I'd like to find the determinant of the matrix $A_n$ given by $(A_n)_{i,j}={n\choose |i-j|}$ for all $n\in\mathbb{Z}_{\ge 1}$ and $i,j\in\{1,2,\ldots,n\}$. Here is what I know so far:
- $\det(A_n)=0$ if and only if $6\mid n$.
- $2^n-1$ is an eigenvalue of all $A_n$, with eigenvector $(1,1,\ldots,1)$
- If $n$ is prime, then $\det(A_n)\equiv 1\pmod n$
- If $n+1$ is prime and $n>2$, then $\det(A_n)\equiv 0\pmod {n+1}$
As you noted, your matrix is circulant. In particular, if we take $P$ to be the permutation matrix described here, then we have $$ A = \sum_{k=0}^{n-1} \binom nk P^k = (I + P)^n - I $$ Thus, we may compute your determinant as the product of all eigenvalues, namely $$ \det(A) = \prod_{j=0}^{n-1} [(1 + e^{(2 \pi j/n) i})^n - 1] $$