We first make a few definitions, seemingly out of the blue (they are introduced/defined in this paper).

Let $F^0_{a}(z) = (1-z)^{-1}$ and define recursively $$ F^{k+1}_{a}(z) = z^{a-1} \frac{d^a}{dz^a} F^{k}_{a}(z), \qquad k\geq 0. $$ Let $G_{a,b}(z)= \left(\prod_{k=0}^{b-1}\frac{k!}{(a+k)!} \right) z^{1-a}(1-z)^{ab+1} F^{b}_{a}(z)$.

Here are the first few values of $G_{a,b}(z)$ as $a=1,2,3$, $b=1,\dotsc,4$. \begin{matrix} 1 & 1 & 1 & 1 \\ 1 & z+1 & z^2+3 z+1 & z^3+6 z^2+6 z+1 \\ 1 & z^2+3 z+1 & z^4+10 z^3+20 z^2+10 z+1 & z^6+22 z^5+113 z^4+190 z^3+113 z^2+22 z+1 \end{matrix}

The combinatorialist might then be inclined to make the following conjecture (verified for $1\leq a,b\leq 4$):
$$
G_{a,b}(z) = \sum_{T \in \mathrm{SYT}(a^b)} z^{des(T)-b+1}.
$$
Here, $\mathrm{SYT}(a^b)$ denotes the set of standard Young tableaux of rectangular shape $(a,a,\dots,a)$, and $des(T)$ is the number of *descents*.

**Question: What is going on?** Can this be proved?

I have never seen anything like this, but the paper above introduces these as generalizations of Eulerian polynomials and Narayana polynomials, so the first two columns (and rows) of the table above agrees (by finding the appropriate pages in Stanley's EC2). If the conjecture is true, then $G_{a,b}(z)$ can be shown to be $h^*$-polynomials for some nice family of polytopes.

Mathematica code for the function $G_{a,b}(z)$ is as follows:

```
GG[a_, b_] := (Product[(k)!/(a + k)!, {k, 0, b - 1}]) z^(1 - a) (1 -
z)^(a b + 1) Nest[Simplify[z^(a - 1) D[#, {z, a}]] &, 1/(1 - z),
b];
```

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