TITLE: Second-order Eulerian numbers, Lambert's W function, and Schröder's fourth problem
QUESTION [6 upvotes]: Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556
have been proved on MSE [  a ,
 b ,  c  ].
A special property unfolds when these numbers are interpreted as coefficients of polynomials:
$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n = 
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$
If we evaluate these polynomials at the point $ x = 1/2 $, we get
a sequence whose exponential generating function is based on
the amazing Lambert W function.
$$ 2^n \left\langle\!\!\left\langle  \frac{1}{2}  \right\rangle\!\!\right\rangle_n
= n! \, [x^n]\, ({2 + 2\operatorname{W}(- \exp((x-1)\,/\,2)\,/\,2))^{-1}} $$
Not only that: the sequence represents the solution of a
famous combinatorial problem, Schröder's fourth problem
(see A000311)!
It would be great if someone could show us a proof of this formula.

REPLY [3 votes]: We seek to show that the following identity holds:
$$2^{n+1} \sum_{k=0}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
\frac{1}{2^k} =
n! [x^n] \frac{1}{1 + W(-\exp((x-1)/2)/2)}.$$
We will be using data from Wikipedia on Lambert
W and work with the
combinatorial branch which is  $W_0(z)$.
Recall that
$$W'(z) \frac{z}{W(z)} = \frac{1}{1+W(z)}.$$
We obtain
$$[z^m] \frac{1}{1+W(z)} =
\frac{1}{2\pi i} 
\int_{|z|=\varepsilon} \frac{1}{z^{m}} 
\frac{1}{W(z)} W'(z) \; dz.$$
Putting $W(z) = v$ we find
$$\frac{1}{2\pi i} 
\int_{|v|=\gamma} \frac{1}{v^{m} \exp(mv)} 
\frac{1}{v} \; dv
= \frac{1}{2\pi i}
\int_{|v|=\gamma} \frac{1}{v^{m+1}} \exp(-mv) \; dv
= \frac{(-1)^m m^m}{m!}.$$
so that
$$\frac{1}{1+W(z)} = \sum_{m\ge 0} (-1)^m m^m \frac{z^m}{m!}.$$
We get for the original RHS
$$n! [x^n] \sum_{m\ge 0} \frac{m^m}{m!}
\exp(m(x-1)/2) \frac{1}{2^m}
\\ = n! [x^n]
\sum_{m\ge 0} \frac{m^m}{m!} \frac{\exp(-m/2)}{2^m} \exp(mx/2)
\\ = \sum_{m\ge 0} \frac{m^{m+n}}{m!} 
\frac{\exp(-m/2)}{2^{m+n}}.$$
First part.
Introduce the tree function $T(z)$  from combinatorics where $T(z) = z
\exp  T(z)$ and  $T(z) =  - W_0(-z).$  Note that  we have  by Cayley's
theorem that $T(z)  = \sum_{m\ge 1} m^{m-1}  \frac{z^m}{m!}.$ We claim
that with $n\ge 1$
$$Q_n(z) = \sum_{m\ge 0} m^{m+n} \frac{z^m}{m!} =
\frac{1}{(1-T(z))^{2n+1}}
\sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k.$$
This  means the  RHS is  $\frac{1}{2^n} Q_n(\exp(-1/2)/2).$  To verify
this last identity  note that $Q_{n+1}(z) = z  \frac{d}{dz} Q_n(z)$ so
we may prove it by induction.
We get for the RHS of the series identity on differentiating and
multiplying by $z$
$$\frac{(2n+1) z T'(z)}{(1-T(z))^{2n+2}}
\sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
\frac{z}{(1-T(z))^{2n+1}}
\sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k-1} T'(z)$$
Extracting the term $z T'(z)/(1-T(z))^{2n+2}$ in front leaves us with
$$(2n+1) \sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
(1-T(z))
\sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k-1}
\\ = (2n+1) \sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k +
\sum_{k=0}^{n-1} 
\left\langle\!\! \left\langle n \atop k+1
\right\rangle\!\! \right\rangle
(k+1) T(z)^{k}
- \sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
k T(z)^{k}
\\ = \sum_{k=1}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
(2n+2-(k+1)) T(z)^k +
\sum_{k=0}^{n-1} 
\left\langle\!\! \left\langle n \atop k+1
\right\rangle\!\! \right\rangle
(k+1) T(z)^{k}.$$
We may include $k=0$ in the first sum and $k=n$ in the second.
