Abstract

We derive some new and interesting identities involving Bernoulli and Euler numbers by using some polynomial identities and p-adic integrals on 𝑝.

1. Introduction and Preliminaries

Let 𝑝 be a fixed odd prime. Throughout this paper, 𝑝,𝑝,𝑝 will, respectively, denote the ring of 𝑝-adic integers, the field of 𝑝-adic rational numbers, and the completion of algebraic closure of 𝑝. The 𝑝-adic absolute value ||𝑝 on 𝑝 is normalized so that |𝑝|𝑝=1/𝑝. Let >0 be the set of natural numbers and 0=>0{0}.

As is well known, the Bernoulli polynomials 𝐵𝑛(𝑥) are defined by the generating function as follows: 𝑡𝐹(𝑡,𝑥)=𝑒𝑡𝑒1𝑥𝑡=𝑒𝐵(𝑥)𝑡=𝑛=0𝐵𝑛(𝑡𝑥)𝑛,𝑛!(1.1) with the usual convention of replacing 𝐵(𝑥)𝑛 by 𝐵𝑛(𝑥).

In the special case, 𝑥=0, 𝐵𝑛(0)=𝐵𝑛 is referred to as the 𝑛th Bernoulli number. That is, the generating function of Bernoulli numbers is given by 𝑡𝐹(𝑡)=𝐹(𝑡,0)=𝑒𝑡=1𝑛=0𝐵𝑛𝑡𝑛𝑛!=𝑒𝐵𝑡,(1.2) with the usual convention of replacing 𝐵𝑛 by 𝐵𝑛, (cf. [123]).

From (1.2), we see that the recurrence formula for the Bernoulli numbers is (𝐵+1)𝑛𝐵𝑛=𝛿1,𝑛,for𝑛0,(1.3) where 𝛿𝑘,𝑛 is the Kronecker symbol.

By (1.1) and (1.2), we easily get the following: 𝐵𝑛(𝑥)=(𝐵+𝑥)𝑛=𝑛𝑙=0𝑛𝑙𝐵𝑙𝑥𝑛𝑙=𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝑥𝑙,for𝑛0.(1.4) Let 𝑈𝐷(𝑝) be the space of uniformly differentiable 𝑝-valued functions on 𝑝. For 𝑓𝑈𝐷(𝑝), the bosonic 𝑝-adic integral on 𝑝 is defined by 𝐼(𝑓)=𝑝𝑓(𝑥)𝑑𝜇(𝑥)=lim𝑁1𝑝𝑁𝑝𝑁1𝑥=0𝑓(𝑥),(1.5)(cf. [12]). Then it is easy to see that 𝐼𝑓1=𝐼(𝑓)+𝑓(0),(1.6) where 𝑓1(𝑥)=𝑓(𝑥+1) and 𝑓(0)=𝑑𝑓(𝑥)/𝑑𝑥|𝑥=0.

By (1.6), we have the following: 𝑝𝑒(𝑥+𝑦)𝑡𝑡𝑑𝜇(𝑦)=𝑒𝑡𝑒1𝑥𝑡=𝑛=0𝐵𝑛(𝑡𝑥)𝑛,𝑛!(1.7)(cf. [1214]). From (1.7), we can derive the Witt's formula for the 𝑛th Bernoulli polynomial as follows: 𝑝(𝑥+𝑦)𝑛𝑑𝜇(𝑦)=𝐵𝑛(𝑥),for𝑛0.(1.8)

