Abstract

We develop methods for computing the product of several Bernoulli and Euler polynomials by using Bernoulli basis for the vector space of polynomials of degree less than or equal to 𝑛.

1. Introduction

It is well known that, the 𝑛th Bernoulli and Euler numbers are defined by 𝑛𝑙=0𝑛𝑙𝐵𝑙𝐵𝑛=𝛿1,𝑛,𝑛𝑙=0𝑛𝑙𝐸𝑙+𝐸𝑛=2𝛿0,𝑛,(1.1) where 𝐵0=𝐸0=1 and 𝛿𝑘,𝑛 is the Kronecker symbol (see [120]).

The Bernoulli and Euler polynomials are also defined by 𝐵𝑛(𝑥)=𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝑥𝑙,𝐸𝑛(𝑥)=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝑥𝑙.(1.2) Note that {𝐵0(𝑥),𝐵1(1),,𝐵𝑛(𝑥)} forms a basis for the space 𝑛={𝑝(𝑥)[𝑥]deg𝑝(𝑥)𝑛}.

So, for a given 𝑝(𝑥)𝑛, we can write 𝑝(𝑥)=𝑛𝑘=0𝑎𝑘𝐵𝑘(𝑥),(1.3) (see [812]) for uniquely determined 𝑎𝑘.

Further, 𝑎𝑘=1𝑝𝑘!(𝑘1)(1)𝑝(𝑘1)(0),where𝑝(𝑘)𝑑(𝑥)=𝑘𝑝(𝑥)𝑑𝑥𝑘,𝑎0=10𝑝(𝑡)𝑑𝑡,where𝑘=1,2,,𝑛.(1.4) Probably, {1,𝑥,,𝑥𝑛} is the most natural basis for the space 𝑛. But {𝐵0(𝑥),𝐵1(𝑥),,𝐵𝑛(𝑥)} is also a good basis for the space 𝑛, for our purpose of arithmetical and combinatorial applications.

What are common to 𝐵𝑛(𝑥), 𝐸𝑛(𝑥), 𝑥𝑛? A few proportion common to them are as follows: (i)they are all monic polynomials of degree 𝑛 with rational coefficients; (ii)(𝐵𝑛(𝑥))=𝑛𝐵𝑛1(𝑥), (𝐸𝑛(𝑥))=𝑛𝐸𝑛1(𝑥), (𝑥𝑛)=𝑛𝑥𝑛1; (iii)𝐵𝑛(𝑥)𝑑𝑥=𝐵𝑛+1(𝑥)/(𝑛+1)+𝑐, 𝐸𝑛(𝑥)𝑑𝑥=𝐸𝑛+1(𝑥)/(𝑛+1)+𝑐, 𝑥𝑛𝑑𝑥=𝑥𝑛+1/(𝑛+1)+𝑐.

In [5, 6], Carlitz introduced the identities of the product of several Bernoulli polynomials: 𝐵𝑚(𝑥)𝐵𝑛(𝑥)=𝑟=0𝑚𝑛𝑚𝐵2𝑟𝑛+2𝑟2𝑟𝐵𝑚+𝑛2𝑟(𝑥)𝑚+𝑛2𝑟+(1)𝑚+1×𝑚!𝑛!𝐵(𝑚+𝑛)!𝑚+𝑛(𝑚+𝑛2),10𝐵𝑚(𝑥)𝐵𝑛(𝑥)𝐵𝑝(𝑥)𝐵𝑞(𝑥)𝑑𝑥=(1)𝑚+𝑛+𝑝+𝑞𝑟,𝑠=0𝑚𝑛𝑚𝑝𝑞𝑝×2𝑟𝑛+2𝑟2𝑠𝑞+2𝑠(𝑚+𝑛2𝑟1)!(𝑝+𝑞2𝑠1)!𝐵(𝑚+𝑛+𝑝+𝑞2𝑟2𝑠)!𝑟𝐵𝑠𝐵𝑚+𝑛+𝑝+𝑞2𝑟2+(1)𝑚+𝑝𝑚!𝑛!(𝑚+𝑛)!𝑝!𝑞!(𝐵𝑝+𝑞)!𝑚+𝑛𝐵𝑝+𝑞.(1.5) In this paper, we will use (1.4) to derive the identities of the product of several Bernoulli and Euler polynomials.

