# On bi-variate poly-Bernoulli polynomials

Claudio Pita-Ruiz  Claudio Pita-Ruiz – Universidad Panamericana. Facultad de
Ingeniería. Augusto Rodin 498, Ciudad de México, 03920, México.
<EMAIL_ADDRESS>
###### Abstract

We introduce poly-Bernoulli polynomials in two variables by using a
generalization of Stirling numbers of the second kind that we studied in a
previous work. We prove the bi-variate poly-Bernoulli polynomial version of
some known results on standard Bernoulli polynomials, as the addition formula
and the binomial formula. We also prove a result that allows us to obtain
poly-Bernoulli polynomial identities from polynomial identities, and we use
this result to obtain several identities involving products of poly-Bernoulli
and/or standard Bernoulli polynomials. We prove two generalized recurrences
for bi-variate poly-Bernoulli polynomials, and obtain some corollaries from
them.

###### keywords:

poly-Bernoulli polynomial, poly-Bernoulli number, Generalized Stirling number,
Generalized recurrence.

11B68, 11B73. 31 2023 1 https://doi.org/10.46298/cm.10327

##  1 Introduction

Bernoulli numbers are one of the most important mathematical objects that have
been studied by mathemathicians since they appeared in the 18-th century (see
[Ma]). A recent important generalization of the Bernoulli numbers $B_{n}$ and
Bernoulli polynomials $B_{n}(x)$ is about the so-called poly-Bernoulli numbers
and poly-Bernoulli polynomials. Poly-Bernoulli numbers
$B_{n}^{\left(k\right)}$, where $k$ is a given positive integer, were
introduced by M. Kaneko [Ka] in 1997, by means of the generating function

$\frac{\text{{Li}}_{k}(1-e^{-t})}{1-e^{-t}}=\sum_{n=0}^{\infty}B_{n}^{\left(k\right)}\frac{t^{n}}{n!},$

where Li${}_{k}(z)=\sum_{j=1}^{\infty}z^{j}/j^{k}$ is the polylogarithm
function. The case $k=1$ corresponds to the standard Bernoulli numbers $B_{n}$
(except the sign of $B_{1}^{\left(1\right)}$). Poly-Bernoulli polynomials
$B_{n}^{\left(k\right)}(x)$ can be defined by the generating function

$\frac{\text{{Li}}_{k}(1-e^{-t})}{1-e^{-t}}e^{-tx}=\sum_{n=0}^{\infty}B_{n}^{\left(k\right)}(x)\frac{t^{n}}{n!},$

(see [Ce-Ko]). The case $k=1$ corresponds to $(-1)^{n}B_{n}(x)$, and the case
$x=0$ corresponds to the poly-Bernoulli numbers $B_{n}^{\left(k\right)}$
mentioned before. Some slightly different definitions of poly-Bernoulli
polynomials $B_{n}^{\left(k\right)}(x)$, with $x$ replaced by $-x$, and/or
with an additional factor $(-1)^{n}$, can be found in some related papers (see
[Ba-Ha], [Ce-Ko], [Co-Ca], [Ha]). In this work we use the following explicit
formula

$B_{n}^{\left(k\right)}(x)=\sum_{l=0}^{n}\frac{1}{(l+1)^{k}}\sum_{j=0}^{l}(-1)^{j}\binom{l}{j}(j+x)^{n},$
(1)

as our definition of poly-Bernoulli polynomials (see formula (1.8) in [Ba-
Ha]). It is important to mention that the notation $B_{n}^{\left(k\right)}(x)$
is also used for a different kind of mathematical objects, namely, Bernoulli
polynomials of $k$-th order (see [Ca1]). A different generalization of
Bernoulli polynomials, studied in the past few years, is about considering
Bernoulli polynomials in several variables
$B_{p_{1},\ldots,p_{t}}(x_{1},\ldots,x_{t})$, that is, polynomials of degree
$p_{i}$ in the variable $x_{i}$, with

$B_{0,\dotsc,p_{i},\dotsc,0}(x_{1},\ldots,x_{t})=B_{p_{i}}\left(x_{i}\right)\quad\textup{
for each }i\in\\{1,\dotsc,t\\},$

seeking that reasonable generalizations of the known properties in the one-
variable case, remain valid. This kind of work is done in [Shib], with a
flavor of multivariable analysis and working with Jack polynomials. A
different approach is presented in [Shishk] (see also [Bre], [DiNar]). In this
work we study poly-Bernoulli polynomials in two variables (bi-variate poly-
Bernoulli polynomials). We define the bi-variate poly-Bernoulli polynomials by
using a generalization of Stirling numbers of the second kind we studied in
[Pi], and then we use the results in [Pi] to obtain results for the bi-variate
poly-Bernoulli polynomials considered in this work. We present now the
definitions and results in [Pi] that we will use in the remaining sections.
The generalized Stirling numbers of the second kind (GSN, for short), denoted
as $S_{a_{1},b_{1}}^{\left(a_{2},b_{2},p_{2}\right)}\left(p_{1},k\right)$,
where $a_{j},b_{j}\in\mathbb{C}$, $a_{j}\neq 0$, $j=1,2$, and $p_{1},p_{2}$
non-negative integers, are defined by means of the expansion

$(a_{1}n+b_{1})^{p_{1}}(a_{2}n+b_{2})^{p_{2}}=\sum_{k=0}^{p_{1}+p_{2}}k!S_{a_{1},b_{1}}^{a_{2},b_{2},p_{2}}(p_{1},k)\binom{n}{k},$
(2)

($S_{a_{1},b_{1}}^{a_{2},b_{2},p_{2}}(p_{1},k)=0$ if $k<0$ or $k>p_{1}+p_{2}$
). An explicit formula for these numbers is

$S_{a_{1},b_{1}}^{a_{2},b_{2},p_{2}}(p_{1},k)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}\binom{k}{j}(a_{1}(k-j)+b_{1})^{p_{1}}(a_{2}(k-j)+b_{2})^{p_{2}}.$
(3)

If $p_{2}=0$, we write the GSN $S_{a,b}^{a_{2},b_{2},0}(p,k)$ as
$S_{a,b}(p,k)$. We have

$S_{a,b}(p,k)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}\binom{k}{j}\left(a\left(k-j\right)+b\right)^{p}.$
(4)

In the case $a=1,b=0$, the corresponding GSN
$S_{1,0}(p,k)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}\binom{k}{j}\left(k-j\right)^{p}$
are the known Stirling numbers of the second kind. We will refer to them as
“standard Stirling numbers”, and in this case we use the known notation
$S(p,k)$. From (3) it is clear that
$S_{a,b}^{a,b,p_{2}}\left(p_{1},k\right)=S_{a,b}\left(p_{1}+p_{2},k\right)$.
We can see directly from (4) that

$\displaystyle S_{1,1}(p,k)$ $\displaystyle=$ $\displaystyle S(p+1,k+1),$ (5)
$\displaystyle S_{1,2}(p,k)$ $\displaystyle=$ $\displaystyle
S(p+2,k+2)-S(p+1,k+2).$ (6)

In this work we will use GSN of the form
$S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$. Some important facts about the GSN
$S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ are the following:

  * •

Some values of the GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ are

$\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},0)$ $\displaystyle=$
$\displaystyle\\!\\!x_{1}^{p_{1}}\\!x_{2}^{p_{2}}\\!,$ (7) $\displaystyle
S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},1)$ $\displaystyle=$
$\displaystyle(x_{1}+1)^{p_{1}}\\!(x_{2}+1)^{p_{2}}\\!-x_{1}^{p_{1}}\\!x_{2}^{p_{2}}\\!,$
$\displaystyle\vdots$ $\displaystyle
S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},p_{1}+p_{2})$ $\displaystyle=$
$\displaystyle 1.$

  * •

The GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ can be written in terms of the
GSN $S_{1,y_{1}}^{1,y_{2},p_{2}}(p_{1},k)$ as follows

$S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-y_{1})^{p_{1}-j_{1}}(x_{2}-y_{2})^{p_{2}-j_{2}}S_{1,y_{1}}^{1,y_{2},j_{2}}(j_{1},k).$
(8)

  * •

The GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ can be written in terms of
standard Stirling numbers as follows

$\displaystyle k!S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ $\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-m)^{p_{1}-j_{1}}(x_{2}-m)^{p_{2}-j_{2}}$
(9) $\displaystyle\times\sum_{i=0}^{m}\binom{m}{i}(k+i)!S(j_{1}+j_{2},k+i),$

where $m$ is an arbitrary non-negative integer.