By (1.1), we have the following: 𝐵𝑛(1𝑥)=(1)𝑛𝐵𝑛(𝑥),for𝑛0.(1.9) Thus, from (1.3), (1.4), and (1.9), we have the following: 𝐵𝑛(1)=𝐵𝑛+𝛿1,𝑛=(1)𝑛𝐵𝑛,for𝑛0.(1.10) By (1.4), we have the following: 𝐵𝑛(𝑥+𝑦)=𝑛𝑘=0𝑛𝑘𝐵𝑘(𝑥)𝑦𝑛𝑘,for𝑛0.(1.11) Especially, for 𝑥=1 and 𝑦=1, 𝐵𝑛(2)=𝑛𝑘=0𝑛𝑘𝐵𝑘(1)=𝑛𝑘=0𝑛𝑘𝐵𝑘+𝛿1,𝑘,for𝑛0.(1.12) Therefore, from (1.9), (1.10), and (1.12), we can derive the following relation. For 𝑛0, (1)𝑛𝐵𝑛(1)=𝐵𝑛(2)=𝑛+𝐵𝑛(1)=𝑛+𝐵𝑛+𝛿1,𝑛=𝑛+(1)𝑛𝐵𝑛.(1.13) Let 𝑓(𝑦)=(𝑥+𝑦)𝑛+1. By (1.6), we have the following: 𝑝(𝑥+𝑦+1)𝑛+1𝑑𝜇(𝑦)𝑝(𝑥+𝑦)𝑛+1𝑑𝜇(𝑦)=(𝑛+1)𝑥𝑛,for𝑛0.(1.14) By (1.8) and (1.14), we have the following: 𝐵𝑛+1(𝑥+1)𝐵𝑛+1(𝑥)=(𝑛+1)𝑥𝑛,for𝑛0.(1.15) Thus, by (1.11) and (1.15), we have the following identity. 𝑥𝑛=1𝑛+1𝑛𝑙=0𝑙𝐵𝑛+1𝑙(𝑥),for𝑛0.(1.16)

As is well known, the Euler polynomials 𝐸𝑛(𝑥) are defined by the generating function as follows: 2𝐺(𝑡,𝑥)=𝑒𝑡𝑒+1𝑥𝑡=𝑒𝐸(𝑥)𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛,𝑛!(1.17) with the usual convention of replacing 𝐸(𝑥)𝑛 by 𝐸𝑛(𝑥).

In the special case, 𝑥=0, 𝐸𝑛(0)=𝐸𝑛 is referred to as the 𝑛th Euler number. That is, the generating function of Euler numbers is given by 2𝐺(𝑡)=𝐺(𝑡,0)=𝑒𝑡=+1𝑛=0𝐸𝑛𝑡𝑛𝑛!=𝑒𝐸𝑡,(1.18) with the usual convention of replacing 𝐸𝑛 by 𝐸𝑛, (cf. [123]).

From (1.18), we see that the recurrence formula for the Euler numbers is (𝐸+1)𝑛+𝐸𝑛=2𝛿0,𝑛,for𝑛0.(1.19) By (1.17) and (1.18), we easily get the following: 𝐸𝑛(𝑥)=(𝐸+𝑥)𝑛=𝑛𝑙=0𝑛𝑙𝐸𝑙𝑥𝑛𝑙=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝑥𝑙,for𝑛0.(1.20) Let 𝐶(𝑝) be the space of continuous 𝑝-valued functions on 𝑝. For 𝑓𝐶(𝑝), the fermionic 𝑝-adic integral on 𝑝 is defined by Kim as follows: 𝐼1(𝑓)=𝑝𝑓(𝑥)𝑑𝜇1(𝑥)=lim𝑝𝑁𝑁1𝑥=0𝑓(𝑥)(1)𝑥,(1.21)(cf. [9]). Then it is easy to see that 𝐼1𝑓1+𝐼1(𝑓)=2𝑓(0),(1.22) where 𝑓1(𝑥)=𝑓(𝑥+1).

By (1.22), we have the following: 𝑝𝑒(𝑥+𝑦)𝑡𝑑𝜇1(2𝑦)=𝑒𝑡𝑒+1𝑥𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛.𝑛!(1.23) From (1.23), we can derive the Witt's formula for the 𝑛-th Euler polynomial as follows: 𝑝(𝑥+𝑦)𝑛𝑑𝜇1(𝑦)=𝐸𝑛(𝑥),for𝑛0.(1.24) By (1.17), we have the following: 𝐸𝑛(1𝑥)=(1)𝑛𝐸𝑛(𝑥),for𝑛0.(1.25) Thus, from (1.19), (1.20), and (1.25), we have the following: 𝐸𝑛(1)=𝐸𝑛+2𝛿0,𝑛=(1)𝑛𝐸𝑛,for𝑛0.(1.26) By (1.20), we have the following: 𝐸𝑛(𝑥+𝑦)=𝑛𝑘=0𝑛𝑘𝐸𝑘(𝑥)𝑦𝑛𝑘,for𝑛0.(1.27) Especially, for 𝑥=1 and 𝑦=1, 𝐸𝑛(2)=𝑛𝑘=0𝑛𝑘𝐸𝑘(1)=𝑛𝑘=0𝑛𝑘𝐸𝑛+2𝛿0,𝑘,for𝑛0.(1.28) Therefore, from (1.25), (1.26), and (1.28), we can derive the following relations. For 𝑛0, (1)𝑛𝐸𝑛(1)=𝐸𝑛(2)=2𝐸𝑛(1)=2+𝐸𝑛2𝛿0,𝑛=2(1)𝑛𝐸𝑛.(1.29) Let 𝑓(𝑦)=(𝑥+𝑦)𝑛. By (1.22), we have the following: 𝑝(𝑥+𝑦+1)𝑛𝑑𝜇1(𝑦)+𝑝(𝑥+𝑦)𝑛𝑑𝜇1(𝑦)=2𝑥𝑛,for𝑛0.(1.30) By (1.24) and (1.30), we have the following: 𝐸𝑛(𝑥+1)+𝐸𝑛(𝑥)=2𝑥𝑛,for𝑛0.(1.31) Thus, by (1.27) and (1.31), we get the following identity. 𝑥𝑛=12𝑛1𝑙=0𝑛𝑙𝐸𝑙(𝑥)+𝐸𝑛(𝑥),for𝑛0.(1.32)

