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### Article

Kyungpook Mathematical Journal 2024; 64(1): 133-159

Published online March 31, 2024 https://doi.org/10.5666/KMJ.2024.64.1.133

### Bernoulli and Euler Polynomials in Two Variables

Claudio Pita-Ruiz

e-mail : cpita@up.edu.mx

Received: August 11, 2022; Revised: October 23, 2022; Accepted: November 8, 2022

In a previous work we studied generalized Stirling numbers of the second kind Sa1,b1(a2,b2,p2)(p1,k), where a1, a2, b1, b2 are given complex numbers, a1, a2 ̸= 0, and p1, p2 are non-negative integers given. In this work we use these generalized Stirling numbers to define Bernoulli polynomials in two variables Bp1,p2 (x1, x2), and Euler polynomials in two variables Ep1,p2(x1, x2). By using results for S1,x1(1,x2,p2)(p1,k), we obtain generalizations, to the bivariate case, of some well-known properties from the standard case, as addition formulas, difference equations and sums of powers. We obtain some identities for bivariate Bernoulli and Euler polynomials, and some generalizations, to the bivariate case, of several known identities for Bernoulli and Euler numbers and polynomials of the standard case.

Keywords: Bernoulli number, Euler number, Bernoulli polynomial, Euler polynomial, Generalized Stirling number

Bernoulli polynomials Bnx and Euler polynomials Enx can be defined by the corresponding generating functions

textet-1=n=0Bnxtnn! and 2extet+1=n=0Enxtnn!,

or by the explicit formulas [3, p. 48]

Bpx=k=0ppkBkxp-k and Epx=k=0ppkEk2kx-12p-k,

where Bk=Bk0 is the k-th Bernoulli number, and Ek=2kEk12 is the k-th Euler number.

In a previous work [10] we studied a generalization of Stirling Numbers of the second kind (GSN, for short), denoted as Sa1,b1a2,b2,p2p1,k, where aj,bjC, aj0, j=1,2, and p1,p2 are non-negative integers, involved in the expansion

a1n+b1p1a2n+b2p2=k=0p1+p2k!Sa1,b1a2,b2,p2p1,knk,

(Sa1,b1a2,b2,p2p1,k=0 if k<0 or k>p1+p2). An explicit formula for these numbers is

Sa1,b1a2,b2,p2p1,k=1k!j=0k-1jkja1k-j+b1p1a2k-j+b2p2.

If p2=0, we write the corresponding GSN as Sa,bp,k. That is, we have

Sa,bp,k=1k!j=0k-1jkjak-j+bp.

In particular, if a1=a2=a, b1=b2=b, expression (1.2) beomes

Sa,ba,b,p2p1,k=Sa,bp1+p2,k.

Two trivial examples are S1,0p,k=Sp,k (the standard Stirling numbers of the second kind), and S1,1p,k=Sp+1,k+1.

We summarize some facts about GSN S1,x11,x2,p2p1,k (contained in [10]) that we will use in this work.

• Some values

S1,x11,x2,p2p1,0=x1p1x2p2,S1,x11,x2,p2p1,1=x1+1p1x2+1p2-x1p1x2p2,S1,x11,x2,p2p1,2=12x1+2p1x2+2p2-x1+1p1x2+1p2+12x1p1x2p2,S1,x11,x2,p2(p1,p1+p2)=1.

• The GSN S1,x11,x2,p2p1,k can be written in terms of the GSN S1,y11,y2,p2p1,k as follows

S1,x11,x2,p2p1,k=j1=0p1j2=0p2p1j1p2j2x1-y1p1-j1x2-y2p2-j2S1,y11,y2,j2j1,k.

• The GSN S1,x11,x2,p2p1,k can be written in terms of standard Stirling numbers as follows

S1,x11,x2,p2p1,k=1k!j1=0p1j2=0p2p1j1p2j2x1-mp1-j1x2-mp2-j2×i=0mmik+i!Sj1+j2,k+i,

where m is a non-negative integer.

• The GSN S1,x11,x2,p2p1,k can be written in terms of standard Stirling numbers as follows

S1,x11,x2,p2p1,k=j1=0p1j2=0p2p1j1p2j2x1-np1-j1x2-np2-j2×i=0n-1-1isn,n-iSj1+j2+n-i,k+n,

where n is a positive integer, and s·,· are the Stirling numbers of the first kind (with recurrence sq+1,k=sq,k-1+qs(q,k)). In particular, from (1.6) with m=0, and (1.7) with n=1, we have

S1,x11,x2,p2p1,k=j1=0p1j2=0p2p1j1p2j2x1p1-j1x2p2-j2S(j1+j2,k)
=j1=0p1j2=0p2p1j1p2j2(x1-1)p1-j1(x2-1)p2-j2S(j1+j2+1,k+1).

• The GSN S1,x11,x2,p2p1,k satisfy the identity

S1,x1+11,x2+1,p2p1,k=S1,x11,x2,p2p1,k+k+1S1,x11,x2,p2p1,k+1.

