Lonesum decomposable matrices

TL;DR

This paper introduces lonesum decomposable matrices, characterizes their properties, and provides a generating function for counting such matrices, along with congruence relations modulo primes.

Contribution

It defines lonesum decomposable matrices, establishes their key properties, and derives a generating function for counting them, including congruence results.

Findings

Derived a necessary and sufficient condition for lonesum decomposability.

Provided a generating function for counting lonesum decomposable matrices.

Proved congruences for the number of such matrices modulo primes.

Abstract

A lonesum matrix is a $(0,1)$-matrix that is uniquely determined by its row and column sum vectors. In this paper, we introduce lonesum decomposable matrices and study their properties. We provide a necessary and sufficient condition for a matrix $A$ to be lonesum decomposable, and give a generating function for the number $D_k(m,n)$ of $m\times n$ lonesum decomposable matrices of order $k$. Moreover, by using this generating function we prove some congruences for $D_k(m,n)$ modulo a prime.

Tables

Table 1. Table 1: D 1 ​ ( m , n ) subscript 𝐷 1 𝑚 𝑛 D_{1}(m,n)

$m\backslash n$	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	1	3	7	15	31
2	0	3	13	45	145	453
3	0	7	45	229	1065	4717
4	0	15	145	1065	6901	41505
5	0	31	453	4717	41505	329461

Table 2. Table 2: D 2 ​ ( m , n ) subscript 𝐷 2 𝑚 𝑛 D_{2}(m,n)

$m\backslash n$	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	0	0	0	0	0
2	0	0	2	12	50	180
3	0	0	12	108	660	3420
4	0	0	50	660	5714	40860
5	0	0	180	3420	40860	391500

Table 3. Table 3: D ​ ( m , n ) 𝐷 𝑚 𝑛 D(m,n)

$m\backslash n$	0	1	2	3	4	5
0	1	1	1	1	1	1
1	1	2	4	8	16	32
2	1	4	16	58	196	634
3	1	8	58	344	1786	8528
4	1	16	196	1786	13528	90946
5	1	32	634	8528	90446	833432

Equations

$$\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\end{array}\right).$$

$$\{\begin{array}{ll}{a}_{ij}=0\Rightarrow a_{kj}=0& (k\ge i),\\ a_{ij}=0\Rightarrow a_{il}=0& (l\ge j).\end{array}$$

$$\frac{{\mathrm{Li}}_k(1-e^{-t})}{1-e^{-t}}=\sum _{n=0}^{\infty }{B}_n^{(k)}\frac{t^n}{n!},$$

$$L(m,n)=B_n^{(-m)}(m,n\ge 0).$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }L(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\frac{e^{x+y}}{e^x+e^y-e^{x+y}}.$$

$$A\sim \left(\begin{array}{cccc}{L}_1& & & O\\ & L_2& & \\ & & \ddots & \\ O& & & L_k\end{array}\right),$$

$$A\sim \left(\begin{array}{ccccc}{F}_1& & & O& \\ & F_2& & & \\ & & \ddots & & \\ O& & & F_k& \\ & & & & O\end{array}\right),$$

$$U:=\left(\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\end{array}\right).$$

$$A\sim\left(\begin{array}[]{cccc|c}F_{1}&&O&&\boldsymbol{b}_{1}\\ &\ddots&&&\vdots\\ O&&F_{k}&&\boldsymbol{b}_{k}\\ &&&O&\boldsymbol{c}\end{array}\right),$$

$$\left(\begin{array}[]{cc|c}F_{k}&&\boldsymbol{b}_{k}\\ &O&\boldsymbol{c}\end{array}\right)\sim\left(\begin{array}[]{ccc}\boldsymbol{1}&F_{k}&\\ \boldsymbol{c}&&O\end{array}\right).$$

$$\left(\begin{array}[]{cc|c}F_{k}&&\boldsymbol{b}_{k}\\ &O&\boldsymbol{c}\end{array}\right)\sim\left(\begin{array}[]{ccc}F_{k}&\boldsymbol{0}&\\ &\boldsymbol{c}&O\end{array}\right).$$

