Convolutions for Bernoulli and Euler–Genocchi Polynomials of Order (r,m) and Their Probabilistic Interpretation ^†

Abstract

The main purpose of this article is to derive several convolutions for generalized Bernoulli and Euler–Genocchi polynomials of order $(r,m)$, $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$, respectively. These polynomials have been introduced recently and contain the generalized Bernoulli, Euler and Genocchi polynomials as special members. Some of our results extend the results of M. Merca and others concerning Bernoulli numbers and polynomials. Probabilistic interpretations of the presented results are also given.

1. Introduction

Bernoulli numbers ${\left(B_n\right)}_{n\ge 0}$ are defined by

$$B\left(z\right)=\sum _{n=0}^{\infty }{B}_n\frac{z^n}{n!}=\frac{z}{e^z-1}\left(\right|z|<2\pi ).$$

A generalization of Bernoulli numbers are Bernoulli polynomials $B_n\left(x\right),x\in \mathbb{C}$, defined by

$$B(x,z)=\sum _{n=0}^{\infty }{B}_n\left(x\right)\frac{z^n}{n!}=\frac{z}{e^z-1}{e}^{xz}\left(\right|z|<2\pi ).$$

These numbers (polynomials) are fascinating objects, appearing in many mathematical branches such as number theory, combinatorics and analysis. The basic properties of Bernoulli numbers and polynomials are discussed in [1,2].

Closely related to Bernoulli polynomials are the Euler and Genocchi polynomials. These polynomials are defined for $\left|z\right|<\pi$ by

$$\sum _{n=0}^{\infty }{E}_n\left(x\right)\frac{z^n}{n!}=\frac{2}{e^z+1}{e}^{xz}\mathrm{a}\mathrm{n}\mathrm{d}\sum _{n=0}^{\infty }{G}_n\left(x\right)\frac{z^n}{n!}=\frac{2z}{e^z+1}{e}^{xz}.$$

Finding recurrences and convolutions for these polynomials is still an active field of research. Many interesting identities for Bernoulli, Euler and Genocchi polynomials can be found in the articles [3,4,5,6,7,8,9,10] for instance. See [11,12,13,14] for some properties of generalizations of these polynomials.

The popularity and importance of Bernoulli numbers and polynomials in number theory comes also from their connection to the Riemann zeta function

$$\zeta \left(2n\right)={(-1)}^{n+1}\frac{{\left(2\pi \right)}^{2n}}{2·\left(2n\right)!}{B}_{2n},$$

where

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{n^s},\Re \left(s\right)>1,$$

is the Riemann zeta function [15]. A great deal of proof for this relation has been provided over the years. See [16] for references. Recently, Merca [16] proved the following relation between Bernoulli numbers:

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k=(2-2^n)B_n.$$

(1)

This relation can be used to derive a recurrence relation for $\zeta \left(2n\right)$. Moreover, in his next article on the topic, Merca [17] used recurrence relations for Bernoulli polynomials $B_n\left(x\right)$ to derive two new infinite families of linear recurrence relations for the Riemann zeta function at positive even integer arguments. Merca’s elegant results are based on the following relations (Theorems 2.1 and 3.1 in [17]): Let n be a positive integer and $x,\alpha \in \mathbb{C}$. Then

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k\left(x\right)({\alpha }^{n-k}-{(-1)}^k{(2x-1+\alpha )}^{n-k})=0,$$

(2)

and

$$\sum _{k=0}^{⌊\frac{n}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k}\right){\alpha }^{2k}{B}_{n-2k}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}\frac{{(2x-1+\alpha )}^k+{(2x-1-\alpha )}^k}{2}{B}_{n-k}\left(x\right).$$

(3)

In fact, identities Equations (1) and (2) have been widely known and used already in the 19th century as immediate consequences of the simple functional relations

$$B\left(2z\right)(e^z+1)=2B\left(z\right)\mathrm{a}\mathrm{n}\mathrm{d}B(x,z)e^{yz}=B(x,-z)e^{(2x-1+y)z}.$$

It may be very difficult to identify the first pioneers who discovered them after all this time. We refer to the notable books by Saalschütz [18], Nielsen [19] and Hansen [20] as a resource in which a large number of classical identities, mainly developed in the 18th and the 19th centuries, can be found. In addition, Equation (1) can be extended to [21]

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)m^k{B}_k\sum _{i=0}^{m-1}{i}^{n-k}=m{B}_n,$$

valid for all integers $m\ge 1$. Thus, Equation (1) is the special case for $m=2$. Next, we can define the function ${\Phi }_n^{\left(m\right)}(x,y),m\ge 1,x,y\in \mathbb{C},$ by

$${\Phi }_n^{\left(m\right)}(x,y)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)m^k{B}_k\left(x\right)y^{n-k}.$$

Then, the following identity has been known for a long time:

$${\Phi }_n^{\left(m\right)}(x,y)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-m)}^k{B}_k\left(x\right){(2mx-m+y)}^{n-k},m\ge 1.$$

(4)

The identity Equation (4) reduces to Equation (2) when $m=1$ and $y=\alpha$. It is easy to see that ${\Phi }_n^{\left(m\right)}(x,y)$ satisfies the following functional equation.

Proposition 1.

