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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 protected]
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.
\msc

11B68, 11B73. \VOLUME31 \YEAR2023 \NUMBER1 \DOIhttps://doi.org/10.46298/cm.10327 {paper}

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 BnB_{n} and Bernoulli polynomials Bn(x)B_{n}(x) is about the so-called poly-Bernoulli numbers and poly-Bernoulli polynomials. Poly-Bernoulli numbers Bn(k)B_{n}^{\left(k\right)}, where kk is a given positive integer, were introduced by M. Kaneko [Ka] in 1997, by means of the generating function

Lik(1et)1et=n=0Bn(k)tnn!,\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(z)k=j=1zj/jk{}_{k}(z)=\sum_{j=1}^{\infty}z^{j}/j^{k} is the polylogarithm function. The case k=1k=1 corresponds to the standard Bernoulli numbers BnB_{n} (except the sign of B1(1)B_{1}^{\left(1\right)}). Poly-Bernoulli polynomials Bn(k)(x)B_{n}^{\left(k\right)}(x) can be defined by the generating function

Lik(1et)1etetx=n=0Bn(k)(x)tnn!,\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=1k=1 corresponds to (1)nBn(x)(-1)^{n}B_{n}(x), and the case x=0x=0 corresponds to the poly-Bernoulli numbers Bn(k)B_{n}^{\left(k\right)} mentioned before. Some slightly different definitions of poly-Bernoulli polynomials Bn(k)(x)B_{n}^{\left(k\right)}(x), with xx replaced by x-x, and/or with an additional factor (1)n(-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

Bn(k)(x)=l=0n1(l+1)kj=0l(1)j(lj)(j+x)n,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 Bn(k)(x)B_{n}^{\left(k\right)}(x) is also used for a different kind of mathematical objects, namely, Bernoulli polynomials of kk-th order (see [Ca1]). A different generalization of Bernoulli polynomials, studied in the past few years, is about considering Bernoulli polynomials in several variables Bp1,,pt(x1,,xt)B_{p_{1},\ldots,p_{t}}(x_{1},\ldots,x_{t}), that is, polynomials of degree pip_{i} in the variable xix_{i}, with

B0,,pi,,0(x1,,xt)=Bpi(xi) for each i{1,,t},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 Sa1,b1(a2,b2,p2)(p1,k)S_{a_{1},b_{1}}^{\left(a_{2},b_{2},p_{2}\right)}\left(p_{1},k\right), where aj,bja_{j},b_{j}\in\mathbb{C}, aj0a_{j}\neq 0, j=1,2j=1,2, and p1,p2p_{1},p_{2} non-negative integers, are defined by means of the expansion

(a1n+b1)p1(a2n+b2)p2=k=0p1+p2k!Sa1,b1a2,b2,p2(p1,k)(nk),(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)

(Sa1,b1a2,b2,p2(p1,k)=0S_{a_{1},b_{1}}^{a_{2},b_{2},p_{2}}(p_{1},k)=0 if k<0k<0 or k>p1+p2k>p_{1}+p_{2} ). An explicit formula for these numbers is

Sa1,b1a2,b2,p2(p1,k)=1k!j=0k(1)j(kj)(a1(kj)+b1)p1(a2(kj)+b2)p2.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 p2=0p_{2}=0, we write the GSN Sa,ba2,b2,0(p,k)S_{a,b}^{a_{2},b_{2},0}(p,k) as Sa,b(p,k)S_{a,b}(p,k). We have

Sa,b(p,k)=1k!j=0k(1)j(kj)(a(kj)+b)p.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=0a=1,b=0, the corresponding GSN S1,0(p,k)=1k!j=0k(1)j(kj)(kj)pS_{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)S(p,k). From (3) it is clear that Sa,ba,b,p2(p1,k)=Sa,b(p1+p2,k)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

S1,1(p,k)\displaystyle S_{1,1}(p,k) =\displaystyle= S(p+1,k+1),\displaystyle S(p+1,k+1), (5)
S1,2(p,k)\displaystyle S_{1,2}(p,k) =\displaystyle= S(p+2,k+2)S(p+1,k+2).\displaystyle S(p+2,k+2)-S(p+1,k+2). (6)

In this work we will use GSN of the form S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k). Some important facts about the GSN S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) are the following:

  • Some values of the GSN S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) are

    S1,x11,x2,p2(p1,0)\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},0) =\displaystyle= x1p1x2p2,\displaystyle\!\!x_{1}^{p_{1}}\!x_{2}^{p_{2}}\!, (7)
    S1,x11,x2,p2(p1,1)\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},1) =\displaystyle= (x1+1)p1(x2+1)p2x1p1x2p2,\displaystyle(x_{1}+1)^{p_{1}}\!(x_{2}+1)^{p_{2}}\!-x_{1}^{p_{1}}\!x_{2}^{p_{2}}\!,
    \displaystyle\vdots
    S1,x11,x2,p2(p1,p1+p2)\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},p_{1}+p_{2}) =\displaystyle= 1.\displaystyle 1.
  • The GSN S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) can be written in terms of the GSN S1,y11,y2,p2(p1,k)S_{1,y_{1}}^{1,y_{2},p_{2}}(p_{1},k) as follows

    S1,x11,x2,p2(p1,k)=j1=0p1j2=0p2(p1j1)(p2j2)(x1y1)p1j1(x2y2)p2j2S1,y11,y2,j2(j1,k).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 S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) can be written in terms of standard Stirling numbers as follows

    k!S1,x11,x2,p2(p1,k)\displaystyle k!S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) =\displaystyle= j1=0p1j2=0p2(p1j1)(p2j2)(x1m)p1j1(x2m)p2j2\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)
    ×i=0m(mi)(k+i)!S(j1+j2,k+i),\displaystyle\times\sum_{i=0}^{m}\binom{m}{i}(k+i)!S(j_{1}+j_{2},k+i),

    where mm is an arbitrary non-negative integer.