• The GSN S1,x11,x2,p2p1,k satisfy the recurrence

S1,x11,x2,p2p1,k=S1,x11,x2,p2p1-1,k-1+k+x1S1,x11,x2,p2p1-1,k.

### 2. Definitions and Preliminary Results

The relation between Bernoulli numbers and Stirling numbers of the second kind, is an old known story, that dates back to Worpitzky [13]. In fact, we have

Bp=k=0pSp,k(-1)kk!k+1

(see [7]). In the Euler case, formula (3.3) in [6] shows Euler polynomials written in terms of Stirling numbers of the second kind. In [8] and [11] we used the GSN S1,x(p,k) described above, to write, respectively, Bernoulli polynomials Bp(x) and Euler polynomials Ep(x) as

Bp(x)=k=0pS1,xp,k(-1)kk!k+1,
Ep(x)=k=0pS1,xp,k(-1)kk!2k.

(Formula (2.2) was inspired by (2.1), and (2.3) was inspired by formula (3.3) in [6] mentioned before.)

In this work we use the GSN S1,x11,x2,p2p1,k to define bivariate Bernoulli polynomials Bp1,p2(x1,x2) and bivariate Euler polynomials Ep1,p2(x1,x2), as

Bp1,p2(x1,x2)=k=0p1+p2S1,x11,x2,p2p1,k(-1)kk!k+1,
Ep1,p2(x1,x2)=k=0p1+p2S1,x11,x2,p2p1,k(-1)kk!2k.

Explicitly, we have

Bp1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1p1-j1x2p2-j2S(j1+j2,k)(-1)kk!k+1,
Ep1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1p1-j1x2p2-j2S(j1+j2,k)(-1)kk!2k.

Clearly we have that

B0,0(x1,x2)=1,Bp1,0(x1,x2)=Bp1x1,B0,p2(x1,x2)=Bp2x2,Bp1,p2x,x=Bp1+p2x,

and similar results for E replacing B.

Some examples are the following

B1,1(x1,x2)=x1x2-12x1-12x2+16,E1,1(x1,x2)=x1x2-12x1-12x2,B1,2(x1,x2)=x1x22-x1x2+16x1-12x22+13x2,E1,2(x1,x2)=x1x22-x1x2-12x22+14.

From (1.6) and (1.7) we can write the following families of formulas for bivariate Bernoulli and Euler polynomials

Bp1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1-mp1-j1x2-mp2-j2×i=0mmiSj1+j2,k+i(-1)kk+i!k+1,
Ep1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1-mp1-j1x2-mp2-j2×i=0mmiSj1+j2,k+i(-1)kk+i!2k,

where m is a non-negative integer (the case m=0 of (2.8) and (2.9) are (2.6) and (2.7), respectively), and

Bp1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1-np1-j1x2-np2-j2×i=0n-1-1isn,n-iSj1+j2+n-i,k+n(-1)kk!k+1,
Ep1,p2(x1,x2)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2x1-np1-j1x2-np2-j2×i=0n-1-1isn,n-iSj1+j2+n-i,k+n(-1)kk!2k,

where n is a positive integer.

We begin this section with the bivariate version of addition formulas for Bernoulli and Euler polynomials.

Proposition 3.1. We have

Pp1,p2(x1,x2)=j1=0p1j2=0p2p1j1p2j2(x1-y1)p1-j1(x2-y2)p2-j2Pj1,j2(y1,y2),

where P is B or E.

Proof. In the case P=B, by using (1.5) we have

Bp1,p2(x1,x2)=k=0p1+p2S1,x11,x2,p2p1,k(-1)kk!k+1=j1=0p1j2=0p2p1j1p2j2x1-y1p1-j1x2-y2p2-j2k=0j1+j2S1,y11,y2,j2j1,k(-1)kk!k+1=j1=0p1j2=0p2p1j1p2j2(x1-y1)p1-j1(x2-y2)p2-j2Bj1,j2(y1,y2),

as desired. The proof of the Euler case is similar.

We have the following particular case of (3.1), corresponding to y1=y2=y,

Pp1,p2(x1,x2)=j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2Pj1+j2y.

By setting y=12 in the case P=E of (3.2), we have also the following expression for Ep1,p2(x1,x2) involving Euler numbers

Ep1,p2(x1,x2)=2-p1-p2j1=0p1j2=0p2p1j1p2j22x1-1p1-j12x2-1p2-j2Ej1+j2.

From the case P=B of (3.2) with y=x1,x2, we see that

Bp1,p2(x1,x2)=j2=0p2p2j2x2-x1p2-j2Bp1+j2x1=j1=0p1p1j1x1-x2p1-j1Bj1+p2x2.

In particular, with x1=1 and x2=0 we obtain from (3.4) that

Bp1,p21,0=-1p1+p2j2=0p2p2j2Bp1+j2=j1=0p1p1j1Bj1+p2.

That is, we have the identity

-1p2j2=0p2p2j2Bp1+j2=-1p1j1=0p1p1j1Bj1+p2.

Formula (3.6) is the well-known Carlitz identity [1].