For all $m,n\in \mathbb{N}$ the following functional equation holds:

$${\Phi }_n^{\left(m\right)}(x,-(2mx-m+y))={(-1)}^n{\Phi }_n^{\left(m\right)}(x,y).$$

(5)

In addition, we can examine a calculation of the sum ${\Phi }_n^{\left(m\right)}(x,y)+{\Phi }_n^{\left(m\right)}(x,-y)$. We have

$${\Phi }_n^{\left(m\right)}(x,y)+{\Phi }_n^{\left(m\right)}(x,-y)=\left\{\begin{array}{cc}2\sum _{l=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0pt}{}{n}{2l}\right)m^{n-2l}{B}_{n-2l}\left(x\right)y^{2l},\hfill & if n-k=2l;\hfill \\ 0,\hfill & \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\hfill \end{array}\right.$$

In view of such an identity, we recognize that Equation (3) is an obvious consequence of Equation (2). Finally, we remark that if $n-k$ is odd, then ${\Phi }_n^{\left(m\right)}(x,y)=-{\Phi }_n^{\left(m\right)}(x,-y)$, i.e., the function ${\Phi }_n^{\left(m\right)}(x,y)$ is an odd or asymmetric function with respect to y.

We conclude this section by recalling the definition of $B_n^{(r,m)}\left(x\right)$, which comes from [22].

Definition 1.

For integers $r,m\ge 1$ the generalized Bernoulli polynomials of order $(r,m)$ are defined by the generating function

$$B(r,m;x,z)=\sum _{n=0}^{\infty }{B}_n^{(r,m)}\left(x\right)\frac{z^n}{n!}={\left(\frac{z^r}{e^z-1}\right)}^m{e}^{xz}\left(\right|z|<2\pi ).$$

(6)

The numbers $B_n^{(r,m)}\left(0\right)=B_n^{(r,m)}$ are called generalized Bernoulli numbers of order r and m.

The polynomials $B_n^{(r,m)}\left(x\right)$ belong to the family of Appell polynomials. We mention that they are defined for $n\ge m(r-1)$, i.e., $B_j^{(r,m)}\left(x\right)=0$ for $j<m(r-1)$. From the definition, it is obvious that $B_n^{(1,1)}\left(x\right)=B_n\left(x\right)$ and $B_n^{(1,m)}\left(x\right)=B_n^{\left(m\right)}\left(x\right)$ are the generalized Bernoulli polynomials of order m. Moreover, $B_n^{(1,1)}=B_n$ are the Bernoulli numbers.

The goal of the present article is to derive several convolutions for generalized Bernoulli polynomials of order $(r,m)$, $B_n^{(r,m)}\left(x\right)$. First, we will generalize Merca’s results for Bernoulli polynomials to the more general class of polynomials. This will be performed in Section 2. In Section 3, we will prove the analogue identities for generalized Euler–Genocchi polynomials $A_n^{(r,m)}\left(x\right)$. A range of additional convolutions for $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$ will be given in Section 4. Among other things, we will rediscover identity Equation (1) as a special case of our findings. In Section 5 and Section 6 we will state some additional remarks concerning applications and future work.

2. Notes on Merca’s Identities

Our first result is an extension of Theorem 2.1 of [17].

Theorem 1.

Let r and m be positive integers and $x,y\in \mathbb{C}$. Then,

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)(y^{n-k}-{(-1)}^{m(r-1)+k}{(2x-m+y)}^{n-k})=0.$$

(7)

Especially, with $y=m$, we have

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)(m^{n-k}-{(-1)}^{m(r-1)+k}{\left(2x\right)}^{n-k})=0.$$

(8)

Proof. 

For $\left|z\right|<2\pi$ we have from Equation (6)

$$\begin{array}{ccc}& & \left(\sum _{n=0}^{\infty }{B}_n^{(r,m)}\left(x\right)\frac{z^n}{n!}\right)\left(\sum _{n=0}^{\infty }{y}^n\frac{z^n}{n!}\right)={\left(\frac{z^r}{e^z-1}\right)}^m{e}^{xz}{e}^{yz}\hfill \\ & & =\frac{z^{rm}}{{(-1)}^m{(e^{-z}-1)}^m}{e}^{-xz}{e}^{(y+2x-m)z}\hfill \\ & & ={(-1)}^{m(r-1)}{\left(\frac{{(-z)}^r}{e^{-z}-1}\right)}^m{e}^{-xz}{e}^{(y+2x-m)z}\hfill \\ & & ={(-1)}^{m(r-1)}B(r,m;x,-z)e^{(y+2x-m)z}\hfill \\ & & ={(-1)}^{m(r-1)}\left(\sum _{n=0}^{\infty }{B}_n^{(r,m)}\left(x\right){(-1)}^n\frac{z^n}{n!}\right)\left(\sum _{n=0}^{\infty }{(y+2x-m)}^n\frac{z^n}{n!}\right)\hfill \\ & & =\sum _{n=0}^{\infty }\left(\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{m(r-1)+k}{B}_k^{(r,m)}\left(x\right){(y+2x-m)}^{n-k}\right)\frac{z^n}{n!}.\hfill \end{array}$$

On the other hand, using Cauchy’s rule, it is obvious that

$$\left(\sum _{n=0}^{\infty }{B}_n^{(r,m)}\left(x\right)\frac{z^n}{n!}\right)\left(\sum _{n=0}^{\infty }{y}^n\frac{z^n}{n!}\right)=\sum _{n=0}^{\infty }\left(\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)y^{n-k}\right)\frac{z^n}{n!}.$$

Comparing the coefficients for $z^n$ in the two power series proves the formula. □

For $(r,m)=(1,1)$ Theorem 1 reduces to Merca’s Theorem 2.1. For $(r,m)=(1,m)$ our theorem gives a convolutional relation for generalized Bernoulli polynomials:

Corollary 1.

For $x,y\in \mathbb{C}$ the following relation holds for generalized Bernoulli polynomials:

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{\left(m\right)}\left(x\right)(y^{n-k}+{(-1)}^{k+1}{(2x-m+y)}^{n-k})=0.$$

(9)

Corollary 2.

