The value of the second-order Eulerian polynomials at $x = \frac{-1}{2}$.

Question

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 = \frac{1}{2} $ 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 = -\frac{1}{2}$ (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: $$ \left\langle\! \left\langle x \right\rangle\! \right\rangle_n = (n+1)!\, (1-t)^{2n + 1}\, [x^{n+1}]\, \operatorname{Reversion}_{x}(x + t - t \exp(x)) \quad (n \ge 0) $$

Answer 1

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.