I am trying to evaluate the following sum: \begin{equation} \sum_{n=0}^\infty \frac{(-ix)^n}{(n+2)!} T_n \end{equation}
where $T_n$ represent the central trinomial coefficients , with the following convetion: $T_0 = 1, T_1 =1, T_2 = 3, T_3 = 7 ...$. I know that this sum converges, however I do not know if there is a way to evaluate it. If it is not possible (or too hard) it would be nice if I could obtain at least the asymptotic behavior when $x \to \infty$ .
We can use the exponential generating function (see OEIS A002426) $$\sum_{n=0}^\infty T_n \frac{z^n}{n!}=e^z I_0(2z)$$ to obtain an integral representation for the sum. Here $I_0(x)$ is the zeroth modified Bessel function, but the substitution $z=-i x$ converts it to
$$\sum_{n=0}^\infty T_n \frac{(-i x)^n}{n!}=e^{-i x} I_0(-2i x)=e^{-i x}J_0(2x)$$ where $J_0(x)$ is the regular zeroth Bessel function. Integrating both sides twice (with base point $0$) and dividing by $x^2$ produces
\begin{align} \sum_{n=0}^\infty T_n \frac{(-i x)^{n}}{(n+2)!} &=\frac{1}{x^2}\int_0^x\int_0^{x''} e^{-i x''}J_0(2x'')\,dx''\,dx'\\ &=\frac{1}{x^2}\int_0^x (x-t)e^{-i t}J_0(2t)\,dt. \end{align} where in the second line I've made use of the Cauchy formula for repeated integration. This is as far as I know how to go, unfortunately: Mathematica does not yield a closed form. But this may be sufficient for asymptotics or numerical exploration.