For $x\in \mathbb{C}$ the following relation holds:

$$B_n^{(r,m)}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{m(r-1)+k}{B}_k^{(r,m)}\left(x\right){(2x-m)}^{n-k}.$$

(10)

Especially for the generalized Bernoulli polynomials we have the identity

$$B_n^{\left(m\right)}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k{B}_k^{\left(m\right)}\left(x\right){(2x-m)}^{n-k}.$$

(11)

Proof. 

Set $y=0$ in Theorem 1. □

Next, for $x,y\in \mathbb{C}$, we define the function

$$F_n^{(r,m)}(x,y)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)y^{n-k},r,m\ge 1.$$

(12)

Then, we note the following functional equation:

Proposition 2.

For all $r,m,n\in \mathbb{N}$ the following functional equation holds:

$$F_n^{(r,m)}(x,-(2mx-m+y))={(-1)}^{n+m(r-1)}{F}_n^{(r,m)}(x,y).$$

(13)

Proof. 

Replacing y with $-(2x-m+y)$ in Equation (12), using Equation (7), we get the result. □

We can also calculate

$$F_n^{(r,m)}(x,y)+F_n^{(r,m)}(x,-y)=\left\{\begin{array}{cc}2\sum _{l=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0pt}{}{n}{2l}\right)B_{n-2l}^{(r,m)}\left(x\right)y^{2l},\hfill & if n-k=2l;\hfill \\ 0,\hfill & \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e},\hfill \end{array}\right.$$

and

$$F_n^{(r,m)}(x,y)-F_n^{(r,m)}(x,-y)=\left\{\begin{array}{cc}2\sum _{l=0}^{\lfloor (n-1)/2\rfloor }\left(\genfrac{}{}{0pt}{}{n}{2l+1}\right)B_{n-2l-1}^{(r,m)}\left(x\right)y^{2l+1},\hfill & if n-k=2l+1;\hfill \\ 0,\hfill & \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\hfill \end{array}\right.$$

These calculations confirm the following facts.

(i)

If $n-k$ is odd, then $F_n^{(r,m)}(x,y)=-F_n^{(r,m)}(x,-y)$, i.e., the function $F_n^{(r,m)}(x,y)$ is an odd or asymmetric function with respect to y.

(ii)

If $n-k$ is even, then $F_n^{(r,m)}(x,y)=F_n^{(r,m)}(x,-y)$, i.e., the function $F_n^{(r,m)}(x,y)$ is an even or symmetric function with respect to y.

The above observations lead to the next corollary, which provides an extension of Theorem 3.1 of [17] to the class $B_n^{(r,m)}\left(x\right)$.

Corollary 3.

Let r and m be positive integers and $x,y\in \mathbb{C}$. Then,

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k}\right)y^{2k}{B}_{n-2k}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}\frac{{(2x-m+y)}^k+{(2x-m-y)}^k}{2}{B}_{n-k}^{(r,m)}\left(x\right)\hfill \end{array}$$

(14)

and

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n-1}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k+1}\right)y^{2k+1}{B}_{n-2k-1}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}\frac{{(2x-m+y)}^k-{(2x-m-y)}^k}{2}{B}_{n-k}^{(r,m)}\left(x\right).\hfill \end{array}$$

(15)

We conclude this section with the following results.

Corollary 4.

Let r and m be positive integers and $x,y\in \mathbb{C}$. Then,

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k}\right){(-1)}^k{y}^{2k}{B}_{n-2k}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{B}_{n-k}^{(r,m)}\left(x\right)\sum _{p=0}^k\left(\genfrac{}{}{0pt}{}{k}{p}\right)y^{p}cos\left(\frac{\pi p}{2}\right){(2x-m)}^{k-p}\hfill \end{array}$$

(16)

and

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n-1}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k+1}\right){(-1)}^k{y}^{2k+1}{B}_{n-2k-1}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{B}_{n-k}^{(r,m)}\left(x\right)\sum _{p=0}^k\left(\genfrac{}{}{0pt}{}{k}{p}\right)y^{p}sin\left(\frac{\pi p}{2}\right){(2x-m)}^{k-p}.\hfill \end{array}$$

(17)

Proof. 

Replace y by $iy$ with $i=\sqrt{-1}$ in Corollary 3 and simplify. □

Corollary 5.

Let r and m be positive integers and $x\in \mathbb{C}$. Then,

$$\begin{array}{ccc}& & \sum _{k=0}^{⌊\frac{n}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k}\right){(-1)}^k{m}^{2k}{B}_{n-2k}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{2}^{k-1}(x^k+{(x-m)}^k)B_{n-k}^{(r,m)}\left(x\right)\hfill \end{array}$$

(18)

and

$$\begin{array}{ccc}& & \sum _{k=0}^{⌊\frac{n-1}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k+1}\right){(-1)}^k{m}^{2k+1}{B}_{n-2k-1}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{2}^{k-1}(x^k-{(x-m)}^k)B_{n-k}^{(r,m)}\left(x\right).\hfill \end{array}$$

(19)

Proof. 

Set $y=m$ in Corollary 3 and simplify. □

3. Analogue Relations for Generalized Euler–Genocchi Polynomials

The definition of generalized Euler–Genocchi polynomials of order $(r,m)$ also comes from the paper [22], where many basic properties of the polynomials are discussed.

Definition 2.