  • The GSN S1,x11,x2,p2(p1,k)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

    S1,x11,x2,p2(p1,k)\displaystyle S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) =\displaystyle= j1=0p1j2=0p2(p1j1)(p2j2)(x1n)p1j1(x2n)p2j2\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)
    ×i=0n1(1)is(n,ni)S(j1+j2+ni,k+n),\displaystyle\times\sum_{i=0}^{n-1}(-1)^{i}s(n,n-i)S(j_{1}+j_{2}+n-i,k+n),

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

  • The GSN S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) satisfy the identity

    S1,x1+11,x2+1,p2(p1,k)=S1,x11,x2,p2(p1,k)+(k+1)S1,x11,x2,p2(p1,k+1).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 S1,x11,x2,p2(p1,k)S_{1,x_{1}}^{1,x_{2},p_{2}}(p_{1},k) satisfy the recurrence

    S1,x11,x2,p2(p1,k)=S1,x11,x2,p2(p11,k1)+(k+x1)S1,x11,x2,p2(p11,k).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

Bp=l=0pS(p,l)(1)ll!l+1,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

Bp(x)=l=0pj=0p(pj)xpjS(j,l)(1)ll!l+1.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 Bp(k)(x)B_{p}^{\left(k\right)}(x) can be written in terms of the GSN as

Bp(k)(x)=l=0pS1,x(p,l)(1)ll!(l+1)k.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 x1,x2x_{1},x_{2}, denoted by Bp1,p2(k)(x1,x2)B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}), as

Bp1,p2(k)(x1,x2)=l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!(l+1)k.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 p2=0p_{2}=0, formula (16) becomes (15). By using (9) we can write Bp1,p2(k)(x1,x2)B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) in terms of standard Stirling numbers as

Bp1,p2(k)(x1,x2)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= l=0p1+p2j1=0p1j2=0p2(p1j1)(p2j2)(x1m)p1j1(x2m)p2j2\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)
×i=0m(mi)S(j1+j2,l+i)(1)l(l+i)!(l+1)k,\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 mm is an arbitrary non-negative integer. Similarly, by using (10) we have that

Bp1,p2(k)(x1,x2)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= l=0p1+p2j1=0p1j2=0p2(p1j1)(p2j2)(x1n)p1j1(x2n)p2j2\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)
×i=0n1(1)is(n,ni)S(j1+j2+ni,l+n)(1)ll!(l+1)k,\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 nn is an arbitrary positive integer. The simplest cases of (17) and (18) are

Bp1,p2(k)(x1,x2)=l=0p1+p2j1=0p1j2=0p2(p1j1)(p2j2)x1p1j1x2p2j2S(j1+j2,l)(1)ll!(l+1)k,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

Bp1,p2(k)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)} (x1,x2)\displaystyle(x_{1},x_{2}) (20)
=l=0p1+p2j1=0p1j2=0p2(p1j1)(p2j2)(x11)p1j1(x21)p2j2S(j1+j2+1,l+1)(1)ll!(l+1)k,\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

B1,1(k)(x1,x2)\displaystyle B_{1,1}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= 23k12k(x1+x2+1)+x1x2,\displaystyle\frac{2}{3^{k}}-\frac{1}{2^{k}}\left(x_{1}+x_{2}+1\right)+x_{1}x_{2},
B1,2(k)(x1,x2)\displaystyle B_{1,2}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= 13k(2x1+4x2+6)12k(x1(2x2+1)+(x2+1)2)+x1x2264k.\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

B0,0(k)(x1,x2)=1.B_{0,0}^{\left(k\right)}(x_{1},x_{2})=1. (21)

Observe also that

Bp1,p2(k)(x,x)=Bp1+p2(k)(x).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

Bp1,p2(k)(0,0)=Bp1+p2(k),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

Bp1,p2(k)(x1,x2)=l=0p1+p2j=0l(1)j(lj)(lj+x1)p1(lj+x2)p2(1)l(l+1)k,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

x1Bp1,p2(k)(x1,x2)\displaystyle\frac{\partial}{\partial x_{1}}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= p1Bp11,p2(k)(x1,x2),\displaystyle p_{1}B_{p_{1}-1,p_{2}}^{\left(k\right)}(x_{1},x_{2}),
x2Bp1,p2(k)(x1,x2)\displaystyle\frac{\partial}{\partial x_{2}}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= p2Bp1,p21(k)(x1,x2).\displaystyle p_{2}B_{p_{1},p_{2}-1}^{\left(k\right)}(x_{1},x_{2}).

We can use (8) to write Bp1,p2(k)(x1,x2)B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) in terms of Bj1,j2(k)(y1,y2)B_{j_{1},j_{2}}^{\left(k\right)}\left(y_{1},y_{2}\right), 0jipi0\leq j_{i}\leq p_{i}, i=1,2i=1,2, as

Bp1,p2(k)(x1,x2)=j1=0p1j2=0p2(p1j1)(p2j2)(x1y1)p1j1(x2y2)p2j2Bj1,j2(k)(y1,y2),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 Bp(k)(x)=j=0p(pj)(xy)pjBj(k)(y)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

Bp1,p2(k)(x1,x2)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) =l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!(l+1)k\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}}
=j1=0p1j2=0p2(p1j1)(p2j2)(x1y1)p1j1(x2y2)p2j2l=0j1+j2S1,y11,y2,j2(j1,l)(1)ll!(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}}
=j1=0p1j2=0p2(p1j1)(p2j2)(x1y1)p1j1(x2y2)p2j2Bj1,j2(k)(y1,y2),\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 y1=y2=yy_{1}=y_{2}=y in (25) we obtain an expression for the bi-variate poly-Bernoulli polynomial Bp1,p2(k)(x1,x2)B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) in terms of one-variable poly-Bernoulli polynomials Bj(k)(y)B_{j}^{\left(k\right)}\left(y\right), 0jp1+p20\leq j\leq p_{1}+p_{2}, namely

Bp1,p2(k)(x1,x2)=j1=0p1j2=0p2(p1j1)(p2j2)(x1y)p1j1(x2y)p2j2Bj1+j2(k)(y),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 Bp1,p2(k)(x1,x2)B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2}) in terms of poly-Bernoulli numbers Bj(k)B_{j}^{\left(k\right)}, 0jp1+p20\leq j\leq p_{1}+p_{2}, as

Bp1,p2(k)(x1,x2)=j1=0p1j2=0p2(p1j1)(p2j2)x1p1j1x2p2j2Bj1+j2(k).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=0k=0 and k=1k=-1 of (27) are

Bp1,p2(0)(x1,x2)=(x11)p1(x21)p2,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

