The value of the second-order Eulerian polynomials at x = -1/2. Recently, the second-order Eulerian polynomials
$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n $
have been discussed on MSE [ a ,
 b ].
$$ \left\langle\!\left\langle x \right\rangle\!\right\rangle_n = 
\sum_{k=0}^n \left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle \, x^k $$
Here $\left\langle\!\!\left\langle n\atop k \right\rangle\!\!\right\rangle $
are the second-order Eulerian numbers A340556.
The values of these polynomials at $ x = \frac12 $ generate a sequence
that represents the solution of Schröder's fourth problem
(see MSE and
A000311).
That's a pretty nice result that suggests the question:
Do the values of the polynomials at $ x = -\frac12$ (times $ 2^n $) also have
a combinatorial meaning? That this might indeed be the case is a conjecture
suggested by many examples of combinatorial polynomials. The reader can get
an impression for himself while browsing through the OEIS.




Polynomials
P(x)
2^nP(1/2)
2^nP(-1/2)




Laguerre
A021009
A103194
A000262


Motzkin
A064189
A330796
A000244


BigSchröder
A080247
A065096
A239204


StirlingSet
A048993
A005493
A000110


StirlingCycle
A132393
A000254
A000774


Eulerian 1st
A173018
A180119
A001710




However, this question may not be easy to answer, so we ask a more concrete question.
The sequence $ 2^n \left\langle \!\left\langle - 1/2 \right\rangle \!\right\rangle_n $
can also be generated without reference to the second-order Eulerian polynomials by
series reversion.
$$ 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n 
= (n + 1)!\, [x^{n+1}]\, \text{Reversion}\left(\frac{6x + \exp(3x) - 1}{9}\right) \quad (n \ge 0). $$
Can someone confirm this equation?
Addendum:
The general form of the reversion is described by:
$$ \left\langle\! \left\langle x \right\rangle\! \right\rangle_n = 
(n+1)!\, (1-t)^{2n + 1}\, [x^{n+1}]\, 
\operatorname{Reversion}_{x}(x + t - t \exp(x)) \quad (n \ge 0) $$

Here are some comments. Seeking to invert
$$-\frac{1}{9} + \frac{2}{3} x + \frac{1}{9} \exp(3x) = z$$
we consult Wikipedia on
LambertW
to find  that the closed form solution to
$$x = a + b \exp(cx)$$
is given by
$$x = a - \frac{1}{c} W(-bc \exp(ac)).$$
We thus write
$$x = \frac{3}{2} z + \frac{1}{6} - \frac{1}{6} \exp(3x).$$
to obtain
$$\frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W((1/2)\times \exp((9/2)z+1/2))
\\ = \frac{1}{6} + \frac{3}{2} z
- \frac{1}{3} W(\exp((1+9z)/2)/2).$$
We have
$$W(z) = \sum_{m\ge 1} (-1)^{m-1} m^{m-1} \frac{z^m}{m!}$$
We then obtain for $n\ge 1$
$$-\frac{1}{3} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
\frac{\exp(m/2)}{2^m} \exp(9mx/2)
\\ = \frac{1}{3}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m+n+1}} 9^{n+1}
\\ = \frac{3^{2n+1}}{2^{n+1}}
\sum_{m\ge 1} (-1)^{m} \frac{m^{m+n}}{m!}
\frac{\exp(m/2)}{2^{m}}
.$$
Using the notation from the cited post we get for our closed form
$$\frac{3^{2n+1}}{2^{n+1}} Q_n(-\exp(1/2)/2).$$
Now with $T(-\exp(1/2)/2) = -1/2$ we get with the cited identity
$$\frac{3^{2n+1}}{2^{n+1}}
\frac{1}{(3/2)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}
\\ = 2^n \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle
\left(-\frac{1}{2}\right)^{k}.$$
This is the claim. Note that with $n=0$ we get $Q_0(-\exp(1/2)/2) - 1$
from the series for a total of $\frac{3}{2} +
\frac{3}{2} (Q_0(-\exp(1/2)/2)-1) = \frac{3}{2} +
\frac{3}{2} \frac{1}{3/2} - \frac{3}{2} = 1$ which also agrees with
the polynomial. 
Concerning the addendum. Inverting $x+t-t\exp(x)$ with respect to
$x$ we obtain
$$-W(- t \exp ( -t + z )) -t + z.$$
We then get for the proposed closed form with $n\ge 1$
$$- (1-t)^{2n+1} (n+1)! [x^{n+1}]
\sum_{m\ge 1} (-1)^{m-1} \frac{m^{m-1}}{m!}
(-1)^m t^m \exp(-tm) \exp(mx)
\\ = (1-t)^{2n+1}
\sum_{m\ge 1} \frac{m^{m+n}}{m!}
t^m \exp(-tm).$$
This is $(1-t)^{2n+1} Q_n(t\exp(-t)).$ Now we have $T(t\exp(-t)) = t$
so we obtain
$$(1-t)^{2n+1} \frac{1}{(1-t)^{2n+1}}
\sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k
\\ = \sum_{k=0}^n \left\langle\!\! \left\langle n\atop k
\right\rangle\!\! \right\rangle t^k$$
as claimed. We get for $n=0$, that the coefficient on the singleton $z$
is $(1-t)$ for a total of $1-t+ (1-t) (Q_0(t\exp(-t))-1) = 1-t + (1-t)
t/(1-t) = 1$ which is the correct value.