  * •

The GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}\left(p_{1},k\right)$ can be written in
terms of standard Stirling numbers as follows

$\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ $\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-n)^{p_{1}-j_{1}}(x_{2}-n)^{p_{2}-j_{2}}$
(10)
$\displaystyle\times\sum_{i=0}^{n-1}(-1)^{i}s(n,n-i)S(j_{1}+j_{2}+n-i,k+n),$

where $n$ is an arbitrary positive integer, and $s(\cdot,\cdot)$ are the
unsigned Stirling numbers of the first kind.

  * •

The GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ satisfy the identity

$S_{1,x_{1}+1}^{1,x_{2}+1,p_{2}}(p_{1},k)=S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)+(k+1)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k+1).$
(11)

  * •

The GSN $S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)$ satisfy the recurrence

$S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)=S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1}-1,k-1)+(k+x_{1})S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1}-1,k).$
(12)

##  2 Definitions and preliminary results

The relation of Bernoulli (numbers and polynomials) with Stirling (numbers of
the second kind) is an old story, that dates back to Worpitsky [W] (see also
[G-K-P, p. 560] and [Ke, p. 5]). We have the following formula for Bernoulli
numbers

$B_{p}=\sum_{l=0}^{p}S(p,l)\frac{(-1)^{l}l!}{l+1},$ (13)

and in the case of Bernoulli polynomials we have

$B_{p}(x)=\sum_{l=0}^{p}\sum_{j=0}^{p}\binom{p}{j}x^{p-j}S(j,l)\frac{(-1)^{l}l!}{l+1}.$
(14)

An important observation of formula (1) is that poly-Bernoulli polynomial
$B_{p}^{\left(k\right)}(x)$ can be written in terms of the GSN as

$B_{p}^{\left(k\right)}(x)=\sum_{l=0}^{p}S_{1,x}(p,l)\frac{(-1)^{l}l!}{(l+1)^{k}}.$
(15)

The generalization of (15) to the case of two variables comes through the GSN:
we define poly-Bernoulli polynomial in the variables $x_{1},x_{2}$, denoted by
$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$, as

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}}.$
(16)

If $p_{2}=0$, formula (16) becomes (15). By using (9) we can write
$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ in terms of standard Stirling
numbers as

$\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(x_{1}-m\right)^{p_{1}-j_{1}}\left(x_{2}-m\right)^{p_{2}-j_{2}}$
(17)
$\displaystyle\times\sum_{i=0}^{m}\binom{m}{i}S\left(j_{1}+j_{2},l+i\right)\frac{(-1)^{l}(l+i)!}{(l+1)^{k}},$

where $m$ is an arbitrary non-negative integer. Similarly, by using (10) we
have that

$\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(x_{1}-n\right)^{p_{1}-j_{1}}\left(x_{2}-n\right)^{p_{2}-j_{2}}$
(18)
$\displaystyle\times\sum_{i=0}^{n-1}(-1)^{i}s\left(n,n-i\right)S\left(j_{1}+j_{2}+n-i,l+n\right)\frac{(-1)^{l}l!}{(l+1)^{k}},$

where $n$ is an arbitrary positive integer. The simplest cases of (17) and
(18) are

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}S\left(j_{1}+j_{2},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}},$
(19)

and

$\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}$ $\displaystyle(x_{1},x_{2})$
(20)
$\displaystyle=\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(x_{1}-1\right)^{p_{1}-j_{1}}\left(x_{2}-1\right)^{p_{2}-j_{2}}S\left(j_{1}+j_{2}+1,l+1\right)\frac{(-1)^{l}l!}{(l+1)^{k}},$

respectively. Two examples are the following

$\displaystyle B_{1,1}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle\frac{2}{3^{k}}-\frac{1}{2^{k}}\left(x_{1}+x_{2}+1\right)+x_{1}x_{2},$
$\displaystyle B_{1,2}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{3^{k}}\left(2x_{1}+4x_{2}+6\right)-\frac{1}{2^{k}}\left(x_{1}\left(2x_{2}+1\right)+\left(x_{2}+1\right)^{2}\right)+x_{1}x_{2}^{2}-\frac{6}{4^{k}}.$

Clearly we have

$B_{0,0}^{\left(k\right)}(x_{1},x_{2})=1.$ (21)

Observe also that

$B_{p_{1},p_{2}}^{\left(k\right)}\left(x,x\right)=B_{p_{1}+p_{2}}^{\left(k\right)}(x).$
(22)

In particular, we have

$B_{p_{1},p_{2}}^{\left(k\right)}\left(0,0\right)=B_{p_{1}+p_{2}}^{\left(k\right)},$
(23)

From (3) and (16) we have

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{l=0}^{p_{1}+p_{2}}\sum_{j=0}^{l}(-1)^{j}\binom{l}{j}\left(l-j+x_{1}\right)^{p_{1}}\left(l-j+x_{2}\right)^{p_{2}}\frac{(-1)^{l}}{(l+1)^{k}},$
(24)

from where we see that

$\displaystyle\frac{\partial}{\partial
x_{1}}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle p_{1}B_{p_{1}-1,p_{2}}^{\left(k\right)}(x_{1},x_{2}),$
$\displaystyle\frac{\partial}{\partial
x_{2}}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle p_{2}B_{p_{1},p_{2}-1}^{\left(k\right)}(x_{1},x_{2}).$

We can use (8) to write $B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ in
terms of $B_{j_{1},j_{2}}^{\left(k\right)}\left(y_{1},y_{2}\right)$, $0\leq
j_{i}\leq p_{i}$, $i=1,2$, as

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(x_{1}-y_{1}\right)^{p_{1}-j_{1}}\left(x_{2}-y_{2}\right)^{p_{2}-j_{2}}B_{j_{1},j_{2}}^{\left(k\right)}\left(y_{1},y_{2}\right),$
(25)

that generalizes the known addition formula
$B_{p}^{\left(k\right)}(x)=\sum_{j=0}^{p}\binom{p}{j}\left(x-y\right)^{p-j}B_{j}^{\left(k\right)}\left(y\right)$
for one-variable poly-Bernoulli polynomials. In fact, we have

$\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-y_{1})^{p_{1}-j_{1}}(x_{2}-y_{2})^{p_{2}-j_{2}}\sum_{l=0}^{j_{1}+j_{2}}S_{1,y_{1}}^{1,y_{2},j_{2}}\left(j_{1},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-y_{1})^{p_{1}-j_{1}}(x_{2}-y_{2})^{p_{2}-j_{2}}B_{j_{1},j_{2}}^{\left(k\right)}(y_{1},y_{2}),$

as claimed. In particular, if we set $y_{1}=y_{2}=y$ in (25) we obtain an
expression for the bi-variate poly-Bernoulli polynomial
$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ in terms of one-variable poly-
Bernoulli polynomials $B_{j}^{\left(k\right)}\left(y\right)$, $0\leq j\leq
p_{1}+p_{2}$, namely

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-y)^{p_{1}-j_{1}}(x_{2}-y)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k\right)}\left(y\right),$
(26)

and then we can write the bi-variate poly-Bernoulli polynomial
$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})$ in terms of poly-Bernoulli
numbers $B_{j}^{\left(k\right)}$, $0\leq j\leq p_{1}+p_{2}$, as

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k\right)}.$
(27)

The cases $k=0$ and $k=-1$ of (27) are

$B_{p_{1},p_{2}}^{\left(0\right)}(x_{1},x_{2})=\left(x_{1}-1\right)^{p_{1}}\left(x_{2}-1\right)^{p_{2}},$
(28)

and

$B_{p_{1},p_{2}}^{(-1)}(x_{1},x_{2})=\left(x_{1}-2\right)^{p_{1}}\left(x_{2}-2\right)^{p_{2}},$
(29)

respectively. In fact, according to (16), formulas (28) and (29) are the
particular cases $r=0$ and $r=1$ of the identity

$\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}\left(p_{1},l\right)(-1)^{l}\left(l+r\right)!=r!\left(x_{1}-r-1\right)^{p_{1}}\left(x_{2}-r-1\right)^{p_{2}},$
(30)

where $r$ is a non-negative integer. We leave the proof of (30) to the reader.
Observe also that (25) implies that

$B_{p_{1},p_{2}}^{\left(k\right)}(x_{1}+1,x_{2}+1)=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{j_{1},j_{2}}^{\left(k\right)}(x_{1},x_{2}),$
(31)

which generalizes the known binomial formula for standard Bernoulli
polynomials

$B_{p}(x+1)=\sum_{j=0}^{p}\binom{p}{j}B_{j}(x).$ (32)

If we set $x_{1}=x_{2}=x$ in (25), we obtain a formula for the standard poly-
Bernoulli polynomial $B_{p_{1}+p_{2}}^{\left(k\right)}(x)$ in terms of the bi-
variate poly-Bernoulli polynomials
$B_{j_{1},j_{2}}^{\left(k\right)}(y_{1},y_{2})$, $0\leq j_{i}\leq p_{i}$,
$i=1,2$, namely

$B_{p_{1}+p_{2}}^{\left(k\right)}(x)=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x-y_{1})^{p_{1}-j_{1}}(x-y_{2})^{p_{2}-j_{2}}B_{j_{1},j_{2}}^{\left(k\right)}(y_{1},y_{2}).$
(33)

Some additional simple observations are the following

$\displaystyle B_{p_{1},0}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$
$\displaystyle B_{p_{1}}^{\left(k\right)}(x_{1}),$ (34) $\displaystyle
B_{0,p_{2}}^{\left(k\right)}(x_{1},x_{2})$ $\displaystyle=$ $\displaystyle
B_{p_{2}}^{\left(k\right)}(x_{2}),$ (35)

and

$\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}\left(0,x_{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}x_{2}^{p_{2}-j_{2}}B_{p_{1}+j_{2}}^{\left(k\right)},$
(36) $\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}\left(x_{1},0\right)$
$\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}x_{1}^{p_{1}-j_{1}}B_{j_{1}+p_{2}}^{\left(k\right)}.$
(37)

##  3 Some identities

In this section we obtain some identities involving poly-Bernoulli
polynomials, by using the following result:

######  Theorem 3.1.

The polynomial identity

$\sum_{r=0}^{n}a_{n,r}(x+\alpha)^{r}=\sum_{r=0}^{n}b_{n,r}(x+\beta)^{r}.$ (38)

implies the poly-Bernoulli polynomial identity

$\sum_{r=0}^{n}a_{n,r}B_{r}^{\left(k\right)}(x+\alpha)=\sum_{r=0}^{n}b_{n,r}B_{r}^{\left(k\right)}(x+\beta).$
(39)

######  Proof 3.2.

Observe that the hypothesis of the polynomial identity (38), comes together
with the identity of its derivatives:

$\sum_{r=0}^{n}\binom{r}{j}a_{n,r}(x+\alpha)^{r-j}=\sum_{r=0}^{n}\binom{r}{j}b_{n,r}(x+\beta)^{r-j}$

where $j$ is a non-negative integer. We have

$\displaystyle\sum_{r=0}^{n}a_{n,r}B_{r}^{\left(k\right)}(x+\alpha)$
$\displaystyle=$
$\displaystyle\sum_{r=0}^{n}a_{n,r}\sum_{l=0}^{r}S_{1,x+\alpha}(r,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{r=0}^{n}a_{n,r}\sum_{l=0}^{r}\sum_{j=0}^{r}\binom{r}{j}(x+\alpha)^{r-j}S(j,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{n}\sum_{j=0}^{n}\left(\sum_{r=0}^{n}\binom{r}{j}a_{n,r}(x+\alpha)^{r-j}\right)S(j,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{n}\sum_{j=0}^{n}\left(\sum_{r=0}^{n}\binom{r}{j}b_{n,r}(x+\beta)^{r-j}\right)S(j,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{r=0}^{n}b_{n,r}\sum_{l=0}^{r}\sum_{j=0}^{r}\binom{r}{j}(x+\beta)^{r-j}S(j,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{r=0}^{n}b_{n,r}\sum_{l=0}^{r}S_{1,x+\beta}(r,l)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=$
$\displaystyle\sum_{r=0}^{n}b_{n,r}B_{r}^{\left(k\right)}(x+\beta),$

as desired.

######  Remark 3.3.

The case $k=1$ of Theorem 3.1 is an old result: based on [Ko-Pi], we obtained
Theorem 3.1 in the case $k=1$ and we used it to generate several identities in
[Pi2].

For example, by using Theorem 3.1 in the trivial identity
$x^{p}=\sum_{j=0}^{p}\binom{p}{j}(x-y)^{j}y^{p-j}$ we obtain the addition
formula for poly-Bernoulli polynomials

$B_{p}^{\left(k\right)}(x)=\sum_{j=0}^{p}\binom{p}{j}(x-y)^{p-j}B_{j}^{\left(k\right)}(y),$
(40)

that we can write as

$\sum_{j=0}^{p}\binom{p}{j}x^{p-j}B_{j}^{\left(k\right)}=\sum_{j=0}^{p}\binom{p}{j}(x-y)^{p-j}B_{j}^{\left(k\right)}(y).$
(41)

We can use again Theorem 3.1 to obtain from (41) that

$\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}(x)B_{j}^{\left(k\right)}=\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}(x-y)B_{j}^{\left(k\right)}(y).$
(42)

Set $y=x$ in (42) to get the identity

$\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}(x)B_{j}^{\left(k\right)}=\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}B_{j}^{\left(k\right)}(x).$
(43)

If we set $k=1$ in (43), and replace $x$ by $x+1$, we obtain

$\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}(x+1)B_{j}=\sum_{j=0}^{p}\binom{p}{j}B_{p-j}^{\left(k_{1}\right)}B_{j}(x+1).$
(44)

By using that $B_{j}(x+1)=B_{j}(x)+jx^{j-1}$ together with (43), we obtain
from (44), after some elementary algebraic steps, the curious identity

$\sum_{j=0}^{p}\binom{p}{j}\left(B_{p-j}^{\left(k_{1}\right)}(x+1)-B_{p-j}^{\left(k_{1}\right)}(x)\right)B_{j}=pB_{p-1}^{\left(k_{1}\right)}(x).$
(45)

From (17), (18) and (19) we have that

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}$
$\displaystyle\binom{p_{2}}{j_{2}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}S(j_{1}+j_{2},l)\frac{(-1)^{l}l!}{(l+1)^{k_{0}}}$
(46)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-m)^{p_{1}-j_{1}}(x_{2}-m)^{p_{2}-j_{2}}$
$\displaystyle\qquad\times\sum_{i=0}^{m}\binom{m}{i}S(j_{1}+j_{2},l+i)\frac{(-1)^{l}(l+i)!}{(l+1)^{k_{0}}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-n)^{p_{1}-j_{1}}(x_{2}-n)^{p_{2}-j_{2}}$
$\displaystyle\qquad\times\sum_{i=0}^{n-1}(-1)^{i}s(n,n-i)S(j_{1}+j_{2}+n-i,l+n)\frac{(-1)^{l}l!}{(l+1)^{k_{0}}}.$

where $m$, $n$ are arbitrary integers, $m\geq 0$, $n>0$. Now we use Theorem
3.1 in (46) and then set $x_{1}=x_{2}=m+n$, to obtain the identities

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}$
$\displaystyle
B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(m+n)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(m+n)S(j_{1}+j_{2},l)\frac{(-1)^{l}l!}{(l+1)^{k_{0}}}$
(47)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(n)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(n)$
$\displaystyle\qquad\times\sum_{i=0}^{m}\binom{m}{i}S(j_{1}+j_{2},l+i)\frac{(-1)^{l}(l+i)!}{(l+1)^{k_{0}}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(m)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(m)$
$\displaystyle\qquad\times\sum_{i=0}^{n-1}(-1)^{i}s(n,n-i)S(j_{1}+j_{2}+n-i,l+n)\frac{(-1)^{l}l!}{(l+1)^{k_{0}}}.$

From (26) we see that

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}$
$\displaystyle(x_{1}-y)^{p_{1}-j_{1}}(x_{2}-y)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(y)$
(48)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-z)^{p_{1}-j_{1}}(x_{2}-z)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(z).$

We can use Theorem 3.1 to get from (48), the identity

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}$
$\displaystyle
B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-y)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-y)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(y)$
(49)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-z)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-z)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(z).$

Set $x_{1}=x_{2}=y=x$ to obtain from (49) that

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}$
$\displaystyle
B_{p_{1}-j_{1}}^{\left(k_{1}\right)}B_{p_{2}-j_{2}}^{\left(k_{2}\right)}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(x)$
(50)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x-z)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x-z)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(z).$

With $z=0$, $z=1-\left(q-1\right)x$, and $z=qx-1$, where $q$ is an arbitrary
parameter, we obtain from (50) the identities

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}$
$\displaystyle\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}B_{p_{2}-j_{2}}^{\left(k_{2}\right)}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(x)$
(51)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(qx-1)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(qx-1)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(1-\left(q-1\right)x)$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(1-(q-1)x)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(1-(q-1)x)B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(qx-1).$

In the case $q=2$, if some (or all) of the parameters $k_{0},k_{1},k_{2}$ are
equal to $1$, we can use the known property
$B_{r}\left(1-x\right)=(-1)^{r}B_{r}(x)$ to simplify the corresponding
expression in (51). For example, if $k_{0}=k_{1}=k_{2}=1$ and $q=2$, we have
the following identities involving standard Bernoulli numbers and polynomials

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}$
$\displaystyle\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}B_{p_{2}-j_{2}}B_{j_{1}+j_{2}}(x)$
(52)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}(x)B_{p_{2}-j_{2}}(x)B_{j_{1}+j_{2}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(-1)^{j_{1}+j_{2}}B_{p_{1}-j_{1}}(2x-1)B_{p_{2}-j_{2}}(2x-1)B_{j_{1}+j_{2}}(x)$
$\displaystyle=(-1)^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(-1)^{j_{1}+j_{2}}B_{p_{1}-j_{1}}(x)B_{p_{2}-j_{2}}(x)B_{j_{1}+j_{2}}(2x-1).$

In particular, by setting $x=1$ in (52), we see that if $p_{1}+p_{2}$ is odd,
then

$\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(-1)^{j_{1}+j_{2}}B_{p_{1}-j_{1}}B_{p_{2}-j_{2}}B_{j_{1}+j_{2}}=0.$
(53)

From (30) together with (9) (with $m=0$) and (10) (with $n=1$), we have

$\displaystyle\sum_{j_{1}=0}^{p_{1}}$
$\displaystyle\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}S(j_{1}+j_{2},l)(-1)^{l}(l+r)!$
(54)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-1)^{p_{1}-j_{1}}(x_{2}-1)^{p_{2}-j_{2}}S(j_{1}+j_{2}+1,l+1)(-1)^{l}(l+r)!$
$\displaystyle=r!(x_{1}-r-1)^{p_{1}}(x_{2}-r-1)^{p_{2}},$

and then, applying Theorem 3.1 in (54) we get

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1})B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2})S(j_{1}+j_{2},l)(-1)^{l}(l+r)!$
(55)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-1)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-1)S(j_{1}+j_{2}+1,l+1)(-1)^{l}(l+r)!$
$\displaystyle=r!B_{p_{1}}^{\left(k_{1}\right)}(x_{1}-r-1)B_{p_{2}}^{\left(k_{2}\right)}(x_{2}-r-1),$

where $r$ is an arbitrary non-negative integer. Now let us consider the
difference

$B_{p_{1},p_{2}}^{\left(k_{0}\right)}(x_{1}+r,x_{2}+r)-B_{p_{1},p_{2}}^{\left(k_{0}\right)}(x_{1},x_{2}),$
(56)

where $r$ is an arbitrary positive integer. In the case $k_{0}=1$, we know
that (56) is equal to (see [Shib])

$p_{1}\sum_{t=0}^{r-1}(x_{2}+t)^{p_{2}}(x_{1}+t)^{p_{1}-1}+p_{2}\sum_{t=0}^{r-1}(x_{1}+t)^{p_{1}}(x_{2}+t)^{p_{2}-1}.$
(57)

We can use (26) to write (56) as

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}$
$\displaystyle\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left((x_{1}+r)^{p_{1}-j_{1}}(x_{2}+r)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}-x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}\right)$
(58)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left((x_{1}+r-y)^{p_{1}-j_{1}}(x_{2}+r-y)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(y)\right.$
$\displaystyle\left.\qquad-(x_{1}-z)^{p_{1}-j_{1}}(x_{2}-z)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k_{0}\right)}(z)\right),$

where $y$ and $z$ are arbitrary parameters. If $k_{0}=1$, we have from (57)
and (58) that

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}$
$\displaystyle\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left((x_{1}+r)^{p_{1}-j_{1}}(x_{2}+r)^{p_{2}-j_{2}}-x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}\right)B_{j_{1}+j_{2}}$
(59)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left((x_{1}+r-y)^{p_{1}-j_{1}}(x_{2}+r-y)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}(y)\right.$
$\displaystyle\left.\qquad-(x_{1}-z)^{p_{1}-j_{1}}(x_{2}-z)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}(z)\right)$
$\displaystyle=p_{1}\sum_{t=0}^{r-1}(x_{2}+t)^{p_{2}}(x_{1}+t)^{p_{1}-1}+p_{2}\sum_{t=0}^{r-1}(x_{1}+t)^{p_{1}}(x_{2}+t)^{p_{2}-1}.$

By using Theorem 3.1 in (59) we get

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}$
$\displaystyle\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}+r)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}+r)-B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1})B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2})\right)B_{j_{1}+j_{2}}$
(60)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}+r-y)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}+r-y)B_{j_{1}+j_{2}}(y)\right.$
$\displaystyle\left.\qquad-
B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-z)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-z)B_{j_{1}+j_{2}}(z)\right)$
$\displaystyle=p_{1}\sum_{t=0}^{r-1}B_{p_{1}-1}^{\left(k_{1}\right)}(x_{1}+t)B_{p_{2}}^{\left(k_{2}\right)}(x_{2}+t)+p_{2}\sum_{t=0}^{r-1}B_{p_{1}}^{\left(k_{1}\right)}(x_{1}+t)B_{p_{2}-1}^{\left(k_{2}\right)}(x_{2}+t).$