Similarly, we have from the case P=E of (3.2) with y=x1,x2

Ep1,p2(x1,x2)=j2=0p2p2j2x2-x1p2-j2Ep1+j2x1=j1=0p1p1j1x1-x2p1-j1Ej1+p2x2.

By setting x1=1,x2=0, we obtain the following Euler versions of Carlitz identity (3.6):

(-1)p2j2=0p2p2j2Ep1+j20=(-1)p1j1=0p1p1j1Ej1+p20,
j2=0p2p2j2(-1)j2Ep1+j21=j1=0p1p1j1(-1)j1Ej1+p21.

Combining formulas for Bp1,p2(x1,x2) and Ep1,p2(x1,x2) in (3.4) and (3.7), we can write the following expressions for the difference Bp1,p2(x1,x2)-Ep1,p2(x1,x2):

Bp1,p2(x1,x2)-Ep1,p2(x1,x2)=j2=0p2p2j2x2-x1p2-j2Bp1+j2x1-Ep1+j2x1=j1=0p1p1j1x1-x2p1-j1Bj1+p2x2-Ej1+p2x2.

Moreover, we can use Cheon's formula

Bn(x)-En(x)=k=2nnkBkEn-k(x),

where n>0 (see [2]), to write (3.10) as

Bp1,p2(x1,x2)-Ep1,p2(x1,x2)=j2=0p2p2j2x2-x1p2-j2k=2p1+j2p1+j2kBkEp1+j2-kx1=j1=0p1p1j1x1-x2p1-j1k=2j1+p2j1+p2kBkEj1+p2-kx2,

where p1,p2>0.

Next we consider properties related to symmetries, difference equations and sums of powers for bivariate Bernoulli and Euler polynomials. We will need some preliminary results, contained in the following lemma.

Lemma 3.2. (a) For non-negative integer p and xR, we have

k=0p-1S1,xp,k+1(-1)kk!=pxp-1.

(b) For non-negative integers p1,p2, and x1,x2R, we have

k=0p1+p2S1,x11,x2,p2p1,k+1(-1)kk!=p1x1p1-1x2p2+p2x1p1x2p2-1.

Proof. (a) We proceed by induction on p. For p=0 the result is clear. If the formula works for a given pN, then, by using the recurrence (1.11) and the value S1,xp,0=xp, we have

k=0pS1,xp+1,k+1(-1)kk!=k=0pS1,xp,k+k+1+xS1,xp,k+1(-1)kk!=k=0pS1,xp,k(-1)kk!+k=0p-1S1,xp,k+1(-1)kk+1!+xk=0p-1S1,xp,k+1(-1)kk!=k=0pS1,xp,k(-1)kk!-k=1pS1,xp,k(-1)kk!+pxp=S1,xp,0+pxp=p+1xp,

as desired.

(b) We proceed by induction on p2. If p2=0 the result is true by (3.13). If formula (3.14) is true for a given p2N, then, by using the recurrence (1.11) and the value S1,x11,x2,p2p1,0=x1p1x2p2, we have that

k=0p1+p2S1,x11,x2,p2+1p1,k+1(-1)kk!=k=0p1+p2S1,x21,x1,p1p2,k+k+1+x2S1,x21,x1,p1p2,k+1(-1)kk!=k=0p1+p2S1,x21,x1,p1p2,k(-1)kk!+k=0p1+p2-1S1,x21,x1,p1p2,k+1(-1)kk+1!+x2k=0p1+p2-1S1,x21,x1,p1p2,k+1(-1)kk!=k=0p1+p2S1,x21,x1,p1p2,k(-1)kk!-k=1p1+p2S1,x21,x1,p1p2,k(-1)kk!+x2p1x2p2x1p1-1+p2x1p1x2p2-1=S1,x11,x2,p2p1,0+x2p1x2p2x1p1-1+p2x1p1x2p2-1=x1p1x2p2+p1x2p2+1x1p1-1+p2x1p1x2p2=p1x2p2+1x1p1-1+p2+1x1p1x2p2,

as desired.

Proposition 3.3. The bivariate Bernoulli and Euler polynomials have the following properties (when possible, we write P to denote B or E).

(a) (Symmetry) We have

Pp1,p21-x1,1-x2=-1p1+p2Pp1,p2(x1,x2).

(b) (Difference equation) We have

Bp1,p2x1+1,x2+1-Bp1,p2(x1,x2)=p1x1p1-1x2p2+p2x1p1x2p2-1,
Ep1,p2x1+1,x2+1+Ep1,p2(x1,x2)=2x1p1x2p2.

(c) (Sum of powers) If r is a positive integer, then

Bp1,p2x1+r,x2+r-Bp1,p2(x1,x2)=p1l=0r-1x1+lp1-1x2+lp2+p2l=0r-1x1+lp1x2+lp2-1,
Ep1,p2x1+r,x2+r--1rEp1,p2(x1,x2)=2l=0r-1-1r-1-lx1+lp1x2+lp2.

