In Pascal's triangle, denote $S_n=\prod\limits_{k=0}^n\binom{n}{k}$. It can be shown that
$$\lim_{n\to\infty}\frac{S_{n-1}S_{n+1}}{{S_n}^2}=e$$
What is the analogous result for the trinomial triangle?
That is, denote $T_n=\prod\limits_{k=-n}^n\binom{n}{k}'$ where $\binom{n}{k}'$ are the numbers in the $n$th row of the trinomial triangle*.
$$\lim_{n\to\infty}\frac{T_{n-1}T_{n+1}}{{T_n}^2}=\space ?$$
Numerical investigation using A027907 gives:
$\dfrac{T_{0}T_{2}}{{T_{\color{red}{1}}}^2}=12$
$\dfrac{T_{1}T_{3}}{{T_{\color{red}{2}}}^2}=15.75$
$\dfrac{T_{2}T_{4}}{{T_{\color{red}{3}}}^2}=18.1555$
$\dfrac{T_{9}T_{11}}{{T_{\color{red}{10}}}^2}=22.4713$
$\dfrac{T_{49}T_{51}}{{T_{\color{red}{50}}}^2}=24.2722$
$\dfrac{T_{73}T_{75}}{{T_{\color{red}{74}}}^2}=24.4284$
$\dfrac{T_{97}T_{99}}{{T_{\color{red}{98}}}^2}=24.5087$
suggesting a limit of approximately $24.7$.
I found that the trinomial coefficients have a closed form expression involving Gegenbauer polynomials, but I am unable to work out the limit.
*The article uses the notation $\binom{n}{k}_2$ for trinomial coefficients, but I thought that may be confusing ($2$ for trinomial?), so I decided to use $\binom{n}{k}'$.