Motivated by [Question 402249](https://mathoverflow.net/questions/402249) of [Zhi-Wei Sun](https://mathoverflow.net/users/124654/zhi-wei-sun), I consider the permanent of matrices


$$e(n)=\mathrm{per}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1},$$
where $n$ is an odd integer greater than 1 and $ \operatorname{sgn} $ is the sign-function.


When $n=3,5,7,$ the matrices are
$$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array}
 \right] ,
$$


$$
\left[ \begin {array}{cccc} 1&-1&-1&0\\ -1&-1&0&1
\\ -1&0&1&1\\ 0&1&1&-1\end {array}
 \right], 
$$


$$
\left[ \begin {array}{cccccc} 1&1&-1&-1&-1&0\\ 1&-1
&-1&-1&0&1\\ -1&-1&-1&0&1&1\\ -1&-
1&0&1&1&1\\ -1&0&1&1&1&-1\\ 0&1&1&
1&-1&-1\end {array} \right].
$$


Numerical computation indicates that


$$e(n)=E(n-1)$$


for $3 \leq n \leq 23 $, where $E(n)$ are the [Euler numbers](https://encyclopediaofmath.org/wiki/Euler_numbers).



```
                 [e(3) = -1, E(2) = -1]

                  [e(5) = 5, E(4) = 5]

                [e(7) = -61, E(6) = -61]

               [e(9) = 1385, E(8) = 1385]

            [e(11) = -50521, E(10) = -50521]

           [e(13) = 2702765, E(12) = 2702765]

        [e(15) = -199360981, E(14) = -199360981]

       [e(17) = 19391512145, E(16) = 19391512145]

    [e(19) = -2404879675441, E(18) = -2404879675441]

   [e(21) = 370371188237525, E(20) = 370371188237525]

[e(23) = -69348874393137901, E(22) = -69348874393137901]

```

**Conjecture.** For any odd integer $n>1$, $e(n)=E(n-1)$.


**Question 1.** How to prove this conjecture?




---


**Edit.**


By [T. Amdeberhan's recent MO post](https://mathoverflow.net/questions/402596),


$$\mathrm{det}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}=(-1)^{\frac{n-1}{2}}=\sin(\frac{n\pi}{2})\neq 0,$$


we can consider the permanent of inverse matrices


$$e'(n)=\mathrm{per}(A^{-1}),$$


where $A=\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}$ and $n$ is an odd integer greater than 1.


When $n=3,5,7,$ $A^{-1}$ equals


$$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array}
 \right], 
$$


$$
\left[ \begin {array}{cccc} 1&-1&1&0\\ -1&-1&0&-1
\\ 1&0&1&1\\ 0&-1&1&-1\end {array}
 \right], 
$$
$$
\left[ \begin {array}{cccccc} 1&1&-1&1&1&0\\ 1&-1&-
1&1&0&-1\\ -1&-1&-1&0&-1&-1\\ 1&1&0
&1&1&1\\ 1&0&-1&1&1&-1\\ 0&-1&-1&1
&-1&-1\end {array} \right] 
$$
respectively.


Numerical computation indicates that all elements of $A^{-1}$ are $-1, 0, 1 $ and


$$e'(n)=E(n-1)$$


for $3 \leq n \leq 23 $.


**Added Conjecture 1.** For any odd integer $n>1$, $e'(n)=E(n-1)$.


The following are some properties of the $A$ and $A^{-1}$ that may be easy to prove:


**Added Conjecture 2.** For any odd integer $n>1$, all elements of $A^{-1}$ are $-1, 0, 1 $.


**Added Conjecture 3.** For any odd integer $n>1$, the Characteristic Polynomials of $A$ and $A^{-1}$ are
$$
\mathrm{Char}(A)=\mathrm{Char}(A^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}},
$$
where $i^2=-1$ and $\mathrm{Char}(A)=\mathrm{det}(\lambda I-A)$.