(d) (Derivatives) We have

x1Pp1,p2(x1,x2)=p1Pp1-1,p2(x1,x2),
x2Pp1,p2(x1,x2)=p2Pp1,p2-1(x1,x2).

(e) (Integrals) We have

t01t01Pp1,p2(x1,x2)dx1dx2=1+(-1)p1+p2(p1+1)(p2+1)Pp1+1,p2+1(1,1)-Pp1+1,p2+1(1,0).

Proof. (a) Let us consider the case P=B. In (3.2) replace x1 by 1-x1 and x2 by 1-x2, and set y=1, to obtain

Bp1,p21-x1,1-x2=j1=0p1j2=0p2p1j1p2j2(-x1)p1-j1(-x2)p2-j2Bj1+j2(1),

from where (3.15) follows. The proof in the case P=E is similar.

(b) By using (1.10) and (3.14) we have

Bp1,p2x1+1,x2+1=k=0p1+p2S1,x1+11,x2+1,p2p1,k(-1)kk!k+1=k=0p1+p2S1,x11,x2,p2p1,k+k+1S1,x11,x2,p2p1,k+1(-1)kk!k+1=k=0p1+p2S1,x11,x2,p2p1,k(-1)kk!k+1+k=0p1+p2S1,x11,x2,p2p1,k+1(-1)kk!=Bp1,p2(x1,x2)+p1x2p2x1p1-1+p2x1p1x2p2-1,

as desired. Similarly, by using (1.10) we have

Ep1,p2x1+1,x2+1=k=0p1+p2S1,x11,x2,p2p1,k+k+1S1,x11,x2,p2p1,k+1(-1)kk!2k=Ep1,p2(x1,x2)-2k=1p1+p2S1,x11,x2,p2p1,k(-1)kk!2k=Ep1,p2(x1,x2)-2Ep1,p2(x1,x2)-x1p1x2p2=-Ep1,p2(x1,x2)+2x1p1x2p2,

as claimed.

(c) The case r=1 of (3.18) is (3.16). If we suppose that (3.18) is valid for a positive integer r>1, then by using (3.16) we have

Bp1,p2x1+r+1,x2+r+1Bp1,p2(x1,x2)=Bp1,p2x1+r+1,x2+r+1Bp1,p2x1+1,x2+1+Bp1,p2x1+1,x2+1Bp1,p2(x1,x2)=p1 l=0 r1x1+l+1p11 x 2 +l+1p2+p2 l=0 r1x1+l+1p1 x 2 +l+1p21+p1x1p11x2p2+p2x1p1x2p21=p1 l=0rx1+lp11 x 2 +lp2+p2 l=0rx1+lp1 x 2 +lp21,

as desired. The proof of (3.19) is similar.

(d) Formulas (3.20) and (3.21) say that for p1 or p2 fixed, the sequences of bivariate Bernoulli and Euler polynomials are Appel sequences, and this fact is equivalent to addition formula (3.1). However, a direct proof follows from the following easy observation:

x1S1,x11,x2,p2p1,k=x1j1=0p1j2=0p2p1j1p2j2x1p1-j1x2p2-j2S(j1+j2,k)=p1j1=0p1-1j2=0p2p1-1j1p2j2x1p1-1-j1x2p2-j2S(j1+j2,k)=p1S1,x11,x2,p2p1-1,k.

Similarly, we have x2S1,x11,x2,p2p1,k=p2S1,x11,x2,p2-1p1,k. Then (3.20) and (3.21) follows directly from the definitions (2.4) and (2.5), and the observation above.

(e) We use (3.20) and (3.21) to write

t01t01Pp1,p2(x1,x2)dx1dx2=1p1+1t01Pp1+1,p2(1,x2)-Pp1+1,p2(0,x2)dx2=1(p1+1)(p2+1)Pp1+1,p2+11,1-Pp1+1,p2+11,0  -Pp1+1,p2+10,1+Pp1+1,p2+10,0=1+(-1)p1+p2(p1+1)(p2+1)Pp1+1,p2+1(1,1)-Pp1+1,p2+1(1,0),

which proves (3.22). In the last step we used (3.15) to write Pp1+1,p2+10,1=(-1)p1+p2Pp1+1,p2+11,0 and Pp1+1,p2+11,1=(-1)p1+p2Pp1+1,p2+10,0.

An immediate consequence of (3.22) is the following:

Corollary 3.4. If p1+p2 is odd, we have

t01t01Bp1,p2(x1,x2)dx1dx2=t01t01Ep1,p2(x1,x2)dx1dx2=0.

In this section we obtain some identities involving bivariate Bernoulli and/or Euler polynomials.

Proposition 4.1. We have the following identities

(1)

12(p1Ep1-1,p2(x1,x2)+p2Ep1,p2-1(x1,x2))=Bp1,p2(x1,x2)-2p1+p2Bp1,p2x12,x22.

(2)

p1Ep1-1,p2(x1,x2)+p2Ep1,p2-1(x1,x2)=2p1+p2Bp1,p2x1+12,x2+12-Bp1,p2x12,x22.

