# What are the prime divisors of $\det(A_n)$, where $A_n$ is the $n\times n$ matrix given by $(A_n)_{i,j}={n\choose|i-j|}$?

For $$n\in\mathbb{Z}_{\ge 1}$$, let $$A_n$$ be the $$n\times n$$ matrix given by $$(A_n)_{i,j}={n\choose |i-j|}$$. From this post it is clear that $$\det(A_n)=\prod_{k=0}^{n-1}\left[\left(\exp\left(\frac{2\pi k i}{n}\right)+1\right)^n-1\right]=\prod_{k=0}^{n-1}\left(2^n(-1)^k\cos^n\left(\frac{\pi k}n\right)-1\right).$$ Also, $$\det(A_n)$$ is obviously integer for all $$n$$, and $$\det(A_n)=0$$ iff $$6\mid n$$.

Question: What is known about the (prime) divisors of $$\det(A_n)$$?

If this is too broad, I am particularly interested in pairs $$(p,d)$$ where $$p$$ is prime with $$d\mid p-1$$ and $$p\nmid\det(A_{(p-1)/d})$$.

Edit: It seems like the product with cosines in it is integer independent of the exponent, see this question of mine.

Edit 2: The first few values of $$\det A_n$$ are: \begin{align*} \det(A_1) &= 1\\ \det(A_2) &= -3=-(2^2-1)\cdot 1^2\\ \det(A_3) &= 28 = (2^3-1)\cdot2^2\\ \det(A_4) &= -375 = (2^4-1)\cdot5^2\\ \det(A_5) &= 3751 = (2^5-1)\cdot11^2\\ \det(A_6) &= 0\\ \det(A_7) &= 6835648 = (2^7-1)\cdot232^2\\ \det(A_8) &=-1343091375 = -(2^8-1)\cdot2295^2\\ \det(A_9) &= 364668913756 = (2^9-1)\cdot26714^2 \end{align*}

Edit 3: It looks like this is called 'Wendt's determinant'. My entire motivation for asking this question had to do with a possible proof for Fermat's last theorem and this link is apparently known as 'Wendt's theorem'

• It looks like $\det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:\mathbb{Z}\to\mathbb{Z}$. – Mastrem Dec 20 '18 at 21:21
• $m(n)$ must be the product over $0<2k<n$. – metamorphy Dec 20 '18 at 22:03
• @metamorhpy. Oh, nevermind, I got it. Also, if $d\mid k$, then $\det A_d\mid \det A_k$ – Mastrem Dec 20 '18 at 22:16
• – Mastrem Dec 21 '18 at 10:27
• – lhf Dec 21 '18 at 10:28