To start we recall the combinatorial class specification for labeled
trees
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T)}$$
(trees are sets of trees attached to a root node) which gives the
functional equation
$$T(z) = z \exp T(z).$$
We know by Cayley's theorem that
$$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}.$$
We will just do the first term of the asymptotics.
Tree function $T(z)$
We will take as our starting point the result by Flajolet and Odlyzko on
the nature of the singularity of the tree function at $1/e.$
Specifically they establish on page 6 proposition 1 of the landmark
paper Random Mapping Statistics INRIA RR 1114 that the tree function
has Pusieux series at a square root singularity
$$\sqrt{1-ez}.$$
We now compute the series as in the paper A sequence of series for the
Lambert W function by Corless, Jeffrey and Knuth. The Pusieux series
is where we put $P=\sqrt{1-ez}$ so that $z=(1-P^2)/e$ and
$$T(z) = \sum_{k\ge 0} Q_k P^k$$
With $T(1/e) = 1$ and hence $Q_0 = 1$ we re-write the functional equation
as
$$T(z) \exp (-T(z)) = z$$ to get
$$(T-1) \exp (1-T) \exp(-1) + \exp (1-T) \exp(-1) =
\frac{1-P^2}{e}.$$
We are ready to compute the expansion in $P$ which we do by equating
the coefficients on the powers of $P$ on the left and the right
side. Note that $(T-1)^k = Q_1^k P^k + \cdots$ so to compute the range
$Q_0$ to $Q_m$ we can work with the initial segment $T_m =
\sum_{k=1}^m Q_k P^k.$ In particular we also truncate the exponential
at $m.$ This is quick and easy, the reader may want to establish a
recurrence instead. We obtain two branches, of which we choose
numerically the one that gives the combinatorial branch of the tree
function and get
$$T(z) = 1 -\sqrt{2} \sqrt{1-ez}
+ {\frac{2}{3}} (1-ez)
-\frac{11 \sqrt{2}}{36} (1-ez)^{3/2}
+ {\frac{43}{135}} (1-ez)^2
\\ -\frac{769 \sqrt{2}}{4320} (1-ez)^{5/2}
+ {\frac{1768}{8505}} (1-ez)^3
- \frac{680863 \sqrt{2}}{5443200} (1-ez)^{7/2}
+ \cdots.$$
We recall from Wilf's generatingfunctionolgy page 180
the well-known asymptotic
$${n-\alpha-1\choose n} \sim \frac{n^{-\alpha-1}}{\Gamma(-\alpha)}$$
This does not hold when $\alpha$ is a non-negative integer
due to the poles of the Gamma function. It does hold for
$\alpha$ a negative integer where we get
$$\frac{n^{(-\alpha)-1}}{((-\alpha)-1)!}$$
which matches the Gamma function.
We should now obtain Stirling's formula from the first
term. We find
$$\frac{n^{n-1}}{n!}
\sim -\sqrt{2} {1/2\choose n} (-1)^n \exp(n)
\\ = -\sqrt{2} {n-1/2-1\choose n} \exp(n)
\sim -\sqrt{2} \frac{n^{-3/2}}{\Gamma(-1/2)} \exp(n)
= \frac{n^{-3/2}}{\sqrt{2\pi}} \exp(n).$$
This yields on rearranging
$$\frac{n^n}{n!} \sim \frac{\exp(n)}{\sqrt{2\pi n}}$$
as desired.
