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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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  • $\begingroup$ @SamHopkins Ah, it does not! It is relevant later, as the number of times one applies the recursion.... $\endgroup$ – Per Alexandersson Sep 3 '19 at 14:40
  • $\begingroup$ Could you recall what a descent is. $\endgroup$ – Abdelmalek Abdesselam Sep 3 '19 at 14:53
  • $\begingroup$ Your definition of $F^0_a$ is missing $a$ $\endgroup$ – Marcel Sep 3 '19 at 15:02
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    $\begingroup$ @SamHopkins, yes one can indeed use MacMahons formula, although I have not tried to get anything explicit with it yet. $\endgroup$ – Per Alexandersson Sep 3 '19 at 18:52
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    $\begingroup$ @AbdelmalekAbdesselam We say that $j$ is a descent if $j+1$ appear in a row with strictly larger index than $j$ (a strictly lower row if you are not French). Des count the number of descents. $\endgroup$ – Per Alexandersson Sep 3 '19 at 18:52
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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 $$z^{a-1}\frac{d^a}{dz^a} z^{m+a-(b-1)-1} = \prod_{j=1}^a(m+j-b) z^{m+a-1-b}.$$

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  • $\begingroup$ That is very neat! This gives a combinatorial interpretation of the coefficients in this OEIS entry, if I am not mistaken. oeis.org/A245173 $\endgroup$ – Per Alexandersson Sep 5 '19 at 7:51

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