Let r and m be integers with $r\ge 0$ and $m\ge 1$. The generalized Euler–Genocchi polynomials of order $(r,m)$, $A_n^{(r,m)}\left(x\right)$, $x\in \mathbb{C}$, are defined by the generating function

$$A(r,m;x,z)=\sum _{n=0}^{\infty }{A}_n^{(r,m)}\left(x\right)\frac{z^n}{n!}={\left(\frac{2z^r}{e^z+1}\right)}^m{e}^{xz}\left(\right|z|<\pi )$$

(20)

with $A_j^{(r,m)}\left(x\right)=0$ for $j<rm$. The numbers $A_n^{(r,m)}\left(0\right)=A_n^{(r,m)}$ are called the generalized Euler–Genocchi numbers of order r and m.

We see that $A_n^{(0,m)}\left(x\right)=E_n^{\left(m\right)}\left(x\right)$ and $A_n^{(1,m)}\left(x\right)=G_n^{\left(m\right)}\left(x\right)$ are the generalized Euler and Genocchi polynomials, respectively, where

$$\sum _{n=0}^{\infty }{E}_n^{\left(m\right)}\left(x\right)\frac{z^n}{n!}={\left(\frac{2}{e^z+1}\right)}^m{e}^{xz}\left(\right|z|<\pi )$$

(21)

and

$$\sum _{n=0}^{\infty }{G}_n^{\left(m\right)}\left(x\right)\frac{z^n}{n!}={\left(\frac{2z}{e^z+1}\right)}^m{e}^{xz}\left(\right|z|<\pi ).$$

(22)

Finally, we mention that the degenerated case $m=0$ gives $A_n^{(r,0)}\left(x\right)=x^n$ for all $r\ge 0$.

The first analogue result of Merca’s identities is stated in the next theorem.

Theorem 2.

Let r and m be integers with $r\ge 0$ and $m\ge 1$, and $x,y\in \mathbb{C}$. Then,

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}\left(x\right)(y^{n-k}-{(-1)}^{mr+k}{(2x-m+y)}^{n-k})=0.$$

(23)

Especially, with $y=m$, we have

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}\left(x\right)(m^{n-k}-{(-1)}^{mr+k}{\left(2x\right)}^{n-k})=0.$$

(24)

Proof. 

Due to the high degree of similarity in the proofs, we only sketch the proofs. The identity basically follows from

$$A(r,m;x,z)e^{yz}={\left(\frac{2z^r}{e^z+1}\right)}^m{e}^{xz}{e}^{yz}={(-1)}^{mr}{\left(\frac{2{(-z)}^r}{e^{-z}+1}\right)}^m{e}^{-xz}{e}^{(2x+y-m)z}.$$

□

Corollary 6.

For $x,y\in \mathbb{C}$ the following relations hold:

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)E_k^{\left(m\right)}\left(x\right)(y^{n-k}+{(-1)}^{k+1}{(2x-m+y)}^{n-k})=0$$

(25)

and

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)G_k^{\left(m\right)}\left(x\right)(y^{n-k}+{(-1)}^{m+k+1}{(2x-m+y)}^{n-k})=0.$$

(26)

Proof. 

Set $r=0$ and $r=1$ in Theorem 2. □

Corollary 7.

For $x\in \mathbb{C}$ the following relation holds:

$$A_n^{(r,m)}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{mr+k}{A}_k^{(r,m)}\left(x\right){(2x-m)}^{n-k}.$$

(27)

Proof. 

Set $y=0$ in Theorem 2. □

Theorem 3.

Let r and m be integers with $r\ge 0$ and $m\ge 1$, and $x,y\in \mathbb{C}$. Then,

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k}\right)y^{2k}{A}_{n-2k}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{mr}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}\frac{{(2x-m+y)}^k+{(2x-m-y)}^k}{2}{A}_{n-k}^{(r,m)}\left(x\right)\hfill \end{array}$$

(28)

and

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^{⌊\frac{n-1}{2}⌋}\left(\genfrac{}{}{0pt}{}{n}{2k+1}\right)y^{2k+1}{A}_{n-2k-1}^{(r,m)}\left(x\right)\hfill \\ & & ={(-1)}^{mr}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}\frac{{(2x-m+y)}^k-{(2x-m-y)}^k}{2}{A}_{n-k}^{(r,m)}\left(x\right).\hfill \end{array}$$

(29)

Proof. 

The proof follows the same arguments as the proof of Theorem 3. □

The special cases where y is replaced by $iy$ and $y=m$ are obvious. We continue skipping the presentation of these explicit results.

4. More Convolutions for $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$

In this section, several other convolutions for $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$ are derived. The first two theorems contain convolutions involving $B_n^{(r,m)}\left(x\right)$ and powers of 2.

Theorem 4.

Let r and m be positive integers and $x,y\in \mathbb{C}$. Then

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k^{(r,m)}\left(x\right)y^{n-k}=2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}(2x+y)E_{n-k}^{\left(m\right)}\left(0\right),$$

(30)

where $E_n^{\left(m\right)}\left(x\right)$ is the generalized Euler polynomial of order m.

Proof. 

For $\left|z\right|<2\pi$ we have from Equation (6)

$$\begin{array}{ccc}\hfill B(r,m;x,2z)e^{yz}& =& \left(\sum _{n=0}^{\infty }{2}^n{B}_n^{(r,m)}\left(x\right)\frac{z^n}{n!}\right)\left(\sum _{n=0}^{\infty }{y}^n\frac{z^n}{n!}\right)\hfill \\ & =& \sum _{n=0}^{\infty }\left(\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k^{(r,m)}\left(x\right)y^{n-k}\right)\frac{z^n}{n!}.\hfill \end{array}$$

On the other hand, we observe that

$$\begin{array}{ccc}\hfill B(r,m;x,2z)e^{yz}& =& 2^{rm}{\left(\frac{z^r}{e^{2z}-1}\right)}^m{e}^{(2x+y)z}\hfill \\ & =& 2^{m(r-1)}{\left(\frac{z^r}{e^z-1}\right)}^m{e}^{(2x+y)z}{\left(\frac{2}{e^z+1}\right)}^m\hfill \\ & =& 2^{m(r-1)}\left(\sum _{n=0}^{\infty }{B}_n^{(r,m)}(2x+y)\frac{z^n}{n!}\right)\left(\sum _{n=0}^{\infty }{E}_n^{\left(m\right)}\left(0\right)\frac{z^n}{n!}\right)\hfill \end{array}$$

Comparing the coefficients for $z^n$ in the two power series proves the formula. □

Remark 1.

