Is $(m,n)=(2,3)$ the only solution to $\sum_\limits{k \in\Bbb Z}\frac1{(W_k(x)+1)^m}=\sum_\limits{k \in\Bbb Z}\frac1{(W_k(x)+1)^n}$? From discussions 1, 2, @HenriCohen wrote a paper on Lambert $W$-Function Branch Identities which includes identities such as $$\sum_\limits{k
\in\Bbb Z}\frac1{(W_k(x)+1)^2}=\sum_\limits{k
\in\Bbb Z}\frac1{(W_k(x)+1)^3}=\frac1{xe+1}$$ by repeated differentiation of $\prod_{k\in\Bbb Z}(1-t/W_k(x))=e^{-t/2}-te^{t/2}/x$.

Naturally, we ask: is $(m,n)=(2,3)$ the only solution to $$\sum_\limits{k
\in\Bbb Z}\frac1{(W_k(x)+1)^m}=\sum_\limits{k
\in\Bbb Z}\frac1{(W_k(x)+1)^n}$$ for positive integers $m<n$ and for all $x>-1/e$?

E.g. When $m>2$ and $n=m+1$, it suffices to show that equality does not hold when $x=2$ for instance, and it then suffices to show that $$\frac{d^m}{ds^m}\left(2m\log(e^{-2s}-s)+\frac{1+2s}{e^{-2s}-s}\right)\bigg\vert_{s=-1/2}\ne0.$$
 A: The answer is "yes":
Define the generating function
$$
G(x,t) = \sum_{k \in \Bbb Z}\log\left[1- \frac{t}{W_k(x)}\right].\tag{1}
$$
Then,
$$
\sum_{k \in \Bbb Z} \frac 1{(W_k(x)+1)^m}
= -\frac{1}{\Gamma(m)} \, G^{(0,m)}(x,-1),\tag{2}
$$
where $G^{(0,m)}(x,-1)$ denotes the $m$-th derivative w.r.t. $t$ at $t=-1$.
From the cited paper, (1) equals
$$
G(x,t) = \log(e^{-t/2} - t x^{-1} e^{t/2}),\tag{3}
$$
such that we need to calculate the $m$-th derivative of (3) at $t=-1$.
While this generates complicated rational polynomials, it is sufficient to expand them to first order around $x=0$ and look at the linear term, note that $G^{(0,m)}(0,-1)/\Gamma(m)=-1$ for $m>1$. We find
\begin{align}
\frac{1}{e\,\Gamma(m)}\,G^{(1,m)}(0,-1) 
&= \frac{!m}{\Gamma(m)} = \frac{\Gamma(m+1,-1)}{e\,\Gamma(m)}\tag{4a}\\
&= 0, 1, 1, \tfrac{3}{2}, \tfrac{11}{6}, \tfrac{53}{24},
\tfrac{103}{40}, \tfrac{2119}{720}, \ldots,\tag{4b}
\end{align}
where $!m$ denotes the subfactorial of $m$, and $\Gamma$ is the incomplete gamma function.
Note that
$$
\frac{!m}{m!}=\sum_{k=0}^{m}\frac{(-1)^k}{k!}.\tag{5}
$$
So, only $m=2,3$ have the same constant and linear order near $x=0$.