Bp1,p2(1)(x1,x2)=(x12)p1(x22)p2,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=0r=0 and r=1r=1 of the identity

l=0p1+p2S1,x11,x2,p2(p1,l)(1)l(l+r)!=r!(x1r1)p1(x2r1)p2,\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 rr is a non-negative integer. We leave the proof of (30) to the reader. Observe also that (25) implies that

Bp1,p2(k)(x1+1,x2+1)=j1=0p1j2=0p2(p1j1)(p2j2)Bj1,j2(k)(x1,x2),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

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

If we set x1=x2=xx_{1}=x_{2}=x in (25), we obtain a formula for the standard poly-Bernoulli polynomial Bp1+p2(k)(x)B_{p_{1}+p_{2}}^{\left(k\right)}(x) in terms of the bi-variate poly-Bernoulli polynomials Bj1,j2(k)(y1,y2)B_{j_{1},j_{2}}^{\left(k\right)}(y_{1},y_{2}), 0jipi0\leq j_{i}\leq p_{i}, i=1,2i=1,2, namely

Bp1+p2(k)(x)=j1=0p1j2=0p2(p1j1)(p2j2)(xy1)p1j1(xy2)p2j2Bj1,j2(k)(y1,y2).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

Bp1,0(k)(x1,x2)\displaystyle B_{p_{1},0}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= Bp1(k)(x1),\displaystyle B_{p_{1}}^{\left(k\right)}(x_{1}), (34)
B0,p2(k)(x1,x2)\displaystyle B_{0,p_{2}}^{\left(k\right)}(x_{1},x_{2}) =\displaystyle= Bp2(k)(x2),\displaystyle B_{p_{2}}^{\left(k\right)}(x_{2}), (35)

and

Bp1,p2(k)(0,x2)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}\left(0,x_{2}\right) =\displaystyle= j2=0p2(p2j2)x2p2j2Bp1+j2(k),\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)
Bp1,p2(k)(x1,0)\displaystyle B_{p_{1},p_{2}}^{\left(k\right)}\left(x_{1},0\right) =\displaystyle= j1=0p1(p1j1)x1p1j1Bj1+p2(k).\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

r=0nan,r(x+α)r=r=0nbn,r(x+β)r.\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

r=0nan,rBr(k)(x+α)=r=0nbn,rBr(k)(x+β).\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:

r=0n(rj)an,r(x+α)rj=r=0n(rj)bn,r(x+β)rj\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 jj is a non-negative integer. We have

r=0nan,rBr(k)(x+α)\displaystyle\sum_{r=0}^{n}a_{n,r}B_{r}^{\left(k\right)}(x+\alpha) =\displaystyle= r=0nan,rl=0rS1,x+α(r,l)(1)ll!(l+1)k\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= r=0nan,rl=0rj=0r(rj)(x+α)rjS(j,l)(1)ll!(l+1)k\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= l=0nj=0n(r=0n(rj)an,r(x+α)rj)S(j,l)(1)ll!(l+1)k\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= l=0nj=0n(r=0n(rj)bn,r(x+β)rj)S(j,l)(1)ll!(l+1)k\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= r=0nbn,rl=0rj=0r(rj)(x+β)rjS(j,l)(1)ll!(l+1)k\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= r=0nbn,rl=0rS1,x+β(r,l)(1)ll!(l+1)k\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= r=0nbn,rBr(k)(x+β),\displaystyle\sum_{r=0}^{n}b_{n,r}B_{r}^{\left(k\right)}(x+\beta),

as desired.

Remark 3.3.

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

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

Bp(k)(x)=j=0p(pj)(xy)pjBj(k)(y),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

j=0p(pj)xpjBj(k)=j=0p(pj)(xy)pjBj(k)(y).\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

j=0p(pj)Bpj(k1)(x)Bj(k)=j=0p(pj)Bpj(k1)(xy)Bj(k)(y).\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=xy=x in (42) to get the identity

j=0p(pj)Bpj(k1)(x)Bj(k)=j=0p(pj)Bpj(k1)Bj(k)(x).\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=1k=1 in (43), and replace xx by x+1x+1, we obtain

j=0p(pj)Bpj(k1)(x+1)Bj=j=0p(pj)Bpj(k1)Bj(x+1).\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 Bj(x+1)=Bj(x)+jxj1B_{j}(x+1)=B_{j}(x)+jx^{j-1} together with (43), we obtain from (44), after some elementary algebraic steps, the curious identity

j=0p(pj)(Bpj(k1)(x+1)Bpj(k1)(x))Bj=pBp1(k1)(x).\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

j1=0p1j2=0p2l=0j1+j2(p1j1)\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}} (p2j2)x1p1j1x2p2j2S(j1+j2,l)(1)ll!(l+1)k0\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)
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)(x1m)p1j1(x2m)p2j2\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}}
×i=0m(mi)S(j1+j2,l+i)(1)l(l+i)!(l+1)k0\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}}}
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)(x1n)p1j1(x2n)p2j2\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}}
×i=0n1(1)is(n,ni)S(j1+j2+ni,l+n)(1)ll!(l+1)k0.\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 mm, nn are arbitrary integers, m0m\geq 0, n>0n>0. Now we use Theorem 3.1 in (46) and then set x1=x2=m+nx_{1}=x_{2}=m+n, to obtain the identities

j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)\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}} Bp1j1(k1)(m+n)Bp2j2(k2)(m+n)S(j1+j2,l)(1)ll!(l+1)k0\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)
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)Bp1j1(k1)(n)Bp2j2(k2)(n)\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)
×i=0m(mi)S(j1+j2,l+i)(1)l(l+i)!(l+1)k0\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}}}
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)Bp1j1(k1)(m)Bp2j2(k2)(m)\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)
×i=0n1(1)is(n,ni)S(j1+j2+ni,l+n)(1)ll!(l+1)k0.\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

j1=0p1j2=0p2(p1j1)(p2j2)\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}} (x1y)p1j1(x2y)p2j2Bj1+j2(k0)(y)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)(x1z)p1j1(x2z)p2j2Bj1+j2(k0)(z).\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

j1=0p1j2=0p2(p1j1)(p2j2)\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}} Bp1j1(k1)(x1y)Bp2j2(k2)(x2y)Bj1+j2(k0)(y)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(x1z)Bp2j2(k2)(x2z)Bj1+j2(k0)(z).\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 x1=x2=y=xx_{1}=x_{2}=y=x to obtain from (49) that

