Iterated derivative and rectangular standard Young tableaux

Question

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];

Answer 1

The formula follows from a result in EC2 (Stanley's "enumerative Combinatorics" Vol. 2) -- Chapter 7, equation (7.96), which is a result from the expansion of Schur functions in terms of fundamental quasisymmetric functions. The equation reads: $$\sum_{m\geq 0} s_{\lambda/\mu} (1^m) z^m = \frac{\sum_T z^{des(T)+1}}{(1-z)^{n+1}},$$ where $T$ ranges over all SYTs of shape $\lambda/\mu$. In the case of rectangular shape $(a^b)$, the hook-content formula (equivalent to MacMahon's here) gives $$s_{a^b} (1^m) = \prod_{i=1}^{b}\prod_{j=1}^a \frac{m+j-i}{i+j-1},$$ and we notice that the product of hook-lengths in the denominator is exactly the constant $\prod_{k=0}^{b-1} \frac{k!}{(a+k)!}$. So after canceling the common factors on both sides, Per's identity becomes equivalent to: $$\sum_{m\geq 0} \prod_{i=1}^b \prod_{j=1}^a (m+j-i) z^{m+a-b-1} = F_a^b(z)$$ Now this follows easily by induction on $b$, since $$z^{a-1}\frac{d^a}{dz^a} z^{m+a-(b-1)-1} = \prod_{j=1}^a(m+j-b) z^{m+a-1-b}.$$