2. The Product of Several Bernoulli and Euler Polynomials

Let us consider the following polynomials of degree 𝑛: 𝑝(𝑥)=𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥),(2.1) where the sum runs over all nonnegative integers 𝑖1,,𝑖𝑟,  𝑗1,𝑗𝑠 satisfying 𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛,  𝑟+𝑠=1, 𝑟,𝑠0.

Thus, from (2.1), we have 𝑝(𝑘)×(𝑥)=(𝑛+𝑟+𝑠1)(𝑛+𝑟+𝑠2)(𝑛+𝑟+𝑠𝑘)𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥).(2.2) For 𝑘=1,2,,𝑛, by (1.4), we get 𝑎𝑘=1𝑝𝑘!(𝑘1)(1)𝑝(𝑘1)=(0)𝑘𝑛+𝑟+𝑠𝑛+𝑟+𝑠𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠=𝑘𝑛+𝑟+𝑠𝑛+𝑟+𝑠0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐×𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛+𝑎+1𝑘𝑟𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠.(2.3) From (2.3), we note that 𝑎𝑛=(𝑛𝑛+𝑟+𝑠)𝑛+𝑟+𝑠𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠=(𝑛𝑛+𝑟+𝑠)1𝑛+𝑟+𝑠21+1𝑟21𝑠2=((𝑟+𝑠)𝑛𝑛+𝑟+𝑠)𝑛,𝑎𝑛+𝑟+𝑠(𝑟+𝑠)=𝑛+𝑟+𝑠1𝑛1=1×𝑛+𝑟+𝑠𝑛+𝑟+𝑠𝑛1𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=2𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠=11𝑛+𝑟+𝑠𝑛+𝑟+𝑠𝑛161𝑟+21221𝑟+𝑠61𝑟2122𝑎𝑟+𝑠=0,0=10=𝑝(𝑡)𝑑𝑡𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠×𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠𝑙1++𝑙𝑟+𝑝1++𝑝𝑠.+1(2.4) Therefore, by (1.3), (2.1), (2.3), and (2.4), we obtain the following theorem.

Theorem 2.1. For 𝑛 with 𝑛2, we have 𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(=1𝑥)𝑛+𝑟+𝑠𝑛2𝑘=1𝑘×𝑛+𝑟+𝑠0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑛+𝑎+1𝑘𝑟𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑘1𝐸𝑗𝑠𝐵𝑘𝑛𝐵(𝑥)+𝑛+𝑟+𝑠1𝑛+(𝑥)𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠×𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+1(2.5)

Let us take the polynomial 𝑝(𝑥) of degree 𝑛 as follows: 𝑝(𝑥)=𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡,(2.6) Then, from (2.6), we have 𝑝(𝑘)×(𝑥)=(𝑛+𝑟+𝑠)(𝑛+𝑟+𝑠1)(𝑛+𝑟+𝑠𝑘+1)𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝑘𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡,(2.7) By (1.4) and (2.7), we get, for 𝑘=1,2,,𝑛, 𝑎𝑘=1𝑝𝑘!(𝑘1)(1)𝑝(𝑘1)=1(0)𝑘×𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠+1𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝑘+1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=𝑘𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠+10𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐×𝑛+𝑎+1𝑘𝑟𝑡=0𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑡𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠,(2.8) Now, we look at 𝑎𝑛 and 𝑎𝑛1. 𝑎𝑛=𝑛𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠+1𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=𝑛𝑛+𝑟+𝑠+11𝑛+𝑟+𝑠+121(𝑟+𝑠)+12=(𝑟+𝑠)𝑟+𝑠+1𝑛=𝑛,𝑎𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠𝑛1=𝑛+𝑟+𝑠+1𝑛1𝑛+𝑟+𝑠+1𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=2𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=𝑛+𝑟+𝑠+1𝑛11𝑛+𝑟+𝑠+161𝑟+1+2122+1𝑟+𝑠21(𝑟+𝑠)61𝑟2122=1𝑟+𝑠𝑛+𝑟+𝑠+1𝑛+𝑟+𝑠+1𝑛1𝑟+𝑠+22=12,𝑛+𝑟+𝑠𝑛1(2.9) From (2.6), we note that 𝑎0=10=𝑝(𝑡)𝑑𝑡𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖+𝑡=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠×𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠1𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+𝑡+1(2.10) Therefore, by (1.3), (2.6), (2.8), (2.9), and (2.10), we obtain the following theorem.