**Question 2.** How to prove these added conjectures?




---


**Edit.**


**The connection between characteristic polynomials and recurrences of Euler numbers:**


Let


$$\mathrm{Char}(A)|\_{\lambda=iE}=\mathrm{det}(\lambda I-A)|\_{\lambda=iE}={\frac { \left( iE+i \right) ^{n-1} + \left( iE-i \right) ^{n-1}}{2}}=0,$$


we deduce


$$(E+1)^{n-1}+(E-1)^{n-1}=0,$$


which is [the recurrence formula for the Euler numbers](https://encyclopediaofmath.org/wiki/Euler_numbers) ($E^n≡E\_n≡E(n)$ in symbolic notation).


**Question 3.** Is this a coincidence?




---


**Edit.**


Let $B=\left[\operatorname{sgn} \left(\sin\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}$ where $n>1$ is an integer.


[Zhi-Wei Sun](https://mathoverflow.net/users/124654/zhi-wei-sun) and [Nemo](https://mathoverflow.net/users/82588/nemo) guessed in the comments that


$$\mathrm{per}(B)=E(n-1)$$
for any integer $n>1$.


Numerical calculations show that this is correct for $1<n\leq 23$.


It is easy to see that $\mathrm{det}(B)=0$ for any even number $n$ and it seems that
$$\mathrm{det}(B)=\mathrm{det}(A)=(-1)^{\frac{n-1}{2}}\neq 0$$
for any odd integer $n>1$.


When $n=3,5,7,$ $B$ and $B^{-1}$ are equal to


$$\left[ \begin {array}{cc} 1&0\\ 0&-1\end {array}
 \right] , \left[ \begin {array}{cc} 1&0\\ 0&-1
\end {array} \right] ,$$


$$ \left[ \begin {array}{cccc} 1&1&1&0\\ 1&1&0&-1
\\ 1&0&-1&-1\\ 0&-1&-1&-1
\end {array} \right] , \left[ \begin {array}{cccc} 1&-1&1&0
\\ -1&1&0&-1\\ 1&0&-1&1
\\ 0&-1&1&-1\end {array} \right] ,$$


$$ 
\left[ \begin {array}{cccccc} 1&1&1&1&1&0\\ 1&1&1&
1&0&-1\\ 1&1&1&0&-1&-1\\ 1&1&0&-1&
-1&-1\\ 1&0&-1&-1&-1&-1\\ 0&-1&-1&
-1&-1&-1\end {array} \right] , \left[ \begin {array}{cccccc} 1&-1&1&-1
&1&0\\ -1&1&-1&1&0&-1\\ 1&-1&1&0&-1&1\\ -1&1&0&-1&1&-1\\ 1&0&-1&1&-1
&1\\ 0&-1&1&-1&1&-1\end {array} \right] 
$$
respectively.


A natural question is: Does $B$ and $B^{-1}$ have the same properties as $A$ and $A^{-1}$ for any odd integer $n>1$.


**Added Conjecture 4.** For any odd integer $n>1$,
$$\mathrm{per}(B)=\mathrm{per}(B^{-1})=E(n-1)$$ and $$
\mathrm{Char}(B)=\mathrm{Char}(B^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}}.
$$


**Question 4.** If $$\mathrm{per}(C)=\mathrm{per}(C^{-1})=E(n-1)$$
and
$$\mathrm{Char}(C)=\mathrm{Char}(C^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}},$$
where $n$ is an odd number greater than $1,$ $C$ is an $(n-1)\times (n-1)$ matrix and all elements of $C$ are $-1, 0, 1 ,$ what can we say about $C$?



Euler numbers and permanent of matrices
The conjecture and the added Conjectures 1 and 2 have been proved.
See the preprint *Proof of five conjectures relating permanents to combinatorial sequences* by Fu, Lin and me available from <http://arXiv.org/abs/2109.11506>.