j1=0p1j2=0p2(p1j1)(p2j2)\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{1}}{j_{1}}\binom{p_{2}}{j_{2}} Bp1j1(k1)Bp2j2(k2)Bj1+j2(k0)(x)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(xz)Bp2j2(k2)(xz)Bj1+j2(k0)(z).\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=0z=0, z=1(q1)xz=1-\left(q-1\right)x, and z=qx1z=qx-1, where qq is an arbitrary parameter, we obtain from (50) the identities

j1=0p1j2=0p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}} (p1j1)(p2j2)Bp1j1(k1)Bp2j2(k2)Bj1+j2(k0)(x)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(x)Bp2j2(k2)(x)Bj1+j2(k0)\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)}
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(qx1)Bp2j2(k2)(qx1)Bj1+j2(k0)(1(q1)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)}(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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(1(q1)x)Bp2j2(k2)(1(q1)x)Bj1+j2(k0)(qx1).\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=2q=2, if some (or all) of the parameters k0,k1,k2k_{0},k_{1},k_{2} are equal to 11, we can use the known property Br(1x)=(1)rBr(x)B_{r}\left(1-x\right)=(-1)^{r}B_{r}(x) to simplify the corresponding expression in (51). For example, if k0=k1=k2=1k_{0}=k_{1}=k_{2}=1 and q=2q=2, we have the following identities involving standard Bernoulli numbers and polynomials

j1=0p1j2=0p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}} (p1j1)(p2j2)Bp1j1Bp2j2Bj1+j2(x)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(x)Bp2j2(x)Bj1+j2\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}}
=j1=0p1j2=0p2(p1j1)(p2j2)(1)j1+j2Bp1j1(2x1)Bp2j2(2x1)Bj1+j2(x)\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)
=(1)p1+p2j1=0p1j2=0p2(p1j1)(p2j2)(1)j1+j2Bp1j1(x)Bp2j2(x)Bj1+j2(2x1).\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=1x=1 in (52), we see that if p1+p2p_{1}+p_{2} is odd, then

j1=0p1j2=0p2(p1j1)(p2j2)(1)j1+j2Bp1j1Bp2j2Bj1+j2=0.\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=0m=0) and (10) (with n=1n=1), we have

j1=0p1\displaystyle\sum_{j_{1}=0}^{p_{1}} j2=0p2l=0j1+j2(p1j1)(p2j2)x1p1j1x2p2j2S(j1+j2,l)(1)l(l+r)!\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)
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)(x11)p1j1(x21)p2j2S(j1+j2+1,l+1)(1)l(l+r)!\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)!
=r!(x1r1)p1(x2r1)p2,\displaystyle=r!(x_{1}-r-1)^{p_{1}}(x_{2}-r-1)^{p_{2}},

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

j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)Bp1j1(k1)(x1)Bp2j2(k2)(x2)S(j1+j2,l)(1)l(l+r)!\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)
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)Bp1j1(k1)(x11)Bp2j2(k2)(x21)S(j1+j2+1,l+1)(1)l(l+r)!\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)!
=r!Bp1(k1)(x1r1)Bp2(k2)(x2r1),\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 rr is an arbitrary non-negative integer. Now let us consider the difference

Bp1,p2(k0)(x1+r,x2+r)Bp1,p2(k0)(x1,x2),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 rr is an arbitrary positive integer. In the case k0=1k_{0}=1, we know that (56) is equal to (see [Shib])

p1t=0r1(x2+t)p2(x1+t)p11+p2t=0r1(x1+t)p1(x2+t)p21.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

j1=0p1j2=0p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}} (p1j1)(p2j2)((x1+r)p1j1(x2+r)p2j2Bj1+j2(k0)x1p1j1x2p2j2Bj1+j2(k0))\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)
=j1=0p1j2=0p2(p1j1)(p2j2)((x1+ry)p1j1(x2+ry)p2j2Bj1+j2(k0)(y)\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.
(x1z)p1j1(x2z)p2j2Bj1+j2(k0)(z)),\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 yy and zz are arbitrary parameters. If k0=1k_{0}=1, we have from (57) and (58) that

j1=0p1j2=0p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}} (p1j1)(p2j2)((x1+r)p1j1(x2+r)p2j2x1p1j1x2p2j2)Bj1+j2\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)
=j1=0p1j2=0p2(p1j1)(p2j2)((x1+ry)p1j1(x2+ry)p2j2Bj1+j2(y)\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.
(x1z)p1j1(x2z)p2j2Bj1+j2(z))\displaystyle\left.\qquad-(x_{1}-z)^{p_{1}-j_{1}}(x_{2}-z)^{p_{2}-j_{2}}B_{j_{1}+j_{2}}(z)\right)
=p1t=0r1(x2+t)p2(x1+t)p11+p2t=0r1(x1+t)p1(x2+t)p21.\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

j1=0p1j2=0p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\sum_{j_{2}=0}^{p_{2}} (p1j1)(p2j2)(Bp1j1(k1)(x1+r)Bp2j2(k2)(x2+r)Bp1j1(k1)(x1)Bp2j2(k2)(x2))Bj1+j2\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)
=j1=0p1j2=0p2(p1j1)(p2j2)(Bp1j1(k1)(x1+ry)Bp2j2(k2)(x2+ry)Bj1+j2(y)\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.
Bp1j1(k1)(x1z)Bp2j2(k2)(x2z)Bj1+j2(z))\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)
=p1t=0r1Bp11(k1)(x1+t)Bp2(k2)(x2+t)+p2t=0r1Bp1(k1)(x1+t)Bp21(k2)(x2+t).\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 (y,z)=(r,r),(r,0),(0,r),(2r,r)\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