$$\left(\begin{array}[]{cc|c}F_{k}&&\boldsymbol{b}_{k}\\ &O&\boldsymbol{c}\end{array}\right)\sim\left(\begin{array}[]{ccc}F_{k}&\boldsymbol{b}_{k}&\\ &\boldsymbol{0}&O\end{array}\right).$$

$$A\sim \left(\begin{array}{cc}{L}_1& O\\ O& L_2\end{array}\right),$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{D}_k(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\frac{e^{x+y}}{k!}{\left(\frac{1}{e^x+e^y-e^{x+y}}-1\right)}^k(k\ge 0).$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }D(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\exp \left(\frac{1}{e^x+e^y-e^{x+y}}+x+y-1\right).$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\tilde{L}(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\frac{1}{e^x+e^y-e^{x+y}}.$$

$$L(m,n)=\sum _{i=0}^m\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{m}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{j}\right)\tilde{L}(i,j),$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\tilde{D}}_1(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\frac{1}{e^x+e^y-e^{x+y}}-1.$$

$$\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\tilde{D}}_k(m,n)\frac{x^m}{m!}\frac{y^n}{n!}=\frac{1}{k!}{\left(\frac{1}{e^x+e^y-e^{x+y}}-1\right)}^k(k\ge 0).$$

$$D_k(m+1,n)$$

$$=D_k(m,n)+\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left((k-1)D_k(m,l)+D_{k-1}(m,l)+D_k(m,l+1)\right).$$

$$\frac{\partial }{\partial x}{G}_k=G_k+(e^y-1)\left((k-1)G_k+G_{k-1}+\frac{\partial }{\partial y}{G}_k\right).$$

$$e^{-xt}\frac{{\text{Li}}_{k_1,\dots ,k_r}^{\star }(1-e^{-t})}{1-e^{-t}}=\sum _{n=0}^{\infty }{B}_{n,\star }^{(k_1,\dots ,k_r)}(x)\frac{t^n}{n!},$$

$${\text{Li}}_{k_1,\dots ,k_r}^{\star }(z):=\sum _{1\le m_1\le \cdots \le m_r}\frac{z^{m_r}}{m_1^{k_1}\cdots m_r^{k_r}}.$$

$$e^{-xt}\frac{{\text{Li}}_k(1-e^{-t})}{1-e^{-t}}=\sum _{n=0}^{\infty }{B}_n^{(k)}(x)\frac{t^n}{n!}$$

$$D_k(m,n)=\frac{(-1{)}^k}{k!}\left(1+\sum _{i=1}^k\left(\genfrac{}{}{0ex}{}{k}{i}\right)(-1{)}^i{B}_{n,\star }^{(\stackrel{i-1}{\overbrace{0,\dots ,0}},-m)}(i-1)\right).$$

$$\frac{e^{x+y}}{(e^x+e^y-e^{x+y}{)}^i}$$

$$=e^{x+y}{e}^{-iy}\sum _{l_1,\dots ,l_i\ge 0}{e}^{(l_1+\cdots +l_i)x}(1-e^{-y}{)}^{l_1+\cdots +l_i}$$

$$=e^{-(i-1)y}\sum _{l_1,\dots ,l_i\ge 0}{e}^{(l_1+\cdots +l_i+1)x}(1-e^{-y}{)}^{l_1+\cdots +l_i+1}\frac{1}{1-e^{-y}}$$

$$=e^{-(i-1)y}\sum _{m=0}^{\infty }\sum _{l_1,\dots ,l_i\ge 0}\frac{(1-e^{-y}{)}^{l_1+\cdots +l_i+1}}{(l_1+\cdots +l_i+1{)}^{-m}}\frac{1}{1-e^{-y}}\frac{x^m}{m!}$$

