The theory of the $r$-Lambert function is more-or-less well known. The main property that is still missing is the asymptotic behavior .
The classical Lambert function satisfies the well-known asymptotic expansion:
$$ W_k(z)=\log k z-\log \log k z+\sum{m=1}^{\infty} \sum{n=0}^{\infty} \frac{(-1)^n}{m!}\binom{n+m}{n+1} \frac{\left(\log _{\left.\log _k z\right)^m}^{\log _k^{n+m} z} . . . ~\right.}{\text { n }}\qquad(1) $$
The $r$-Lambert function (which satisfies $W_r e^{a W_r}=z$ ) modifies this by introducing an additional perturbation in $r$.
To determine the asymptotic expansion for large $z$, we proceed as follows:
Assume an expansion of the form:
$$ W_r(z)=W(z)+\sum_{k=1}^{\infty} c_k r^k $$
Substitute into the defining equation and solve order-by-order.
The leading-order term remains the classical asymptotic form:
$$ W_r(z) \approx \log z-\log \log z $$
To find the first correction term in $r$, we substitute the expansion into:
$$ W_r e^{a W_r}=z $$
Using a perturbative approach, the first-order correction is:
$$ W_r(z) \approx W(z)-\frac{W^2(z)}{\downarrow(1+W(z))} r $$
For higher-order corrections, we recursively substitute into the equation.Thus, a generalized formula like Equation (1) for $W_r$ should contain similar asymptotics but with $r$ dependent terms correcting the standard Lambert function expansion.
The series for $W_r(w)$ is:
$$ W_r(w)=W(w)-\frac{W^2(w)}{w(1+W(w))} r-\frac{1}{2} \frac{W^3(w)\left(W^2(w)-2\right)}{w^2(W(w)+1)^3} r^2-\cdots\qquad(2) $$
To analyze the radius of convergence, we consider: