Recall that the permanent of a matrix $A=[a_{j,k}]_{1\le j,k\le n}$ is given by
$$\mathrm{per}(A)=\sum_{\tau\in S_n}\prod_{j=1}^na_{j,\tau(j)}.$$

Let $n$ be an odd integer greater than one. In 2019 I studied
$$t(n):=\frac{(-1)^{(n-1)/2}}n\mathrm{per}\left[\tan\pi\frac{j+k}n\right]_{1\le j,k\le n-1}.$$ 
I have proved that $t(n)\in\mathbb Q$, and that 
$$t(p)\in\mathbb Z\ \ \ \text{and}\ \ \ t(p)\equiv(-1)^{(p+1)/2}2\pmod p$$
for every odd prime $p$. 

**Question.** Whether for any odd integer $n>1$ the number $t(n)$ is always a positive integer?

My numerical computation inducates that
$$t(3)=1,\ t(5)=13,\ t(7)=285,\ t(9)=16569,
\ t(11)=1218105,\ t(13)=164741445.$$
I guess the question has a positive answer.