$$=e^{-(i-1)y}\sum _{m=0}^{\infty }\frac{{\text{Li}}_{0,\dots ,0,-m}^{\star }(1-e^{-y})}{1-e^{-y}}\frac{x^m}{m!}$$

$$=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{B}_{n,\star }^{(\stackrel{i-1}{\overbrace{0,\dots ,0}},-m)}(i-1)\frac{x^m}{m!}\frac{x^n}{n!}.$$

$$\frac{e^{x+y}}{k!}{\left(\frac{1}{e^x+e^y-e^{x+y}}-1\right)}^k$$

Full text

MSC2010: Primary 05A15, Secondary 11B68; 15B36

Keywords: Lonesum matrices; poly-Bernoulli numbers

1 Introduction

A $(0,1)$-matrix (resp. vector) is a matrix (resp. vector) in which each entry is zero or one. A $(0,1)$-matrix $A$ is called a lonesum matrix if $A$ is uniquely determined by its row and column sum vectors. For example, a $(0,1)$-matrix with a row sum vector ${}^{t}(3,1)$ and a column sum vector $(1,2,1)$ is uniquely determined as the following:

[TABLE]

Hence, the matrix $\begin{pmatrix}1&1&1\\ 0&1&0\end{pmatrix}$ is a lonesum matrix. Because $\begin{pmatrix}1&0&1\\ 0&1&0\end{pmatrix}$ and $\begin{pmatrix}0&1&1\\ 1&0&0\end{pmatrix}$ have the same row and column sum vectors, they are not lonesum matrices. We denote by $L(m,n)$ the number of $m\times n$ lonesum matrices. For simplicity, we set $L(m,0)=L(0,m)=1$ for any non-negative integer $m$. It is known that lonesum matrices are related to certain combinatorial objects. For example, the number $L(m,n)$ is equal to the number of acyclic orientations of the complete bipartite graph $K_{m,n}$ ([5, Theorem 2.1]).

An $m\times n$ $(0,1)$-matrix $A=(a_{ij})$ is called a Ferrers matrix if $A$ satisfies the condition

[TABLE]

This condition means that all 1 entries of $A$ are placed to the upper left of $A$. For example, the matrix $\begin{pmatrix}1&1&1\\ 1&0&0\end{pmatrix}$ is a Ferrers matrix. Ryser [12] investigated matrices that have fixed row and column sum vectors. In our setting, his result can be written as follows:

Proposition 1.1**.**

Let $A$ be a $(0,1)$-matrix. Then, the following conditions are equivalent:

• (i)

$A$* is a lonesum matrix.*

• (ii)

$A$* does not contain $\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$ or $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$ as a submatrix.*

• (iii)

$A$* is obtained from a Ferrers matrix by permutations of rows and columns.*

For an integer $k$, Kaneko [10] introduced poly-Bernoulli numbers $B_{n}^{(k)}$ of index $k$ as

[TABLE]

where ${\rm Li}_{k}(z)$ denotes the $k$-th polylogarithm, defined by ${\rm Li}_{k}(z):=\sum_{n=1}^{\infty}z^{n}/n^{k}$. Brewbaker [4] proved that the numbers $L(m,n)$ are equal to the poly-Bernoulli numbers of negative indices:

[TABLE]

The generating function of poly-Bernoulli numbers of negative indices has been given by Kaneko [10], hence the numbers of lonesum matrices have the following generating function:

[TABLE]

The present author, Ohno, and Yamamoto [9] introduced “weighted” lonesum matrices and a simple proof of (3) was given (see [9, Proof of Theorem 1]).

For $m\times n$ matrices $A$ and $B$, we write $A\sim B$ if $A$ is obtained from $B$ by row or column exchanges. We call a $(0,1)$-matrix $A$ is lonesum decomposable if $A$ satisfies the condition

[TABLE]

where $L_{i}$ ($1\leq i\leq k$) are lonesum matrices. A lonesum matrix is clearly lonesum decomposable. Since a lonesum matrix can be obtained from a Ferrers matrix, a lonesum decomposable matrix $A$ can be transformed as

[TABLE]

where $F_{i}$ ($1\leq i\leq k$) are Ferrers matrices with no zero rows or zero columns. We call the right-hand side of (4) the decomposition matrix of $A$ and $k$ the decomposition order of $A$.