From the above proof, it is clear that we can also write

$$\begin{array}{ccc}\hfill \sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k^{(r,m)}\left(x\right)y^{n-k}& =& 2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}(x+y)E_{n-k}^{\left(m\right)}\left(x\right)\hfill \\ & =& 2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)E_{n-k}^{\left(m\right)}(x+y)\hfill \\ & =& 2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(y\right)E_{n-k}^{\left(m\right)}\left(2x\right).\hfill \end{array}$$

Corollary 8.

Let r and m be positive integers and $x\in \mathbb{C}$. Then,

$$2^n{B}_n^{(r,m)}\left(x\right)=2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(2x\right)E_{n-k}^{\left(m\right)}\left(0\right),$$

(31)

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{2}^k{B}_k^{(r,m)}\left(x\right)x^{n-k}=2^{m(r-1)}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)E_{n-k}^{\left(m\right)}\left(0\right)$$

(32)

and

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{B}_k^{(r,m)}\left(x\right)x^{n-k}=2^{m(r-1)-n}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(0\right)E_{n-k}^{\left(m\right)}\left(0\right).$$

(33)

Proof. 

Set $y=0,y=-x$ and $y=-2x$ in (30), respectively. □

Evidently, the sums in the Corollary contain some interesting special cases. The evaluations with $(r,m)=(1,1)$, and $x=0$ and $x=1/2$, respectively, yield the following identities for Bernoulli numbers:

$$\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{2-2^{n+2-k}}{n+1-k}{B}_k{B}_{n+1-k}=(2^n-1)B_n,$$

(34)

$$\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k\frac{2-2^{n+2-k}}{n+1-k}{B}_k{B}_{n+1-k}=(2-2^n-{(-1)}^n)B_n,$$

(35)

$$\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}(1-2^{k-1})B_k=(2^n-1)B_n$$

(36)

and

$${\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{n-k}{2}^{2k}(2^{1-k}-1)B_k=-2^{n+1}{\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{(2^{1-k}-1)(2^{n+1-k}-1)}{n+1-k}{B}_k{B}_{n+1-k},$$

(37)

where we have employed the following relations

$$E_{n-1}\left(x\right)=\frac{2}{n}(B_n\left(x\right)-2^n{B}_n(x/2)),$$

and

$$B_n(1/2)=(2^{1-n}-1)B_n.$$

It is difficult to say whether the Bernoulli identities Equations (34)–(37) are original. We could not find them in the book [20]. Hence, they are maybe not classical. However, they may have appeared elsewhere before. Furthermore, setting $r=m=1$, $x=0$ and $y=1$ in Equation (30) gives

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k\left(1\right)E_{n-k}\left(0\right)=2^n{B}_n(1/2),$$

(38)

where we have used Equation (31). Hence, we rediscover Merca’s identity Equation (1).

Theorem 5.

Let r and m be positive integers and $x,y\in \mathbb{C}$. Then,

$${\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k^{(r,m)}\left(x\right)y^{n-k}=2^{m(r-1)}{\sum }_{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{2}^j{\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(0\right)E_{n-k}^{(m-j)}(2x-m+y),$$

(39)

where $E_n^{\left(m\right)}\left(x\right)$ is the generalized Euler polynomial of order m.

Proof. 

The identity follows from

$$\begin{array}{ccc}\hfill B(r,m;x,2z)e^{yz}& =& 2^{rm}{\left(\frac{z^r}{e^{2z}-1}\right)}^m{e}^{(2x+y)z}\hfill \\ & =& 2^{rm}{e}^{(2x-m+y)z}{\left(\frac{z^r}{e^z-1}-\frac{z^r}{e^{2z}-1}\right)}^m\hfill \\ & =& 2^{rm}{e}^{(2x-m+y)z}\left(\sum _{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{\left(\frac{z^r}{e^z-1}\right)}^j{\left(\frac{z^r}{e^{2z}-1}\right)}^{m-j}\right)\hfill \\ & =& 2^{rm}{\left(\frac{z^r}{e^z-1}\right)}^m{e}^{(2x-m+y)z}\left(\sum _{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{\left(\frac{1}{e^z+1}\right)}^{m-j}\right)\hfill \\ & =& 2^{rm}{\left(\frac{z^r}{e^z-1}\right)}^m\left(\sum _{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{2}^{j-m}{\left(\frac{2}{e^z+1}\right)}^{m-j}{e}^{(2x-m+y)z}\right)\hfill \\ & =& 2^{r(m-1)}\sum _{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{2}^j{\left(\frac{z^r}{e^z-1}\right)}^m{\left(\frac{2}{e^z+1}\right)}^{m-j}{e}^{(2x-m+y)z}.\hfill \end{array}$$

□

Corollary 9.

