For $n\in\mathbb{N}$ and $m=\lfloor\frac{n}2\rfloor$, consider the $n\times n$ skew-symmetric matrix $A\_n$ where each entry in the first $m$
sub-diagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Let $I\_n$ be the $n\times n$ identity matrix. 


Next, construct the matrix $M\_n=A\_n+xI\_n$. For example, we have
$$M\_3=\begin{pmatrix} x&-1&1 \\ 1&x&-1 \\ -1&1&x
\end{pmatrix} \qquad \text{and} \qquad
M\_4=\begin{pmatrix} x&-1&-1&1 \\ 1&x&-1&-1 \\ 1&1&x&-1 \\ -1&1&1&x
\end{pmatrix}.$$



> 
> **QUESTION.** Is the following true? Experiments suggest to be so.
>  $$\det(M\_n)=\sum\_{k=0}^m\binom{n}{2k}x^{n-2k}.$$
> 
> 
> 



Determinant of "skew-symmetric" matrices
For $n$ odd, $M\_n$ is an $n\times n$ circulant matrix, and so Theorem 17 in Krattenthaler's marvellous [text](https://arxiv.org/pdf/math/9902004.pdf) applies. Denoting by $w$ a primitive $n$th root of unity, it gives 
$$\det M\_n=\prod\_{i=0}^{n-1} (x-w^i-w^{2i}-\dots -w^{mi}+w^{(m+1)i}+\dots +w^{(n-1)i}),$$
something that should not be hard to simplify.


For $n$ even, you still have a special Hankel matrix, for which again there are general methods in [loc.cit.].


