A Cauchy type determinant $\det ((x_i+x_j)^{n-1})$ Let $x_1$, $\ldots$, $x_n$ $n$ variables. The $n\times n$ determinant
$$ \det \left( (x_i+x_j)^{n-1}\right)$$
is a symmetric homogenous polynomial in $x_1$, $\ldots$, $x_n$ of degree $n(n-1)$, $0$ when $x_i = x_j$, so divisible by $\prod_{i< j} (x_i-x_j)^2$, so it must be an (integer) constant times $\prod_{i< j} (x_i-x_j)^2$.  I wonder what that constant for a general $n$. I know that it is $\ne 0$, and its sign is  $(-1)^{\left[\frac{n}{2}\right]}$.
$\bf{Added}$ The quick solution is to notice (Newton binomial) that the matrix is the product of two matrices, one a Vandermonde, the other a modified Vandermonde.  This in fact appears in Faddeev & Sominsky -- Problems in Higher Algebra #293 a).
 A: The absolute value of the constant is $\prod_{k=0}^{n-1} \binom{n-1}{k}$, which is A001142 up to an off-by-one shift. Here's one proof; maybe someone can find a better one.
Let $\vec{v}_j = ((x_i+x_j)^{n-1})_{i=1}^n$. By the binomial theorem, $\vec{v}_j = \sum_{k=0}^{n-1} \binom{n-1}{k} x_j^{n-1-k} (x_i^k)_{i=1}^n$. By multilinearity of the determinant,
\begin{align*}
  \det(\vec{v}_1, \ldots, \vec{v}_n)
  &= \sum_{0 \leq k_1, \ldots, k_n \leq n-1} \binom{n-1}{k_1} \cdots \binom{n-1}{k_n} x_1^{n-1-k_1} \cdots x_n^{n-1-k_n} \det(x_i^{k_j})_{i,j=1}^n.
\end{align*}
Let $x_i = t^{i-1}$ be the principal specialization. The $t$-degree of the summand is
\begin{align*}
  &(n-1-k_1) \cdot 0 + (n-1-k_2) \cdot 1 + \cdots + (n-1-k_n) \cdot (n-1) \\
  &+(1-1) \cdot k_{[1]} + (2-1) \cdot k_{[2]} + \cdots + (n-1) \cdot k_{[n]}
\end{align*}
where $k_{[1]} \leq \cdots \leq k_{[n]}$ is the weakly increasing rearrangement of $k_1, \ldots, k_n$. This is uniquely maximized when $k_n=0, k_{n-1}=1, \ldots, k_1=n-1$ so $k_{[1]} = 0, k_{[2]} = 1, \ldots, k_{[n]} = n-1$. Hence the top-degree coefficient is $\binom{n-1}{n-1} \binom{n-1}{n-2} \cdots \binom{n-1}{0} = \prod_{k=0}^{n-1} \binom{n-1}{k}$. The result follows since $\prod_{1 \leq i < j \leq n} (t^{j-1} - t^{i-1})$ is monic.
