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.


2 Answers 2


$$\sum_{n=0}^\infty T_n\frac{x^n}{n!}=\frac{1}{2\pi i}\sum_{n=0}^\infty\frac{x^n}{n!}\oint\frac{(1+z+z^2)^n}{z^{n+1}}\,dz=\frac{e^x}{2\pi i}\oint e^{x(z+1/z)}\frac{dz}{z}=e^x I_0(2x),$$ 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 $I_0 (2x) = \sum_{j=0}^\infty \frac{x^{2j}}{(j!)^2}$

\begin{equation} e^x I_{0}(2x) = \sum_{i = 0}^\infty \sum_{j=0}^\infty \frac{x^{i+2j}}{(i!)(j!)^2} \end{equation} We can group the factors with the same exponent $i+2j $, obtaining: \begin{equation} 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 \end{equation}

  • 2
    $\begingroup$ Very direct. Nice answer. $\endgroup$
    – K.defaoite
    Dec 30, 2020 at 22:56

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