Inspired by [Question 402572][1], I consider the permanent of matrices
$$f(n)=\mathrm{per}(A)=\mathrm{per}\left[\operatorname{sgn} \left(\sin\pi\frac{j+2k}{n+1} \right)\right]_{1\le j,k\le n},$$

where $n$ is  a positive integer and $\operatorname{sgn}$ is the sign-function.

When $n=1,2,3,4,5,6,$ the matrices are
$$\left[ \begin {array}{c} -1\end {array} \right],$$
$$\left[ \begin {array}{cc} 0&-1\\ -1&0\end {array}\right],$$
$$\left[ \begin {array}{ccc} 1&-1&-1\\ 0&-1&0
\\ -1&-1&1\end {array} \right]
,$$
$$\left[ \begin {array}{cccc} 1&0&-1&-1\\ 1&-1&-1&0
\\ 0&-1&-1&1\\ -1&-1&0&1
\end {array} \right]
,$$
$$\left[ \begin {array}{ccccc} 1&1&-1&-1&-1\\ 1&0&-1&
-1&0\\ 1&-1&-1&-1&1\\ 0&-1&-1&0&1
\\ -1&-1&-1&1&1\end {array} \right]
,$$
$$\left[ \begin {array}{cccccc} 1&1&0&-1&-1&-1\\ 1&1&
-1&-1&-1&0\\ 1&0&-1&-1&-1&1\\ 1&-1
&-1&-1&0&1\\ 0&-1&-1&-1&1&1\\ -1&-
1&-1&0&1&1\end {array} \right]
.$$
Numerical computation indicates that
\begin{equation}
   \mathrm{per}(A)=
   \begin{cases}
   -T_n&\mbox{if $n$ is odd}\\
   E_n&\mbox{if $n$ is even}
   \end{cases}
  \end{equation}
for $3\leq n \leq 21$, where $T_n$ and $E_n$ are [tangent numbers][2] and [secant numbers][3], i.e.

$$\tan{x}=\sum_{2\nmid n}T_n\frac{x^n}{n!}=1\frac{x}{1!}+2\frac{x^3}{3!}+16\frac{x^5}{5!}+272\frac{x^7}{7!}+7936\frac{x^9}{9!}+\cdots$$
and 

$$\sec{x}=\sum_{2\mid n}E_n\frac{x^n}{n!}=1+1\frac{x^2}{2!}+5\frac{x^4}{4!}+61\frac{x^6}{6!}+1385\frac{x^8}{8!}+50521\frac{x^{10}}{10!}+\cdots$$
respectively.
```
f(1) = -1

f(2) = 1

f(3) = -2

f(4) = 5

f(5) = -16

f(6) = 61

f(7) = -272

f(8) = 1385

f(9) = -7936

f(10) = 50521

f(11) = -353792

f(12) = 2702765

f(13) = -22368256

f(14) = 199360981

f(15) = -1903757312

f(16) = 19391512145

f(17) = -209865342976

f(18) = 2404879675441

f(19) = -29088885112832

f(20) = 370371188237525

f(21) = -4951498053124096
```

Thus, we obtain the following

**Conjecture.**  For any positive integer $n$,

\begin{equation}
   \mathrm{per}(A)=
   \begin{cases}
   -T_n&\mbox{if $n$ is odd}\\
   E_n&\mbox{if $n$ is even}
   \end{cases}.
  \end{equation}

**Question.** Is this identity correct? How to prove it?

___
**EDIT**

It is easy to see that (switch some rows of $A$ and multiply some rows by $-1$)

$\mathrm{per}(A)=E_n$ for any even integer $n$ $\iff $ initial conjecture of [Question 402572][1].

It is still open.


  [1]: https://mathoverflow.net/questions/402572
  [2]: https://mathworld.wolfram.com/TangentNumber.html
  [3]: https://mathworld.wolfram.com/SecantNumber.html