Set
$\left(y,z\right)=\left(r,r\right),\left(r,0\right),\left(0,-r\right),\left(2r,r\right)$
in (60), to obtain the following identities involving poly-Bernoulli
polynomials, standard Bernoulli polynomials and Bernoulli numbers

$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}+r)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}+r)-B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1})B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2})\right)B_{j_{1}+j_{2}}$
(61)
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\left(B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1})B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2})-B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-r)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-r)\right)B_{j_{1}+j_{2}}(r)$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1})B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2})\left(B_{j_{1}+j_{2}}(r)-B_{j_{1}+j_{2}}\right)$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}+r)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}+r)\left(B_{j_{1}+j_{2}}-B_{j_{1}+j_{2}}(-r)\right)$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}B_{p_{1}-j_{1}}^{\left(k_{1}\right)}(x_{1}-r)B_{p_{2}-j_{2}}^{\left(k_{2}\right)}(x_{2}-r)\left(B_{j_{1}+j_{2}}(2r)-B_{j_{1}+j_{2}}(r)\right)$
$\displaystyle=p_{1}\sum_{t=0}^{r-1}B_{p_{1}-1}^{\left(k_{1}\right)}(x_{1}+t)B_{p_{2}}^{\left(k_{2}\right)}(x_{2}+t)+p_{2}\sum_{t=0}^{r-1}B_{p_{1}}^{\left(k_{1}\right)}(x_{1}+t)B_{p_{2}-1}^{\left(k_{2}\right)}(x_{2}+t).$

##  4 Generalized recurrences

In this section we show two generalized recurrences for bi-variate poly-
Bernoulli polynomials, and obtain some consequences of them.

######  Proposition 4.1.

We have

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}B_{p_{1},p_{2}+l}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=$
$\displaystyle-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

where $\mathcal{R}_{-1,k}(y)=-1$, and for $\mu\geq 0$ the functions
$\mathcal{R}_{\mu,k}(y)$ are defined recursively by

$\mathcal{R}_{\mu,k}(y)=y(y+\mu+2)^{k}\mathcal{R}_{\mu-1,k}(y)-(y+1)^{k+1}\mathcal{R}_{\mu-1,k}(y+1).$

######  Proof 4.2.

We prove that

$\sum_{l=0}^{q}\binom{q}{k}(-x_{1})^{q-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}$
(63)

by induction on $q$. The case $q=0$ of (63) is the definition (16). Let us
suppose formula (63) is true for a given $q\in\mathbb{N}$. Then

$\displaystyle\sum_{l=0}^{q+1}$
$\displaystyle\binom{q+1}{l}(-x_{1})^{q+1-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=-x_{1}\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})+\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}B_{p_{1}+1+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=x_{1}\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}-\sum_{l=0}^{p_{1}+1}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1}+1,l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

Now we use the recurrence (12) to write

$\displaystyle\sum_{l=0}^{q+1}\binom{q+1}{l}$
$\displaystyle(-x_{1})^{q+1-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=x_{1}\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}$
$\displaystyle\qquad-\sum_{l=0}^{p_{1}+1}\left(S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l-1)+(l+x_{1})S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\right)\frac{(-1)^{l}l!\mathcal{R}_{q-1}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

Some further simplifications give us

$\displaystyle\sum_{l=0}^{q+1}\binom{q+1}{l}(-x_{1})^{q+1-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=-\sum_{l=0}^{p_{1}+1}\left(S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l-1)+lS_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\right)\frac{(-1)^{l}l!\mathcal{R}_{q-1}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}$
$\displaystyle=\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}(l+1)!\mathcal{R}_{q-1}(l+1)}{\left(\prod_{i=1}^{q+1}(l+i+1)\right)^{k}}-\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!l\mathcal{R}_{q-1}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}$
$\displaystyle=-\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}\left(p_{1},l\right)\left(l(l+q+2)^{k}\mathcal{R}_{q-1}(l)-(l+1)^{k+1}\mathcal{R}_{q-1}(l+1)\right)\frac{(-1)^{l}l!}{\left(\prod_{i=1}^{q+2}(l+i)\right)^{k}}$
$\displaystyle=-\sum_{l=0}^{p_{1}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q}(l)}{\left(\prod_{i=1}^{q+2}(l+i)\right)^{k}},$

as desired. The proof of

$\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}B_{p_{1},p_{2}+l}^{\left(k\right)}(x_{1},x_{2})=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}},$

is similar.

For example, we have

$\mathcal{R}_{0,k}(y)=(y+1)^{k+1}-y(y+2)^{k},$
$\mathcal{R}_{1,k}(y)=(2y+1)(y+1)^{k+1}(y+3)^{k}-y^{2}(y+2)^{k}(y+3)^{k}-(y+1)^{k+1}(y+2)^{k+1},$
(64)

and then, formula (4.1) with $q=1$ is

$\displaystyle-x_{1}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})+$
$\displaystyle
B_{p_{1}+1,p_{2}}^{\left(k\right)}(x_{1},x_{2})=-x_{2}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})+B_{p_{1},p_{2}+1}^{\left(k\right)}(x_{1},x_{2})$
(65)
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k)(-1)^{l}l!\left(\frac{l+1}{(l+2)^{k}}-\frac{l}{(l+1)^{k}}\right),$

and with $q=2$ is

$\displaystyle x_{1}^{2}B_{p_{1},p_{2}}^{\left(k\right)}$
$\displaystyle(x_{1},x_{2})-2x_{1}B_{p_{1}+1,p_{2}}^{\left(k\right)}(x_{1},x_{2})+B_{p_{1}+2,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
(66)
$\displaystyle=x_{2}^{2}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})-2x_{2}B_{p_{1},p_{2}+1}^{\left(k\right)}(x_{1},x_{2})+B_{p_{1},p_{2}+2}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)(-1)^{l}l!\left(\frac{(2l+1)(l+1)}{(l+2)^{k}}-\frac{l^{2}}{(l+1)^{k}}-\frac{(l+1)(l+2)}{(l+3)^{k}}\right).$

We can write (4.1) by using (26) as

$\displaystyle\sum_{l=0}^{q}$
$\displaystyle\binom{q}{l}(-x_{1})^{q-l}\sum_{j_{1}=0}^{p_{1}+l}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}+l}{j_{1}}\binom{p_{2}}{j_{2}}(x_{1}-y)^{p_{1}+l-j_{1}}(x_{2}-y)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}^{\left(k\right)}(y)$
(67)
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}+l}\binom{p_{1}}{j_{1}}\binom{p_{2}+l}{j_{2}}(x_{1}-z)^{p_{1}-j_{1}}(x_{2}-z)^{p_{2}+l-j_{2}}B_{j_{1}+j_{2}}^{\left(k\right)}(z)$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

