# Deriving exponential generating function for central trinomial coefficients

A recent question (link) asked for a derivation of the (ordinary) generating function for the central trinomial coefficients $$\{T_n\}$$. But the OEIS page (A002426) also lists an exponential generating function

$$\sum_{n=0}^\infty T_n \frac{x^n}{n!}=e^x I_0(2x)$$ where $$I_0(x)$$ is the zeroth Bessel function. How is this derived? I'll take a stab myself at showing this using the tools of analytic combinatorics, but I wanted to open this up to more knowledgeable folks as well.

$$\sum_{n=0}^\infty T_n\frac{x^n}{n!}=\frac{1}{2\pi i}\sum_{n=0}^\infty\frac{x^n}{n!}\oint\frac{(1+z+z^2)^n}{z^{n+1}}\,dz=\frac{e^x}{2\pi i}\oint e^{x(z+1/z)}\frac{dz}{z}=e^x I_0(2x),$$ using $$T_n=[z^n](1+z+z^2)^n$$, then the exponential series, then the contour integral representation of $$I_0$$ based on the generating function $$e^{z(t+1/t)/2}=\sum_{n\in\mathbb{Z}}I_n(z)t^n$$.
Using the fact that $$I_0 (2x) = \sum_{j=0}^\infty \frac{x^{2j}}{(j!)^2}$$
$$$$e^x I_{0}(2x) = \sum_{i = 0}^\infty \sum_{j=0}^\infty \frac{x^{i+2j}}{(i!)(j!)^2}$$$$ We can group the factors with the same exponent $$i+2j$$, obtaining: $$$$e^x I_{0}(2x) = \sum_{n = 0}^\infty \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{x^n}{(k!)^2 (n-2k)!} = \sum_{n=0}^\infty \frac{x^n}{n!}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} \binom{2k}{k} = \sum_{n=0}^\infty \frac{x^n}{n!} T_n$$$$