Orthogonal : which weights do you use?

Thank you. (1) The formal series $\sum_{n\geq 1}\frac{n^n}{n!}p(n) z^n$ can be expressed as a rational function of $T(z)$ if $p$ is a polynomial in $n$ and $\tfrac{1}{n}$. My first guess is that it will not be easy to treat the case $p(n)=H_n$ via Lagrange inversion. (2) $T(z)$ is certainly not "my" function. $T(z)=-W(-z)$ where $W$ is "Lambert's W-function", and $T(z)=z e^{T(z)}$ goes back to Eisenstein.

\@ Alexey Ustinov: unfortunately not my formula. I think that the formula allows to complete the proof, sketch (steps): (1) show that the series on the rhs defines a holomorphic function for $|y|<1$ (check that the convergence is locally uniform) (2) $q(y):=\frac{e^{y}}{c(y)}$ is holomorphic and $\neq 0$ in a small disc punctured at $0$, and the same holds for $d(y):=1/q(y)$ (3) putting $d(0):=0$ makes $d$ holomorphic in that disc (4) thus $d$ is given in that disc by its Taylor series (which you showed to compute formally above)

In addition it can be shown (following the lines of Flajolet\&Odlyzko, see hal.archives-ouvertes.fr/inria-00075445/document , p.15) that for $y\in (0,1)$ $${e^y \over c_n(y)}=\frac{1}{2}\frac{y(1-y^n)}{1-y}+\frac{1}{1-e^{-y}}+y\sum_{j=0}^{n-1} w(\delta_j(y))y^j$$ and $${e^y \over c(y)}=\frac{1}{2}\frac{y}{1-y}+\frac{1}{1-e^{-y}}+y\sum_{j=0}^\infty w(\delta_j(y))y^j$$ where $w(z)=\frac{1}{1-e^{-z}}-\frac{1}{z}-\frac{1}{2}$, $\delta_{-1}(y)=y$ and $\delta_n(y)=y(1-e^{-\delta_{n-1}(y)})$ for $n \geq 0$

You're right (and my assumption above probably false). The branching process perspective suggests that the limiting distribution of the no. of nodes in the highest level should be a discrete distribution concentrated on positive integers (maybe even degenerate). This might be known, if so, it can possibly be found in Pavlov's book "Random Forests". Else you'll have to analyse the g.f. yourself (using the saddle point method). I would first try to find the asymptotic distribution of the number of nodes in height $k$ in a random tree of height $k$ for fixed $k$ and then try to extend.

(1) True, but it shows that the order of the no. of nodes at maximal height is not higher than $\sqrt{n}$. (2) It is plausible that $\sqrt{n}$ is the correct order, since by the results of Stepanov (see epubs.siam.org/doi/10.1137/1114007) and Meir&Moon (see cms.math.ca/10.4153/CJM-1978-085-0) asymptotically each layer (stratum) at height $x\sqrt{n}$ of a rooted random tree with $n$ nodes contains of order $\sqrt{n}$ nodes, moreover asymptotically a randomly chosen node lies at a height of order $\sqrt{n}$. Question: how precise do you need this information to be made?

The height $H_n$ of a random rooted Cayley tree with $n$ nodes is of order $\sqrt{n}$, more precisely: the distribution of $H_n/\sqrt{n}$ converges to the Kolmogorov-Smirnov distribution as $n\longrightarrow \infty$. This is a special case of the results here epubs.siam.org/doi/10.1137/1128044.