By setting $y,z=x_{1},x_{2}$ in (67), we get the identities

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}$
$\displaystyle(-x_{1})^{q-l}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}(x_{2}-x_{1})^{p_{2}-j_{2}}B_{p_{1}+l+j_{2}}^{\left(k\right)}(x_{1})$
(68)
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}\sum_{j_{1}=0}^{p_{1}+l}\binom{p_{1}+l}{j_{1}}(x_{1}-x_{2})^{p_{1}+l-j_{1}}B_{j_{1}+p_{2}}^{\left(k\right)}(x_{2})$
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}\sum_{j_{2}=0}^{p_{2}+l}\binom{p_{2}+l}{j_{2}}(x_{2}-x_{1})^{p_{2}+l-j_{2}}B_{p_{1}+j_{2}}^{\left(k\right)}(x_{1})$
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}(x_{1}-x_{2})^{p_{1}-j_{1}}B_{j_{1}+p_{2}+l}^{\left(k\right)}(x_{2})$
$\displaystyle=-\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}S\left(j_{1}+j_{2},l\right)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

If we set $x_{2}=0$ in (68), we get

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}$
$\displaystyle(-x_{1})^{p_{2}-j_{2}}B_{p_{1}+l+j_{2}}^{\left(k\right)}(x_{1})$
(69)
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}\sum_{j_{1}=0}^{p_{1}+l}\binom{p_{1}+l}{j_{1}}x_{1}^{p_{1}+l-j_{1}}B_{j_{1}+p_{2}}^{\left(k\right)}$
$\displaystyle=\sum_{j_{2}=0}^{p_{2}+q}\binom{p_{2}+q}{j_{2}}(-x_{1})^{p_{2}+q-j_{2}}B_{p_{1}+j_{2}}^{\left(k\right)}(x_{1})$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}x_{1}^{p_{1}-j_{1}}B_{j_{1}+p_{2}+q}^{\left(k\right)}$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}}^{1,0,p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}.$

The case $q=0,x_{1}=1,k=1$ of (69) reduces to

$\displaystyle(-1)^{p_{1}+p_{2}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+j_{2}}$
$\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}B_{j_{1}+p_{2}}$
$\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{l=0}^{j_{1}+p_{2}}\binom{p_{1}}{j_{1}}S(j_{1}+p_{2},l)\frac{(-1)^{l}l!}{l+1}.$

Formula (4) is the famous Carlitz identity [Ca]. In terms of bi-variate
Bernoulli polynomials, Carlitz identity is written as

$B_{p_{1},p_{2}}(1,0)=(-1)^{p_{1}+p_{2}}B_{p_{1},p_{2}}(0,1).$ (71)

For example, we can use (71) to write the following version of (4.1) in the
case $k=1$, when $x_{1}=0,x_{2}=1$

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-1)^{q-l}B_{p_{1}+l,p_{2}}(1,0)$
$\displaystyle=$
$\displaystyle(-1)^{p_{1}+p_{2}+q}\sum_{l=0}^{q}\binom{q}{l}B_{p_{1}+l,p_{2}}(0,1)$
$\displaystyle=$ $\displaystyle
B_{p_{1},p_{2}+q}(1,0)=(-1)^{p_{1}+p_{2}+q}B_{p_{1},p_{2}+q}(0,1)$
$\displaystyle=$
$\displaystyle-\sum_{l=0}^{p_{1}+p_{2}}S_{1,1}^{1,0,p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}l!\mathcal{R}_{q-1,1}\left(l\right)}{\prod_{i=1}^{q+1}(l+i)},$

or, explicitly

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-1)^{q-l}\sum_{j_{1}=0}^{p_{1}+l}$
$\displaystyle\binom{p_{1}+l}{j_{1}}B_{j_{1}+p_{2}}$
$\displaystyle=(-1)^{p_{1}+p_{2}+q}\sum_{l=0}^{q}\binom{q}{l}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+l+j_{2}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}B_{j_{1}+p_{2}+q}=(-1)^{p_{1}+p_{2}+q}\sum_{j_{2}=0}^{p_{2}+q}\binom{p_{2}+q}{j_{2}}B_{p_{1}+j_{2}}$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}S(j_{1}+p_{2},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,1}(l)}{\prod_{i=1}^{q+1}(l+i)}.$

It is easy to check that in the case $k=0$, the functions
$\mathcal{R}_{\mu,k}\left(y\right)$ of Proposition 4.1 are
$\mathcal{R}_{\mu,0}(y)=(-1)^{\mu}$. Thus, the case $k=0$ of (4.1) is the case
$r=0$ of (30). Also, in the case $k=-1$, we have that
$\mathcal{R}_{\mu,-1}(y)=\frac{(-1)^{\mu}2^{\mu+1}}{\prod_{i=2}^{\mu+2}(y+i)}$,
and then the case $k=-1$ of (4.1) is the case $r=1$ of (30).

######  Proposition 4.3.

For non-negative integers $p_{1},p_{2},q$ we have

$\displaystyle\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}$
$\displaystyle\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)$ (72)
$\displaystyle=\sum_{l=0}^{q}(-1)^{l}B_{p_{1},p_{2}+l}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{2}^{l}}\prod_{i=0}^{q-1}(x_{2}+i)$
$\displaystyle=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}}.$

######  Proof 4.4.

We prove that

$\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}\left(l+q\right)!}{\left(l+q+1\right)^{k}},$
(73)

by induction on $q$. The case $q=0$ of (73) is the definition (16). If we
suppose that (73) is true for a given $q\in\mathbb{N}$, then

$\displaystyle\sum_{l=0}^{q+1}(-1)^{l}$ $\displaystyle
B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q}(x_{1}+i)$
$\displaystyle=\sum_{l=0}^{q+1}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\left((x_{1}+q)\prod_{i=0}^{q-1}(x_{1}+i)\right)$
$\displaystyle=\sum_{l=0}^{q+1}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\left((x_{1}+q)\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)+l\frac{d^{l-1}}{dx_{1}^{l-1}}\prod_{i=0}^{q-1}(x_{1}+i)\right)$
$\displaystyle=(x_{1}+q)\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)$
$\displaystyle\qquad+\sum_{l=1}^{q+1}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{\left(l-1\right)!}\frac{d^{l-1}}{dx_{1}^{l-1}}\prod_{i=0}^{q-1}(x_{1}+i)$
$\displaystyle=(x_{1}+q)\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)$
$\displaystyle\qquad-\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+1+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)$
$\displaystyle=(x_{1}+q)\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}\left(l+q\right)!}{\left(l+q+1\right)^{k}}$
$\displaystyle\qquad-\sum_{l=0}^{p_{1}+p_{2}+1}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}\left(p_{1}+1,l\right)\frac{(-1)^{l}\left(l+q\right)!}{\left(l+q+1\right)^{k}}.$

Now we use the recurrence (12) and formula (11) to write

$\displaystyle\sum_{l=0}^{q+1}(-1)^{l}$ $\displaystyle
B_{p_{1}+l,p_{2}}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q}(x_{1}+i)$
$\displaystyle=(x_{1}+q)\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}}$
$\displaystyle\qquad-\sum_{l=0}^{p_{1}+p_{2}+1}\left(S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l-1)+(l+x_{1}+q)S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l)\right)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}}$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}+1}\left(S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l-1)+lS_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l)\right)\frac{(-1)^{l}(l+q)!}{\left(l+q+1\right)^{k}}$
$\displaystyle=-\sum_{l=1}^{p_{1}+p_{2}+1}S_{1,x_{1}+q+1}^{1,x_{2}+q+1,p_{2}}(p_{1},l-1)\frac{(-1)^{l}(l+q)!}{\left(l+q+1\right)^{k}}$
$\displaystyle=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q+1}^{1,x_{2}+q+1,p_{2}}(p_{1},l)\frac{(-1)^{l}(l+q+1)!}{(l+q+2)^{k}},$

as desired. The proof of

$\sum_{l=0}^{q}(-1)^{l}B_{p_{1},p_{2}+l}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!}\frac{d^{l}}{dx_{2}^{l}}\prod_{i=0}^{q-1}(x_{2}+i)=\sum_{l=0}^{p_{1}+p_{2}}S_{1,x_{1}+q}^{1,x_{2}+q,p_{2}}(p_{1},l)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}},$

is similar.