The Bernstein polynomials are defined by 𝐵𝑘,𝑛𝑛𝑘𝑥(𝑥)=𝑘(1𝑥)𝑛𝑘,for𝑘,𝑛0,(1.33) with 0𝑘𝑛 (cf. [14]).

By the definition of 𝐵𝑘,𝑛(𝑥), we note that 𝐵𝑘,𝑛(𝑥)=𝐵𝑛𝑘,𝑛(1𝑥).(1.34)

In this paper, we derive some new and interesting identities involving Bernoulli and Euler numbers from well-known polynomial identities. Here, we note that our results are “complementary” to those in [6], in the sense that we take a fermionic 𝑝-adic integral where a bosonic 𝑝-adic integral is taken and vice versa, and we use the identity involving Euler polynomials in (1.32) where that involving Bernoulli polynomials in (1.16) is used and vice versa. Finally, we report that there have been a lot of research activities on this direction of research, namely, on derivation of identities involving Bernoulli and Euler numbers and polynomials by exploiting bosonic and fermionic 𝑝-adic integrals (cf. [68]).

2. Identities Involving Bernoulli Numbers

Taking the bosonic 𝑝-adic integral on both sides of (1.16), we have the following: 𝑝𝑥𝑚𝑑𝜇(𝑥)=𝑝1𝑚+1𝑚𝑘=0𝑘𝐵𝑚+1𝑘=1(𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1𝑝𝐵𝑘(=1𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝑝𝑥𝑗=1𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗.(2.1) Therefore, we obtain the following theorem.

Theorem 2.1. Let 𝑚0. Then on has the following: 𝐵𝑚=1𝑚+1𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝐵𝑚+1𝑘𝑗𝐵𝑗.(2.2)

Let us apply (1.9) to the bosonic 𝑝-adic integral of (1.16). 𝑝𝑥𝑚1𝑑𝜇(𝑥)=𝑚+1𝑚𝑘=0𝑘𝑚+1𝑝𝐵𝑘=1(𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘(𝑚+11)𝑘𝑝𝐵𝑘(=11𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝑝(1𝑥)𝑗=1𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗(1)𝑗𝐵𝑗(1).(2.3) Then, we can express (2.3) in three different ways.

By (1.13), (2.3) can be written as 𝑝𝑥𝑚1𝑑𝜇(𝑥)=𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝑗+𝐵𝑗+𝛿1,𝑗=1𝑚+1𝑚𝑘=0𝑘(𝑚+11)𝑘𝑘𝐵𝑘1(1)+𝑘𝐵𝑘1+𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗=𝑚1𝑘=0𝑚𝑘𝐵𝑘+(1)𝑘𝐵𝑘+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗=𝑚1𝑘=0𝑚𝑘𝐵𝑘+𝐵𝑘+𝛿1,𝑘+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗𝐵=2𝑚(1)𝐵𝑚𝑚𝛿1,𝑚+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗=𝛿1,𝑚1𝑚+𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝐵𝑗.(2.4) Thus, we have the following theorem.