Cycle of trees $\log\frac{1}{1-T(z)}$
These are the connected components of random mappings. We seek
$$n! [z^n] \log\frac{1}{1-T(z)}
= (n-1)! [z^{n-1}] \frac{T'(z)}{1-T(z)}.$$
Working with the initial segment of the Pusieux series we get
$$(n-1)! [z^{n-1}]
\left[ \sqrt{2}\sqrt{1-ez} - \frac{2}{3}(1-ez) + \cdots
\right]^{-1} \times
\left[ \frac{\sqrt{2}e}{2} \frac{1}{\sqrt{1-ez}}
- \frac{2e}{3} 1\times (1-ez)^0
+ \frac{\sqrt{2}e11}{24} \sqrt{1-ez} + \cdots \right].$$
This becomes
$$(n-1)! [z^{n-1}] \frac{\sqrt{2}e}{2} \frac{1}{\sqrt{1-ez}}
\frac{1}{\sqrt{2}\sqrt{1-ez}}
\left[1 - \frac{\sqrt{2}}{3}\sqrt{1-ez} + \cdots
\right]^{-1} \times
\left[ 1
- \frac{2\sqrt{2}}{3} \sqrt{1-ez}
+ \frac{11}{12} (1-ez) + \cdots \right].$$
Now expanding the inverse term into a geometric series
gives a constant one plus higher order terms in
$\sqrt{1-ez}$ where a power
$(\frac{\sqrt{2}}{3}\sqrt{1-ez} + \cdots)^k$ starts with
$\sqrt{1-ez}^k$ which means fixed powers of the square
root term are only contributed to a finite number of
times. Extracting the asymptotic we thus have
$$(n-1)! [z^{n-1}] \frac{1}{2} \frac{1}{1-ez}
= \frac{1}{2} (n-1)! \exp(n)
= \frac{\exp(n)}{2n} n!.$$
This matches the result from the Flajolet paper. For the formula in the
book we use Stirling to get
$$\frac{\exp(n)}{2n} \frac{n^n}{\exp(n)} \sqrt{2\pi n}
= \frac{1}{2} \sqrt{2\pi} n^{n-1/2}$$
again as claimed.
A square root of the tree function $\frac{1}{\sqrt{1-T(z)}}$
This time we seek
$$n! [z^n] \frac{1}{\sqrt{1-T(z)}}.$$
Once again we have from the intial segment
$$n! [z^n]
\left[ \sqrt{2}\sqrt{1-ez} - \frac{2}{3}(1-ez) + \cdots
\right]^{-1/2}
\\ = n! [z^n] \frac{1}{2^{1/4} (1-ez)^{1/4}}
\left[1 - \frac{\sqrt{2}}{3}\sqrt{1-ez} + \cdots\right]^{-1/2}.$$
We are justified in applying the generalized binomial theorem to the
square root term as explained in the previous section. This means the
dominant asymptotic originates with
$$n! [z^n] \frac{1}{2^{1/4} (1-ez)^{1/4}}
= n! \frac{1}{2^{1/4}} {-1/4\choose n} (-1)^n \exp(n)
\\ = n! \frac{1}{2^{1/4}} {n+1/4-1\choose n} \exp(n)
\sim n! \frac{1}{2^{1/4}} \frac{n^{-3/4}}{\Gamma(1/4)} \exp(n)$$
Now use Stirling to get
$$\frac{n^n}{\exp(n)} \sqrt{2\pi n}
\frac{1}{2^{1/4}} \frac{n^{-3/4}}{\Gamma(1/4)} \exp(n)
= \frac{\sqrt{\pi} 2^{1/4}}{\Gamma(1/4)} n^{n-1/4}.$$
This matches the book, moreover we have computed the constant, which
was not given.
A sequence of $m$ maps $\frac{1}{(1-T(z))^m}$
This time we seek with $m$ a positive integer.
$$n! [z^n] \frac{1}{(1-T(z))^m}.$$
Do the case $m$ odd first. Once again we have from the
intial segment
$$n! [z^n]
\left[ \sqrt{2}\sqrt{1-ez} - \frac{2}{3}(1-ez) + \cdots
\right]^{-m}
\\ = n! [z^n] \frac{1}{2^{m/2} (1-ez)^{m/2}}
\left[1 - \frac{\sqrt{2}}{3}\sqrt{1-ez} + \cdots\right]^{-m}.$$
Extracting the dominant asymptotic we find
$$\frac{1}{2^{m/2}} n! [z^n] \frac{1}{(1-ez)^{m/2}}
= \frac{1}{2^{m/2}} n! {-m/2\choose n} (-1)^n \exp(n)
\\ = \frac{1}{2^{m/2}} n! {n+m/2-1\choose n} \exp(n)
\\ \sim \frac{1}{2^{m/2}} n! \frac{n^{m/2-1}}{\Gamma(m/2)} \exp(n)
\sim \frac{1}{2^{m/2}} \frac{n^n}{\exp(n)} \sqrt{2\pi n}
\frac{n^{m/2-1}}{\Gamma(m/2)} \exp(n)
\\ = \frac{\sqrt{\pi}}{2^{(m-1)/2}\Gamma(m/2)} n^{n+(m-1)/2}.$$
This also matches the book where the constant was not given.