For $x,y\in \mathbb{C}$ the generalized Bernoulli polynomials of order $(r,m)$ satisfy the following relation:

$$\begin{array}{ccc}\hfill & & \sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{k-m(r-1)}{2}^k{B}_k^{(r,m)}\left(x\right)y^{n-k}\hfill \\ & & =2^{m(r-1)}\sum _{j=0}^m\left(\genfrac{}{}{0pt}{}{m}{j}\right){(-1)}^{m-j}{2}^j\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(0\right)E_{n-k}^{(m-j)}(m+y-2x).\hfill \end{array}$$

(40)

Proof. 

Replace x with $m-x$ and use the reciprocal relation (see [22])

$$B_n^{(r,m)}(m-x)={(-1)}^{n-m(r-1)}{B}_n^{(r,m)}\left(x\right).$$

(41)

□

Setting $r=m=1$ and using the fact that $E_n^{\left(0\right)}\left(x\right)=x^n$, we obtain:

Corollary 10.

For $x,y\in \mathbb{C}$ the Bernoulli polynomials satisfy the following relation:

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k\left(x\right)y^{n-k}=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k(2{(2x-1+y)}^{n-k}-E_{n-k}(2x-1+y))$$

(42)

and

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k{2}^k{B}_k\left(x\right)y^{n-k}=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k(2{(1+y-2x)}^{n-k}-E_{n-k}(1+y-2x)).$$

(43)

Inserting $x=(1-y)/2$ in Equation (42) or $x=(1+y)/2$ in Equation (43) yields

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k{B}_k\left(\frac{1-y}{2}\right)y^{n-k}=(2-2^n)B_n$$

(44)

and

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k{2}^k{B}_k\left(\frac{1+y}{2}\right)y^{n-k}=(2-2^n)B_n$$

(45)

for each $y\in \mathbb{C}$. We can see that $y=1$ or $y=-1$ produce Merca’s identity Equation (1). It is also worth mentioning the special cases of the above results for y being a power of 2.

We now focus on presenting other types of convolutions. Some types follow straightforwardly from the definitions Equations (6) and (20). For instance, it is fairly easy to deduce that for each integer $p\ge 0$ and $x,y\in \mathbb{C}$

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)B_{n-k}^{(r,p)}\left(y\right)=B_n^{(r,m+p)}(x+y)$$

and

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}\left(x\right)A_{n-k}^{(r,p)}\left(y\right)=A_n^{(r,m+p)}(x+y).$$

Setting $p=x=0$ corresponds to the representation

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}{x}^{n-k}=B_n^{(r,m)}\left(x\right)and\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}{x}^{n-k}=A_n^{(r,m)}\left(x\right).$$

Theorem 6.

For $m,p\ge 0$ and $x,y\in \mathbb{C}$ we have

$$\begin{array}{c}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)B_{n-k}^{(r,p)}\left(y\right)=\hfill \\ \sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{(r-1)(m+p)+k}{B}_k^{(r,m+p)}\left(x\right){(2x+y-(m+p))}^{n-k}\hfill \end{array}$$

(46)

and

$$\begin{array}{c}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}\left(x\right)A_{n-k}^{(r,p)}\left(y\right)=\hfill \\ \sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{r(m+p)+k}{A}_k^{(r,m+p)}\left(x\right){(2x+y-(m+p))}^{n-k}.\hfill \end{array}$$

(47)

Proof. 

The first identity follows from

$${\left(\frac{z^r}{e^z-1}\right)}^m{e}^{xz}·{\left(\frac{z^r}{e^z-1}\right)}^p{e}^{yz}={(-1)}^{(r-1)(m+p)}{\left(\frac{{(-z)}^r}{e^{-z}-1}\right)}^{m+p}{e}^{-xz}{e}^{(2x+y-(m+p\left)\right)z}.$$

The other one can be proved similarly. □

Theorem 7.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$, we have the following convolution

$$\begin{array}{ccc}\hfill & & m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}\left(x\right)B_{n-k}^{(r,1)}(y+1-m)=(x+y-m){\left(n\right)}_r{B}_{n-r}^{(r,m)}(x+y-m)\hfill \\ & & -(n+1-r(m+1)){\left(n\right)}_{r-1}{B}_{n+1-r}^{(r,m)}(x+y-m),\hfill \end{array}$$

(48)

where ${\left(n\right)}_r$ denotes the falling factorial defined by

$${\left(n\right)}_r=\left\{\begin{array}{cc}1,\hfill & r=0;\hfill \\ n(n-1)\cdots (n-r+1),\hfill & r\ge 1.\hfill \end{array}\right.$$

(49)

Proof. 

Since

$$\frac{d}{dz}\frac{e^{(x+y-m)z}}{{(e^z-1)}^m}=(x+y-m)\frac{e^{(x+y-m)z}}{{(e^z-1)}^m}-m\frac{e^{(x+y+1-m)z}}{{(e^z-1)}^{m+1}},$$

we have the relation

$$m{\left(\frac{z^r}{e^z-1}\right)}^{m+1}{e}^{(x+y+1-m)z}=(x+y-m)z^r{\left(\frac{z^r}{e^z-1}\right)}^m{e}^{(x+y-m)z}-z^{r(m+1)}\frac{d}{dz}\frac{e^{(x+y-m)z}}{{(e^z-1)}^m}.$$

□

Replacing x with $m-x$ and using (41), we immediately get the alternating version of Theorem 7.

