Finding $\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ for a recursively defined sequence $(a_n)$ with $a_{n}=\frac13(a_{n-1}+a_{n-2}+a_{n-1}a_{n-2})$ Question.

Let $(a_n)$ be a complex sequence defined recursively:
$$
a_0 = 0,\quad  a_1=1, \quad a_{n}=\frac13(a_{n-1}+a_{n-2}+a_{n-1}a_{n-2})\quad (n>1)
$$
What is $\displaystyle\limsup_{n\to\infty}\sqrt[n]{|a_n|}$?

Remarks.
This question is motivated by an unanswered question on the site that I am not able to solve. By the Cauchy-Hadamard theorem, the problem of finding the radius of convergence for the power series $f(z)=\sum_{n=0}^\infty a_nz^n$ reduces to the limit problem above. One can probably find some other way to figure out the radius of convergence, but I would like to focus on Cauchy-Hadamard and thus particularly the limit in the question.
One straightforward attempt is to find a closed-form formula for the sequence so that one may able to apply asymptotic techniques to analyze $|a_n|^{1/n}$. While there are systematic ways to handle the linear recurrence, I don't know how to handle this particular nonlinear case. The usual idea of "linearizing the nonlinear problem" seems not helpful here.
With help of the short Python script, I plotted the sequence $(|a_n|^{1/n})$. The result seems to suggest that the limit is $1$.

 A: Hint:
$$\frac{a_n}{a_{n-1}}=\frac13\Big(1+\frac{a_{n-2}}{a_{n-1}}+a_{n-2}\Big)$$
Show that $a_n\xrightarrow{n\rightarrow\infty}0$ and deduce that $x_n:=\frac{a_n}{a_{n-1}}$ converges to some $x>0$ satisfying
$$x=\frac{1}{3}\Big(1+\frac{1}{x}\Big)$$

Recall that
$$\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}|}{|a_n|}$$

Another approach, is to consider the discrete dynamical system on $\mathbb{R}^2$ given as
\begin{align}
\boldsymbol{x}_{n+1}&=G(\boldsymbol{x}_n)\\
\boldsymbol{x}_0 &= \begin{pmatrix} 0\\1\end{pmatrix}
\end{align}
where
$$
G(x,y)=\begin{pmatrix} 0 & 1\\
\frac13 & \frac13
\end{pmatrix} \begin{pmatrix} x\\ y
\end{pmatrix}
+ \frac13\begin{pmatrix}  0\\ xy \end{pmatrix}
$$
The eigenvalues of the linear part of $G$ around the critical point $\boldsymbol{x}^*=[0\;0]^\intercal$ are $\frac{1\pm\sqrt{13}}{6}$.
From this, one can show that with initial conditions closed enough to $\boldsymbol{x}^*$, $\boldsymbol{x}_n$ converges to $\boldsymbol{x^*}$ exponentially fast (the rate of convergence being $\frac{1+\sqrt{13}}{6}$ which is happened to be $\lim_n\frac{a_{n+1}}{a_n}$ where
$\boldsymbol{x}_n:=[a_{n-1}\,a_n]^\intercal $.
A: I finally figure out the details to follow Oliver's hint and I would like to write out an answer here.
The two essential steps are proving that the sequences $(a_n)$ and $(\frac{a_n}{a_{n-1}})$ are both convergent. These two steps are nontrivial (!) and they are answered in the following two follow-up posts:

*

*Convergence of a recursively defined sequence of complex numbers: $3z_{n}=z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}$

*Finding $\lim_{n\to\infty} \frac{z_{n}}{z_{n-1}}$ where $3z_{n}=z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}$
It then follows by straightforward calculations that
$$
\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}} = \frac16(1+\sqrt{13}).
$$
Consequently, by a well-known theorem (see, e.g., Rudin's Principle of Mathematical Analysis) which compares the ratio and root test:
$$
\liminf _{n} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|} \leq \liminf _{n} \sqrt[n]{\left|a_{n}\right|} \leq \lim _{n} \sup \sqrt[n]{\left|a_{n}\right|} \leq \limsup _{n} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}
$$
the desired result follows.

Remark.
It is worthing knowing that, in Python, one needs a good algorithm to calculate the $n$-th root for very big $n$. In the original attempt mentioned in the post, the total iteration number was taken too big that the plot is actually not accurate because a**(1/n) was used in the code to find the $n$-th root.  However, if one only iterates 100 times, one gets $b_{100}\approx 0.76759187924+\pm 10^{-11}$ which is an approximate value of $\frac{1}{6}(1+\sqrt{13})$.

