A functional power series equation

I am interested in solving the following functional equation:

$$(1-w-zw)F(z,w)+zw^nF(z,w^2)=z-zw\,.$$

Here $n\geq2$ is a fixed integer and $F(z,w)$ is a power series in two variables with complex coefficients, which you can assume that converges in a neighborhood of the origin in $\mathbb C^2$.

My thoughts: Defining $G(j)=F(z,w^j)$ for $j\geq1$ and using the functional equation we obtain

$$G(j)-\underbrace{\,\frac{-zw^{jn}}{1-w^j-zw^j}\,\\[4mm]}_{a_j}\,G(2j)=\underbrace{\,\frac{z-zw^j}{1-w^j-zw^j}\,\\[4mm]}_{b_j}\,.$$

Dividing both sides by $\prod_{k=0}^\infty a_{2^kj}=c_j$ and defining $H(j)=G(j)/c_j$ we get

$$H(j)-H(2j)=\frac{b_j}{c_j}\,,$$

which implies

$$\begin{align*} F(z,w)=&\,G(1)=c_1H(1)\\ =&\,c_1\sum_{r=0}^\infty\bigl[H(2^r)-H(2^{r+1})\bigr]+c_1\lim_{r\to\infty}H(2^r)\\ =&\,\sum_{r=0}^\infty b_{2^r}\frac{c_1}{c_{2^r}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ %=&\,\sum_{r=0}^\infty \frac{z-zw^{2^r}}{1-w^{2^r}-zw^{2^r}}\,\prod_{k=0}^{r-1}\frac{zw^{2^kn}}{1-w^{2^k}-zw^{2^k}}+\lim_{r\to\infty}\frac{c_1}{c_{2^r}}G(2^r)\\ \end{align*}$$

Since $F$ is holomorphic, then for $|w|<1$ we have $\lim\limits_{r\to\infty}G(2^r)=F(z,0)=z$ by the functional equation. Defining

$$d_r=\frac{c_1}{c_{2^r}}=\prod_{k=0}^{r-1}\frac{-zw^{2^kn}}{1-w^{2^k}-zw^{2^k}}$$

we arrive to the following "explicit" formula, provided that the sequence $\boldsymbol{(d_r)_{r\geq1}}$ converges for $\boldsymbol{|z|,|w|}$ small:

$$F(z,w)=\sum_{r=0}^\infty b_{2^r}d_r+z\lim_{r\to\infty}d_r\,.$$

I personally believe that in fact $(d_r)_{r\geq1}$ converges. Ironically, I wasn't lazy to think the partial solution above, but I am not feeling like solving the system of infinite linear equations satisfied by the coefficients of $F$. Who knows? some seasoned complex analyst can help me and obtain an explicit formula for both the series and the infinite product above.