Proposition 1.2**.**

Let $A$ be a lonesum decomposable matrix. Then the decomposition matrix of $A$ is uniquely determined up to the order of $F_{i}$ ($1\leq i\leq k$). In particular, the decomposition order of $A$ is uniquely determined.

Proof.

For a lonesum decomposable matrix $A=(a_{ij})$, it follows from Proposition 1.1 that two elements $a_{ij}=1$ and $a_{i^{\prime}j^{\prime}}=1$ belong to the same Ferrers block if and only if $a_{ij}$ and $a_{i^{\prime}j^{\prime}}$ do not form a submatrix $\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$ or $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$. Because the type of Ferrers matrix is uniquely determined, a decomposition matrix of $A$ is also uniquely determined up to the order of its Ferrers blocks. ∎

The outline of this paper is as follows. In Section 2, we show that a $(0,1)$-matrix $A$ is lonesum decomposable if and only if $A$ does not contain certain matrices as submatrices. In Section 3, we give a generating function for the number of lonesum decomposable matrices. In Section 4, we derive some congruences for the numbers of lonesum decomposable matrices of order $k$ by using the generating function given in Section 3.

2 Lonesum decomposable matrices

Let us define a $2\times 3$ matrix $U$ as

[TABLE]

It can be easily checked that $U$ is not lonesum decomposable. Let $\mathcal{N}$ be a set of all matrices obtained from $U$ or ${}^{t}U$ by permutations of rows and columns. Namely, the elements of $\mathcal{N}$ are the following twelve matrices:

$\begin{pmatrix}1&1&0\\ 1&0&1\end{pmatrix}$,  $\begin{pmatrix}1&0&1\\ 1&1&0\end{pmatrix}$,  $\begin{pmatrix}1&1&0\\ 0&1&1\end{pmatrix}$,  $\begin{pmatrix}0&1&1\\ 1&1&0\end{pmatrix}$,  $\begin{pmatrix}1&0&1\\ 0&1&1\end{pmatrix}$,  $\begin{pmatrix}0&1&1\\ 1&0&1\end{pmatrix}$,

$\begin{pmatrix}1&1\\ 1&0\\ 0&1\end{pmatrix}$,  $\begin{pmatrix}1&1\\ 0&1\\ 1&0\end{pmatrix}$,  $\begin{pmatrix}1&0\\ 1&1\\ 0&1\end{pmatrix}$,  $\begin{pmatrix}0&1\\ 1&1\\ 1&0\end{pmatrix}$,  $\begin{pmatrix}1&0\\ 0&1\\ 1&1\end{pmatrix}$,  $\begin{pmatrix}0&1\\ 1&0\\ 1&1\end{pmatrix}$.

The following is the first main result of this paper.

Theorem 2.1**.**

Let $A$ be a $(0,1)$-matrix. Then, the following two conditions are equivalent.

1. $(i)$

$A$* is lonesum decomposable.* 2. $(ii)$

$A$* does not contain an element of $\mathcal{N}$ as a submatrix.*

Proof.

It is clear that (i) $\Rightarrow$ (ii), and we show (ii) $\Rightarrow$ (i). This statement clearly holds for $0\leq m,n\leq 2$, where $m$ and $n$ are the numbers of rows and columns of $A$, respectively. A transpose of a lonesum decomposable matrix is also lonesum decomposable, hence we only have to prove that if the statement holds for all $m\times n$ matrices, then it holds for any $m\times(n+1)$ matrix for $m,n\geq 2$.