(3) For mN,

Bp1,p2mx1,mx2=mp1+p2-1k=0m-1Bp1,p2x1+km,x2+km.

(4) For m=1,3,5,

Ep1,p2(mx1,mx2)=mp1+p2k=0m-1(-1)kEp1,p2x1+km,x2+km.

(5) For m=2,4,6,,

p1Ep1-1,p2mx1,mx2+p2Ep1,p2-1mx1,mx2=-2mp1+p2-1k=0m-1(-1)kBp1,p2x1+km,x2+km.

Proof. We show the proofs of (4.1), (4.3) and (4.5).

(1) Beginning with the right-hand side of (4.1), we have

Bp1,p2(x1,x2)-2p1+p2Bp1,p2x12,x22=j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2Bj1+j2y-2p1+p2j1=0p1j2=0p2p1j1p2j2x12-y2p1-j1x22-y2p2-j2Bj1+j2y2=j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2×  ×Bj1+j2y-2j1+j2Bj1+j2y2.

By using the identity En-1y=2nBny-2nBny2 (see [9, Eq. 24.4.22]), we have

(4.6)=12j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2j1+j2Ej1+j2-1y=p12j1=0p1-1j2=0p2p1-1j1p2j2x1-yp1-1-j1x2-yp2-j2Ej1+j2y+p22j1=0p1j2=0p2-1p1j1p2-1j2x1-yp1-j1x2-yp2-1-j2Ej1+j2y=12p1Ep1-1,p2(x1,x2)+p2Ep1,p2-1(x1,x2),

as desired

(2) Beginning with the left-hand side of (4.3), we have

Bp1,p2mx1,mx2=j1=0p1j2=0p2p1j1p2j2mx1-myp1-j1mx2-myp2-j2Bj1+j2my.

By using the identity Bnmy=mn-1k=0m-1Bny+km (see [9, Eq. 24.4.18]) we obtain

(4.7)=mp1+p2-1k=0m-1j1=0p1j2=0p2p1j1p2j2x1+km-y+kmp1-j1×  ×x2+km-y+kmp2-j2Bj1+j2y+km=mp1+p2-1k=0m-1Bp1,p2x1+km,x2+km,

as desired.

(3) Beginning with the right-hand side of (4.5), we have

-2mp1+p2-1k=0m-1(-1)kBp1,p2x1+km,x2+km=-2mp1+p2-1j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2×  ×k=0m-1(-1)kBj1+j2y+km.

By using the identity Enmy=-2mnn+1k=0m-1(-1)kBn+1y+km for m=2,4,6, (see [9, Eq. 24.4.19]) we obtain

(4.8)=-2mp1+p2-1j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2××-j1+j22mj1+j2-1Ej1+j2-1my=p1j1=0p1-1j2=0p2p1-1j1p2j2mx1-myp1-1-j1mx2-myp2-j2Ej1+j2my+p2j1=0p1j2=0p2-1p1j1p2-1j2mx1-myp1-j1mx2-myp2-1-j2Ej1+j2my=p1Ep1-1,p2mx1,mx2+p2Ep1,p2-1mx1,mx2,

as desired.

The proofs of (4.2) and (4.4) are similar, using the identities

En-1y=2nnBny+12-Bny2

(see [9, Eq. 24.4.23])

and

Enmy=mnk=0m-1(-1)kEny+km for m=1,3,5,

(see [9, Eq. 24.4.20]).

Now we consider a different kind of identities involving Bernoulli and Euler numbers and polynomials, which are based in the following result.

Proposition 4.2. The polynomial identity

l=0nan,lx+αl=l=0nbn,lx+βl,

implies the Bernoulli and Euler polynomials identities

l=0nan,lBlx+α=l=0nbn,lBlx+β,
l=0nan,lElx+α=l=0nbn,lElx+β.

Proof. Observe that the polynomial identity (4.9) comes together with the identities of the corresponding derivatives

l=0nan,llrx+αl-r=l=0nbn,llrx+βl-r,

for 0rln. Beginning with the left-hand side of (4.10), and using (4.12) in the third step, we have

l=0nan,lBlx+α=l=0nan,lk=0lS1,x+αl,k(-1)kk!k+1=k=0nj=0nl=0nan,lljx+αl-jSj,k(-1)kk!k+1=k=0nj=0nl=0nbn,lljx+βl-jSj,k(-1)kk!k+1=l=0nbn,lk=0lS1,x+βl,k(-1)kk!k+1=l=0nbn,lBlx+β,

which shows (4.10). The proof of (4.11) is similar.

For example, from (3.2) we can write

j1=0p1j2=0p2p1j1p2j2x1-yp1-j1x2-yp2-j2Bj1+j2y=j1=0p1j2=0p2p1j1p2j2x1-zp1-j1x2-zp2-j2Bj1+j2z,

where y,z are arbitrary parameters. By using Proposition 4.2 in (4.13) we obtain identities of the form

j1=0p1j2=0p2p1j1p2j2Pp1-j1x1-yQp2-j2x2-yBj1+j2y=j1=0p1j2=0p2p1j1p2j2Pp1-j1x1-zQp2-j2x2-zBj1+j2z,

where P,Q are Bernoulli or Euler polynomials.

