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Considering the Jacobi theta: $\theta_3(z) = \sum_{n\in\mathbb{Z}} q^{n^2}$, we can invert $\theta_3-1$ in a small enough neighbourhood of 0.

Routine computation with Lagrange-Burmann inversion gives that the inverse have expansion starting by:

$\frac{q}{2}-\frac{q^4}{16}+\frac{q^7}{32}-\frac{q^9}{512}-\frac{11 q^{10}}{512}+\frac{13 q^{12}}{4096}+ O(q^{13})$

Is there any known closed form (or recursive formula) giving the terms of this series?

(it is very closely related to OEIS A259938 suite: https://oeis.org/A259938, corresponding to $\theta_3(q/2)$ but which doesn't seem very studied as well)

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    $\begingroup$ By closely related, one should understand related by the substitution $q \mapsto q/2$. $\endgroup$
    – F. C.
    Commented Jul 5, 2021 at 18:00
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    $\begingroup$ Morse-Feshbach (see bottom of page) gives a sum over partitions of $n-1$ into numbers one less than a square; it's not a closed form or an elegant recurrence, but if your concern is practical evaluation then it may be of interest. $\endgroup$
    – Peter Taylor
    Commented Jul 6, 2021 at 7:38
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    $\begingroup$ With a bit of simplification, $$A_n = n2^{-n} \sum_{s,t,u,\ldots} (-1)^{s+t+u+\cdots} \binom{n-1+s+t+u+\cdots}{s,t,u,\ldots}$$ where the sum is over non-negative integer solutions to $s + 2t + 3u + \cdots = n-1$. $\endgroup$
    – Peter Taylor
    Commented Jul 6, 2021 at 7:47
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    $\begingroup$ To avoid confusion it would be better to call it reversion rather than inversion (the latter can be more likely understood as multiplicative inverse). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 6, 2021 at 9:18
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    $\begingroup$ Apologies, I seem to have lost a reciprocal along the way. Change $n2^{-n}$ to $n^{-1} 2^{-n}$. $\endgroup$
    – Peter Taylor
    Commented Jul 6, 2021 at 10:13

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Perhaps, the form given by Lagrange inversion theorem cannot be much simplified here. It expresses the $n$-th coefficient of a series reversion as the sum of $n-1$ values of exponential Bell polynomials.

From the practical perspective, since $\theta_3-1$ contains nonzero coefficients only at square powers, computation of the $n$-th reversion coefficient amounts to iterating over the partitions of $n-1$ into squares $>1$ decreased by $1$ (almost OEIS A243148 if we fix a number of squares). This yields the formula that Peter Taylor gave in the comments (modulo the corrections) for $n>1$:

$$A_n = \frac{1}{n!2^n} \sum_{(2^2-1)j_2 + (3^2-1)j_3 + \dots = n-1} (-1)^{j_2+j_3+\dots}\cdot \frac{(n-1+j_2+j_3+\dots)!}{j_2!j_3!\dots}.$$

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  • $\begingroup$ Seems indeed to be the best we can do. I'll accept it as an answer to the question, thanks ! $\endgroup$
    – user70925
    Commented Jul 7, 2021 at 9:08

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