About the high-order derivatives of Lambert function

Question

In the mid seventies, in my former research group, we found that the $n^{\text{th}}$ derivative of $W_0(x)$ could write $$\frac {d^n\,W_0(x)}{dx^n}=(-1)^{n+1}\,\,\frac{\,P_n(w)}{ e^{nw}\,(1+w)^{2n-1}}\qquad \text{with}\qquad w=W_0(x)$$ where, with $P_1(w)=1$, the polynomials

$$P_n(w)=\sum_{k=0}^{n-1} a_{n,k}\,w^k$$ satisfy the simple recurrence relation $$P_{n+1}(w)=\big(n (w + 3) - 1\big)\,P_{n}(w)-(1+w)\,P'_{n}(w)$$

The very first of them $$\left( \begin{array}{cc} n & P_n(w) \\ 1 & 1 \\ 2 & w+2 \\ 3 & 2 w^2+8 w+9 \\ 4 & 6 w^3+36 w^2+79 w+64 \\ 5 & 24 w^4+192 w^3+622 w^2+974 w+625 \\ 6 & 120 w^5+1200 w^4+5126 w^3+11758 w^2+14543 w+7776 \\ 7 & 720 w^6+8640 w^5+45756 w^4+137512 w^3+248250 w^2+255828 w+117649 \\ \end{array} \right)$$ show some interesting patterns such that $$a_{n,0}=n^{n-1}$$ $$a_{n,1}= 3n^n-(n+1)^n-n^{n-1}$$ $$a_{n,n-2}=2(n-1)(n-1)!$$ $$a_{n,n-1}=(n-1)!$$

We were not be able to generate the general recurrence relation for the $a_{n,k}$ and this should be my first question.

My second question is : how to prove that $a_{n,k}$ are all positve, possibly unimodal and log-concave with respect to $k$.

Any help, idea or suggestion would be more than welcome.

Answer 1

Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin, see also arXiv:1011.5940

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials.

In section 4.2.3 Kalugin proves that the coefficients are positive, unimodal, and log-concave.