In the case P=Q=B, with x1=x2=z=1 and y=0, we obtain that

(-1)p1+p2j1=0p1j2=0p2p1j1p2j2(-1)j1+j2Bp1-j1Bp2-j2Bj1+j2=j1=0p1j2=0p2p1j1p2j2(-1)j1+j2Bp1-j1Bp2-j2Bj1+j2,

from where we conclude that, if p1+p2 is odd, then

j1=0p1j2=0p2p1j1p2j2(-1)j1+j2Bp1-j1Bp2-j2Bj1+j2=0.

With similar arguments it is possible to show that if p1+p2 is odd, then

j1=0p1j2=0p2p1j1p2j2Bp1-j1Bp2-j2Ej1+j21=0,j1=0p1j2=0p2p1j1p2j2(-1)j1+j2Ep1-j10Ep2-j20Bj1+j2=0,j1=0p1j2=0p2p1j1p2j2Ep1-j1Ep2-j2Ej1+j2=0,j1=0p1j2=0p2p1j1p2j2-1j1+j2Ep1-j1Ep2-j2Ej1+j2=0.

Now let us consider the difference equations (3.16) and (3.17). Write (3.16) as

j1=0p1j2=0p2p1j1p2j2x1+1p1-j1x2+1p2-j2-x1p1-j1x2p2-j2Bj1+j2=p1x1p1-1x2p2+p2x2p2-1x1p1.

By using Proposition 4.2, we obtain from (4.18) the identities

j1=0p1j2=0p2p1j1p2j2××Bp1-j1x1+1Bp2-j2x2+1-Bp1-j1x1Bp2-j2x2Bj1+j2=p1Bp1-1x1Bp2x2+p2Bp1x1Bp2-1x2,
j1=0p1j2=0p2p1j1p2j2××Ep1-j1x1+1Ep2-j2x2+1-Ep1-j1x1Ep2-j2x2Bj1+j2=p1Ep1-1x1Ep2x2+p2Ep1x1Ep2-1x2,

and

j1=0p1j2=0p2p1j1p2j2××Bp1-j1x1+1Ep2-j2x2+1-Bp1-j1x1Ep2-j2x2Bj1+j2=p1Bp1-1x1Ep2x2+p2Bp1x1Ep2-1x2.

Similarly, by writing (3.17) as

j1=0 p1j2=0p2 p 1 j 1 p2 j2 x 1+1 p 1 j 1 x 2+1 p2 j2 +x1 p 1 j 1 x2 p2 j2 Ej1+j2(0)=2x1p1x2p2,

and using Proposition 4.2, we obtain from (4.21) identities of the form

j1=0p1j2=0p2p1j1p2j2××Pp1-j1x1+1Qp2-j2x2+1+Pp1-j1x1Qp2-j2x2Ej1+j2(0)=2Pp1x1Qp2x2,

where P and Q are Bernoulli or Euler polynomials.

To end this section, we show some particular cases of the previous results.

By setting p2=2 and x1=x2=0 in (4.19), we obtain (after some simplifications) the following identity for p>0:

j=1p+12p+12j-1B2p-2j+2B2j=-(p+1/2)B2p+2.

Similarly, from (4.20) with p2=1 and x1=x2=0, we get the following identity for p>0:

2j=1p2p2j-1E2p-2j+11B2j=-B2p-pE2p-10.

By using Proposition 4.2 in the case m=2 of (4.3), we obtain that

21-p1-p2j1=0p1j2=0p2p1j1p2j22p1+p2-j1-j2Bp1-j1(x1)Bp2-j2(x2)Bj1+j2=j1=0p1j2=0p2p1j1p2j2××Bp1-j1(x1)Bp2-j2(x2)+Bp1-j1(x1+1/2)Bp2-j2(x2+1/2Bj1+j2.

Set x1=x2=0 and use that Bn12=21-n-1Bn, to get from (4.23) that

2j1=0p1j2=0p2p1j1p2j22-j1-j2Bp1-j1Bp2-j2Bj1+j2=j1=0p1j2=0p2p1j1p2j21+21+j1-p1-121+j2-p2-1Bp1-j1Bp2-j2Bj1+j2.

The case p2=1 of (4.24) produces the identity

j=0p-12p2j1-21-2jB2jB2p-2j=21-2p-12pB2p,

which can be written (by using the Euler's identity j=0p2p2jB2jB2p-2j=-(2p-1)B2p for p>1, see [4]) as

j=0p2p2j2-2jB2jB2p-2j=1-2p2-2pB2p.

With similar procedures it is possible to obtain also the identities

j=0p-12p2j22jB2jE2p-2j=21-22pB2p,
j=0p-12p-12j1-21-2jB2jE2p-1-2j(0)=21-2p(2p-1)E2p-2.

### 5. Generalized Recurrences

In this section we show some recurrences for bivariate Bernoulli and Euler polynomials.

