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)



Closed formula for reversion of Jacobi theta series
Perhaps, the form given by [Lagrange inversion theorem](https://en.wikipedia.org/wiki/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](https://oeis.org/A243148) if we fix a number of squares). This yields the formula that Peter Taylor gave in the comments (modulo the corrections) for $n>1$:


$$A\_n = \frac{1}{n!2^n} \sum\_{(2^2-1)j\_2 + (3^2-1)j\_3 + \dots = n-1} (-1)^{j\_2+j\_3+\dots}\cdot \frac{(n-1+j\_2+j\_3+\dots)!}{j\_2!j\_3!\dots}.$$


