Recently, several of my conjectures in Question 402572 and Question 403336 were proved by Fu, Lin and Sun available from http://arXiv.org/abs/2109.11506.

These questions relate to the permanents of the matrices $$\left[\operatorname{sgn} \left(\sin\pi\frac{j+k}{n+1} \right)\right]_{1\le j,k\le n}, \left[\operatorname{sgn} \left(\sin\pi\frac{j+2k}{n+1} \right)\right]_{1\le j,k\le n} , \left[\operatorname{sgn} \left(\tan\pi\frac{j+k}{2n+1} \right)\right]_{1\le j,k\le 2n}.$$

It seems natural to ask: what is the value of $\mathrm{per}(A)$ where $$A=\left[\operatorname{sgn} \left(\cos\pi\frac{j+k}{n+1} \right)\right]_{1\le j,k\le n}.$$

When $n=1,2,3,4,5,6,$ $A=$ $$\left[ \begin {array}{c} -1\end {array} \right] ,$$

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

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

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

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

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

Definition. Given a sequence $\{a_k\}$, define a new sequence $\{b_m\}$ by

$$b_m=T_m(a_k)=\sum_{k=0}^{m}\binom{m}{k}a_k,$$

The sequence $\{b_m\}$ is the binomial transform of $\{a_k\}$.

Let $E_n$ satisfy $$\sec{x}+\tan{x}=\sum_{n\geq 0}E_n\frac{x^n}{n!}=1+1\frac{x}{1!}+1\frac{x^2}{2!}+2\frac{x^3}{3!}+5\frac{x^4}{4!}+16\frac{x^5}{5!}+61\frac{x^6}{6!}+272\frac{x^7}{7!}+1385\frac{x^8}{8!}+7936\frac{x^9}{9!}+\cdots$$

The binomial transform of $\{E_{2k+1}\}$ and $\{E_{2k}\}$ are $$T_m(E_{2k+1})=\sum_{k=0}^{m}\binom{m}{k}E_{2k+1}$$ and $$T_m(E_{2k})=\sum_{k=0}^{m}\binom{m}{k}E_{2k}$$ respectively.

We have the following

Conjecture 1. For any positive integer $n$,

\begin{align} \mathrm{per}(A)&= \begin{cases} -T_m(E_{2k+1})&\mbox{if $n=2m+1$ }\\ T_m(E_{2k})&\mbox{if $n=2m$ } \end{cases}. \end{align}

Numerical computation indicates that it is ture for $1≤n≤21$.

per(A)= -1, 2, -3, 8, -21, 80, -327, 1664, -9129, 58112, -396363, 3027968, -24615741, 219392000, -2068052367, 21065007104, -225742096209, 2586813857792, -31048132997523, 395317106966528, -5252064083753061.

Denote $a(n)=\mathrm{per}(A)$.

Motivated by Question 404530, especially Daniel Barsky's congruence for Fubini numbers, I discovered the following interesting arithmetic properties of $a(n)$

Conjecture 2. For any prime $p$ and positive integer $n>1$, $$a(n) \equiv a(n+2p-2) \pmod p.$$ Noting that $a(2)=2$, we have

Conjecture 3. For any prime $p$, $$a(2p) \equiv 2 \pmod p.$$

Numerical computation indicates that it is ture for $2≤p≤173$.

Moreover

Conjecture 4. For any prime $p$ and positive integer $n>h≥1$, $$a(n) \equiv a(n+2(p-1)p^{h-1}) \pmod{p^h}.$$

I believe that for some "nice matrices" $B_n$, $b(n)=\mathrm{per}(B_n)$ will have congruence like Daniel Barsky's congruence, e.g. matrices in Question 402572 and Question 403336.

Question. Are these results correct? How to prove them?