In a previous [MO question](https://mathoverflow.net/questions/154250/a-binomial-determinant-fomula), the OP asks a proof for $\det\_{1\leq i,j\leq n}\left(\binom{i}{2j}+\binom{-i}{2j}\right)=1$. Subsequently, Gjergji Zaimi generalized the problem to
$$\det\_{1\le i,j\le n}\left( \binom{x\_i}{2j}+ \binom{-x\_i}{2j}\right)=\prod\_{i=1}^n x\_i^2 \prod\_{i<j} (x\_j^2-x\_i^2) \prod\_{j=1}^n \frac{2}{(2j)!}.$$
Let's define
$$\Psi\_n:=\prod\_{i=1}^n x\_i^2 \prod\_{i<j} (x\_j^2-x\_i^2) \prod\_{j=1}^{\mathbf{n+1}} \frac{2}{(2j)!}.$$
Recall the [elementary symmetric functions](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) $e\_k=e\_k(x\_1^2,\dots,x\_n^2)$. Now, I would like ask:



> 
> **Question 1.** Is it true that there exist positive integers $a\_0, a\_1, \dots, a\_n$ such that
>  $$\det\_{1\le i,j\le n}\left( \binom{x\_i}{2j+2}+ \binom{-x\_i}{2j+2}\right)=
> (a\_0+a\_1e\_1+\cdots+a\_{n-1}e\_{n-1}+a\_ne\_n)\cdot\Psi\_n.$$
> **Question 2.** What are these coefficients $a\_0, a\_1, \dots, a\_n$?
> 
> 
> 


**NOTE.** For any $n$, we observe that $a\_n=1, a\_{n-1}=11, a\_{n-2}=661, a\_{n-3}=151451$. These sequence does not appear on [OEIS](https://oeis.org/).



A binomial determinant formula: a new variant
Let $M$ be the matrix in question. The entry $M\_{ij}$ is of the form $\frac{2x\_i^2}{(2j+2)!}p\_j(x\_i)$ for some even polynomial $p\_j$ of degree $2j$. After factoring out the $2x\_i^2$ terms from each row and the $\frac{1}{(2j+2)!}$ terms from each column, we are left with the matrix $(p\_j(x\_i))\_{i,j=1}^n$. This can be column reduced to a matrix $P$ with rows $(x\_i^2+a\_1,x\_i^4-a\_2,x\_i^6+a\_3,\dots)$, where $a\_1=11$, $a\_2=661$, $a\_3=151451$, etc. It remains to show that $\det P$ is $(a\_ne\_0+a\_{n-1}e\_1+\cdots+a\_1e\_{n-1}+e\_n)\prod\_{i<j}(x\_j^2-x\_i^2)$.


$P$ breaks into the sum of a Vandermonde matrix $V$ in the indeterminates $x\_i^2$ and a rank-$1$ matrix $uv^\top$, where $u=(1,\dots,1)^\top$ and $v=(a\_1,-a\_2,a\_3,\dots,(-1)^{n-1}a\_n)^\top$. By the matrix determinant lemma, $\det P=\det V(1+v^\top V^{-1} u)$. Since $\det V=e\_n\prod\_{i<j}(x\_j^2-x\_i^2)$, it suffices to show that $V^{-1}u=(e\_{n-1},-e\_{n-2},\dots,(-1)^{n-1} e\_0)^\top/e\_n$. Indeed, the $i$th entry of $V(e\_{n-1},-e\_{n-2},\dots,(-1)^{n-1} e\_0)^\top/e\_n$ is
$$(e\_n-e\_n+x\_i^2e\_{n-1}-\cdots\pm x\_i^{2n}e\_0)/e\_n=(e\_n\pm(x\_i^2-x\_1^2)\cdots(x\_i^2-x\_n^2))/e\_n=1.$$


