# A binomial determinant formula: a new variant

In a previous MO question, 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 $$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i Let's define $$\Psi_n:=\prod_{i=1}^n x_i^2 \prod_{i Recall the elementary symmetric functions $$e_k=e_k(x_1^2,\dots,x_n^2)$$. 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.

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 $$(a_ne_0+a_{n-1}e_1+\cdots+a_1e_{n-1}+e_n)\prod_{i.
$$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, 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 $$(e_n-e_n+x_i^2e_{n-1}-\cdots\pm x_i^{2n}e_0)/e_n=(e_n\pm(x_i^2-x_1^2)\cdots(x_i^2-x_n^2))/e_n=1.$$