Let $A$ be an $m\times(n+1)$ $(0,1)$-matrix not containing an element of $\mathcal{N}$. The matrix obtained by removing the $(n+1)$-st column from $A$ is $m\times n$ matrix. Hence, by the induction assumption, the matrix $A$ can be transformed as

[TABLE]

where $F_{i}$ ($1\leq i\leq k$) are Ferrers matrices with no zero rows or columns, and $\boldsymbol{b}_{i}$ ($1\leq i\leq k$) and $\boldsymbol{c}$ are $(0,1)$-vectors. If there exist two non-zero vectors $\boldsymbol{b}_{i}$ and $\boldsymbol{b}_{j}$ ($i\neq j$), then the submatrix $\begin{pmatrix}F_{i}&O&\boldsymbol{b}_{i}\\ O&F_{j}&\boldsymbol{b}_{j}\end{pmatrix}$ contains a matrix $\begin{pmatrix}1&0&1\\ 0&1&1\end{pmatrix}$, and this contradicts the assumption that $A$ does not contain any element of $\mathcal{N}$. Therefore, there is at most one non-zero vector in $\boldsymbol{b}_{i}$ ($1\leq i\leq k$), and we can set $\boldsymbol{b}_{1}=\cdots=\boldsymbol{b}_{k-1}=\boldsymbol{0}$ without loss of generality.

We consider the two cases where (i) $\boldsymbol{c}$ has $1$’s and (ii) $\boldsymbol{c}$ has no $1$’s.

(i). The case that $\boldsymbol{c}$ has $1$’s.

If the vector $\boldsymbol{b}_{k}$ has both [math]’s and $1$’s, then $\begin{pmatrix}F_{k}&\boldsymbol{b}_{k}\\ O&\boldsymbol{c}\end{pmatrix}$ contains a matrix $\begin{pmatrix}1&1\\ 1&0\\ 0&1\end{pmatrix}$ or $\begin{pmatrix}1&0\\ 1&1\\ 0&1\end{pmatrix}$, and this contradicts the assumption that $A$ does not contain an element of $\mathcal{N}$. Therefore, $\boldsymbol{b}_{k}=\boldsymbol{1}$ or $\boldsymbol{0}$. If $\boldsymbol{b}_{k}=\boldsymbol{1}$, then

[TABLE]

Because the right-hand side of (5) is a lonesum matrix, the statement holds. If $\boldsymbol{b}_{k}=\boldsymbol{0}$, then

[TABLE]

The right-hand side of (6) is lonesum decomposable of order $2$, and hence the statement again holds.

(ii). The case that $\boldsymbol{c}$ has no $1$’s.

We have

[TABLE]

By Proposition 1.1, if the matrix $(F_{k}\,\boldsymbol{b}_{k})$ is not a lonesum matrix then it contains $\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$ or $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$ as a submatrix. Because $F_{k}$ has no zero columns, the matrix $(F_{k}\,\boldsymbol{b}_{k})$ also contains $\begin{pmatrix}1&1&0\\ 1&0&1\end{pmatrix}$ or $\begin{pmatrix}1&0&1\\ 1&1&0\end{pmatrix}$ as a submatrix, and this contradicts the assumption that $A$ does not contain an element of $\mathcal{N}$. Therefore, the matrix $(F_{k}\,\boldsymbol{b}_{k})$ is a lonesum matrix and the statement also holds in this case. ∎

For a $(0,1)$-matrix $A$, we define $\overline{A}$ as the matrix in which the [math] and $1$ entries of $A$ are inverted. If $A$ is a lonesum matrix, then $\overline{A}$ is also a lonesum matrix. However, lonesum decomposable matrices do not have this property. For example, the matrix $V=\begin{pmatrix}1&0&0\\ 0&1&0\end{pmatrix}$ is lonesum decomposable, but $\overline{V}=\begin{pmatrix}0&1&1\\ 1&0&1\end{pmatrix}\in\mathcal{N}$ is not lonesum decomposable. The following theorem determines when a matrix $A$ satisfies that both $A$ and $\overline{A}$ are lonesum decomposable.