Theorem 2.2. For 𝑛 with 𝑛2, one has 𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡=1𝑛+𝑟+𝑠+1𝑛2𝑘=1𝑘×𝑛+𝑟+𝑠+10𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐𝑛+𝑎+1𝑘𝑟𝑡=0𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑡𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝐵𝑘+1(𝑥)2𝐵𝑛+𝑟+𝑠𝑛1𝑛1(𝑛𝐵𝑥)+𝑛+𝑟+𝑠𝑛(+𝑥)𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖+𝑡=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠×𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠×1𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+𝑡+1(2.11)

Consider the following polynomial of degree 𝑛: 𝑝(𝑥)=𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛1𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥).(2.12) Then, from (2.12), one has 𝑝(𝑘)(𝑥)=(𝑟+𝑠)𝑘𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!.(2.13) By (1.4) and (2.13), one gets, for 𝑘=1,2,,𝑛, 𝑎𝑘=1𝑝𝑘!(𝑘1)(1)𝑝(𝑘1)=(0)(𝑟+𝑠)𝑘1𝑘!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝑘+1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!=(𝑟+𝑠)𝑘1𝑘!0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑛+𝑎+1𝑘𝑟𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1!𝑖2!𝑖𝑎!𝑗1!𝑗𝑐!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+11𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠.(2.14) Now look at 𝑎𝑛 and 𝑎𝑛1: 𝑎𝑛=(𝑟+𝑠)𝑛1𝑛!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=1𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!=(𝑟+𝑠)𝑛11𝑛!21(𝑟+𝑠)2=(𝑟+𝑠)(𝑟+𝑠)𝑛,𝑎𝑛!𝑛1=(𝑟+𝑠)𝑛2(𝑛1)!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=2𝐵𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!=(𝑟+𝑠)𝑛21(𝑛1)!2161𝑟+21221𝑟+𝑠2161𝑟2122𝑟+𝑠=0.(2.15)

It is easy to show that 𝑎0=10𝑝(𝑡)𝑑𝑡=𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛1𝑖1!𝑖𝑟!𝑗1!𝑗𝑠!×𝑖1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠𝑙1++𝑙𝑟+𝑝1+𝑝𝑠𝑖+11𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠.(2.16) Therefore, by (1.3), (2.14), and (2.15), one obtains the following theorem.

Theorem 2.3. For 𝑛 with 𝑛2, one has 𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛1𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠=(𝑥)𝑛2𝑘=1(𝑟+𝑠)𝑘1𝑘!0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐×𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑛+𝑎+1𝑘𝑟𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1!𝑖2!𝑖𝑎!𝑗1!𝑗𝑐!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+11𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!×𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝐵𝑘(𝑥)+(𝑟+𝑠)𝑛𝐵𝑛!𝑛+(𝑥)𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠×𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+1(2.17)

Take the polynomial 𝑝(𝑥) of degree 𝑛 as follows: 𝑝(𝑥)=𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛1𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠𝐵!𝑡!𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡.(2.18) Then, from (2.18), one gets 𝑝(𝑘)(𝑥)=(𝑟+𝑠+1)𝑘×𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝑘1𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠𝐵!𝑡!𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡.(2.19) By (1.4) and (2.19), one gets, for 𝑘=1,,𝑛, 𝑎𝑘=1𝑝𝑘!(𝑘1)(1)𝑝(𝑘1)=(0)(𝑟+𝑠+1)𝑘1𝑘!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛𝑘+11𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠×𝐵!𝑡!𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=(𝑟+𝑠+1)𝑘1𝑘!0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐𝑛+𝑎+1𝑘𝑟𝑡=01×(𝑛+𝑎+1𝑘𝑟𝑡)!𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑡1𝑖1!𝑖2!𝑖𝑎!𝑗1!𝑗𝑐!𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠.(2.20)

