<s>[INST] Deriving Generating function for centered trinomial coefficients Let $c_n$ denote the $n$-th center trinomial coefficient (OEIS sequence here).
It appears they cannot be generated by a linear recurrence relation, so how should I go about to find the generating function $G(x)$ for the sequence?
$$G(x)=\sum_{n=0}^{∞}c_nx^n=1+x+3x^2+7x^3+19x^4+51x^5+141x^6+...$$
The geometric ratio appears to have a limit close to $$\lim_{n\to ∞}\frac{c_{n+1}}{c_n}=2.95...$$ (these are the successive ratios of the last two listed terms in the OEIS sequence).
Also, what is the interval of convergence (and divergence)? Based on the geometric limit, it seems that $G(1/3)$ will converge.
Edit: The generating function is $$G(x)=\frac{1}{\sqrt{1-2x-3x^2}}$$
Any idea as to how this answer is derived?
 [/INST] Here is a variation based upon G. P. Egorychev's Classic: Integral Representation and the Computation of Combinatorial Sums. We start with the central trinomial coefficients:
\begin{align*}
[x^n](1+x+x^2)^n\qquad\qquad n\geq 0
\end{align*}
We consider the function
\begin{align*}
f(x)=1+x+x^2\tag{1}
\end{align*}
and derive a function $y=y(x)$:
\begin{align*}
y(x)=\frac{x}{f(x)}=\frac{x}{1+x+x^2}\qquad\qquad y^{\prime}(x)=\frac{1-x^2}{(1+x+x^2)^2 }\tag{2}
\end{align*}
With $f(x)$ and $y(x)=\frac{x}{f(x)}$ we can now apply the substitution rule (Rule 5, one-dimensional case) from section 1.2.2 in G. P. Egorychev's book as follows:
\begin{align*}
\color{blue}{[x^n](f(x))^n=[y^n]\left.\left(\frac{1}{f(x)y^{\prime}(x)}\right)\right|_{x=g(y)}}\tag{3}
\end{align*}
with $g(y)$ the inverted function given by $y=y(x)$ in (2).

We obtain from (1) - (3):
\begin{align*}
\color{blue}{[x^n]}&\color{blue}{\left(1+x+x^2\right)^n}\\
&=[y^n]\left.\left(\frac{1}{\left(1+x+x^2\right)\frac{d}{dx}\left(\frac{x}{1+x+x^2}\right)}\right)\right|_{x=g(y)}\\
&=[y^n]\left.\frac{1+x+x^2}{1-x^2}\right|_{x=g(y)}\\
&\,\,\color{blue}{=[y^n]\frac{1}{\sqrt{1-2y-3y^2}}}\tag{4}
\end{align*}
and the claim follows.

In (4) we use the identity
\begin{align*}
2y=\frac{2x}{1+x+x^2}&=1-3\left(\frac{x}{1+x+x^2}\right)^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
&=1-3y^2-\left(\frac{1-x^2}{1+x+x^2}\right)^2\\
\frac{1+x+x^2}{1-x^2}&=\left(1-2y-3y^2\right)^{-\frac{1}{2}}
\end{align*}
</s>