Now the Eulerian number recurrence (second order) according to OEIS
A349556 is
$$\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
=
\left\langle\!\! \left\langle n-1 \atop k
\right\rangle\!\! \right\rangle k
+ \left\langle\!\! \left\langle n-1 \atop k-1
\right\rangle\!\! \right\rangle (2n-k)$$
We have shown that
$$Q_{n+1}(z) = \frac{z T'(z)}{(1-T(z))^{2n+2}}
\sum_{k=0}^n \left\langle\!\! \left\langle n+1 \atop k+1
\right\rangle\!\! \right\rangle T(z)^k
\\ = \frac{z T'(z)}{T(z) (1-T(z))^{2n+2}}
\sum_{k=1}^{n+1} \left\langle\!\! \left\langle n+1 \atop k
\right\rangle\!\! \right\rangle T(z)^k.$$
Now we just have to verify that
$$\frac{z T'(z)}{T(z) (1-T(z))^{2n+2}}
= \frac{1}{(1-T(z))^{2n+3}}
\quad\text{or}\quad
z T'(z) (1-T(z)) = T(z).$$
The functional equation tells us that $T'(z) = \exp T(z) + z \exp T(z) 
T'(z)$ so that $T'(z) (1 - T(z)) = \exp T(z) = T(z) / z$ which is just
what we need. It remains to verify the base case so the induction
starts properly. We seek
$$Q_1(z) = \sum_{m\ge 0} m^{m+1} \frac{z^m}{m!} =
\frac{T(z)}{(1-T(z))^3}.$$
We verify this by coefficient extraction. We get
$$m! [z^m] Q_1(z) =
\frac{m!}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^{m+1}} \frac{T(z)}{(1-T(z))^3} \; dz.$$
With $T(z) = z + \cdots$ this integral will produce the correct value
zero for $m=0.$ For $m\ge 1$,  we put $T(z) = w$ so that $z = w \exp(-w)$
and $dz = \exp(-w) (1-w) \;  dw$ and obtain
$$\frac{m!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp((m+1)w)}{w^{m+1}} \frac{w}{(1-w)^3} 
\exp(-w) (1-w) \; dw
\\ = \frac{m!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(mw)}{w^{m}} \frac{1}{(1-w)^2} 
\; dw.$$
This is
$$m! \sum_{q=0}^{m-1} \frac{m^q}{q!} (m-q)
= m! \sum_{q=0}^{m-1} \frac{m^{q+1}}{q!}
- m! \sum_{q=1}^{m-1} \frac{m^q}{(q-1)!}
\\ = m! \sum_{q=0}^{m-1} \frac{m^{q+1}}{q!}
- m! \sum_{q=0}^{m-2} \frac{m^{q+1}}{q!}
= m! \frac{m^m}{(m-1)!} = m^{m+1}$$
as desired. 
Sequel. Note  that in  the identity  for $Q_n(z)$  we have  by the
definition of the Eulerian numbers that $\left\langle\!\! \left\langle
n  \atop 0  \right\rangle\!\! \right\rangle$  is zero  when $n\ge  1.$
Therefore we  may extend $k$  to include zero  (with $n\ge 1$  for the
moment) which yields
$$\bbox[5px,border:2px solid #00A000]{
Q_n(z) = \sum_{m\ge 0} m^{m+n} \frac{z^m}{m!} =
\frac{1}{(1-T(z))^{2n+1}}
\sum_{k=0}^n 
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle
T(z)^k.}$$
Now  observe  that this  will  produce  $Q_0(z)  = \sum_{m\ge  0}  m^m
\frac{z^m}{m!}    =   \frac{1}{1-T(z)}$   due   to   $\left\langle\!\!
\left\langle 0 \atop  0 \right\rangle\!\! \right\rangle =  1$ which is
in fact correct because unlike $Q_n(z)$  with $n\ge 1$, $Q_0(z)$ has a
constant term, which  is one (this is because $m^{m+n}  = 0$ for $m=0$
and  $n\ge 1$  and  $m^{m+n} =  1$ for  $m=0$  and $n=0$).   Therefore
$$Q_0(z) = 1 + z T'(z)  = 1 + \frac{T(z)}{1-T(z)} = \frac{1}{1-T(z)}$$
as obtained from the boxed version of the main identity, which is seen
to hold for all $n\ge 0.$ 
Conclusion. We are now ready  to answer the original question.  We
have shown that  the RHS is $\frac{1}{2^n}  Q_n(\exp(-1/2)/2).$ By our
formula for  $Q_n(z)$ in  terms of  the tree  function we  obtain with
$T(\exp(-1/2)/2) = \frac{1}{2}$ at last the closed form
$$\frac{1}{2^n} \frac{1}{(1-1/2)^{2n+1}} 
\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle \frac{1}{2^k}
= 2^{n+1}
\sum_{k=0}^n
\left\langle\!\! \left\langle n \atop k
\right\rangle\!\! \right\rangle \frac{1}{2^k}$$
which is the LHS and hence the claim.