j1=0p1j2=0p2(p1j1)(p2j2)(Bp1j1(k1)(x1+r)Bp2j2(k2)(x2+r)Bp1j1(k1)(x1)Bp2j2(k2)(x2))Bj1+j2\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)
=j1=0p1j2=0p2(p1j1)(p2j2)(Bp1j1(k1)(x1)Bp2j2(k2)(x2)Bp1j1(k1)(x1r)Bp2j2(k2)(x2r))Bj1+j2(r)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(x1)Bp2j2(k2)(x2)(Bj1+j2(r)Bj1+j2)\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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(x1+r)Bp2j2(k2)(x2+r)(Bj1+j2Bj1+j2(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}+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)
=j1=0p1j2=0p2(p1j1)(p2j2)Bp1j1(k1)(x1r)Bp2j2(k2)(x2r)(Bj1+j2(2r)Bj1+j2(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}-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)
=p1t=0r1Bp11(k1)(x1+t)Bp2(k2)(x2+t)+p2t=0r1Bp1(k1)(x1+t)Bp21(k2)(x2+t).\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

l=0q(ql)(x1)qlBp1+l,p2(k)(x1,x2)\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= l=0q(ql)(x2)qlBp1,p2+l(k)(x1,x2)\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= l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!q1,k(l)(i=1q+1(l+i))k.\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 1,k(y)=1\mathcal{R}_{-1,k}(y)=-1, and for μ0\mu\geq 0 the functions μ,k(y)\mathcal{R}_{\mu,k}(y) are defined recursively by

μ,k(y)=y(y+μ+2)kμ1,k(y)(y+1)k+1μ1,k(y+1).\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

l=0q(qk)(x1)qlBp1+l,p2(k)(x1,x2)=l=0p1+p2S1,x11,x2,p2(p1,k)(1)ll!q1,k(l)(i=1q+1(l+i))k\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 qq. The case q=0q=0 of (63) is the definition (16). Let us suppose formula (63) is true for a given qq\in\mathbb{N}. Then

l=0q+1\displaystyle\sum_{l=0}^{q+1} (q+1l)(x1)q+1lBp1+l,p2(k)(x1,x2)\displaystyle\binom{q+1}{l}(-x_{1})^{q+1-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})
=x1l=0q(ql)(x1)qlBp1+l,p2(k)(x1,x2)+l=0q(ql)(x1)qlBp1+1+l,p2(k)(x1,x2)\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})
=x1l=0p1S1,x11,x2,p2(p1,l)(1)ll!q1,k(l)(i=1q+1(l+i))kl=0p1+1S1,x11,x2,p2(p1+1,l)(1)ll!q1,k(l)(i=1q+1(l+i))k.\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

l=0q+1(q+1l)\displaystyle\sum_{l=0}^{q+1}\binom{q+1}{l} (x1)q+1lBp1+l,p2(k)(x1,x2)\displaystyle(-x_{1})^{q+1-l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})
=x1l=0p1S1,x11,x2,p2(p1,l)(1)ll!q1(l)(i=1q+1(l+i))k\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}}
l=0p1+1(S1,x11,x2,p2(p1,l1)+(l+x1)S1,x11,x2,p2(p1,l))(1)ll!q1(l)(i=1q+1(l+i))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

l=0q+1(q+1l)(x1)q+1lBp1+l,p2(k)(x1,x2)\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})
=l=0p1+1(S1,x11,x2,p2(p1,l1)+lS1,x11,x2,p2(p1,l))(1)ll!q1(l)(i=1q+1(l+i))k\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}}
=l=0p1S1,x11,x2,p2(p1,l)(1)l(l+1)!q1(l+1)(i=1q+1(l+i+1))kl=0p1S1,x11,x2,p2(p1,l)(1)ll!lq1(l)(i=1q+1(l+i))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}}
=l=0p1S1,x11,x2,p2(p1,l)(l(l+q+2)kq1(l)(l+1)k+1q1(l+1))(1)ll!(i=1q+2(l+i))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}}
=l=0p1S1,x11,x2,p2(p1,l)(1)ll!q(l)(i=1q+2(l+i))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

l=0q(ql)(x2)qlBp1,p2+l(k)(x1,x2)=l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!q1,k(l)(i=1q+1(l+i))k,\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

0,k(y)=(y+1)k+1y(y+2)k,\mathcal{R}_{0,k}(y)=(y+1)^{k+1}-y(y+2)^{k},
1,k(y)=(2y+1)(y+1)k+1(y+3)ky2(y+2)k(y+3)k(y+1)k+1(y+2)k+1,\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=1q=1 is

x1Bp1,p2(k)(x1,x2)+\displaystyle-x_{1}B_{p_{1},p_{2}}^{\left(k\right)}(x_{1},x_{2})+ Bp1+1,p2(k)(x1,x2)=x2Bp1,p2(k)(x1,x2)+Bp1,p2+1(k)(x1,x2)\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)
=l=0p1+p2S1,x11,x2,p2(p1,k)(1)ll!(l+1(l+2)kl(l+1)k),\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=2q=2 is

x12Bp1,p2(k)\displaystyle x_{1}^{2}B_{p_{1},p_{2}}^{\left(k\right)} (x1,x2)2x1Bp1+1,p2(k)(x1,x2)+Bp1+2,p2(k)(x1,x2)\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)
=x22Bp1,p2(k)(x1,x2)2x2Bp1,p2+1(k)(x1,x2)+Bp1,p2+2(k)(x1,x2)\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})
=l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!((2l+1)(l+1)(l+2)kl2(l+1)k(l+1)(l+2)(l+3)k).\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

l=0q\displaystyle\sum_{l=0}^{q} (ql)(x1)qlj1=0p1+lj2=0p2(p1+lj1)(p2j2)(x1y)p1+lj1(x2y)p2j2Bj1+j2(k)(y)\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)
=l=0q(ql)(x2)qlj1=0p1j2=0p2+l(p1j1)(p2+lj2)(x1z)p1j1(x2z)p2+lj2Bj1+j2(k)(z)\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)
=l=0p1+p2S1,x11,x2,p2(p1,l)(1)ll!q1,k(l)(i=1q+1(l+i))k.\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=x1,x2y,z=x_{1},x_{2} in (67), we get the identities

l=0q(ql)\displaystyle\sum_{l=0}^{q}\binom{q}{l} (x1)qlj2=0p2(p2j2)(x2x1)p2j2Bp1+l+j2(k)(x1)\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)
=l=0q(ql)(x1)qlj1=0p1+l(p1+lj1)(x1x2)p1+lj1Bj1+p2(k)(x2)\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})
=l=0q(ql)(x2)qlj2=0p2+l(p2+lj2)(x2x1)p2+lj2Bp1+j2(k)(x1)\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})
=l=0q(ql)(x2)qlj1=0p1(p1j1)(x1x2)p1j1Bj1+p2+l(k)(x2)\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})
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)x1p1j1x2p2j2S(j1+j2,l)(1)ll!q1,k(l)(i=1q+1(l+i))k.\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 x2=0x_{2}=0 in (68), we get

