# A novel identity connecting permanents to Bernoulli numbers

For a matrix $$[a_{j,k}]_{1\le j,k\le n}$$ over a field, its permanent is defined by $$\mathrm{per}[a_{j,k}]_{1\le j,k\le n}:=\sum_{\pi\in S_n}\prod_{j=1}^n a_{j,\pi(j)}.$$ In a recent preprint of mine, I investigated arithmetic properties of some permanents.

Let $$n$$ be a positive integer. I have proved that $$\mathrm{per}\left[\left\lfloor\frac{j+k}n\right\rfloor\right]_{1\le j,k\le n}=2^{n-1}+1$$ and that $$\det\left[\left\lfloor\frac{j+k}n\right\rfloor\right]_{1\le j,k\le n}=(-1)^{n(n+1)/2-1} \qquad\text{if}\ \ n>1.$$ where $$\lfloor \cdot\rfloor$$ is the floor function. The proofs are relatively easy.

Recall that the Bernoulli numbers $$B_0,B_1,\ldots$$ are given by $$\frac x{e^x-1}=\sum_{n=0}^\infty B_n\frac{x^n}{n!}\ \ \ \ (|x|<2\pi).$$ Those $$G_n=2(1-2^n)B_n\ (n=1,2,3,\ldots)$$ are sometimes called Genocchi numbers.

Based on my numerical computation, here I pose the following two conjectures.

Conjecture 1. For any positive integer $$n$$, we have $$\mathrm{per}\left[\left\lfloor\frac{2j-k}n\right\rfloor\right]_{1\le j,k\le n}=2(2^{n+1}-1)B_{n+1}.\tag{1}$$

Conjecture 2. For any positive integer $$n$$, we have $$\det\left[\left\lfloor\frac{2j-k}n\right\rfloor\right]_{1\le j,k\le n}=\begin{cases}(-1)^{(n^2-1)/8}&\text{if}\ 2\nmid n,\\0&\text{if}\ 2\mid n.\end{cases}\tag{2}$$

Conjecture 2 seems easier than Conjecture 1.

QUESTION. Are the identities $$(1)$$ and $$(2)$$ correct? How to prove them?

• Some computations suggest that the matrix $A$ is totally unimodular (all minors are $0,\pm 1$). If so, the entries of $A^{-1}$ when $n$ is odd are $0,\pm 1$. Can one describe $A^{-1}$ explicitly? For any $n$, evidence suggests a simple explicit description of the LUD decomposition of $A$. Sep 7, 2021 at 14:44
• The inverse $A^{-1}$ of size $2n+1$ seems to have a very nice structure: The first $n$ columns start all with a $0$ and the remaining rows are very simple (with a unique $1$ or $-1$ in every row) and form a block $B$ of size $2n\times n$. Permuting the initial $0$ row with the simple matrix $B$ yields the next $n$ columns. The final column starts and ends with $0$ and is $1,-1,1,-1,\ldots,1$ otherwise. It should be not too hard to show that this is indeed the inverse matrix and that it has the required determinant. For matrices of even size, there is perhaps an obvious element in the kernel. Sep 7, 2021 at 15:27
• I also conjecture that $$\det\left[x^{\lfloor (2j-k)/n\rfloor}\ \right]_{1\le j,k\le n}=\begin{cases}(-1)^{(n^2-1)/8}\ (x-1)^{n-1}x^{(3-n)/2}&\text{if}\ 2\nmid n,\\0&\text{if}\ 2\mid n. \end{cases}\tag{3}$$ Sep 8, 2021 at 2:18
• Conjecture 2 has the following extension involving $q$-integers: $$\det\left[\left[\left\lfloor\frac{2j-k}n\right\rfloor\right]_q\right]_{1\le j,k\le n}=\begin{cases}(-1)^{(n^2-1)/8}\ q^{(1-n)/2}&\text{if}\ 2\nmid n,\\0&\text{if}\ 2\mid n,\end{cases}\tag{4}$$ where $[m]_q:=(1-q^m)/(1-q).$ Sep 8, 2021 at 2:37
• Conjecture 2 as well as $(3)$ and $(4)$ have been confirmed fully by my graduate student Chen-Kai Ren. Sep 9, 2021 at 3:04