Theorem 2.2. Let 𝑚0. Then one has the following: 𝐵𝑚=𝛿1,𝑚1𝑚+𝑚+1𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝑚+1(1)𝑘𝐵𝑘𝑗𝐵𝑗.(2.5)

Corollary 2.3. Let 𝑚 be an integer 2. Then one has the following: 𝐵𝑚1+𝑚=𝑚+1𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝑚+1(1)𝑘𝐵𝑘𝑗𝐵𝑗.(2.6)

Especially, for an odd integer 𝑚 with 𝑚3, we obtain the following corollary.

Corollary 2.4. Let 𝑚 be an odd integer with 𝑚3. Then one has the following: 𝑚(𝑚+1)=𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝑚+1(1)𝑘𝐵𝑘𝑗𝐵𝑗.(2.7)

By (1.13), (2.3) can be written as 𝑝𝑥𝑚1𝑑𝜇(𝑥)=𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗𝐵𝑘𝑗𝑗+(1)𝑗𝐵𝑗=1𝑚+1𝑚𝑘=0𝑘(𝑚+11)𝑘𝑘𝐵𝑘1(1)+𝑘𝑗=0𝑘𝑗(1)𝑗𝐵𝑘𝑗𝐵𝑗=𝑚1𝑘=0𝑚𝑘𝐵𝑘+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗(1)𝑗𝐵𝑘𝑗𝐵𝑗=𝐵𝑚(1)+𝐵𝑚+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗(1)𝑗𝐵𝑘𝑗𝐵𝑗.(2.8) By (1.10), (2.8) can be written as 𝑝𝑥𝑚𝑑𝜇(𝑥)=(1)𝑚+1𝐵𝑚+𝐵𝑚+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗(1)𝑗𝐵𝑘𝑗𝐵𝑗.(2.9) So, we get the following theorem.

Theorem 2.5. Let 𝑚0. Then one has the following: 𝐵𝑚=1𝑚+1𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝑚+1(1)𝑚+𝑘+𝑗𝐵𝑘𝑗𝐵𝑗.(2.10)

By (1.10), (2.8) can also be written as 𝑝𝑥𝑚𝑑𝜇(𝑥)=𝛿1,𝑚+1𝑚+1𝑚𝑘=0𝑘𝑚+1(1)𝑘𝑘𝑗=0𝑘𝑗(1)𝑗𝐵𝑘𝑗𝐵𝑗.(2.11) Thus, we have the following theorem.

Theorem 2.6. Let 𝑚0. Then one has the following: 𝐵𝑚=𝛿1,𝑚+1𝑚+1𝑚𝑘𝑘=0𝑗=0𝑘𝑘𝑗𝑚+1(1)𝑘+𝑗𝐵𝑘𝑗𝐵𝑗.(2.12)

3. Identities Involving Euler Numbers

Taking the fermionic 𝑝-adic integral on both sides of (1.32), we have the following: 𝑝𝑥𝑚𝑑𝜇1(𝑥)=𝑝𝐸𝑚1(𝑥)+2𝑚1𝑘=0𝑚𝑘𝐸𝑘(𝑥)𝑑𝜇1=(𝑥)𝑚𝑙=0𝑚𝑙𝐸𝑚𝑙𝑝𝑥𝑙𝑑𝜇1(1𝑥)+2𝑚1𝑘=0𝑚𝑘𝑘𝑗=0𝑘𝑗𝐸𝑘𝑗𝑝𝑥𝑗𝑑𝜇1(=𝑥)𝑚𝑙=0𝑚𝑙𝐸𝑚𝑙𝐸𝑙+12𝑚1𝑘=0𝑚𝑘𝑘𝑗=0𝑘𝑗𝐸𝑘𝑗𝐸𝑗.(3.1) So, we obtain the following theorem.

Theorem 3.1. Let 𝑚0. Then one has the following: 𝐸𝑚=𝑚𝑙=0𝑚𝑙𝐸𝑚𝑙𝐸𝑙+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗𝐸𝑘𝑗𝐸𝑗.(3.2)

Let us apply (1.25) to the fermionic 𝑝-adic integral of (1.32). 𝑝𝑥𝑚𝑑𝜇1(𝑥)=(1)𝑚𝑝𝐸𝑚(1𝑥)𝑑𝜇11(𝑥)+2𝑚1𝑘=0𝑚𝑘(1)𝑘𝑝𝐸𝑘(1𝑥)𝑑𝜇1(𝑥)=(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘(1)𝑘𝐸𝑘(+11)2𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗(1)𝑗𝐸𝑗(1).(3.3) Then, we can express (3.3) in two different ways.

