The permanent $\mathrm{per}(A)$ of a matrix $A$ of size $n\times n$ is defined to be:


$$\mathrm{per}(A)=\sum\_{\tau\in S\_n}\prod\_{j=1}^na\_{j,\tau(j)}.$$


Let
$$A=\left[\tan\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$B=\left[\sin\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$C=\left[\cos\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$D=\left[\sec\pi\frac{j+k}n\right]\_{1\le j,k\le n-1}.$$


Motivated by [Question 402249](https://mathoverflow.net/questions/402249), I found the following


**Conjecture 1.** For any odd integer $n>1,$  

$$(-1)^{(n-1)/2}\mathrm{per}(A)=\frac{2(n!!)^2}{n+1}\sum\_{k=0}^{\frac{n-1}{2}}\frac{(-1)^k}{2k+1}. \tag{1}$$


Numerical calculations show that this is correct for $3 \leq n \leq 33$. See [Question 402249](https://mathoverflow.net/questions/402249) for details.


Inspired by [Question 402572](https://mathoverflow.net/questions/402572), I also found the following identities


**Conjecture 2.** For any odd integer $n>1,$


\begin{align}
(-1)^{(n-1)/2}\mathrm{per}(B)&=\frac{n!}{2^{n-2}(n+1)}，\tag{2}
\\ \mathrm{per}(C)&={\frac{(n-1)!}{2^{n-1}}}\sum\_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}},\tag{3}
\\ \mathrm{per}(D)&= (n-2)!!^2\left( (-1)^{\frac{n+1}{2}}+2n\sum\_{k=0}^{\frac{n-1}{2}} {\frac {\left( -1 \right) ^{k}}{2k+1} }
\right) .\tag{4}
\end{align}
Numerical calculations show that it is correct for $3 \leq n \leq 21$.


**Question.** Are these identities correct? How to prove them?



Permanent identities
Let $\zeta$ be a primitive $n$-th root of unity. Then
$$\prod\_{j=1}^{n-1}(x-\zeta^j)=\frac{x^n-1}{x-1}=1+x+\cdots+x^{n-1}$$
and hence
$$\sigma\_k=\sum\_{1\le i\_1<\cdots<i\_k\le n-1}\zeta^{i\_1+\cdots+i\_k}=(-1)^k$$
for all $k=1,\ldots,n-1$.


Observe that
\begin{align\*}\mathrm{per}[1-\zeta^{j+k}]\_{1\le j,k\le n-1}=&\sum\_{\tau\in S\_{n-1}}\ \prod\_{j=1}^{n-1}(1-\zeta^{j+\tau(j)})
\\=&\sum\_{\tau\in S\_{n-1}}1+\sum\_{\tau\in S\_{n-1}}\ \sum\_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum\_{j\in J}\ (j+\tau(j))}
\\=&(n-1)!+\sum\_{\emptyset \not=J\subseteq \{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum\_{j\in J}j}\sum\_{\tau\in S\_{n-1}}\zeta^{\sum\_{j\in J}\ \tau(j)}.
\end{align\*}
For $\emptyset \not=J\subseteq\{1,\ldots,n-1\}$, clearly
\begin{align\*}&\sum\_{\tau\in S\_{n-1}}\zeta^{\sum\_{j\in J}\ \tau(j)}
\\=&\sum\_{1\le i\_1<\cdots<i\_{|J|}\le n-1}\zeta^{i\_1+\cdots+i\_{|J|}}\sum\_{\tau\in S\_{n-1}\atop\{\tau(j):\ j\in J\}
=\{i\_1,\ldots,i\_{|J|}\}}1
\\=&|J|!(n-1-|J|)!\sigma\_{|J|}=(-1)^{|J|}|J|!(n-1-|J|)!.
\end{align\*}
Therefore
\begin{align\*}\mathrm{per}[1-\zeta^{j+k}]\_{1\le j,k\le n-1}=&(n-1)!+\sum\_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}|J|!(n-1-|J|)!\zeta^{\sum\_{j\in J}\ j}
\\=&(n-1)!\sum\_{k=0}^{n-1}\frac{(-1)^k}{\binom{n-1}k}=(1-(-1)^n)\frac{n!}{n+1}.
\end{align\*}
Then the conjectural identity $(2)$ follows from this since
$$\sin\pi\frac{j+k}n=\frac{e^{-\pi i(j+k)/n}}{2i}\left(e^{2\pi i(j+k)/n}-1\right).$$


The identities $(3)$ can be proved similarly, in fact we have
$$\mathrm{per}[1+\zeta^{j+k}x]\_{1\le j,k\le n-1}=(n-1)!\sum\_{k=0}^{n-1}\frac{x^k}{\binom{n-1}k}.$$ The idea here is slight modification of my way to establish Theorem 1.1 in my preprint *Arithmetic properties of some permanents* available from <http://arxiv.org/abs/2108.07723>.


I admit that the identities $(1)$ and $(4)$ remain open.