Proposition 5.1. For non-negative integers p1,p2,q, we have the recurrences

k=0qBp1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j)=k=0qBp1,p2+k(x1,x2)(-1)kk!dkdx2kj=0q-1(x2+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!k+q+1,

and

k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j)=k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=0q-1(x2+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q.

Proof. Formulas in (5.1) are included as particular cases of Proposition 4 in [12]. We prove the identity

k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q,

contained in (5.2) by induction on q. The case q=0 is the definition (2.5). If we suppose that (5.3) is true for a given qN, then

k=0q+1Ep1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q(x1+j)=k=0q+1Ep1+k,p2(x1,x2)(-1)kk!(x1+q)dkdx1kj=0q-1(x1+j)+kdk-1dx1k-1j=0q-1(x1+j)=(x1+q)k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j)-k=0qEp1+1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j).

The induction hypothesis gives us from (5.4), that

k=0q+1Ep1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q(x1+j)=(x1+q)k=0p1+p2S1,x1+q1,x2+q,p2(p1,k)(-1)kk+q!2k+q-k=0p1+p2+1S1,x1+q1,x2+q,p2(p1+1,k)(-1)kk+q!2k+q.

According to the recurrence (1.11), we have from (5.5) that

k=0q+1Ep1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q(x1+j)=x1+qk=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q-k=0p1+p2+1S1,x1+q1,x2+q,p2p1,k-1+(k+x1+q)S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q=-k=0p1+p2+1S1,x1+q1,x2+q,p2p1,k-1+kS1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q=k=0p1+p2S1,x1+q1,x2+q,p2p1,k+(k+1)S1,x1+q1,x2+q,p2p1,k+1(-1)kk+q+1!2k+q+1=k=0p1+p2S1,x1+q+11,x2+q+1,p2p1,k(-1)kk+q+1!2k+q+1,

as desired. In the last step we used (1.10).

The proof of the identity

k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=0q-1(x2+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q.

is similar.

Proposition 5.2. For non-negative integers p1,p2,q, we have the recurrences

k=0qBp1+k,p2(x1,x2)(-1)kk!dkdx1kj=1q(x1-j)=k=0qBp1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q(x2-j)=q!k=0p1+p2S1,x11,x2,p2p1,k-1k+qk!k+q+1,
k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=1q(x1-j)=k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q(x2-j)=k=0p1+p2S1,x11,x2,p2p1,k(-1)k+q(k+q)!2k+q.

Proof. We prove the identity

k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q(x2-j)=k=0p1+p2S1,x11,x2,p2p1,k-1k+qk+q!2k+q,

contained in (5.7) by induction on q. The case q=0 of (5.8) is the definition (2.5). If we suppose (5.8) is true for a given qN, then

k=0q+1Ep1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q+1(x2-j)=k=0q+1Ep1,p2+k(x1,x2)(-1)kk!××(x2-q-1)dkdx2kj=1q(x2-j)+kdk-1dx2k-1j=1q(x2-j)=(x2-q-1)k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q(x2-j)-k=0qEp1,p2+k+1(x1,x2)(-1)kk!dkdx2kj=1q(x2-j).

We use induction hypothesis to write (5.9) as

k=0q+1Ep1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q+1(x2-j)=(x2-q-1)k=0p1+p2S1,x11,x2,p2p1,k-1k+qk+q!2k+q-k=0p1+p2+1S1,x11,x2,p2+1p1,k-1k+qk+q!2k+q.

The recurrence (1.11) gives us

k=0q+1Ep1,p2+k(x1,x2)(-1)kk!dkdx2kj=1q+1(x2-j)=(x2-q-1)k=0p1+p2S1,x11,x2,p2p1,k-1k+qk+q!2k+q-k=0p1+p2+1S1,x11,x2,p2p1,k-1+(k+x2)S1,x11,x2,p2p1,k-1k+qk+q!2k+q=-k=0p1+p2S1,x11,x2,p2p1,k-1k+q+1k+q+1!2k+q+1-k=0p1+p2(k+q+1)S1,x11,x2,p2p1,k-1k+qk+q!2k+q=-k=0p1+p2S1,x11,x2,p2p1,k-1k+q+1k+q+1!2k+q+1.-k=0p1+p2S1,x11,x2,p2p1,k-1k+qk+q+1!2k+q=k=0p1+p2S1,x11,x2,p2p1,k-1k+q+1k+q+1!2k+q+1,

as desired. The proof of the identity

k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=1q(x1-j)=k=0p1+p2S1,x11,x2,p2p1,k-1k+qk+q!2k+q,

and (5.6) are similar.

We can write recurrence (5.7) as

k=0qEp1+k,p2(x1+q,x2+q)(-1)kk!dkdx1kj=0q-1(x1+j)=k=0qEp1,p2+k(x1+q,x2+q)(-1)kk!dkdx2kj=0q-1(x2+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k-1k+qk+q!2k+q.