Theorem 2.2**.**

Let $A$ be a $(0,1)$-matrix. Then, the following conditions are equivalent.

1. $(i)$

Both $A$ and $\overline{A}$ are lonesum decomposable. 2. $(ii)$

$A$* is a lonesum matrix or $A\sim\begin{pmatrix}{\boldsymbol{1}}&{\boldsymbol{O}}\\ {\boldsymbol{O}}&{\boldsymbol{1}}\end{pmatrix}$, where $\boldsymbol{1}$ (resp. $\boldsymbol{O}$) is a matrix whose entries are all $1$ (resp. [math]).*

Proof.

It is clear that (ii) $\Rightarrow$ (i), and we only have to prove that (i) $\Rightarrow$ (ii). Assume that $A$ and $\overline{A}$ are both lonesum decomposable, and let $k$ be the decomposition order of $A$. When $k=0$ or $1$, $A$ is a lonesum matrix. When $k=2$, the matrix $A$ satisfies that

[TABLE]

where $L_{1}$ and $L_{2}$ are non-zero lonesum matrices. If $L_{1}$ or $L_{2}$ has [math]’s, then the matrix $\overline{A}$ contains an element of $\mathcal{N}$ as a submatrix, and $\overline{A}$ is not lonesum decomposable. Therefore, $L_{1}=\boldsymbol{1}$ and $L_{2}=\boldsymbol{1}$. When $k\geq 3$, the matrix $A$ contains a $3\times 3$ submatrix $W$ satisfying $W\sim\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$. This matrix contains $\begin{pmatrix}1&0&0\\ 0&1&0\\ \end{pmatrix}$, and this contradicts the condition that $\overline{A}$ is lonesum decomposable. As a consequence, either $A$ is a lonesum matrix or $A\sim\begin{pmatrix}{\boldsymbol{1}}&{\boldsymbol{O}}\\ {\boldsymbol{O}}&{\boldsymbol{1}}\end{pmatrix}$. ∎

3 Generating function of lonesum decomposable matrices

For a positive integer $k$, let $D_{k}(m,n)$ denote the number of $m\times n$ lonesum decomposable matrices of decomposition order $k$. For simplicity, we set $D_{k}(m,0)=D_{k}(0,m)=0$ for $k\geq 1$ and $m\geq 0$, and $D_{0}(m,n)=1$ for $(m,n)\in\mathbb{Z}_{\geq 0}^{2}$. Moreover, we define $D(m,n):=\sum_{k=0}^{\infty}D_{k}(m,n)$ for $(m,n)\in\mathbb{Z}_{\geq 0}^{2}$. This means that $D(m,n)$ is the number of all $m\times n$ lonesum decomposable matrices. We can see that $D_{k}(m,n)=0$ for $k>\min(m,n)$ and $L(m,n)=D_{0}(m,n)+D_{1}(m,n)$. We present tables showing $D_{1}(m,n)$, $D_{2}(m,n)$, and $D(m,n)$ at the end of this paper.

The generating functions for $D_{k}$ and $D$ are given as follows:

Theorem 3.1**.**

The following equations hold:

[TABLE]

[TABLE]

Proof.

Let $\tilde{L}(m,n)$ be the number of $m\times n$ lonesum matrices with no zero rows or columns. Here, we set $\tilde{L}(0,0)=1$ and $\tilde{L}(m,0)=\tilde{L}(0,m)=0$ for $m>0$. Benyi and Hajnal [3, Theorem 3] mentioned that the generating function of $\tilde{L}(m,n)$ is given by

[TABLE]

By definition, it holds that

[TABLE]

and we can also obtain the generating function (3) of $L(m,n)$ from (10). We note that multiplying the generating function (10) by $e^{x+y}$ means that it allows the lonesum matrices to have zero columns or zero rows.

