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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):

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

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 $n>1$:

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

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 $n-1+k$ into $k$ nonzero squares (OEIS A243148) for $k\in\{1,2,\dots,n-1\}$.

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):

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

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 reversion coefficients amounts to iterating over partitions of $n-1$ into squares (OEIS A001156).

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 $n-1+k$ into $k$ nonzero squares (OEIS A243148) for $k\in\{1,2,\dots,n-1\}$.