Formula (72) with $x_{1}=0,x_{2}=1$ looks as

$\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l)B_{p_{1}+l,p_{2}}^{\left(k\right)}(0,1)$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)B_{p_{1},p_{2}+l}^{\left(k\right)}(0,1)$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{p_{1}+p_{2}}S_{1,q}^{1,q+1,p_{2}}\left(p_{1},l\right)\frac{(-1)^{l}\left(l+q\right)!}{\left(l+q+1\right)^{k}},$

or, explicitly

$\displaystyle\sum_{l=0}^{q}(-1)^{l}$ $\displaystyle
s(q,l)\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+l+j_{2}}^{\left(k\right)}=\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)\sum_{j_{2}=0}^{p_{2}+l}\binom{p_{2}+l}{j_{2}}B_{p_{1}+j_{2}}^{\left(k\right)}$
(75)
$\displaystyle=\sum_{l=0}^{p_{1}+p_{2}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}q^{p_{1}-j_{1}}(q+1)^{p_{2}-j_{2}}S(j_{1}+j_{2},l)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}}.$

If we set $k=1$ in formula (4), we can use Carlitz identity (71) to write the
following enriched version of (4)

$\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l)B_{p_{1}+l,p_{2}}(0,1)$
$\displaystyle=$
$\displaystyle(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q,l)B_{p_{1}+l,p_{2}}(1,0)$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)B_{p_{1},p_{2}+l}(0,1)$
$\displaystyle=$
$\displaystyle(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q+1,l+1)B_{p_{1},p_{2}+l}(1,0)$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{p_{1}+p_{2}}S_{1,q}^{1,q+1,p_{2}}(p_{1},l)\frac{(-1)^{l}(l+q)!}{l+q+1},$

or, explicitly

$\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l)$
$\displaystyle\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+l+j_{2}}$
(77)
$\displaystyle=(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q,l)\sum_{j_{1}=0}^{p_{1}+l}\binom{p_{1}+l}{j_{1}}B_{j_{1}+p_{2}}$
$\displaystyle=\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)\sum_{j_{2}=0}^{p_{2}+l}\binom{p_{2}+l}{j_{2}}B_{p_{1}+j_{2}}$
$\displaystyle=(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q+1,l+1)\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}B_{j_{1}+p_{2}+l}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{l=0}^{j_{1}+j_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}q^{p_{1}-j_{1}}(q+1)^{p_{2}-j_{2}}S(j_{1}+j_{2},l)\frac{(-1)^{l}(l+q)!}{l+q+1}.$

To end this section, let us consider the case $q=1$ of the first two lines of
(72). That is

$B_{p_{1},p_{2}+1}^{\left(k\right)}(x_{1},x_{2})-B_{p_{1}+1,p_{2}}^{\left(k\right)}(x_{1},x_{2})=(x_{2}-x_{1})B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}).$
(78)

Formula (78) is the first step of two results contained in the following
proposition.

######  Proposition 4.5.

We have the following identities:

  1. a)

$\sum_{j=0}^{q}\binom{q}{j}(-1)^{j}B_{p_{1}+j,p_{2}+q-j}^{\left(k\right)}(x_{1},x_{2})=(x_{2}-x_{1})^{q}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}).$
(79)

  2. b)

$\sum_{j=0}^{q}\binom{q}{j}(x_{2}-x_{1})^{j}B_{p_{1}+q-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})=B_{p_{1},p_{2}+q}^{\left(k\right)}(x_{1},x_{2}).$
(80)

######  Proof 4.6.

Let us prove (79) by induction on $q$. The case $q=0$ is a trivial identity.
Let us suppose that (79) is true for a given $q\in\mathbb{N}$. Then

$\displaystyle\sum_{j=0}^{q+1}$
$\displaystyle\binom{q+1}{j}(-1)^{j}B_{p_{1}+j,p_{2}+q+1-j}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=\sum_{j=0}^{q+1}\left(\binom{q}{j}+\binom{q}{j-1}\right)(-1)^{j}B_{p_{1}+j,p_{2}+q+1-j}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=\sum_{j=0}^{q}\binom{q}{j}(-1)^{j}B_{p_{1}+j,p_{2}+q+1-j}^{\left(k\right)}(x_{1},x_{2})+\sum_{j=0}^{q}\binom{q}{j}(-1)^{j+1}B_{p_{1}+1+j,p_{2}+q-j}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=(x_{2}-x_{1})^{q}B_{p_{1},p_{2}+1}^{\left(k\right)}(x_{1},x_{2})-(x_{2}-x_{1})^{q}B_{p_{1}+1,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=(x_{2}-x_{1})^{q}\left(B_{p_{1},p_{2}+1}^{\left(k\right)}(x_{1},x_{2})-B_{p_{1}+1,p_{2}}^{\left(k\right)}(x_{1},x_{2})\right)$
$\displaystyle=(x_{2}-x_{1})^{q+1}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}),$

as desired. In the last step we used (78). Now let us prove (80). Again we
proceed by induction on $q$. The case $q=0$ is a trivial identity. Let us
suppose that (80) is true for a given $q\in\mathbb{N}$. Then

$\displaystyle\sum_{j=0}^{q+1}$
$\displaystyle\binom{q+1}{j}(x_{2}-x_{1})^{j}B_{p_{1}+q+1-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=\sum_{j=0}^{q+1}\left(\binom{q}{j}+\binom{q}{j-1}\right)(x_{2}-x_{1})^{j}B_{p_{1}+q+1-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=\sum_{j=0}^{q}\binom{q}{j}(x_{2}-x_{1})^{j}B_{p_{1}+q+1-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})+\sum_{j=0}^{q}\binom{q}{j}(x_{2}-x_{1})^{j+1}B_{p_{1}+q-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=B_{p_{1}+1,p_{2}+q}^{\left(k\right)}(x_{1},x_{2})+(x_{2}-x_{1})B_{p_{1},p_{2}+q}^{\left(k\right)}(x_{1},x_{2})$
$\displaystyle=B_{p_{1},p_{2}+q+1}^{\left(k\right)}(x_{1},x_{2}),$

as desired. We used (78) (with $p_{2}$ replaced by $p_{2}+q$) in the last
step.

The case $p_{1}=0$ of (80) is

$\sum_{j=0}^{q}\binom{q}{j}(x_{2}-x_{1})^{j}B_{q-j,p_{2}}^{\left(k\right)}(x_{1},x_{2})=B_{p_{2}+q}^{\left(k\right)}(x_{2})\text{.}$
(81)

The case $p_{2}=0$ of (81) is the addition formula for standard poly-Bernoulli
polynomials, namely

$\sum_{j=0}^{q}\binom{q}{j}(x_{2}-x_{1})^{q-j}B_{j}^{\left(k\right)}(x_{1})=B_{q}^{\left(k\right)}(x_{2}).$

The case $p_{1}=p_{2}=0$ of (79) is

$\sum_{j=0}^{q}\binom{q}{j}(-1)^{j}B_{j,q-j}^{\left(k\right)}(x_{1},x_{2})=(x_{2}-x_{1})^{q}.$
(82)

As a final comment, we mention that by considering the GSN
$S_{1,x_{1}}^{(1,x_{2},p_{2}),\ldots,(1,x_{n},p_{n})}\left(p_{1},k\right)$
involved in the expansion

