# A variant of numeric Vandermonde which failed symbolically?

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

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 $$\det\big[(j+f(i))^{i-1}\big]_{i,j=1}^n = c_{1,0}c_{2,1}\cdots c_{n,n-1}\cdot \det\big[j^{i-1}\big]_{i,j=1}^n=\det\big[j^{i-1}\big]_{i,j=1}^n.$$
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 $$B_n$$.