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 $$t(p)\in\mathbb Z\ \ \ \text{and}\ \ \ t(p)\equiv-2\pmod p$$ 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 indicates 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.

Your comments are welcome!