We have when $m$ is even the asymptotic works out the same
as remarked earlier. (Note that when $m$ is odd we have a
branch cut from $1/(1-w)^{m/2}$ and when $m$ is even there
is no cut, only a singularity at $w=1$. Both are analytic
when $|w| \lt 1.$) Just to be sure we can compute an
alternate expression of the coefficient as a sum and check
that against the asymptotic. We get
$$(n-1)! [z^{n-1}] \frac{m}{(1-T(z))^{m+1}} T'(z)$$
which is
$$m (n-1)! \;\underset{z}{\mathrm{res}}\;
\frac{1}{z^n} \frac{1}{(1-T(z))^{m+1}} T'(z)$$
With $T(z) = w$ we get from the functional equation
$z=w\exp(-w)$
$$m (n-1)! \;\underset{w}{\mathrm{res}}\;
\frac{\exp(nw)}{w^n} \frac{1}{(1-w)^{m+1}}
\\ = m (n-1)! n^{n-1} \sum_{q=0}^{n-1}
\frac{n^{-q}}{(n-1-q)!} {q+m\choose q}.$$
This formula was tested against the asymptotic and numeric
equivalence was observed.
Comparison with Theorem 1
Referring to page 6 of the random mapping paper we recall
that $\frac{1}{(1-T(z))^m}$ will produce the dominant asymptotic
$$\frac{1}{2^{m/2}} n! [z^n] \frac{1}{(1-ez)^{m/2}}$$
so that in terms of said theorem we find $\sigma(u) =
u^{m/2}$ so that $\alpha=m/2$, and $s=\exp(-1)$. Applying
this to the closed form we quote
$$[z^n] f(z) \sim s^{-n} \frac{\sigma(n)}{n\Gamma(\alpha)}
= \exp(n) \frac{n^{m/2}}{n\Gamma(m/2)}.$$
On multiplication by $n!$ and the scalar we indeed obtain
the asymptotic from the previous section. We see that once
the Pusieux series is established the asymptotics of a
variety of tree function statistics can be obtained from
the singularity analysis theorem simply by substituting a
set of parameters into the proposed closed form.
Acyclic labeled graphs $\exp(T(z)-T(z)^2/2)$
Here we seek
$$n! [z^n] \exp(T(z)-T(z)^2/2).$$
To get the initial segment of the Pusieux series re-write this as
$$n! [z^n] \sqrt{e} \exp(-1/2 \times (T(z)-1)^2).$$
Expanding the exponential yields
$$\sqrt{e} \sum_{q\ge 0} (-1)^q \frac{1}{2^q}
\frac{1}{q!} (T(z)-1)^{2q}
\\ = \sqrt{e} \sum_{q\ge 0} (-1)^q \frac{1}{2^q}
\frac{1}{q!}
(-\sqrt{2}\sqrt{1-ez} + \frac{2}{3} (1-ez)+\cdots)^{2q}.$$
Observe carefully that the powers $(1-ez)$ and
$(1-ez)^{3/2}$ appear only in the term for $q=1$ with
higher values of $q$ also generating higher
powers. Therefore $0\le q\le 1$ gives the exact first
three terms of the series. We get
$$\sqrt{e}
(1 - \frac{1}{2} (2(1-ez)
- \frac{4\sqrt{2}}{3} (1-ez)^{3/2} + \cdots))
\\ = \sqrt{e}
(1 - (1-ez)
+ \frac{2\sqrt{2}}{3} (1-ez)^{3/2} + \cdots).$$
The dominant asymptotic is therefore
$$n! [z^n] \sqrt{e} \frac{2\sqrt{2}}{3} (1-ez)^{3/2}.$$
We then apply Theorem 1 with $\sigma(u) = u^{-3/2}$ so
that $\alpha = -3/2$ and with $s=\exp(-1)$ to obtain
$$n! \sqrt{e} \frac{2\sqrt{2}}{3}
\exp(n) \frac{n^{-3/2}}{n\Gamma(-3/2)}.$$
With $\Gamma(-3/2) = \frac{4\sqrt{\pi}}{3}$ this becomes
$$n! \sqrt{e} \frac{1}{\sqrt{2\pi}} \exp(n) n^{-5/2}.$$
Use Stirling one last time
$$\sqrt{e} \frac{1}{\sqrt{2\pi}} n^{-5/2}
n^n \sqrt{2\pi n} = \sqrt{e} \times n^{n-2}.$$
Exactly as given in the book. We conclude for the time being.