Corollary 11.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$ we have the following convolution

$$\begin{array}{ccc}\hfill & & m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{k-m(r-1)}{B}_k^{(r,m)}\left(x\right)B_{n-k}^{(r,1)}(y+1-m)=(y-x){\left(n\right)}_r{B}_{n-r}^{(r,m)}(y-x)\hfill \\ & & -(n+1-r(m+1)){\left(n\right)}_{r-1}{B}_{n+1-r}^{(r,m)}(y-x).\hfill \end{array}$$

(50)

For $(r,m)=(1,m)$ the above results reduce to convolutions for generalized Bernoulli polynomials:

$$\begin{array}{ccc}\hfill m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{\left(m\right)}\left(x\right)B_{n-k}(y+1-m)& =& (x+y-m)n{B}_{n-1}^{\left(m\right)}(x+y-m)\hfill \\ & & -(n-m)B_n^{\left(m\right)}(x+y-m),\hfill \end{array}$$

(51)

and

$$m{\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k{B}_k^{\left(m\right)}\left(x\right)B_{n-k}(y+1-m)=(y-x)n{B}_{n-1}^{\left(m\right)}(y-x)-(n-m)B_n^{\left(m\right)}(y-x).$$

(52)

These results generalize the known convolutions for Bernoulli polynomials, which are obtained for $(r,m)=(1,1)$. The analogue convolution for $A_n^{(r,m)}\left(x\right)$ is given next.

Theorem 8.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$, the following convolution result holds:

$$\begin{array}{ccc}\hfill & & m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)A_k^{(r,m)}\left(x\right)A_{n-k}^{(r,1)}(y+1-m)=2(x+y-m){\left(n\right)}_r{A}_{n-r}^{(r,m)}(x+y-m)\hfill \\ & & -2(n+1-r(m+1)){\left(n\right)}_{r-1}{A}_{n+1-r}^{(r,m)}(x+y-m).\hfill \end{array}$$

(53)

For $r=0$, the result becomes

$$m{\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)E_k^{\left(m\right)}\left(x\right)E_{n-k}(y+1-m)=2(x+y-m)E_n^{\left(m\right)}(x+y-m)-2E_{n+1}^{\left(m\right)}(x+y-m).$$

(54)

Proof. 

Use

$$\begin{array}{c}m{\left(\frac{2z^r}{e^z+1}\right)}^{m+1}{e}^{(x+y+1-m)z}=\hfill \\ 2(x+y-m)z^r{\left(\frac{2z^r}{e^z+1}\right)}^m{e}^{(x+y-m)z}-2^{m+1}{z}^{r(m+1)}\frac{d}{dz}\frac{e^{(x+y-m)z}}{{(e^z+1)}^m}.\hfill \end{array}$$

□

Corollary 12.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$ the following convolution result holds:

$$\begin{array}{ccc}\hfill & & m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^{k-rm}{A}_k^{(r,m)}\left(x\right)A_{n-k}^{(r,1)}(y+1-m)=2(y-x){\left(n\right)}_r{A}_{n-r}^{(r,m)}(y-x)\hfill \\ & & -2(n+1-r(m+1)){\left(n\right)}_{r-1}{A}_{n+1-r}^{(r,m)}(y-x).\hfill \end{array}$$

(55)

For $r=0$ the result becomes

$$m\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^k{E}_k^{\left(m\right)}\left(x\right)E_{n-k}(y+1-m)=2(y-x)E_n^{\left(m\right)}(y-x)-2E_{n+1}^{\left(m\right)}(y-x).$$

(56)

Proof. 

This result follows from the reciprocal relation for $A_n^{(r,m)}\left(x\right)$ (see [22])

$$A_n^{(r,m)}(m-x)={(-1)}^{n-mr}{A}_n^{(r,m)}\left(x\right).$$

(57)

□

Theorem 9.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$ the following convolution result holds:

$$2^n{B}_n^{(r,m)}\left(\frac{x+y}{2}\right)=2^{m(r-1)}{\left(n\right)}_{rm}\sum _{k=0}^{n-rm}\left(\genfrac{}{}{0pt}{}{n-rm}{k}\right)B_k^{(r,m)}\left(x\right)A_{n-rm-k}^{(r,m)}\left(y\right).$$

(58)

Proof. 

Use

$${\left(\frac{{\left(2z\right)}^r}{e^{2z}-1}\right)}^m{e}^{(x+y)z}=2^{m(r-1)}{z}^{-rm}{\left(\frac{z^r}{e^z-1}\right)}^m{e}^{xz}{\left(\frac{2z^r}{e^z+1}\right)}^m{e}^{yz}.$$

□

Upon replacing x with $m-x$ and y with $m-y$ and using Equations (41) and (57), we also get the alternating version of the previous result.

Corollary 13.

For $r,m\ge 1$ and $x,y\in \mathbb{C}$ the following convolution result holds:

$$\begin{array}{c}{2}^n{B}_n^{(r,m)}\left(m-\frac{x+y}{2}\right)\hfill \\ ={(-1)}^{n-m(r-1)}{2}^{m(r-1)}{\left(n\right)}_{rm}\sum _{k=0}^{n-rm}\left(\genfrac{}{}{0pt}{}{n-rm}{k}\right)B_k^{(r,m)}\left(x\right)A_{n-rm-k}^{(r,m)}\left(y\right).\hfill \end{array}$$

(59)

The case $(r,m)=(1,1)$ produces relations involving Bernoulli and Genocchi polynomials:

$$2^n{B}_n\left(\frac{x+y}{2}\right)=n\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n-1}{k}\right)B_k\left(x\right)G_{n-1-k}\left(y\right)$$

(60)

and

$$2^n{B}_n\left(1-\frac{x+y}{2}\right)={(-1)}^{n}n\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n-1}{k}\right)B_k\left(x\right)G_{n-1-k}\left(y\right).$$

(61)

5. Probabilistic Interpretation of Bernoulli and Euler–Genocchi Polynomials

In this section, we indicate a possible application of the polynomials discussed in this work.

