Let us define this sum as a function of $z \in \mathbb{C}$ with some positive parameter $a$ $$ f(z; a) = \sum\limits_{n = 0}^{\infty}\frac{z^{2n}}{n!}e^{ian^2}. $$ Probably, it can be expressed (or somehow related) in terms of thetafunction.

2$\begingroup$ It's really $f(z;a)$. That aside, if $a$ is a rational multiple of $\pi$ then $f(z;a)$ is a linear combination of terms $\exp 2cz$ for some roots of unity $c$ (because $e^{ian^2}$ is periodic and can be written as a discrete Fourier series). Irrational multiples of $\pi$ seem much harder. $\endgroup$– Noam D. ElkiesJan 11, 2021 at 19:31

$\begingroup$ Yeh, here $a$  some real parameter. $\endgroup$– MightyPowerJan 11, 2021 at 19:52

2$\begingroup$ There is a question here on MO about this same series (up to substitution) $\endgroup$– მამუკა ჯიბლაძეJan 11, 2021 at 21:30

$\begingroup$ @მამუკაჯიბლაძე Thank you! $\endgroup$– MightyPowerJan 12, 2021 at 10:23
1 Answer
Probably the answer is negative. Your series is a restriction of the analytic function in two complex variables: $$F(\zeta,q)=\sum_{n=0}^\infty\frac{q^{n^2}}{n!}\zeta^n,\quad q\leq1,$$ obtained by setting $q=\exp(ia)$ and $\zeta=z^2>0$. Function $F$ is continuous, entire with respect to $\zeta$ and analytic for $q<1$.
This important function has been studied much recently, see lectures of Alan Sokal for a survey of known results; as Sokal says himself, there are many conjectures and almost no theorems, and there is no indication of its expression in terms of standard special functions. (Except the trivial observation that it is the "Hadamard product" of the "partial thetafunction'' with the exponential, or Borel's transform of the partial theta function, but this partial thetafunction is itself outside of the set of common special functions, and its properties make a current research subject.)
The case when your $a$ is real, that is $q=1$ is especially difficult and mysterious; all these points are singular for $q\mapsto F(\zeta,q)$.

$\begingroup$ I see. Thank you for your answer! And especially for the link. $\endgroup$ Jan 12, 2021 at 10:22