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Inspired by Question 402572, 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 and secant numbers, 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.

It is still open.

Inspired by Question 402572, 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 and secant numbers, 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?

Inspired by Question 402572, 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 and secant numbers, 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.

Inspired by Question 402572, 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 and secant numbers, 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.

It is still open.

Inspired by Question 402572, 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 and secant numbers 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?

Inspired by Question 402572, 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 and secant numbers, 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.