TITLE: Certain matrices of interesting determinant
QUESTION [9 upvotes]: Let $M_n$ be the $n\times n$ matrix with entries
$$\binom{i}{2j}+\binom{j}{2i}, \qquad \text{for $1\leq i,j\leq n$}.$$

QUESTION. Is this true? There is some evidence. The determinant $\det(M_{2n+1})=0$ and
  $$\det(M_{2n})=(-1)^n\binom{2n}n^22^{n(n-3)}.$$

REPLY [2 votes]: As a follow-up on Fedor's wonderful proof, I would like to pen down just a comment. 
Let's replace $c_j+j$ by $u_j$ and treat $N$ as an indeterminate. Then, for non-negative integers $a$ and $b$, we have the determinantal evaluation
$$\det\left(\binom{N+i+a}{u_{j+b}}\right)_{i,j=0}^{n-1}
=\prod_{i=0}^{n-1}\frac{(N+i+a)!}{(N+n-a-1)!}\binom{N+n+a-1}{u_{i+b}}\prod_{i<j}^{0,n-1}(u_{j+b}-u_{i+b}).$$
The proof follows from the method of condensation; see Desnanot–Jacobi identity.
Remark. The introduction (or generalization) of the parameters $a$ and $b$ is what makes above-mentioned method work, effectively.