$\frac{b^{k+1}}{b^k}$$
$$\frac{2^{k+2}z^{k+1}}{2^{k+1}z^{k}}=4z$$
$$q=4z$$ $$W_0=(4z)^2$$ $$A(z)=\frac{4^{z}}{1+(4z)^2}$$
Is this the right answer?
With $\displaystyle\left\vert z\right\vert < {1 \over 2}$, $\displaystyle\quad\sum_{n = 0}^{\infty}a_{n}z^{n} = \sum_{n = 0}^{\infty}2^{3n + 1}\, z^{3n} = 2\sum_{n = 0}^{\infty}\left[\left(2z\right)^{3}\right]^{n} =\ \bbox[#ffd,10px,border:1px groove navy]{2 \over 1 - \left(2z\right)^{3}}$