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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'

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  • $\begingroup$ It looks like $\det(A_n)=(-1)^{n+1}(2^n-1)m(n)^2$ for some $m:\mathbb{Z}\to\mathbb{Z}$. $\endgroup$ – Mastrem Dec 20 '18 at 21:21
  • $\begingroup$ $m(n)$ must be the product over $0<2k<n$. $\endgroup$ – metamorphy Dec 20 '18 at 22:03
  • $\begingroup$ @metamorhpy. Oh, nevermind, I got it. Also, if $d\mid k$, then $\det A_d\mid \det A_k$ $\endgroup$ – Mastrem Dec 20 '18 at 22:16
  • $\begingroup$ I found oeis.org/search?q=232%2C2295&language=english&go=Search $\endgroup$ – Mastrem Dec 21 '18 at 10:27
  • $\begingroup$ See oeis.org/A048954 and oeis.org/A215615 $\endgroup$ – lhf Dec 21 '18 at 10:28

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