By (1.29), (3.3) can be written as 𝑝𝑥𝑚𝑑𝜇1(𝑥)=(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘2+𝐸𝑘2𝛿0,𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗2+𝐸𝑗2𝛿0,𝑗=2𝐸𝑚+(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘𝐸𝑘+2(1)𝑚+1𝐸𝑚+𝑚1𝑘=0𝑚𝑘𝐸𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗𝐸𝑗+𝑚1𝑘=0𝑚𝑘(1)𝑘+1𝐸𝑘=2𝐸𝑚+(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘𝐸𝑘+2(1)𝑚+1𝐸𝑚+𝐸𝑚(1)𝐸𝑚+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗𝐸𝑗+(1)𝑚+1𝐸𝑚(1)𝐸𝑚=2+2𝛿0,𝑚+(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘𝐸𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗𝐸𝑗.(3.4) Thus, we get the following theorem.

Theorem 3.2. Let 𝑚0. Then one has the following: 𝐸𝑚=2+2𝛿0,𝑚+(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘𝐸𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗𝐸𝑗.(3.5)

Corollary 3.3. Let 𝑚>0. Then one has the following: 𝐸𝑚+2=(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘𝐸𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗𝐸𝑗.(3.6)

By (1.29), (3.3) can be written as 𝑝𝑥𝑚𝑑𝜇1(𝑥)=(1)𝑚𝑚𝑘=0𝑚𝑘𝐸𝑚𝑘2(1)𝑘𝐸𝑘+12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘𝐸𝑘𝑗2(1)𝑗𝐸𝑗=2𝐸𝑚+(1)𝑚𝑚+1𝑘=0𝑚𝑘(1)𝑘𝐸𝑚𝑘𝐸𝑘+𝑚1𝑘=0𝑚𝑘𝐸𝑘12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘(1)𝑗𝐸𝑘𝑗𝐸𝑗=2𝐸𝑚+(1)𝑚𝑚+1𝑘=0𝑚𝑘(1)𝑘𝐸𝑚𝑘𝐸𝑘+𝐸𝑚(1)𝐸𝑚12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘(1)𝑗𝐸𝑘𝑗𝐸𝑗=2𝛿0,𝑚+(1)𝑚𝑚+1𝑘=0𝑚𝑘(1)𝑘𝐸𝑚𝑘𝐸𝑘12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘(1)𝑗𝐸𝑘𝑗𝐸𝑗.(3.7) So, we have the following theorem.

Theorem 3.4. Let 𝑚0. Then one has the following: 𝐸𝑚=2𝛿0,𝑚+(1)𝑚𝑚+1𝑘=0𝑚𝑘(1)𝑘𝐸𝑚𝑘𝐸𝑘12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘+𝑗𝐸𝑘𝑗𝐸𝑗.(3.8)

Corollary 3.5. Let 𝑚>1. Then one has the following: 𝐸𝑚=(1)𝑚𝑚+1𝑘=0𝑚𝑘(1)𝑘𝐸𝑚𝑘𝐸𝑘12𝑚1𝑘𝑘=0𝑗=0𝑚𝑘𝑘𝑗(1)𝑘+𝑗𝐸𝑘𝑗𝐸𝑗.(3.9)

4. Identities Involving Bernoulli and Euler Numbers

By (1.16) and (1.32), we have the following: 𝑝𝑥𝑚+𝑛𝑑𝜇(𝑥)=𝑝1𝑚+1𝑚𝑘=0𝑘𝐵𝑚+1𝑘𝐸(𝑥)𝑛1(𝑥)+2𝑛1𝑙=0𝑛𝑙𝐸𝑙=1(𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘=0𝑘𝑚+1𝑝𝐵𝑘(𝑥)𝐸𝑛(+1𝑥)𝑑𝜇(𝑥)2(𝑚+1)𝑚𝑘=0𝑘𝑚+1𝑛1𝑙=0𝑛𝑙𝑝𝐵𝑘(𝑥)𝐸𝑙=1(𝑥)𝑑𝜇(𝑥)𝑚+1𝑚𝑘𝑘=0𝑛𝑗=0𝑙=0𝑘𝑘𝑗𝑛𝑙𝐵𝑚+1𝑘𝑗𝐸𝑛𝑙𝐵𝑗+𝑙+12(𝑚+1)𝑚𝑘=0𝑛1𝑘𝑙=0𝑙𝑗=0𝑖=0𝑘𝑛𝑙𝑘𝑗𝑙𝑖𝐵𝑚+1𝑘𝑗𝐸𝑙𝑖𝐵𝑗+𝑖.(4.1) Therefore, we get the following theorem.

