Recently, several of my conjectures in [Question 402572][1] and [Question 403336][2] were proved by Fu, Lin and Sun available from [Proofs of five conjectures relating permanents to combinatorial sequences](http://arXiv.org/abs/2109.11506).

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][3], especially [Daniel Barsky's congruence][4] 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 $p$,
$$a(2p) \equiv 2 \pmod p.$$
**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 $n>h≥1$,
$$a(n) \equiv a(n+2(p-1)p^{h-1}) \pmod{p^h}.$$

**Proof (Suppose Conjecture 1 holds).**  Similar to the proof of Conjecture 3.

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][1] and [Question 403336][2] (c.f. \[1\]). 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.


  [1]: https://mathoverflow.net/questions/402572
  [2]: https://mathoverflow.net/questions/403336/
  [3]: https://mathoverflow.net/questions/404530
  [4]: http://www.mat.univie.ac.at/~slc/opapers/s05barsky.html