l=0q(ql)(x1)qlj2=0p2(p2j2)\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-x_{1})^{q-l}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}} (x1)p2j2Bp1+l+j2(k)(x1)\displaystyle(-x_{1})^{p_{2}-j_{2}}B_{p_{1}+l+j_{2}}^{\left(k\right)}(x_{1}) (69)
=l=0q(ql)(x1)qlj1=0p1+l(p1+lj1)x1p1+lj1Bj1+p2(k)\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)}
=j2=0p2+q(p2+qj2)(x1)p2+qj2Bp1+j2(k)(x1)\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})
=j1=0p1(p1j1)x1p1j1Bj1+p2+q(k)\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)}
=l=0p1+p2S1,x11,0,p2(p1,l)(1)ll!q1,k(l)(i=1q+1(l+i))k.\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,x1=1,k=1q=0,x_{1}=1,k=1 of (69) reduces to

(1)p1+p2j2=0p2(p2j2)Bp1+j2\displaystyle(-1)^{p_{1}+p_{2}}\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+j_{2}} =\displaystyle= j1=0p1(p1j1)Bj1+p2\displaystyle\sum_{j_{1}=0}^{p_{1}}\binom{p_{1}}{j_{1}}B_{j_{1}+p_{2}}
=\displaystyle= j1=0p1l=0j1+p2(p1j1)S(j1+p2,l)(1)ll!l+1.\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

Bp1,p2(1,0)=(1)p1+p2Bp1,p2(0,1).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=1k=1, when x1=0,x2=1x_{1}=0,x_{2}=1

l=0q(ql)(1)qlBp1+l,p2(1,0)\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-1)^{q-l}B_{p_{1}+l,p_{2}}(1,0) =\displaystyle= (1)p1+p2+ql=0q(ql)Bp1+l,p2(0,1)\displaystyle(-1)^{p_{1}+p_{2}+q}\sum_{l=0}^{q}\binom{q}{l}B_{p_{1}+l,p_{2}}(0,1)
=\displaystyle= Bp1,p2+q(1,0)=(1)p1+p2+qBp1,p2+q(0,1)\displaystyle B_{p_{1},p_{2}+q}(1,0)=(-1)^{p_{1}+p_{2}+q}B_{p_{1},p_{2}+q}(0,1)
=\displaystyle= l=0p1+p2S1,11,0,p2(p1,l)(1)ll!q1,1(l)i=1q+1(l+i),\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

l=0q(ql)(1)qlj1=0p1+l\displaystyle\sum_{l=0}^{q}\binom{q}{l}(-1)^{q-l}\sum_{j_{1}=0}^{p_{1}+l} (p1+lj1)Bj1+p2\displaystyle\binom{p_{1}+l}{j_{1}}B_{j_{1}+p_{2}}
=(1)p1+p2+ql=0q(ql)j2=0p2(p2j2)Bp1+l+j2\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}}
=j1=0p1(p1j1)Bj1+p2+q=(1)p1+p2+qj2=0p2+q(p2+qj2)Bp1+j2\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}}
=l=0p1+p2j1=0p1(p1j1)S(j1+p2,l)(1)ll!q1,1(l)i=1q+1(l+i).\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=0k=0, the functions μ,k(y)\mathcal{R}_{\mu,k}\left(y\right) of Proposition 4.1 are μ,0(y)=(1)μ\mathcal{R}_{\mu,0}(y)=(-1)^{\mu}. Thus, the case k=0k=0 of (4.1) is the case r=0r=0 of (30). Also, in the case k=1k=-1, we have that μ,1(y)=(1)μ2μ+1i=2μ+2(y+i)\mathcal{R}_{\mu,-1}(y)=\frac{(-1)^{\mu}2^{\mu+1}}{\prod_{i=2}^{\mu+2}(y+i)}, and then the case k=1k=-1 of (4.1) is the case r=1r=1 of (30).

Proposition 4.3.

For non-negative integers p1,p2,qp_{1},p_{2},q we have

l=0q(1)lBp1+l,p2(k)(x1,x2)1l!\displaystyle\sum_{l=0}^{q}(-1)^{l}B_{p_{1}+l,p_{2}}^{\left(k\right)}(x_{1},x_{2})\frac{1}{l!} dldx1li=0q1(x1+i)\displaystyle\frac{d^{l}}{dx_{1}^{l}}\prod_{i=0}^{q-1}(x_{1}+i) (72)
=l=0q(1)lBp1,p2+l(k)(x1,x2)1l!dldx2li=0q1(x2+i)\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)
=l=0p1+p2S1,x1+q1,x2+q,p2(p1,l)(1)l(l+q)!(l+q+1)k.\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

l=0q(1)lBp1+l,p2(k)(x1,x2)1l!dldx1li=0q1(x1+i)=l=0p1+p2S1,x1+q1,x2+q,p2(p1,l)(1)l(l+q)!(l+q+1)k,\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 qq. The case q=0q=0 of (73) is the definition (16). If we suppose that (73) is true for a given qq\in\mathbb{N}, then

l=0q+1(1)l\displaystyle\sum_{l=0}^{q+1}(-1)^{l} Bp1+l,p2(k)(x1,x2)1l!dldx1li=0q(x1+i)\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)
=l=0q+1(1)lBp1+l,p2(k)(x1,x2)1l!dldx1l((x1+q)i=0q1(x1+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)
=l=0q+1(1)lBp1+l,p2(k)(x1,x2)1l!((x1+q)dldx1li=0q1(x1+i)+ldl1dx1l1i=0q1(x1+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!}\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)
=(x1+q)l=0q(1)lBp1+l,p2(k)(x1,x2)1l!dldx1li=0q1(x1+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)
+l=1q+1(1)lBp1+l,p2(k)(x1,x2)1(l1)!dl1dx1l1i=0q1(x1+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)
=(x1+q)l=0q(1)lBp1+l,p2(k)(x1,x2)1l!dldx1li=0q1(x1+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)
l=0q(1)lBp1+1+l,p2(k)(x1,x2)1l!dldx1li=0q1(x1+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)
=(x1+q)l=0p1+p2S1,x1+q1,x2+q,p2(p1,l)(1)l(l+q)!(l+q+1)k\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}}
l=0p1+p2+1S1,x1+q1,x2+q,p2(p1+1,l)(1)l(l+q)!(l+q+1)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