Theorem 4.1. Let 𝑚,𝑛0. Then one has the following: 𝐵𝑚+𝑛=1𝑚+1𝑚𝑘𝑘=0𝑛𝑗=0𝑙=0𝑘𝑘𝑗𝑛𝑙𝐵𝑚+1𝑘𝑗𝐸𝑛𝑙𝐵𝑗+𝑙+12(𝑚+1)𝑚𝑘=0𝑛1𝑘𝑙=0𝑙𝑗=0𝑖=0𝑘𝑛𝑙𝑘𝑗𝑙𝑖𝐵𝑚+1𝑘𝑗𝐸𝑙𝑖𝐵𝑗+𝑖.(4.2)

By (1.16) and (1.33), we have the following: 𝑝𝑥𝑚𝐵𝑘,𝑛(𝑥)𝑑𝜇(𝑥)=𝑝1𝑚+1𝑚𝑙=0𝑙𝐵𝑚+1𝑙(𝑥)𝐵𝑘,𝑛=1(𝑥)𝑑𝜇(𝑥)𝑛𝑘𝑚+1𝑚𝑙𝑙=0𝑖=0𝑙𝑙𝑖𝐵𝑚+1𝑙𝑖𝑝𝑥𝑖+𝑘(1𝑥)𝑛𝑘=1𝑑𝜇(𝑥)𝑛𝑘𝑚+1𝑚𝑙𝑙=0𝑖=0𝑛𝑘𝑗=0𝑙𝑙𝑖𝑗𝑚+1𝑛𝑘(1)𝑗𝐵𝑙𝑖𝑝𝑥𝑖+𝑘+𝑗=1𝑑𝜇(𝑥)𝑛𝑘𝑚+1𝑚𝑙𝑙=0𝑖=0𝑛𝑘𝑗=0𝑙𝑙𝑖𝑗𝑚+1𝑛𝑘(1)𝑗𝐵𝑙𝑖𝐵𝑖+𝑘+𝑗.(4.3) By (1.33), we have the following: 𝑝𝑥𝑚𝐵𝑘,𝑛𝑛𝑘(𝑥)𝑑𝜇(𝑥)=𝑝𝑥𝑚+𝑘(1𝑥)𝑛𝑘=𝑛𝑘𝑑𝜇(𝑥)𝑛𝑘𝑗=0𝑗(𝑛𝑘1)𝑗𝑝𝑥𝑚+𝑘+𝑗=𝑛𝑘𝑑𝜇(𝑥)𝑛𝑘𝑗=0𝑗𝑛𝑘(1)𝑗𝐵𝑚+𝑘+𝑗.(4.4) By (4.3) and (4.4), we obtain the following theorem.

Theorem 4.2. Let 𝑚,𝑛,𝑘0. Then one has the following: 𝑛𝑘𝑗=0𝑗𝑛𝑘(1)𝑗𝐵𝑚+𝑘+𝑗=1𝑚+1𝑚𝑙𝑙=0𝑖=0𝑛𝑘𝑗=0𝑙𝑙𝑖𝑗𝑚+1𝑛𝑘(1)𝑗𝐵𝑙𝑖𝐵𝑖+𝑘+𝑗.(4.5) Especially, one has the following: (𝑚+1)𝐵𝑚+𝑛=𝑚𝑙𝑙=0𝑖=0𝑙𝑙𝑖𝐵𝑚+1𝑙𝑖𝐵𝑖+𝑛.(4.6)

By (4.2) and (4.6), we have the following theorem. Note that (4.8) in the following was obtained in [6].

Theorem 4.3. Let 𝑚,𝑛0. Then one has the following: 𝐵𝑚+𝑛=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐵𝑚+𝑙+12𝑛1𝑙𝑙=0𝑖=0𝑛𝑙𝑙𝑖𝐸𝑙𝑖𝐵𝑚+𝑖.(4.7)

In particular, we have the following: 𝐵𝑛=𝑛𝑙=0𝑛𝑙𝐸𝑛1𝐵𝑙+12𝑛1𝑙𝑙=0𝑖=0𝑛𝑙𝑙𝑖𝐸𝑙𝑖𝐵𝑖.(4.8)