Let $\tilde{D}_{k}(m,n)$ be the number of $m\times n$ lonesum decomposable matrices of order $k$ with no zero rows and columns. We set $\tilde{D}_{0}(m,n)=0$ if $(m,n)\neq(0,0)$ and $=1$ if $(m,n)=(0,0)$. When $k=1$, we have $\tilde{D}_{1}(m,n)=\tilde{L}(m,n)$ if $(m,n)\neq(0,0)$ and $=0$ if $(m,n)=(0,0)$. Therefore, we have

[TABLE]

In general, the generating function of $\tilde{D}_{k}$ can be given by

[TABLE]

The generating function of $D_{k}$ can be obtained by multiplying (12) by $e^{x+y}$, hence we obtain (8). Equation (9) follows immediately from (8). ∎

Remark 3.2**.**

Ju and Seo [8] studied generating functions for the number of matrices not including various $2\times 2$ matrices. Theorem 3.1 gives a similar result on matrices that do not include the elements of $\mathcal{N}$.

It is known that the numbers $L(m,n)$ (or the poly-Bernoulli numbers of negative indices) satisfy a recurrence relation (e.g. [2, Prop. 14.3 and 14.4]). Our numbers $D_{k}(m,n)$ also satisfy a recurrence relation.

Proposition 3.3**.**

For $k\geq 1$ and $m,n\geq 0$, we have

[TABLE]

Proof.

Let $G_{k}(x,y):=\dfrac{e^{x+y}}{k!}\left(\dfrac{1}{e^{x}+e^{y}-e^{x+y}}-1\right)^{k}$. By direct calculations, we can verify that

[TABLE]

By comparing the coefficients of both sides of (13), we obtain the proposition. ∎

To conclude this section, we give a relation between $D_{k}(m,n)$ and the poly-Bernoulli polynomials. For any integers $k_{1},\ldots,k_{r}$, we define the multi-poly Bernoulli(-star) polynomials $B_{n,\star}^{(k_{1},\ldots,k_{r})}(x)$ by

[TABLE]

where

[TABLE]

These polynomials have been introduced by Imatomi [7, §6], but they were defined there with $e^{-xt}$ replaced by $e^{xt}$ in (14). When $r=1$, the polynomial $B_{n,\star}^{(k)}(x)$ coincides with the $n$-th poly-Bernoulli polynomial $B_{n}^{(k)}(x)$ defined by

[TABLE]

(see e.g., Coppo-Candelpergher [6]).

Proposition 3.4**.**

For integers $k,m,n\geq 0$, we have

[TABLE]

Proof.

For an integer $i\geq 1$, we have

[TABLE]

From this formula and the binomial expansion, we obtain that

[TABLE]

and this proves the proposition. ∎

Remark 3.5**.**

Kaneko, Sakurai, and Tsumura [11] introduced a sequence $\mathscr{B}_{m}^{(-l)}(n)$ as

[TABLE]

where $\left[n\atop j\right]$ are the Stirling numbers of the first kind. They proved that this sequence has the following simple generating function:

[TABLE]

By using this formula, we can also give an expression for $D_{k}(m,n)$ in terms of poly-Bernoulli polynomials:

[TABLE]

4 Congruences for $D_{k}(m,n)$

It is known that the numbers of $m\times n$ lonesum matrices (or poly-Bernoulli numbers of negative indices) have the following expression:

[TABLE]

where $\left\{m\atop j\right\}$ are the Stirling numbers of the second kind (see e.g., [1] [4]). We note that $\left\{m\atop j\right\}=0$ for $j>m\geq 1$. The following proposition says that the numbers $D_{k}(m,n)$ also have a similar expression.

Proposition 4.1**.**

For integers $k\geq 1$ and $m,n\geq 0$ we have

[TABLE]

Proof.

The generating function for $D_{k}(m,n)$ can be transformed as

[TABLE]

Because

[TABLE]

we have

[TABLE]

Therefore, we obtain (17). ∎

By using this expression, we give some congruences for $D_{k}(m,n)$ modulo a prime. We first recall the following lemma in order to prove them. All of the formulas are deduced from the well-known identities

[TABLE]

and we omit their proofs.

