So I have $$\overset{*}{F} = \overset{*}{F}_{n-1} + \overset{*}{F}_{n-2} + g(n)$$ where $\overset{*}{F}$ is NOT a Fibonacci number for $n \geq 2$.
$g(n)$ is any function $g: \mathbb{N} \to \mathbb{R}$. And $\overset{*}{F}_0 = g(0)$ and $\overset{*}{F}_1 = g(1)$.
I think that $\overset{*}{F}_n$ would be a sequence of $g(n)$s. Actually, I have found that $$\overset{*}{F}_n = f_{n-1}\cdot g(0) + f_n\cdot g(1) + f_{n-1}\cdot g(2) + f_{n-2}\cdot g(3) + \dots + f_1\cdot g(n)$$ Here $f_n$ is the n-th Fibonacci number. So to clarify, let n = 5, then: $$\overset{*}{F}_5 = f_4\cdot g(0) + f_5\cdot g(1) + f_4\cdot g(2) + f_3\cdot g(3) + f_2\cdot g(4) + f_1\cdot g(5)$$ $$\overset{*}{F}_5 = 3\cdot g(0) + 5\cdot g(1) + 3\cdot g(2) + 2\cdot g(3) + 1\cdot g(4) + 1\cdot g(5)$$
My question is how to find the generating function of $\overset{*}{F}_n$ if $$G(x) = \sum^{\infty}_{n=0}g(n)\cdot x^n$$ applies?
I have difficulties representing the generating function of the Fibonacci numbers reversed. Even though that I know that the regular g.f. $F(x) = \frac{1}{1-x-x^2}$
Maybe I should somehow transform this: $$\overset{*}{F}_n = f_{n-1}\cdot g(0)x^0 + f_n\cdot g(1)x^1 + f_{n-1}\cdot g(2)x^2 + f_{n-2}\cdot g(3)x^3 + \dots + f_1\cdot g(n)x^n$$ and use $G(x)$ to get the generating function for all $\overset{*}{F}_n$?
Comment turned answer per request.
Just multiply the $n^{th}$ equation by $z^n$ and start to sum from $n = 2$.
$$ \overbrace{\overset{*}{F}(z) - g(0) - g(1)z}^{ \sum\limits_{n=2}^\infty\overset{*}{F}_n z^n} = \overbrace{z(\overset{*}{F}(z)-g(0))}^{ \sum\limits_{n=2}^\infty \overset{*}{F}_{n-1} z^n } + \overbrace{z^2\overset{*}{F}(z)}^{ \sum\limits_{n=2}^\infty \overset{*}{F}_{n-2} z^n } + \overbrace{(G(z) - g(0) - g(1)z)} ^{ \sum\limits_{n=2}^\infty g(n) z^n } \\ \iff (1 - z - z^2)\overset{*}{F}(z) = G(z) - g(0)z\\ \implies \overset{*}{F}(z) = \frac{G(z) - g(0)z}{1 -z -z^2}$$