# Can the following sum be counted or expressed in terms of special functions?

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 theta-function.

• 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 $\exp 2c|z|$ 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. Jan 11, 2021 at 19:31
• Yeh, here $a$ - some real parameter. Jan 11, 2021 at 19:52
• There is a question here on MO about this same series (up to substitution) Jan 11, 2021 at 21:30
• @მამუკაჯიბლაძე Thank you! Jan 12, 2021 at 10:23

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 $$|q|<1$$.
The case when your $$a$$ is real, that is $$|q|=1$$ is especially difficult and mysterious; all these points are singular for $$q\mapsto F(\zeta,q)$$.