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. This can be used to discover new congruents.

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