"Difficult to please" rabbits and Fibonacci Numbers - A probabilistic variation

A rigorous definition of the random process $(F_n)_{n\geqslant0}$ described in the question is that $F_0=F_1=1$ and, for every $n\geqslant0$, $$ F_{n+2}=F_{n+1}+\sum_{k=1}^{F_n}Z_{n,k}$$ where the doubly-indexed family $(Z_{n,k})_{n\geqslant0,k\geqslant1}$ is i.i.d. Bernoulli with $P(Z_{n,k}=1)=p$ and $P(Z_{n,k}=0)=1-p$.

In turn, this random recursion can be rewritten under the form of the bivariate branching process $$Y_n=\begin{pmatrix}Y^1_n\\ Y^2_n\end{pmatrix}=\begin{pmatrix}F_{n+1}\\ F_n\end{pmatrix}$$ starting from $$Y_0=\begin{pmatrix}1\\1\end{pmatrix}$$ with reproduction mechanism $$Y^1_{n+1}=\sum_{k=1}^{Y^1_n}1+\sum_{k=1}^{Y^2_n}Z_{n,k}\qquad Y^2_{n+1}=\sum_{k=1}^{Y^1_n}1+\sum_{k=1}^{Y^2_n}0$$ Thus, the mean reproduction matrix of $(Y_n)$ is $$\begin{pmatrix}1&p\\1&0\end{pmatrix}$$ whose eigenvalues are $$\lambda=\frac{1+\varrho}2\qquad\mu=\frac{1-\varrho}2$$ where $$\varrho=\sqrt{4p+1}$$ with eigenvectors $$U_\lambda=\begin{pmatrix}\lambda\\1\end{pmatrix}\qquad U_\mu=\begin{pmatrix}\mu\\1\end{pmatrix}$$ A consequence is that, for each $\nu$ in $\{\lambda,\mu\}$, $$M_n^\nu=\frac{F_{n+1}+(\nu-1)F_n}{\nu^{n+1}}$$ defines a martingale $(M_n^\nu)_{n\geqslant0}$ starting from $M_0^\nu=1$. Now, $\lambda>1$ hence $M^\lambda$ is a positive martingale, bounded in $L^2$, in particular, $M^\lambda_n\to W_\lambda$ almost surely, for some nonnegative random variable $W_\lambda$ such that $E(W_\lambda)=1$. From there, follows the almost sure limit $$\lim_{n\to\infty}\frac{F_n}{\lambda^n}= \frac{\lambda}{2\lambda-1}W_\lambda=\frac{\lambda}{\varrho}W_\lambda$$ This convergence is rather sharp since, using the second eigenvalue $\mu$ with $-1<\mu<0$, one can show that, almost surely, $$\varrho\,F_n=\lambda^{n+1}W_\lambda-\mu^{n+1}W_\mu+\mu^nG_n$$ for some second random variable $W_\mu$ such that $E(W_\mu)=1$ and some sequence $(G_n)_{n\geqslant0}$ such that $E(G_n)=0$ and $G_n\to0$ almost surely.

Note finally that this approach also provides the (much easier) result that $$E(F_n)=\frac{\lambda^{n+1}-\mu^{n+1}}\varrho$$