Now look at 𝑎𝑛 and 𝑎𝑛1: 𝑎𝑛=(𝑟+𝑠+1)𝑛1𝑛!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=11𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠×𝐵!𝑡!𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=(𝑟+𝑠+1)𝑛11𝑛!2(1𝑟+𝑠)+12(=𝑟+𝑠)(𝑟+𝑠+1)𝑛1𝑛!(𝑟+𝑠+1)=(𝑟+𝑠+1)𝑛,𝑎𝑛!𝑛1=(𝑟+𝑠+1)𝑛2(𝑛1)!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=21𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠×𝐵!𝑡!𝑖1(1)𝐵𝑖𝑟(1)𝐸𝑗1(1)𝐸𝑗𝑠(1)𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠0𝑡=(𝑟+𝑠+1)𝑛21(𝑛1)!2161𝑟+2+12122+1𝑟+𝑠21(𝑟+𝑠)2161𝑟2122=𝑟+𝑠(𝑟+𝑠+1)𝑛2(𝑛1)!𝑟+𝑠+12=(𝑟+𝑠+1)𝑛1.2(𝑛1)!(2.21) From (2.18), one can derive the following identity: 𝑎0=10=𝑝(𝑡)𝑑𝑡𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛1𝑖1!𝑖𝑟!𝑗1!𝑗𝑠!𝑡!10𝐵𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡=𝑑𝑡𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛1𝑖1!𝑖𝑟!𝑗1!𝑗𝑠!𝑡!𝑖1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0×𝑗𝑠𝑝𝑠=0𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠𝐵𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠1𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+𝑡+1(2.22) Therefore, by (1.3), (2.20), (2.21), and (2.22), one obtains the following theorem.

Theorem 2.4. For 𝑛 with 𝑛2, one has 𝑖1++𝑖𝑟+𝑗1++𝑗𝑠+𝑡=𝑛1𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠𝐵!𝑡!𝑖1(𝑥)𝐵𝑖𝑟(𝑥)𝐸𝑗1(𝑥)𝐸𝑗𝑠(𝑥)𝑥𝑡=𝑛2𝑘=1(𝑟+𝑠+1)𝑘1𝑘!0𝑎𝑟0𝑐𝑠𝑘+𝑟𝑛1𝑎𝑟𝑟𝑎𝑠𝑐(1)𝑐2𝑠𝑐𝑛+𝑎+1𝑘𝑟𝑡=01(×𝑛+𝑎+1𝑘𝑟𝑡)!𝑖1++𝑖𝑎+𝑗1++𝑗𝑐=𝑡1𝑖1!𝑖2!𝑖𝑎!𝑗1!𝑗𝑐!𝐵𝑖1𝐵𝑖𝑎𝐸𝑗1𝐸𝑗𝑐𝑖1++𝑖𝑟+𝑗1++𝑗𝑠=𝑛𝑘+1𝐵𝑖1𝐵𝑖𝑟𝐸𝑗1𝐸𝑗𝑠𝑖1!𝑖2!𝑖𝑟!𝑗1!𝑗𝑠!𝐵𝑘+(𝑥)(𝑟+𝑠+1)𝑛1𝐵2(𝑛1)!𝑛1(𝑥)+(𝑟+𝑠+1)𝑛𝑛!𝑖1++𝑖𝑟+𝑗1++𝑗𝑠𝑖+𝑡=𝑛1𝑙1=0𝑖𝑟𝑙𝑟𝑗=01𝑝1=0𝑗𝑠𝑝𝑠=0×𝑖1𝑙1𝑖𝑟𝑙𝑟𝑗1𝑝1𝑗𝑠𝑝𝑠𝑖1!𝑖𝑟!𝑗1!𝑗𝑠𝐵!𝑡!𝑖1𝑙1𝐵𝑖𝑟𝑙𝑟𝐸𝑗1𝑝1𝐸𝑗𝑠𝑝𝑠1𝑙1++𝑙𝑟+𝑝1+𝑝𝑠.+𝑡+1(2.23)

Acknowledgments

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology 2012R1A1A2003786.