$(m+x_{1})^{p_{1}}\cdots(m+x_{n})^{p_{n}}=\sum_{l=0}^{p_{1}+\cdots+p_{n}}l!S_{1,x_{1}}^{(1,x_{2},p_{2}),\ldots,(1,x_{n},p_{n})}\left(p_{1},l\right)\binom{m}{l},$
(83)

where $p_{1},p_{2},\ldots,p_{n}$ are non-negative integers given, one can
define poly-Bernoulli polynomials in $n$ variables $x_{1},\ldots,x_{n}$,
denoted as $B_{p_{1},\ldots,p_{n}}^{(k)}(x_{1},\ldots,x_{n})$, as

$B_{p_{1},\ldots,p_{n}}^{\left(k\right)}(x_{1},\ldots,x_{n})=\sum_{l=0}^{p_{1}+\cdots+p_{n}}S_{1,x_{1}}^{(1,x_{2},p_{2}),\ldots,(1,x_{n},p_{n})}\left(p_{1},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}},$

or explicitly as

$\displaystyle B_{p_{1},\ldots,p_{n}}^{\left(k\right)}$
$\displaystyle(x_{1},\ldots,x_{n})$
$\displaystyle=\sum_{l=0}^{p_{1}+\cdots+p_{n}}\sum_{j_{1}=0}^{p_{1}}\cdots\sum_{j_{n}=0}^{p_{n}}\binom{p_{1}}{j_{1}}\cdots\binom{p_{n}}{j_{n}}x_{1}^{p_{1}-j_{1}}\cdots
x_{n}^{p_{n}-j_{n}}S\left(j_{1}+\cdots+j_{n},l\right)\frac{(-1)^{l}l!}{(l+1)^{k}}$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\cdots\sum_{j_{n}=0}^{p_{n}}\binom{p_{1}}{j_{1}}\cdots\binom{p_{n}}{j_{n}}x_{1}^{p_{1}-j_{1}}\cdots
x_{n}^{p_{n}-j_{n}}B_{j_{1}+\cdots+j_{n}}^{\left(k\right)}.$

In this more general setting we have natural generalizations of results (4.1),
(72), (79) and (80). We show the corresponding results in the case of poly-
Bernoulli polynomials in 3 variables:

$B_{p_{1},p_{2},p_{3}}^{(k)}(x_{1},x_{2},x_{3})=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{j_{3}=0}^{p_{3}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\binom{p_{3}}{j_{3}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}x_{3}^{p_{3}-j_{3}}B_{j_{1}+j_{2}+j_{3}}^{\left(k\right)},$

  1. (a)

(See (4.1)) We have the generalized recurrences

$\displaystyle\sum_{l=0}^{q}\binom{q}{l}$
$\displaystyle(-x_{1})^{q-l}B_{p_{1}+l,p_{2},p_{3}}^{\left(k\right)}(x_{1},x_{2},x_{3})$
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{2})^{q-l}B_{p_{1},p_{2}+l,p_{3}}^{\left(k\right)}(x_{1},x_{2},x_{3})$
$\displaystyle=\sum_{l=0}^{q}\binom{q}{l}(-x_{3})^{q-l}B_{p_{1},p_{2},p_{3}+l}^{\left(k\right)}(x_{1},x_{2},x_{3})$
$\displaystyle=-\sum_{l=0}^{p_{1}+p_{2}+p_{3}}\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{j_{3}=0}^{p_{3}}\left(\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\binom{p_{3}}{j_{3}}x_{1}^{p_{1}-j_{1}}x_{2}^{p_{2}-j_{2}}x_{3}^{p_{3}-j_{3}}\right.$
$\displaystyle\hskip 170.71652pt\left.\times
S(j_{1}+j_{2}+j_{3},l)\frac{(-1)^{l}l!\mathcal{R}_{q-1,k}(l)}{\left(\prod_{i=1}^{q+1}(l+i)\right)^{k}}\right),$

where $\mathcal{R}_{q-1,k}(l)$ is defined in Proposition 4.1.

  2. (b)

(See (72)) We have the generalized recurrences

$\displaystyle\sum_{l=0}^{q}$
$\displaystyle(-1)^{l}B_{p_{1}+l,p_{2},p_{3}}^{\left(k\right)}(x_{1},x_{2},x_{3})\frac{1}{l!}\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i)$
$\displaystyle=\sum_{l=0}^{q}(-1)^{l}B_{p_{1},p_{2}+l,p_{3}}^{\left(k\right)}(x_{1},x_{2},x_{3})\frac{1}{l!}\frac{d^{l}}{dx_{2}^{l}}\prod_{i=0}^{q-1}(x_{2}+i)$
$\displaystyle=\sum_{l=0}^{q}(-1)^{l}B_{p_{1},p_{2},p_{3}+l}^{\left(k\right)}(x_{1},x_{2},x_{3})\frac{1}{l!}\frac{d^{l}}{dx_{3}^{l}}\prod_{i=0}^{q-1}(x_{3}+i)$
$\displaystyle=\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\sum_{j_{3}=0}^{p_{3}}\left(\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}}\binom{p_{3}}{j_{3}}(x_{1}+q)^{p_{1}-j_{1}}(x_{2}+q)^{p_{2}-j_{2}}(x_{3}+q)^{p_{3}-j_{3}}\right.$
$\displaystyle\hskip 199.16928pt\left.\times
S(j_{1}+j_{2}+j_{3},l)\frac{(-1)^{l}(l+q)!}{(l+q+1)^{k}}\right).$

  3. (c)

(See (79)) We have the identities

$\displaystyle\sum_{j=0}^{q}\binom{q}{j}\left(-1\right)^{j}B_{p_{1},p_{2}+j,p_{3}+q-j}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=$
$\displaystyle\left(x_{3}-x_{2}\right)^{q}B_{p_{1},p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right),$
$\displaystyle\sum_{j=0}^{q}\binom{q}{j}\left(-1\right)^{j}B_{p_{1}+j,p_{2},p_{3}+q-j}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=$
$\displaystyle\left(x_{3}-x_{1}\right)^{q}B_{p_{1},p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right),$
$\displaystyle\sum_{j=0}^{q}\binom{q}{j}\left(-1\right)^{j}B_{p_{1}+j,p_{2}+q-j,p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=$
$\displaystyle\left(x_{2}-x_{1}\right)^{q}B_{p_{1},p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right).$

  4. (d)

(See (80)) We have the identities

$\displaystyle\sum_{j=0}^{q}\binom{q}{j}$
$\displaystyle\left(x_{1}-x_{2}\right)^{j}B_{p_{1},p_{2}+q-j,p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=\sum_{j=0}^{q}\binom{q}{j}\left(x_{1}-x_{3}\right)^{j}B_{p_{1},p_{2},p_{3}+q-j}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)=B_{p_{1}+q,p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right).$
$\displaystyle\sum_{j=0}^{q}\binom{q}{j}$
$\displaystyle\left(x_{2}-x_{1}\right)^{j}B_{p_{1}+q-j,p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=\sum_{j=0}^{q}\binom{q}{j}\left(x_{2}-x_{3}\right)^{j}B_{p_{1},p_{2},p_{3}+q-j}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)=B_{p_{1},p_{2}+q,p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right).$
$\displaystyle\sum_{j=0}^{q}\binom{q}{j}$
$\displaystyle\left(x_{3}-x_{1}\right)^{j}B_{p_{1}+q-j,p_{2},p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)$
$\displaystyle=\sum_{j=0}^{q}\binom{q}{j}\left(x_{3}-x_{2}\right)^{j}B_{p_{1},p_{2}+q-j,p_{3}}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right)=B_{p_{1},p_{2},p_{3}+q}^{\left(k\right)}\left(x_{1},x_{2},x_{3}\right).$

### Acknowledgments

I thank the anonymous referee for his/her helpful observations, that certainly
contributed to improve the final version of this work.

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October 08, 2020November 28, 2021Karl Dilcher