A variant of numeric Vandermonde which failed symbolically?

Question

Given some variables $x_1, x_2, \dots, x_n$, the Vandermonde determinant is given by $$V_n(x_1,\dots,x_n):=\det(x_j^{i-1})_{i,j=1}^n=\prod_{i<j}(x_j-x_i).$$ One can take as special cases: $x_j=j$ or $x_j=q^j$.

I was playing around with the following variation $A_n=((i+j-1)^{i-1})_{i,j=1}^n$. It turns out that (numerically) $\det(A_n)=\det(j^{i-1})_{i,j=1}^n$.

In addition, if $B_n=(q^{(i+j-1)(i-1)})_{i,j=1}^n$ then $\det(B_n)$ and $\det(q^{j(i-1)})_{i,j=1}^n$ differ only by a factor of a power of $q$, that is, $q^c$ for some $c$.

On the other hand, if $C_n=(x_{i+j-1}^{i-1})_{i,j=1}^n$ then (unfortunately) $\det(C_n)$ and $V_n$ are vastly different; in fact, $\det(C_n)$ does not factor as $V_n$.

QUESTION 1. Is this true? $\det(A_n)=\det(j^{i-1})_{i,j=1}^n$.

QUESTION 2. What worked right for $A_n$ andd $B_n$ and failed for $C_n$?

Answer 1

Matrix $A_n$ can be generalized to $\big[(j+f(i))^{i-1}\big]_{i,j=1}^n$ for any function $f(i)$ not just $f(i)=i-1$.

Since $$(j+f(i))^{i-1} = \sum_{k=0}^{i-1} c_{i,k}\cdot j^k,$$ where $c_{i,k} := \binom{i-1}k f(i)^{i-1-k}$, we have $$\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = \begin{bmatrix} c_{1,0} & 0 & 0 & \dots & 0\\ c_{2,0} & c_{2,1} & 0 & \dots & 0\\ c_{3,0} & c_{3,1} & c_{3,2} & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_{n,0} & c_{n,1} & c_{n,2} & \dots & c_{n,n-1} \end{bmatrix}\cdot \big[j^{i-1}\big]_{i,j=1}^n.$$ Taking determinants, we get $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$

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For $B_n$ things work even for a simpler reason: $$q^{(i+j-1)(i-1)} = q^{(i-1)^2} q^{j(i-1)}$$ and we can take out a factor of $q^{(i-1)^2}$ from the $i$-th row of matrix $B_n$.

However for matrix $C_n$ neither binomial expansion work, nor we can take out any common factors from its rows. So, despite visual similarity $C_n$ does not possess the properties of $A_n$ or $B_n$.