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.