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Let ${\left\langle\matrix {n \cr k}\right\rangle}_r$ denote the $r-$Hoggatt binomials defined by $${{\left\langle\matrix {n \cr k}\right\rangle}_r=\frac{\langle n \rangle_r!}{\langle k \rangle_r! \langle n-k \rangle_r!}}$$ with ${\langle n\rangle}_r!= {\langle 1\rangle}_r\dots{\langle n\rangle}_r$, where ${\langle n\rangle}_r=\binom{n+r-1}{r}$ and let $D$ be the differentiation operator.

The well-known formula $$\frac{1}{k!}D^k {\frac{1}{1-x}}=\sum_{n\geq0}\binom{n+k}{k}x^n$$ generalizes to a formula for the generating function of the $r-$Hoggatt binomials

$$\prod_{j=0}^{r-1}\frac{j!}{(k+j)!}(x^{r-1}D^r)^k \frac{1}{1-x}=\sum_{n\geq0}{\left\langle\matrix {n+k \cr k}\right\rangle}_r x^{n+r-1}.$$

An explicit expression of the $r-$Hoggatt binomials is $${\left\langle\matrix {n \cr k}\right\rangle}_r= \prod_{1 \le i \le k} \frac{(n - i + 1)^{\overline{r}}}{(k - i + 1)^{\overline{r}}}$$ where $x^{\overline{r}}$ denotes the rising factorial $x(x + 1) \dots (x + r-1).$

The right-hand side can be interpreted as the number of semistandard Young tableaux with shape $r^k.$

Questions: Does the above formula for the generating function of the $r-$Hoggatt binomials occur in the literature about Young tableaux? Has it a combinatorial interpretation or proof?

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    $\begingroup$ "number of semistandard Young tableaux with shape $r^k$"... and entries in $\{1,2,\ldots,n\}$ I presume? If so, then these are also counting plane partitions in a box, and of course there are many formulas for plane partitions (where maybe your differentiation operator generating function appears). $\endgroup$ Commented Aug 21, 2022 at 15:38
  • $\begingroup$ There are two well-known formulas for counting SSYTs: symmetricfunctions.com/schur.htm#schurSpecializations one is the hook-content, and the other is via the Weyl formula for Schur polynomials. However, I have not seen the term "r-Hoggatt" in any paper mentioning Schur polynynomials (but I do not claim to have read all such papers). $\endgroup$ Commented Aug 21, 2022 at 20:14
  • $\begingroup$ Might be worth mentioning that 2-Hoggatt binomials are the Narayana numbers and that the OEIS also has 3-Hoggatt binomials, 4-Hoggatt binomials, 5-Hoggatt binomials, and 6-Hoggatt binomials. I see that the OP is cited in some of these entries. $\endgroup$ Commented Aug 22, 2022 at 2:12
  • $\begingroup$ @Per Alexandersson: Perhaps I should have formulated my question in the following way: Is it pure chance that Hoggatt binomials count SSYTs or is there some relation between SSYTs and powers of differential operators x^(r-1)D^r ? $\endgroup$ Commented Aug 22, 2022 at 6:17

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This is too long for a comment, but I seem to have asked a similar question here.

The related paper is this one, where some iterated derivative is computed, but no mention of SSYT. There should be a refinement of your question, which also takes the descents of a SSYT into consideration.

If you look at the answer on the MO-question I asked, then there is a proof by Greta Panova. I suspect that one can adapt that proof.

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