l=0q+1(1)l\displaystyle\sum_{l=0}^{q+1}(-1)^{l} Bp1+l,p2(x1,x2)1l!dldx1li=0q(x1+i)\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)
=(x1+q)l=0p1+p2S1,x1+q1,x2+q,p2(p1,l)(1)l(l+q)!(l+q+1)k\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}}
l=0p1+p2+1(S1,x1+q1,x2+q,p2(p1,l1)+(l+x1+q)S1,x1+q1,x2+q,p2(p1,l))(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}}
=l=0p1+p2+1(S1,x1+q1,x2+q,p2(p1,l1)+lS1,x1+q1,x2+q,p2(p1,l))(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}}
=l=1p1+p2+1S1,x1+q+11,x2+q+1,p2(p1,l1)(1)l(l+q)!(l+q+1)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}}
=l=0p1+p2S1,x1+q+11,x2+q+1,p2(p1,l)(1)l(l+q+1)!(l+q+2)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

l=0q(1)lBp1,p2+l(k)(x1,x2)1l!dldx2li=0q1(x2+i)=l=0p1+p2S1,x1+q1,x2+q,p2(p1,l)(1)l(l+q)!(l+q+1)k,\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 x1=0,x2=1x_{1}=0,x_{2}=1 looks as

l=0q(1)ls(q,l)Bp1+l,p2(k)(0,1)\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l)B_{p_{1}+l,p_{2}}^{\left(k\right)}(0,1) =\displaystyle= l=0q(1)ls(q+1,l+1)Bp1,p2+l(k)(0,1)\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)B_{p_{1},p_{2}+l}^{\left(k\right)}(0,1)
=\displaystyle= l=0p1+p2S1,q1,q+1,p2(p1,l)(1)l(l+q)!(l+q+1)k,\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

l=0q(1)l\displaystyle\sum_{l=0}^{q}(-1)^{l} s(q,l)j2=0p2(p2j2)Bp1+l+j2(k)=l=0q(1)ls(q+1,l+1)j2=0p2+l(p2+lj2)Bp1+j2(k)\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)
=l=0p1+p2j1=0p1j2=0p2(p1j1)(p2j2)qp1j1(q+1)p2j2S(j1+j2,l)(1)l(l+q)!(l+q+1)k.\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=1k=1 in formula (4), we can use Carlitz identity (71) to write the following enriched version of (4)

l=0q(1)ls(q,l)Bp1+l,p2(0,1)\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l)B_{p_{1}+l,p_{2}}(0,1) =\displaystyle= (1)p1+p2l=0qs(q,l)Bp1+l,p2(1,0)\displaystyle(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q,l)B_{p_{1}+l,p_{2}}(1,0)
=\displaystyle= l=0q(1)ls(q+1,l+1)Bp1,p2+l(0,1)\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q+1,l+1)B_{p_{1},p_{2}+l}(0,1)
=\displaystyle= (1)p1+p2l=0qs(q+1,l+1)Bp1,p2+l(1,0)\displaystyle(-1)^{p_{1}+p_{2}}\sum_{l=0}^{q}s(q+1,l+1)B_{p_{1},p_{2}+l}(1,0)
=\displaystyle= l=0p1+p2S1,q1,q+1,p2(p1,l)(1)l(l+q)!l+q+1,\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

l=0q(1)ls(q,l)\displaystyle\sum_{l=0}^{q}(-1)^{l}s(q,l) j2=0p2(p2j2)Bp1+l+j2\displaystyle\sum_{j_{2}=0}^{p_{2}}\binom{p_{2}}{j_{2}}B_{p_{1}+l+j_{2}} (77)
=(1)p1+p2l=0qs(q,l)j1=0p1+l(p1+lj1)Bj1+p2\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}}
=l=0q(1)ls(q+1,l+1)j2=0p2+l(p2+lj2)Bp1+j2\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}}
=(1)p1+p2l=0qs(q+1,l+1)j1=0p1(p1j1)Bj1+p2+l\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}
=j1=0p1j2=0p2l=0j1+j2(p1j1)(p2j2)qp1j1(q+1)p2j2S(j1+j2,l)(1)l(l+q)!l+q+1.\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=1q=1 of the first two lines of (72). That is

Bp1,p2+1(k)(x1,x2)Bp1+1,p2(k)(x1,x2)=(x2x1)Bp1,p2(k)(x1,x2).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)
    j=0q(qj)(1)jBp1+j,p2+qj(k)(x1,x2)=(x2x1)qBp1,p2(k)(x1,x2).\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)
    j=0q(qj)(x2x1)jBp1+qj,p2(k)(x1,x2)=Bp1,p2+q(k)(x1,x2).\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 qq. The case q=0q=0 is a trivial identity. Let us suppose that (79) is true for a given qq\in\mathbb{N}. Then

j=0q+1\displaystyle\sum_{j=0}^{q+1} (q+1j)(1)jBp1+j,p2+q+1j(k)(x1,x2)\displaystyle\binom{q+1}{j}(-1)^{j}B_{p_{1}+j,p_{2}+q+1-j}^{\left(k\right)}(x_{1},x_{2})
=j=0q+1((qj)+(qj1))(1)jBp1+j,p2+q+1j(k)(x1,x2)\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})
=j=0q(qj)(1)jBp1+j,p2+q+1j(k)(x1,x2)+j=0q(qj)(1)j+1Bp1+1+j,p2+qj(k)(x1,x2)\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})
=(x2x1)qBp1,p2+1(k)(x1,x2)(x2x1)qBp1+1,p2(k)(x1,x2)\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})
=(x2x1)q(Bp1,p2+1(k)(x1,x2)Bp1+1,p2(k)(x1,x2))\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)
=(x2x1)q+1Bp1,p2(k)(x1,x2),\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 qq. The case q=0q=0 is a trivial identity. Let us suppose that (80) is true for a given qq\in\mathbb{N}. Then

j=0q+1\displaystyle\sum_{j=0}^{q+1} (q+1j)(x2x1)jBp1+q+1j,p2(k)(x1,x2)\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})
=j=0q+1((qj)+(qj1))(x2x1)jBp1+q+1j,p2(k)(x1,x2)\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})
=j=0q(qj)(x2x1)jBp1+q+1j,p2(k)(x1,x2)+j=0q(qj)(x2x1)j+1Bp1+qj,p2(k)(x1,x2)\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})
=Bp1+1,p2+q(k)(x1,x2)+(x2x1)Bp1,p2+q(k)(x1,x2)\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})
=Bp1,p2+q+1(k)(x1,x2),\displaystyle=B_{p_{1},p_{2}+q+1}^{\left(k\right)}(x_{1},x_{2}),

as desired. We used (78) (with p2p_{2} replaced by p2+qp_{2}+q) in the last step.