Thus, from (5.2) and (5.10) we have

k=0qEp1+k,p2(x1+q,x2+q)-1k+qk!dkdx1kj=0q-1(x1+j)=k=0qEp1,p2+k(x1+q,x2+q)-1k+qk!dkdx2kj=0q-1(x2+j)=k=0qEp1+k,p2(x1,x2)(-1)kk!dkdx1kj=0q-1(x1+j)=k=0qEp1,p2+k(x1,x2)(-1)kk!dkdx2kj=0q-1(x2+j)=k=0p1+p2S1,x1+q1,x2+q,p2p1,k(-1)kk+q!2k+q.

From (5.1) with x1=1,x2=0 we have

k=0q(-1)ks(q+1,k+1)Bp1+k,p21,0=k=0q(-1)ks(q,k)Bp1,p2+k1,0=k=0p1+p2S1,q+11,q,p2p1,k(-1)kk+q!k+q+1.

We use Carlitz identity (3.5) to obtain the following explicit enriched version of (5.12),

k=0qj1=0p1+kp1+kj1(-1)ks(q+1,k+1)Bj1+p2=(-1)p1+p2k=0qj2=0p2p2j2s(q+1,k+1)Bp1+k+j2=k=0qj1=0p1p1j1(-1)ks(q,k)Bj1+p2+k=(-1)p1+p2k=0qj2=0p2+kp2+kj2s(q,k)Bp1+j2=k=0p1+p2j1=0p1j2=0p2p1j1p2j2(q+1)p1-j1qp2-j2S(j1+j2,k)(-1)kk+q!k+q+1.

Similarly, from (5.6) we obtain

k=0qj1=0p1+kp1+kj1s(q,k)Bj1+p2=-1p1+p2k=0qj2=0p2p2j2(-1)ks(q,k)Bp1+k+j2=k=0qj1=0p1p1j1s(q+1,k+1)Bj1+p2+k=-1p1+p2k=0qj2=0p2+kp2+kj2(-1)ks(q+1,k+1)Bp1+j2=q!k=0p1+p2j1=0p1p1j1S(j1+p2,k)(-1)kk!k+q+1,

and from (5.11) we get

(-1)q!k=0qj1=0p1+kj2=0p2!!p1+kj1p2j2(-1)ks(q+1,k+1)(q+1)p1+k-j1qp2-j2Ej1+j2(0)=(-1)qk=0qj1=0p1+kj2=0p2p1j1p2+kj2(-1)ks(q,k)(q+1)p1-j1qp2+k-j2Ej1+j2(0)=k=0qj1=0p1+kp1+kj1(-1)ks(q+1,k+1)Ej1+p2(0)=k=0qj1=0p1p1j1(-1)ks(q,k)Ej1+p2+k(0)=k=0p1+p2j1=0p1j2=0p2p1j1p2j2(q+1)p1-j1qp2-j2S(j1+j2,k)(-1)k(k+q)!2k+q.

Proposition 5.3. We have the following identities for bivariate Euler polynomials

j=0qqj(-1)jEp1+j,p2+q-j(x1,x2)=(x2-x1)qEp1,p2(x1,x2),
j=0qqj(x2-x1)jEp1+q-j,p2(x1,x2)=Ep1,p2+q(x1,x2).

Proof. Let us prove (5.14) by induction on q. The case q=0 is a trivial identity. If (5.14) is valid for a given qN, then

j=0q+1q+1j(-1)jEp1+j,p2+q+1-j(x1,x2)=j=0qqj(-1)jEp1+j,p2+1+q-j(x1,x2)-j=0qqj(-1)jEp1+1+j,p2+q-j(x1,x2).

The induction hypothesis gives us

j=0q+1q+1j(-1)jEp1+j,p2+q+1-j(x1,x2)=(x2-x1)qEp1,p2+1(x1,x2)-(x2-x1)qEp1+1,p2(x1,x2)=(x2-x1)qEp1,p2+1(x1,x2)-Ep1+1,p2(x1,x2)=(x2-x1)q+1Ep1,p2(x1,x2),

as desired. In the last step we used the identity

Ep1+1,p2(x1,x2)-Ep1,p2+1(x1,x2)=(x1-x2)Ep1,p2(x1,x2),

which is included in the case q=1 of (5.2) (or (5.7)).

Let us prove (5.15) by induction on q. The case q=0 is a trivial identity. If we suppose (5.15) is valid for a given qN, then

j=0q+1q+1j(x2-x1)jEp1+q+1-j,p2(x1,x2)=j=0qqj(x2-x1)jEp1+q+1-j,p2(x1,x2)+(x2-x1)j=0qqj(x2-x1)jEp1+q-j,p2(x1,x2)=Ep1+1,p2+q(x1,x2)+(x2-x1)Ep1,p2+q(x1,x2)=Ep1,p2+q+1(x1,x2),

as desired. We used (5.16) in the last step.

There are similar results to (5.14) and (5.15) in the case of bivariate Bernoulli polynomials. These results appear in [12] in the context of bivariate poly-Bernoulli polynomials.

I thank the anonymous referee for his/her careful reading of the original version of this work. The many details and observations he/she pointed out, certainly contributed to present this final version.

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