$\def\W{\operatorname W} \def\Li{\operatorname{Li}} $
Interested by $\sum_\limits{n=1}^\infty\frac{\W(n^2)}{n^2}$, here is an example where Lagrange reversion applies to a Lambert W sum:
$$\W(x)=\ln(x)+\sum_{m=1}^\infty\sum_{k=0}^\infty\frac{(-1)^m(m-k+1)_kS_{m-1}^{(k)}}{\ln^{m-1}(x)m!}\ln^{m-k}(\ln(x))$$
for about $|x|>e$ with the Pochhammer symbol $n^{(m)}$ and Stirling S1 $S_n^{(m)}$. Therefore the polylogarithm appears:
$$\sum_{n=0}^\infty\W(e^{e^{an}})x^n=1+\frac1{1-e^ax} +\sum_{m=1}^\infty\sum_{k=0}^\infty\frac{(-1)^m}{m!}(m-k+1)_kS_{m-1}^{(k)}a^{m-k}\Li_{k-m}(e^{(a-1)m}x),a>0$$
Expanding the polylogarithm uses Stirling S2 $s_n^{(m)}$:
$$\sum_{n=0}^\infty\W(e^{e^{an}})x^n= 1+\frac1{1-e^ax}+\sum_{m=1}^\infty\sum_{k=0}^\infty\sum_{j=1}^{m-k+1}\frac{(-1)^m}{m!}a^{m-k}(m-k+1)_k \Gamma(j) S_{m-1}^{(k)}s_{m-k+1}^{(j)}\left(\frac x{e^{(m-1)a}-x}\right)^j$$
Even the $S_n^{(m)}$ limit expansion does not help evaluate the sum over $m$ or $k$, but this method numerically works.
How can one evaluate $\displaystyle\sum_{n=0}^\infty\W(e^{e^{an}})x^n$?