The case p1=0p_{1}=0 of (80) is

j=0q(qj)(x2x1)jBqj,p2(k)(x1,x2)=Bp2+q(k)(x2).\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 p2=0p_{2}=0 of (81) is the addition formula for standard poly-Bernoulli polynomials, namely

j=0q(qj)(x2x1)qjBj(k)(x1)=Bq(k)(x2).\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 p1=p2=0p_{1}=p_{2}=0 of (79) is

j=0q(qj)(1)jBj,qj(k)(x1,x2)=(x2x1)q.\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 S1,x1(1,x2,p2),,(1,xn,pn)(p1,k)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+x1)p1(m+xn)pn=l=0p1++pnl!S1,x1(1,x2,p2),,(1,xn,pn)(p1,l)(ml),(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 p1,p2,,pnp_{1},p_{2},\ldots,p_{n} are non-negative integers given, one can define poly-Bernoulli polynomials in nn variables x1,,xnx_{1},\ldots,x_{n}, denoted as Bp1,,pn(k)(x1,,xn)B_{p_{1},\ldots,p_{n}}^{(k)}(x_{1},\ldots,x_{n}), as

Bp1,,pn(k)(x1,,xn)=l=0p1++pnS1,x1(1,x2,p2),,(1,xn,pn)(p1,l)(1)ll!(l+1)k,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

Bp1,,pn(k)\displaystyle B_{p_{1},\ldots,p_{n}}^{\left(k\right)} (x1,,xn)\displaystyle(x_{1},\ldots,x_{n})
=l=0p1++pnj1=0p1jn=0pn(p1j1)(pnjn)x1p1j1xnpnjnS(j1++jn,l)(1)ll!(l+1)k\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}}
=j1=0p1jn=0pn(p1j1)(pnjn)x1p1j1xnpnjnBj1++jn(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:

Bp1,p2,p3(k)(x1,x2,x3)=j1=0p1j2=0p2j3=0p3(p1j1)(p2j2)(p3j3)x1p1j1x2p2j2x3p3j3Bj1+j2+j3(k),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

    l=0q(ql)\displaystyle\sum_{l=0}^{q}\binom{q}{l} (x1)qlBp1+l,p2,p3(k)(x1,x2,x3)\displaystyle(-x_{1})^{q-l}B_{p_{1}+l,p_{2},p_{3}}^{\left(k\right)}(x_{1},x_{2},x_{3})
    =l=0q(ql)(x2)qlBp1,p2+l,p3(k)(x1,x2,x3)\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})
    =l=0q(ql)(x3)qlBp1,p2,p3+l(k)(x1,x2,x3)\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})
    =l=0p1+p2+p3j1=0p1j2=0p2j3=0p3((p1j1)(p2j2)(p3j3)x1p1j1x2p2j2x3p3j3\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.
    ×S(j1+j2+j3,l)(1)ll!q1,k(l)(i=1q+1(l+i))k),\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 q1,k(l)\mathcal{R}_{q-1,k}(l) is defined in Proposition 4.1.

  2. (b)

    (See (72)) We have the generalized recurrences

    l=0q\displaystyle\sum_{l=0}^{q} (1)lBp1+l,p2,p3(k)(x1,x2,x3)1l!dldx1li=0q1(x1+i)\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)
    =l=0q(1)lBp1,p2+l,p3(k)(x1,x2,x3)1l!dldx2li=0q1(x2+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)
    =l=0q(1)lBp1,p2,p3+l(k)(x1,x2,x3)1l!dldx3li=0q1(x3+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)
    =j1=0p1j2=0p2j3=0p3((p1j1)(p2j2)(p3j3)(x1+q)p1j1(x2+q)p2j2(x3+q)p3j3\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.
    ×S(j1+j2+j3,l)(1)l(l+q)!(l+q+1)k).\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

    j=0q(qj)(1)jBp1,p2+j,p3+qj(k)(x1,x2,x3)\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= (x3x2)qBp1,p2,p3(k)(x1,x2,x3),\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),
    j=0q(qj)(1)jBp1+j,p2,p3+qj(k)(x1,x2,x3)\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= (x3x1)qBp1,p2,p3(k)(x1,x2,x3),\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),
    j=0q(qj)(1)jBp1+j,p2+qj,p3(k)(x1,x2,x3)\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= (x2x1)qBp1,p2,p3(k)(x1,x2,x3).\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

    j=0q(qj)\displaystyle\sum_{j=0}^{q}\binom{q}{j} (x1x2)jBp1,p2+qj,p3(k)(x1,x2,x3)\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)
    =j=0q(qj)(x1x3)jBp1,p2,p3+qj(k)(x1,x2,x3)=Bp1+q,p2,p3(k)(x1,x2,x3).\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).
    j=0q(qj)\displaystyle\sum_{j=0}^{q}\binom{q}{j} (x2x1)jBp1+qj,p2,p3(k)(x1,x2,x3)\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)
    =j=0q(qj)(x2x3)jBp1,p2,p3+qj(k)(x1,x2,x3)=Bp1,p2+q,p3(k)(x1,x2,x3).\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).
    j=0q(qj)\displaystyle\sum_{j=0}^{q}\binom{q}{j} (x3x1)jBp1+qj,p2,p3(k)(x1,x2,x3)\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)
    =j=0q(qj)(x3x2)jBp1,p2+qj,p3(k)(x1,x2,x3)=Bp1,p2,p3+q(k)(x1,x2,x3).\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