I want to solve the following: $$ \begin{align*} G_{T_t}(s) &= s\cdot G_X(G_{T_{t-1}}(s)) \\ G_{T_{t-1}}(s) &= s\cdot G_X(G_{T_{t-2}}(s)) \\ &\vdots \\ G_{T_0}(s) &= s \\ \implies G_{T_t}(s) &= s\cdot G_X(s\cdot G_X(\cdots s\cdot G_X(s))) \end{align*}$$ $G_X(s) = \mathbb{P}(X = 0) + s^2\cdot\mathbb{P}(X = 2) = 1-p + ps^2$.
My goal is to find the result of the above recursion if we iterate it infinitely, i.e. $t\to\infty$ and the result should be the following: $$ G_T(s) = \frac{1 - \sqrt{1 - 4p(1-p)s^{2}}}{2ps}.$$
Maybe someone already realised that $X$ is a Bernoulli kindish random variable, but it takes value $2$ with probability $p$. And $G_X(s)$ is a probability generating function of $X$.
I appreciate your help in advance.
I'll slightly simplify the notation and forget what the expressions come from:
$$g(s) = 1 - p + ps^2$$
and $f_0(s) = s$,
$$f_{n + 1}(s) = sg(f_n(s)).$$
A bit informally (requiring further justification), we can look for a function $f = \lim_{n\to\infty}f_n$ as the solution to
$$f(s) = sg(f(s)).$$
Intuitively, the sequence tends to stabilization. Writing this out, we get
$$f(s) = s(1 - p + pf(s)^2) = spf(s)^2 - ps + s.$$
This is a quadratic equation in $f(s)$, whose solution (one of whose solutions) is the expression in the coefficients that you are looking for.