Consider the standard Vandermonde $V(x\_1, \ldots, x\_n) = \prod\_{i < j} (x\_i - x\_j)$.
I am intersted in the calculation of the following expression for fixed $k$:
$$\sum\_i (x\_i)^k (d/dx\_i)^k V(x\_1 , \ldots , x\_n).$$
My guess is that it equals $c \cdot V(x\_1, \ldots, x\_n)$
where $c$ is an expression depending on $k$ and $n$ but not on the $x\_i$'s. Is it true ? If yes, what is this constant $c$?


I think, if it is true, then it is pretty well-known.
Would you be so kind to provide with the answer and/or proof and/or references?
What can be the context people study it? Symmetric functions? Quantum Calogero-Moser?



A sum involving derivatives of Vandermonde
Perhaps a little Mathematica program will help us form a conjecture. For $k \geq n$ the answer is $0$, so we list your $c$ for $k < n$, as follows:



```
In[1]:= V[x_] := Product[x[[i]] - x[[j]], {i,1,Length[x]},{j,1,i-1}]
In[2]:= V[x/@Range[3]]
Out[2]= (-x[1]+x[2]) (-x[1]+x[3]) (-x[2]+x[3])
In[3]:= s[k_,x_] := Sum[x[[i]]^k*D[V[x],{x[[i]],k}],{i,1,Length[x]}]
In[4]:= Table[s[k,x/@Range[n]]/V[x/@Range[n]], {n,1,6},{k,0,n-1}]//Simplify//TableForm
Out[4]//TableForm=
1
2       1                               
3       3       2                       
4       6       8       6               
5       10      20      30      24      
6       15      40      90      144     120

```

Each row is showing a fixed $n$, and each column a fixed $k$. The first column is easy. The second one is triangular number $n(n+1)/2$, [OEIS](http://www.oeis.org/) tells me the third one is $2 {n \choose 3}$ and the fourth one $n(n+1)(n+2)(n+3)/4$. After a bit of experimentation:



```
In[19]:= c[n_,k_] := Product[(n-k+i),{i,0,k}]/(k+1)
In[20]:= Table[c[n,k],{n,1,6},{k,0,n-1}]//TableForm
Out[20]//TableForm=
1
2       1                               
3       3       2                       
4       6       8       6               
5       10      20      30      24      
6       15      40      90      144     120

```

There we have it, your constant seems to be $n (n-1) \cdots (n-k)/(1+k)$. A true combinatorist should now be able to prove that this is really so. I already believe it.


