Recently, several of my conjectures in Question 402572 and Question 403336 were proved by Fu, Lin and Sun available from Proofs of five conjectures relating permanents to combinatorial sequences.
These questions relate to the permanents of 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 true 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 $n>1$, $$a(n) \equiv a(n+2p-2) \pmod p.$$
Conjecture 3. For any prime $p$, $$a(2p) \equiv 2 \pmod p.$$ Proof (Suppose Conjecture 1 holds).
Noting $E_{2p}\equiv 1\pmod p$ [1], we have $$a(2p) =T_p(E_{2k})= \sum_{k=0}^{p}\binom{p}{k}E_{2k}\equiv 1+E_{2p} \equiv 2\pmod p.$$
Moreover
Conjecture 4. For any prime $p$ and positive integer $n>h≥1$, $$a(n) \equiv a(n+2(p-1)p^{h-1}) \pmod{p^h}.$$
By [1], this is a Kummer-like congruence. I believe that for some "nice matrices" $B_n$, $b(n)=\mathrm{per}(B_n)$ will have congruence like Daniel Barsky'sa Kummer-like congruence, e.g. matrices in Question 402572 and, Question 403336 (c.f. [1]) and Question 404530. This can be used to discover new congruences.
Question. Are these results correct? How to prove them?
[1] Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.
