The [Eulerian numbers](https://en.wikipedia.org/wiki/Eulerian_number) 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.
> 
> 
> 



a new representation for Eulerian numbers?
Quiet a short algebraic proof may go as follows (I cite my own [manuscript](https://arxiv.org/abs/1512.07136)). 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.](http://mathworld.wolfram.com/EulerianNumber.html)


