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Define polynomials $S_{k,n}(x)$ by $$\sum_{j\ge 0}\binom{k+j}{k}^n x^j=\frac{ S_{k,n}(x)}{(1-x)^{k n+1}},$$ which for $k=1$ reduce to the Eulerian polynomials.

Computatios suggest that $$S_{k,n}(1)=\frac{(kn)!}{k!^n}.$$

Any idea how to prove this? Have these polynomials been studied in the literature?

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A simple idea is to use the following "Abelian-type theorem" (with an easy proof):

Suppose that $a(x)=\sum\limits_{j\geqslant 0}a_j x^j$ and $b(x)=\sum\limits_{j\geqslant 0}b_j x^j$ converge when $|x|<1$, with $b_j>0$, and $\lim\limits_{x\to 1^-}b(x)=\infty$. If $\lim\limits_{j\to\infty}\dfrac{a_j}{b_j}=\lambda$ exists then $\lim\limits_{x\to 1^-}\dfrac{a(x)}{b(x)}=\lambda$.

Take $a_j=\binom{k+j}{k}^n$, $b_j=\binom{kn+j}{kn}$, and use $\lim\limits_{j\to\infty}\frac1{j^m}\binom{m+j}{m}=\frac1{m!}$, $\sum\limits_{j\geqslant 0}\binom{m+j}{m}x^j=(1-x)^{-m-1}$.

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  • $\begingroup$ Thank you. This is a nice proof! $\endgroup$ Mar 4 at 9:15

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