Consider the polynomial $P(X) =\prod_{i=1}^k (X-x_i)^s$ and let $M$ be the corresponding confluent Vandermonde matrix. Concretely, here is what I mean by that. Define
$$
M^{(0)} = \begin{pmatrix}
1 & 1 & \cdots & 1\\
x_1 & x_2 & \cdots &x_k\\
x^2_1 & x^2_2 & \cdots &x^2_k\\
\vdots\\
x^{ks-1}_1 & x^{ks-1}_2 & \cdots &x^{ks-1}_k
\end{pmatrix}
$$
which is the $ks\times k$ Vandermonde matrix. Let $D=\sum_i \partial/\partial x_i$ be an operator. Define
$$
M^{(n)} = \frac{1}{n!}D^n M^{(0)}
$$
Then $M=(M^{(0)}, M^{(1)}, \cdots, M^{(s-1)})$ which is $ks\times ks$. With that definition, now here is my question:

> Is there a clean formula for the solution $F$ of the system $MF=B$? Equivalently, is $M^{-1}$ known in terms of, say, symmetric functions? For all intents and purposes we can define $M^{(n)}=D^n M^{(0)}$ instead (as confluent Vandermonde is sometimes defined) if it helps.


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**Extra Detail:** Since I have not managed to find much on this topic online, I'm fearful that there might not be a clean answer in general. In my specific application, however, both $F$ and $B$ have additional structure.  Maybe these additional assumptions make life easier.

*Structure of $F$:* Let $F^{(n)}=(F^{(n)}_1, \cdots, F^{(n)}_k)^T$ so that $F=(F^{(0)}, \cdots, F^{(s-1)})$. Given a function $g(x_1, \cdots, x_k)$, consider the following of Dunkl operator
$$
(Kg)(x_1, \cdots, x_k) = g(x_2, x_3, \cdots, x_k, x_1)
$$
which rotates the variables. Then for each $0\leq n\leq s-1$ and $1\leq m\leq k$ I have
$$
F^{(n)}_m(x_1, \cdots, x_k)  = (K^{m-1} F^{(n)}_1)(x_1, \cdots, x_k)
$$
Moreover, $F_1^{(n)}$ is symmetric (rational function) in variables $x_2, \cdots, x_k$.

*Structure of B:* Let $\Delta = \sum_{i=1}^k  x_i^2\partial /\partial x_i$ and $f(x_1, \cdots, x_k)$ a *symmetric homogeneous polynomial of degree $s$* such that $Df=\Delta^{(k-2)s+1}f=0$. Then $B=(b_1, \cdots, b_{ks})^T$ is obtained via the following
$$
\sum_{l=1}^{ks} b_l t^l = t^{2s}\exp(t\Delta)f
$$

> All I really need is to find a formula for functions $F_1^{(0)},
F_1^{(1)}, \cdots, F_1^{(s-1)}$ in terms of $B$, or even better, $f$.