Consider a probability measure space $(\Omega ,\mathcal{F},P)$, where $\Omega$ is a non-empty space, $\mathcal{F}$ is a $\sigma$-algebra of events and P is a probability measure on $\mathcal{F}$. Let X be a random variable which is a measurable real function on the probability space $(\Omega ,\mathcal{F},P)$, and let $p\left(x\right)$ be a probability density function of X. Then mathematical expectation and moments of order n of X are defined by

$$E\left[X\right]={\int }_{-\infty }^{\infty }xp\left(x\right)dx\mathrm{a}\mathrm{n}\mathrm{d}E\left[X^n\right]={\int }_{-\infty }^{\infty }{x}^{n}p\left(x\right)dx,$$

with $E\left[X^1\right]=E\left[X\right]$. The moment-generating function of X is defined by

$$\mu \left(t\right)=E\left[e^{tX}\right]={\int }_{-\infty }^{\infty }{e}^{tx}p\left(x\right)dx,t\in \mathbb{R}.$$

The relation between the moment generating function $\mu \left(t\right)$ and and moments $E\left[X^n\right]$ is given by

$$\mu \left(t\right)=\sum _{k=0}^{\infty }\frac{t^k}{k!}E\left[X^k\right].$$

Suppose that (6), (20), (21) and (22) are moment-generating functions of random variables $X_1,X_2,\dots ,X_5$. Then,

$$\begin{array}{ccc}\hfill E\left[X_1^n\right]& =& \frac{d^n}{d{z}^n}\left\{{\left(\frac{z^r}{e^z-1}\right)}^m{e}^{xz}\right\}{|}_{z=0}=B_n^{(r,m)}\left(x\right),\hfill \\ \hfill E\left[X_2^n\right]& =& \frac{d^n}{d{z}^n}\left\{{\left(\frac{z}{e^z-1}\right)}^m{e}^{xz}\right\}{|}_{z=0}=B_n^{\left(m\right)}\left(x\right),\hfill \\ \hfill E\left[X_3^n\right]& =& \frac{d^n}{d{z}^n}\left\{{\left(\frac{2z^r}{e^z+1}\right)}^m{e}^{xz}\right\}{|}_{z=0}=A_n^{(r,m)}\left(x\right),\hfill \\ \hfill E\left[X_4^n\right]& =& \frac{d^n}{d{z}^n}\left\{{\left(\frac{2}{e^z+1}\right)}^m{e}^{xz}\right\}{|}_{z=0}=E_n^{\left(m\right)}\left(x\right),\hfill \\ \hfill E\left[X_5^n\right]& =& \frac{d^n}{d{z}^n}\left\{{\left(\frac{2z}{e^z+1}\right)}^m{e}^{xz}\right\}{|}_{z=0}=G_n^{\left(m\right)}\left(x\right).\hfill \end{array}$$

Another interpretation is the next example. Let X be a discrete random variable on $(\Omega ,\mathcal{F},P)$. Then, we can define a probability distribution for X by

$$P(X=k)=\frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}}{B_n^{(r,m)}\left(1\right)},k=0,1,\dots ,n.$$

Note that

$$\sum _{k=0}^{n}P(X=k)=\frac{{\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}}{B_n^{(r,m)}\left(1\right)}=\frac{B_n^{(r,m)}\left(1\right)}{B_n^{(r,m)}\left(1\right)}=1.$$

Next, replacing k by $n-k$ in the identity

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{(r,m)}{x}^{n-k}=B_n^{(r,m)}\left(x\right)$$

we get

$$\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n-k}^{(r,m)}{x}^k=B_n^{(r,m)}\left(x\right),$$

which, upon differentiating with respect to x, yields

$$\sum _{k=0}^{n}k\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n-k}^{(r,m)}{x}^{k-1}=n{B}_{n-1}^{(r,m)}\left(x\right).$$

Inserting $x=1$ gives

$$\sum _{k=0}^{n}k\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n-k}^{(r,m)}=n{B}_{n-1}^{(r,m)}\left(1\right),$$

which shows that

$$E\left[X\right]=n{B}_{n-1}^{(r,m)}\left(1\right).$$

In an analogous fashion, the quantity $A_n^{(r,m)}\left(x\right)$ can be interpreted probabilistically.

6. Conclusions and Future Work

In this paper, mainly focusing on convolutions, we established additional properties of the generalized Bernoulli and Euler–Genocchi polynomials $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$, respectively. These properties provide generalizations of some known facts about generalized Bernoulli and Euler polynomials, respectively. In the future, we intend to work in two different directions. First, it seem desirable to find some new kinds of closed-form expressions for our polynomials (such as combinatorial, integral, hypergeometric and determinantal ones). Such expressions will provide us with new and significant properties of these polynomials. Second, it is possible to study the Apostol-type generalized polynomials ${\mathfrak{B}}_n^{(r,m)}(x;\lambda )$ associated with the complex parameter $\lambda \ne 0$, which are defined by the generating function

$$\mathfrak{B}(r,m;x,z;\lambda )={\left(\frac{z^r}{\lambda e^z-1}\right)}^m{e}^{xz}=\sum _{n=0}^{\infty }{\mathfrak{B}}_n^{(r,m)}(x;\lambda )\frac{z^n}{n!}.$$

Then, it is fairly easy to identify the relations $B(r,m;x,z)=\mathfrak{B}(r,m;x,z;1)$ and $A(r,m;x,z)={(-2)}^m\mathfrak{B}(r,m;x,z;-1)$. Therefore, by means of ${\mathfrak{B}}_n^{(r,m)}(x;\lambda )$ it is possible to discuss both $B_n^{(r,m)}\left(x\right)$ and $A_n^{(r,m)}\left(x\right)$, at once.