Start with a palindromic sequence of integers $(a_0, a_1, \ldots, a_{n+1})$, i.e. $a_j=a_{n+1-j}$, and put $a_j:=0$ for $j<0$ and $j>n+1$. You may readily guess that the choice of the binomial coefficients $a_j=\binom{n+1}j$ has been at the origin for this question.

So we consider the $n \times n$-matrix $B$ whose $\left(i, j\right)$-th entry is $a_{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

E.g. for $n=4$ and the sequence $(1,2,3,3,2,1)$, this would be $$ B = \left(\begin{array}{rrrrr} 2& 3& 1 & 0 \\ 1 & 3& 2& 0 \\ 0 & 2& 3 & 1 \\ 0 & 1 & 3& 2 \end{array}\right).$$

Now what about the * eigenvalues of such a matrix*? In the case of binomial coefficients, they turned out to be integers in the cited question, and more precisely $2^1, 2^2, \ldots, 2^n$. In the general case, one might expect the characteristic polynomial of $B$ to be irreducible, apart from the trivial right eigenvector $(1,\dots,1)$ with eigenvalue $s:=\sum_{j=0}^{n/2} a_j$ for even $n$. But I have found that

*of degrees $\lfloor\frac {n+1}2\rfloor$ and $\lfloor\frac {n-1}2\rfloor$. And of course I wonder why.*

**it always splits into two (generally irreducible) factors**Moreover, it appears that the absolute terms of those two factoring polynomials have an integer ratio $R$, which is in fact just a linear combination of the $a_j$'s: $$R=\sum_{j\in\mathbb Z} (-1)^{j+1}a_{\lfloor\frac {n}2\rfloor+2j}.$$ This is about the easiest way to write this finite sum without having to distinguish between even and odd $n$. In fact, for odd $n=2k+1$, half of the terms cancel out by palindromicity, making this $$R=-a_k +2 a_{k-2}-2 a_{k-4}+2 a_{k-6}-+\cdots,$$ while for even $n=2k$, all the different $a_j$ (thus, say, $a_0,...,a_k$) occur with signs $-++--++-\cdots$, viz. $$R=-a_k + a_{k-1}+ a_{k-2}- a_{k-3}-++-\cdots.$$

Some examples for small $n$, writing the initial sequence as $(a,b,c,b,a)$ etc.: for $n=3$, we have $$P(x)=\Bigl[x-\color{blue}{b}\Bigr]\Bigl[x^2-(b+c)x+\color{blue}{b}\underbrace{(2a-c)}_R\Bigr] $$

For $n=4$, $$P(x)=\overbrace{\Bigl[x-(a+b+c)\Bigr]}^{\text{trivial}}\Bigl[x+\color{blue}{(a-b)}\Bigr]\Bigl[x^2-cx+\color{blue}{(a-b)}\underbrace{(a+b-c)}_R\Bigr] $$

For $n=5$, $$P(x)=\Bigl[x^2+(-a+b+c)x+\color{blue}{(-ab+bc-ad)}\Bigr]\cdot\\ \Bigl[x^3-(a+b+c+d)x^2+(ab-bc+bd+cd)x+\color{blue}{(-ab+bc-ad)}\underbrace{(2b-d)}_R\Bigr] $$

For $n=6$, $$P(x)=\overbrace{\Bigl[x-(a+b+c+d)\Bigr]}^{\text{trivial}}\cdot\Bigl[x^2 + (a - c)x + \color{blue}{(a^2-b^2 + bc-ad)}\Bigr]\cdot\\ \Bigl[x^3-(b+d)x^2+(b^2-c^2+cd-ab)x+\color{blue}{(a^2-b^2 + bc-ad)} \underbrace{(-a+b+c-d)}_R\Bigr] $$

As a by-product it follows that if the determinant of $B$ is divided by $R$ (or, for even $n$, by $Rs$), we remain with a square.

Sadly, there do not seem to exist matrices $M_1$ and $M_2$ of respective sizes $\lfloor\frac {n+1}2\rfloor\times \lfloor\frac {n+1}2\rfloor$ and $\lfloor\frac {n-1}2\rfloor\times \lfloor\frac {n-1}2\rfloor$ with entries depending in an "easy" (i.e. linear) way of the $a_j$ such that for a suitable matrix $V$ $$B=V^{-1}\pmatrix{M_1& 0\\ 0 & M_2}V\quad \text{ resp. (for even $n$)}\quad B=V^{-1}\pmatrix{s&0&0\\ 0 &M_1&0\\0& 0 & M_2}V. $$

So how to prove this reducibility of the characteristic polynomial and the conjectured ratio of the absolute terms?