A difficult determinant

Question

(EDIT: I have removed the denominators I had in a previous version as they were superfluous)

The $N\times N$ determinant $$D(a,\vec{b})=\det\left((2N+a+b_j-i-j)!\right)$$ has the nice form $$D(a,\vec{b})=\prod_{j=1}^N(N+a+b_j-j)!\prod_{i=j+1}^N(b_j-b_i-j+i).$$

Since from the definition it is clear that $D(a,\vec{b})$ is antisymmetric in the variables $x_j=(N+b_j-j)$, it should be proportional to the Vandermonde of the $x$.

I would like to know if the generalization where $a$ is allowed to vary with $i$ has a nice expression as well, $$D(\vec{a},\vec{b})=\det\left((2N+a_i+b_j-i-j)!\right)=?$$

This is antisymmetric in both $x_j=(N+b_j-j)$ and $y_i=(N+a_i-i)$, so it should be proportional to both Vandermondes of the $x$ and the $y$...

I know Krattenthaler has this great paper about determinants, but I was not able to find help there.

Answer 1

I have found a solution myself, at least in the case when $a$ and $b$ are partitions.

The determinant can be written as $$ D=\det((x_i+y_j)!)=\det\left( \int z^{x_i+y_j}e^{-z}dz\right)$$ We resort to the Andreief identity, $$ \int dz \det(f_{i}(z_{j}))\det(g_{i}(z_{j}))=N! \det\left(\int f_{i}(z)g_{j}(z)dz\right).$$ This is usually used from left to right, but I used in reverse to write $$D=\frac{1}{N!}\int dz \det(z_{j}^{x_j})\det(z_{j}^{y_i})\prod_{i=1}^Ne^{-z_i}$$ Now, if $x_i=a_i-i+N$ and $y_i=b_i-i+N$ and if $a\vdash n$ and $b\vdash m$ are partitions, then $\det(z_{j}^{x_j})\det(z_{j}^{y_i})=(V(z))^2s_a(z)s_b(z)$, where $V$ is the Vandermonde and $s$ are the Schur functions. Then $$D=\frac{1}{N!}\int dz (V(z))^2s_a(z)s_b(z)\prod_{i=1}^Ne^{-z_i}$$

The Littlewood-Richardson coefficients are defined by $s_as_b=\sum_{\rho\vdash n+m} c^\rho_{ab}s_\rho$. Using them we have $$D=\frac{1}{N!}\sum_{\rho\vdash n+m} c^\rho_{ab}\int dz (V(z))^2s_\rho(z)\prod_{i=1}^Ne^{-z_i}.$$ This is an integral of the Selberg type, and the result is known: $$\int dz (V(z))^2s_\rho(z)\prod_{i=1}^Ne^{-z_i}=N!\frac{d_\rho}{(n+m)!}\prod_{j=1}^N (\rho_i+N-i)!^2 \quad (\ell(\rho)\le N),$$ where $d_\rho$ is the dimension of the irreducible representation of the permutation group labeled by $\rho$.

Therefore, $$D=\frac{1}{(n+m)!}\sum_{\substack{\rho\vdash n+m\\\ell(\rho)\le N}} d_\rho c^\rho_{ab}\prod_{j=1}^N (\rho_i+N-i)!^2.$$

Curiously, there are no Vandermondes in this solution.