Lemma 4.2**.**

Let $p$ be a prime.

1. $(i)$

For positive integers $m$ and $m^{\prime}$ with $m\equiv m^{\prime}\pmod{p-1}$ and $0\leq i\leq p$, we have $\displaystyle\left\{m\atop i\right\}\equiv\left\{m^{\prime}\atop i\right\}\pmod{p}$. 2. $(ii)$

$\displaystyle\left\{p\atop i\right\}\equiv 0\pmod{p}$* for $2\leq i\leq p-1$.* 3. $(iii)$

$\displaystyle\left\{m\atop 2\right\}=2^{m-1}-1$* for $m\geq 1$.*

Theorem 4.3**.**

Let $k$, $m$, $m^{\prime}$, $n$, and $n^{\prime}$ be positive integers. For any prime $p$, the following congruences hold:

1. $(i)$

If $k\geq p$, then

[TABLE] 2. $(ii)$

If $m\equiv m^{\prime}$ and $n\equiv n^{\prime}\pmod{p-1}$, then

[TABLE] 3. $(iii)$

If $p>k$, then

[TABLE] 4. $(iv)$

[TABLE]

Proof.

1. $(i)$

If $k\geq p$, then $(j!)^{2}/k!\equiv 0\pmod{p}$ in (17), and this proves that $D_{k}(m,n)\equiv 0\pmod{p}$. 2. $(ii)$

By (i), when $k\geq p$ both sides of (19) vanish modulo $p$ and the congruence holds. We may assume that $p>k$. By the symmetric property $D_{k}(m,n)=D_{k}(n,m)$, we only have to show that $D_{k}(m+p-1,n)\equiv D_{k}(m,n)\pmod{p}$. By Proposition 4.1, we have

[TABLE]

The terms for $j\geq m+1$ in (22) vanish modulo $p$. In fact, if $m+1\leq j\leq p-1$ then $\left\{m+p\atop j+1\right\}\equiv\left\{m+1\atop j+1\right\}\equiv 0\pmod{p}$ by Lemma 4.2 (i), and if $j\geq p$ then $j!\equiv 0\pmod{p}$. Consequently, we have

[TABLE]

and this is equal to $D_{k}(m,n)$. 3. $(iii)$

By (ii), we only have to consider the cases with $1\leq n\leq p-1$. By Proposition 4.1, we have

[TABLE]

If $n\leq p-2$, then $\left\{p\atop j+1\right\}\equiv 0\pmod{p}$ by Lemma 4.2 (ii), and $D_{k}(p-1,n)\equiv 0\pmod{p}$. If $n=p-1$, then only the term for $j=p-1$ in (23) remains, and

[TABLE]

The final equivalence is derived from the congruence $\binom{p-2}{k-1}\equiv(-1)^{k-1}k$ and Wilson’s theorem, which states that $(p-1)!\equiv-1\pmod{p}$. 4. $(iv)$

By Proposition 4.1, we have

[TABLE]

When $j=p$, we have $(j!)^{2}/k!\equiv 0\pmod{p}$. When $2\leq j\leq p-1$, we have $\left\{p+1\atop j+1\right\}=\left\{p\atop j\right\}+(j+1)\left\{p\atop j+1\right\}\equiv 0\pmod{p}$ because of Lemma 4.2 (ii). Therefore, the congruence $D_{k}(p,n)\equiv 0\pmod{p}$ holds for $k\geq 2$.

If $k=1$, then the term for $j=1$ in (24) remains, and $D_{k}(p,n)\equiv\left\{p+1\atop 2\right\}\left\{n+1\atop 2\right\}=(2^{p}-1)(2^{n}-1)\equiv 2^{n}-1\pmod{p}$ by Lemma 4.2 (iii) and Fermat’s little theorem.

∎

Acknowledgements

This work was supported by Grant-in-Aid for Young Scientists (B) from JSPS KAKENHI (16K17583).