**Update:** see below

Let $M$ be an $n\times n$ matrix that's constructed as follows. Construct the right-most column of $M$ as $[\alpha_1(x_1),\cdots,\alpha_n(x_n)]^T$ for some class of fixed functions $\alpha_i(x)$. Now, moving left, each column of $M$ is the derivative of the last. So the next column is $[\alpha_1'(x_1),\cdots,\alpha'_n(x_n)]^T$, and the very first column is $[\alpha_1^{(n-1)}(x_1),\cdots,\alpha^{(n-1)}_n(x_n)]^T$. I'm curious if there's a name for such matrices. Specifically, I'm wondering about the determinant of such matrices: $G(x_1,\cdots,x_n)=\det(M(x_1,\cdots,x_n))$. As Jose rightfully pointed out when all variables are set equal we get the usual Wronskian.

I'm particularly curious about $\alpha_i(x)= x^{d_i}/(d_i)!$ for some decreasing positive integer sequence $d_i$. A similar example would be when $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, with the understanding that $M_{ij}$ is zero once we are taking negative factorials. This is starting to look like an offset Vandermonde matrix, or a *confluent* Vandermonde. The determinant isn't quite a Schur polynomial but perhaps a known generalization? Is there a connection of these determinants to some well studied class of polynomials (it seems unlikely they will be symmetric)?

Note that when written as $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, we have that if $|d|=\sum_{i=1}^nd_i$, then when $x_1=x_2=\cdots=x_n=1$, $|d|!\det(M)$ is equal to the number of standard Young tableau of shape $(d_1,d_2,\cdots,d_n)$.

Alternatively, is there a way to write $\det(M)=C\prod_{i=1}^dx_{i}^{d_i}\prod_{i\neq j}(x_i-x_j)^{a_ib_j}?$ (or possibly more complicated). Unfortunately numerical evidence suggests that fixing a variable, say $x_1$ and thinking of the determinant as a polynomial in $x_1$ gives wildly complex expressions for its roots. So the $(x_i-x_j)$ factors aren't quite right either. I've seen "generalized Vandermondes" here but unfortunately it looks like the structure in the link is simpler.

As an example, take $n=4$, and $d_i=9-2i$. Then the resulting matrix is:

\begin{pmatrix} \frac{x_1^4}{4!} & \frac{x_1^5}{5!} & \frac{x_1^6}{6!} & \frac{x_1^7}{7!}\\ \frac{x_2^2}{2!} & \frac{x_2^3}{3!} & \frac{x_2^4}{4!} & \frac{x_2^5}{5!}\\ 1 & x_3 & \frac{x_3^2}{2!} & \frac{x_3^3}{3!}\\ 0 & 0 & 1 & x_4 \end{pmatrix}

The determinant equals:

$$\frac{1}{302400}x_1^4 x_2^2 \left(10 x_1^3 x_2-30 x_1^3 x_3-70 x_1^2 x_2 x_4+210 x_1^2 x_3 x_4-21 x_1 x_2^3+105 x_1 x_2^2 x_4+210 x_1 x_3^3-630 x_1 x_3^2 x_4+105 x_2^3 x_3-525 x_2^2 x_3 x_4-350 x_2 x_3^3+1050 x_2 x_3^2 x_4\right)$$

and it looks like if we single out $x_1$ to make a polynomial in $x_1$, we get complex roots involving $x_2,x_3,x_4$. So, if there is a way to write the determinant as the sum over something nice (not involving signs of permutations), I would also be very interested in see that.

Context: These kinds of determinants come up when counting Young Tableau via integrals over sub-simplices, (specifically (4.1),(4.2)). It's a surprisingly powerful way for calculating the number of non-standard Young tableau shapes. I'm just wondering if there's a neater (representation-theoretic?) interpretation of where these determinants arise from.

**Edit:**

Consider the matrix $\mathcal{M}=\left[\frac{t_i^{x+j-2}}{(x+j-2)!}\right]_{i,j=1^d}$. It's determinant is:

$$\mbox{det}(\mathcal{M})=\frac{1}{\prod_{j=1}^n(x+j-2)}\prod_{i=1}^d t_i^{x-1} \prod_{1\leq i< j\leq d}(t_j-t_i),$$

where we recognize the vandermonde determinant. Then my matrix $M$, and it's determinant can be calculated from this by taking an appropriate amount of derivatives of $t_i$ in each row.

So I think this does have something to do with Schur functions afterall. The Jacobi Trudi identity states that:

$s_{\lambda}=\det(h_{\lambda_i+j-i})_{i,j=1}^d.$

The exponential specialization of $h_k$ to parameter $t$ gives $t^k/k!$. With $t=1$, one gets that the exponential specialization the Schur polynomial above gives the identity:

$$f^{\lambda}=\det\left[\frac{1}{(\lambda_i+j-i)!}\right]_{i,j=1}^d.$$

So the question is, is it sensible to generalize exponential specialization to get an identity of the above form for $t_1,t_2,\cdots,t_d$ instead of just $t_1=\cdots=t_n=t=1$?