Motivated by Question 402249 of 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.
[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,
$$\mathrm{det}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]_{1\le j,k\le n-1}=(-1)^{\frac{n-1}{2}}=\sin(\frac{n\pi}{2})\neq 0,$$
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 ($E^n≡E_n≡E(n)$ 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 and 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 $C$?
Best Answer
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.