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The Eulerian numbers enjoy many different presentations among which I write the two-variable recursive definition: $A(n,0)=1$ and $A(n,k)=0$ for $k<0$ so that $$A(n,k)=(k+1)A(n-1,k)+(n-k)A(n-1,k-1).$$ However, my curiosity is regarding a certain "vanishing-variables" formulation (summation) over the symmetry (permutation) groups $\mathfrak{S}_n$:

QUESTION. For $0\leq k\leq n$ and a set of variables $x_1, x_2, \dots, x_{n+1}$, experiments suggest $$A(n,k)=\sum_{\pi\in\mathfrak{S}_{n+1}}(x_{\pi(1)}+\cdots+x_{\pi(k+1)})^n\prod_{i=1}^n\frac1{x_{\pi(i)}-x_{\pi(i+1)}}.$$ It seems correct to me. Proof or disproof? I am still hoping and waiting.

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  • $\begingroup$ What is $k$? Should it be $i$? $\endgroup$ – Ira Gessel Dec 15 '16 at 0:17
  • $\begingroup$ Thanks, but it is not a typo. The numerator does not depend on $i$. The $k$ reflects the one on the LHS. I rewrote the RHS so as not to confuse anyone. $\endgroup$ – T. Amdeberhan Dec 15 '16 at 0:19
  • $\begingroup$ My mistake—sorry. $\endgroup$ – Ira Gessel Dec 15 '16 at 0:22
  • $\begingroup$ Are you sure about the "rt.representation-theory" tag? That usually refers to linear representations of group, not of the kind of representation you're asking for; it's true that there's a lot of combinatorics bound up in the representation theory of the symmetric group, but as it stands that combinatorics is not visible here. $\endgroup$ – Noam D. Elkies Dec 15 '16 at 1:31
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    $\begingroup$ Shouldn't Theorem 3.1 in arxiv.org/abs/math/0507163v1 (applied to $n = n+1$ and $x_i = \begin{cases} 1, & \text{ if } i \leq k+1; \\ 0, & \text{ if } i > k+1 \end{cases}$) reveal that your right hand side is $n! \operatorname{Vol} P_{n+1}\left(\underbrace{1,1,\ldots,1}_{k+1 \text{ entries}}, \underbrace{0,0,\ldots,0}_{n-k \text{ entries}}\right) = n! \operatorname{Vol} \Delta_{k+1,n+1}$ ? And identifying this with the left hand side is the classical geometric interpretation of Eulerian numbers? $\endgroup$ – darij grinberg Dec 15 '16 at 3:34
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Quiet a short algebraic proof may go as follows (I cite my own manuscript). For a polynomial $f(x_1,\dots,x_{n+1})$ of degree at most $n$ we define a linear operator $$\Phi[f]=\text{Sym}\, \frac{f(x_1,\dots,x_{n+1})}{(x_1-x_2)(x_2-x_3)\dots (x_n-x_{n+1})},$$ where $\text{Sym}\, g(x_1,\dots,x_{n+1})=\sum g(x_{\pi_1},\dots,x_{\pi_{n+1}})$, the summation is over all $(n+1)!$ permutations of the variables. Note that $\Phi[f]$ is actually a polynomial of degree at most 0 (all factors in the denominator are cancelled, see Ilya's comment), in other words, it is a scalar, and the linear operator $\Phi$ is a linear functional.

Several observations about $\Phi$.

1) If $f$ is divisible by $y:=x_1+\dots+x_{n+1}$, then $\Phi[f]=y\Phi[f/y]=0$. (The same holds for any symmetric factor $y$.)

2) Assume that $f=(x_{k}-x_{k+1})h(x_1,\dots,x_k)g(x_{k+1},\dots,x_{n+1})$. Then $$\Phi[f]=\begin{cases}\binom{n+1}k\Phi_k[h]\Phi_{n+1-k}[g],\,\text{if}\deg h=k-1\,\text{and} \deg g=n-k\\ 0,\,\text{otherwise}\end{cases}.$$

3) If $f$ is divisible by $(x_1+\dots+x_k)(x_{k}-x_{k+1})$, then $\Phi[f]=0$. Indeed, it suffices to consider the polynomials $f=(x_1+\dots+x_k)(x_{k+1}-x_k)h(x_1,\dots,x_k)g(x_{k+1},\dots,x_{n+1})$. For them the claim follows from 1) and 2).

Before proving your Euler numbers claim we prove that $c_k:=\Phi[x_k^n]=(-1)^{k-1}\binom{n}{k-1}$. Induction in $n$. Base is straightforward. Induction step. We have $c_0+\dots+c_n=\Phi(x_1^n+\dots+x_{n+1}^n)=0$, by observation in 1). Next, $$ c_k-c_{k+1}=\Phi((x_k-x_{k+1})(x_k^{n-1}+\dots+x_{k+1}^{n-1})= (-1)^{k-1}\binom{n+1}k $$ by 2) and by the induction assumption. We obtain $n+1$ linear relations on $c_1,\dots,c_{n+1}$ which determine them uniquely and also the numbers $c_i=(-1)^{i-1}\binom{n}{i-1}$ satisfy these relations. This finishes the induction step.

Now we want to calculate $\Phi[(x_1+\dots+x_{k+1})^n]$. By 3) we have $$ \Phi[(x_1+\dots+x_{k+1})^n-(x_1+\dots+x_{k}+x_k)^n+x_k^n-x_{k+1}^n]=0, $$ thus $$\Phi[(x_1+\dots+x_{k+1})^n-(x_1+\dots+x_{k}+x_k)^n]=(-1)^{k}\binom{n+1}k.$$ Next, $$\Phi[(x_1+\dots+x_{k-1}+2x_k)^n-(x_1+\dots+x_{k-2}+3x_{k-1})^n]=(-1)^{k-1}2^n\binom{n+1}{k-1}.$$ Proceeding in this way and summing up we get the standard formula for Eulerian number (1) on this MathWorld page.

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