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I recently encountered the following function

$$ f(t) = \sum_{n=0}^\infty \frac{t^{n^2}}{n^2!}. $$

It seems familiar, though I cannot remember where I might have seen it before. I would like to know in what text, if any, it has been studied. Above all, I would like to know its asymptotic properties (how fast it grows) as $t$ tends to infinity ($t$ real and positive).

Thanks,

G.

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  • $\begingroup$ Related to (integral of) the theta function $\theta(0,\tau)$, de.wikipedia.org/wiki/Thetafunktion. $\endgroup$ – Dirk Oct 13 '14 at 7:37
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    $\begingroup$ As with the exponential series, for large $t$ the biggest terms are when $n\approx t$. At this point about one in every $2t$ terms of the exponential series are present, and vary on approximately that scale, so we should have $f(t)\sim c(t)e^t/t$ where $c(t)$ is bounded. I think it will wriggle between two limits indefinitely. $\endgroup$ – Brendan McKay Oct 13 '14 at 7:38
  • $\begingroup$ @Dirk: How it is related to the integral of theta-function? Where will $(n^2)!$ in the denominator will come from? $\endgroup$ – Alexandre Eremenko Oct 13 '14 at 15:19
  • $\begingroup$ Well, its only loosely related. Your f is almost the integral of $\sum t^{n^2}$… $\endgroup$ – Dirk Oct 13 '14 at 15:21
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    $\begingroup$ @Dirk: Why almost? Is $(n^2)!$ "almost" $n^2$? $\endgroup$ – Alexandre Eremenko Oct 13 '14 at 15:41
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This is a slice of the taylor series for $\exp(t)$. The terms that dominate are those near $n=t^{1/2}$. Using Stirling's approximation, as $t\to\infty$ with $q$ more or less bounded, we have $$\frac{t^{(t^{1/2}+q)^2}}{(t^{1/2}+q)^2)!} = \frac{e^{t-2q^2}}{\sqrt{2\pi t}} (1 + O(q^3/t)).$$ So $$f(t) \sim \frac{e^t}{\sqrt{2\pi t}}c(t),$$ where $$c(t) = \sum_q e^{-2q^2},$$ with the sum over all $q$ for which $t^{1/2}+q$ is integer.

It is clear that $c(t)$ depends only on the fractional part of $t^{1/2}$. It is nice curve that oscillates between 1.2353 (when $t^{1/2}$ is midway between two integers) and 1.2679 (when $t^{1/2}$ is an integer).

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