$e$ is hidden in Pascal's (binomial) triangle. What is hidden in the trinomial triangle, in the same way?

Question

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.

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*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 $\binom{n}{k}'$.

Answer 1

A Long Comment. A heuristic argument suggests that the limit is equal to

$$ \exp\left( 2 + \frac{2\pi}{3\sqrt{3}} \right) \approx 24.759260388738970195. $$

Let me demonstrate my heuristic computation that led to this value. (Again, I emphasize that this is by no means a full-fledged solution, but rather a chain of heuristics.)

Heuristic Computation. Assume the limit $L = \lim_{n\to\infty} \frac{T_{n-1}T_{n+1}}{T_n^2}$ exists.¹⁾ Then by Stolz–Cesàro theorem,

$$ \lim_{n\to\infty} \frac{\log T_n}{n^2} = \lim_{n\to\infty} \frac{\log T_{n+1} - 2\log T_n + \log T_{n-1}}{2} = \frac{1}{2}\log L. $$

So it suffices to find the limit of $\frac{1}{n^2}\log T_n$. In doing so, we note that

$$ \binom{n}{k}' = \sum_{j=0}^{n} \frac{n!}{j!(j+k)!(n-2j-k)!}. $$

Substituting $j = sn$ and $k = tn$, and denoting by $s = s_t$ the value of $s$ at which the summand is maximized, we can check that

$$ s_t \sim \frac{1}{6} \left(4-3t-\sqrt{4-3t^2}\right). $$

Plugging this to the asymptotic formula

$$ \log \binom{n}{tn}' = \log \frac{n!}{(s_t n)!((s_t + t)n)!((1 - t - 2s_t)n)!} + \mathcal{O}(\log n) $$

and utilizing Stirling approximation, we get

$$ \lim_{n\to\infty} \frac{\log T_n}{n^2} = \int_{-1}^{1} f(t) \, \mathrm{d}t, $$

where $f(t)$ is given by

\begin{align*} f(t) &= \frac{t}{2} \left[ \log\left(4-3t-\sqrt{4-3t^2}\right) - \log\left(4+3t-\sqrt{4-3t^2}\right) \right] \\ &\quad -\frac{1}{2} \log \left(5-3t^2-2\sqrt{4-3t^2}\right) + \log 3. \end{align*}

Mathematica tells that $\int_{-1}^{1} f(t) \, \mathrm{d}t = 1+\frac{\pi}{3\sqrt{3}}$, hence the claimed value is obtained.

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^{1) I suspect this is probably the hardest part to justify in my heuristic computation.}