Abstract

We give some interesting identities on the Bernoulli numbers and polynomials, on the Genocchi numbers and polynomials by using symmetric properties of the Bernoulli and Genocchi polynomials.

1. Introduction

Let 𝑝 be a fixed odd prime number. Throughout this paper 𝑝,𝑝, and 𝑝 will denote the ring of 𝑝-adic rational integers, the field of 𝑝-adic rational numbers, and the completion of the algebraic closure of 𝑝. Let be the set of natural numbers and +={0}. The 𝑝-adic norm on 𝐶𝑝 is normalized so that |𝑝|𝑝=𝑝1. Let 𝐶(𝑝) be the space of continuous 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(1.1) (see [116]). From (1.1), we have 𝐼1𝑓1=𝐼1(𝑓)+2𝑓(0)(1.2) (see [116]), where 𝑓1(𝑥)=𝑓(𝑥+1).

Let us take 𝑓(𝑥)=𝑒𝑥𝑡. Then, by (1.2), we get 𝑡𝑝𝑒𝑥𝑡𝑑𝜇1(𝑥)=2𝑡𝑒𝑡=+1𝑛=0𝐺𝑛𝑡𝑛,𝑛!(1.3) where 𝐺𝑛 are the 𝑛th ordinary Genocchi numbers (see [8, 15]).

From the same method of (1.3), we can also derive the following equation: 𝑡𝑝𝑒(𝑥+𝑦)𝑡𝑑𝜇1(𝑦)=2𝑡𝑒𝑡𝑒+1𝑥𝑡=𝑛=0𝐺𝑛(𝑡𝑥)𝑛,𝑛!(1.4) where 𝐺𝑛(𝑥) are called the 𝑛th Genocchi polynomials (see [14, 15]).

By (1.3), we easily see that 𝐺𝑛(𝑥)=𝑛𝑙=0𝑛𝑙𝐺𝑙𝑥𝑛𝑙(1.5) (see [15]). By (1.3) and (1.4), we get Witt's formula for the 𝑛th Genocchi numbers and polynomials as follows: 𝑝𝑥𝑛𝑑𝜇1𝐺(𝑥)=𝑛+1,𝑛+1𝑝(𝑥+𝑦)𝑛𝑑𝜇1𝐺(𝑦)=𝑛+1(𝑥)𝑛+1,for𝑛+.(1.6) From (1.2), we have 𝑝(𝑥+1)𝑛𝑑𝜇1(𝑥)+𝑝𝑥𝑛𝑑𝜇1(𝑥)=2𝛿0,𝑛,(1.7) where the symbol 𝛿0,𝑛 is the Kronecker symbol (see [4, 5]).

Thus, by (1.5) and (1.7), we get (𝐺+1)𝑛+𝐺𝑛=2𝛿1,𝑛(1.8)(see [15]). From (1.4), we can derive the following equation: 𝑝(1𝑥+𝑦)𝑛𝑑𝜇1(𝑦)=(1)𝑛𝑝(𝑥+𝑦)𝑛𝑑𝜇1(𝑦).(1.9) By (1.6) and (1.9), we see that 𝐺𝑛+1(1𝑥)𝑛+1=(1)𝑛𝐺𝑛+1(𝑥).𝑛+1(1.10) Thus, by (1.10), we get 𝐺𝑛+1(2)/(𝑛+1)=(1)𝑛(𝐺𝑛+1(1)/(𝑛+1)).

From (1.5) and (1.8), we have 𝐺𝑛+1(2)𝐺𝑛+1=2𝑛+1(1)𝐺𝑛+1=2+𝑛+1𝑛+12𝛿1,𝑛+1.(1.11) The Bernoulli polynomials 𝐵𝑛(𝑥) are defined by 𝑡𝑒𝑡𝑒1𝑥𝑡=𝑒𝐵(𝑥)𝑡=𝑛=0𝐵𝑛(𝑡𝑥)𝑛𝑛!(1.12) (see [6, 9, 12]) with the usual convention about replacing 𝐵𝑛(𝑥) by 𝐵𝑛(𝑥).

In the special case, 𝑥=0, 𝐵𝑛(0)=𝐵𝑛 is called the 𝑛-th Bernoulli number. By (1.12), we easily see that 𝐵𝑛(𝑥)=𝑛𝑙=0𝑛𝑙𝑥𝑛𝑙𝐵𝑙=(𝐵+𝑥)𝑛(1.13) (see [6]). Thus, by (1.12) and (1.13), we get reflection symmetric formula for the Bernoulli polynomials as follows: 𝐵𝑛(1𝑥)=(1)𝑛𝐵𝑛𝐵(𝑥),(1.14)0=1,(𝐵+1)𝑛𝐵𝑛=𝛿1,𝑛(1.15)(see [6, 9, 12]). From (1.14) and (1.15), we can also derive the following identity: (1)𝑛𝐵𝑛(1)=𝐵𝑛(2)=𝑛+𝐵𝑛(1)=𝑛+𝐵𝑛+𝛿1,𝑛.(1.16) In this paper, we investigate some properties of the fermionic 𝑝-adic integrals on 𝑝. By using these properties, we give some new identities on the Bernoulli and the Euler numbers which are useful in studying combinatorics.

2. Identities on the Bernoulli and Genocchi Numbers and Polynomials

Let us consider the following fermionic 𝑝-adic integral on 𝑝 as follows: 𝐼1=𝑝𝐵𝑛(𝑥)𝑑𝜇1(𝑥)=𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝑝𝑥𝑙𝑑𝜇1=(𝑥)𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐺𝑙+1𝑙+1,for𝑛+={0}.(2.1) On the other hand, by (1.14) and (1.15), we get 𝐼1=(1)𝑛𝑝𝐵𝑛(1𝑥)𝑑𝜇1(𝑥)=(1)𝑛𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝑝(1𝑥)𝑙𝑑𝜇1(𝑥)=(1)𝑛𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙(1)𝑙𝐺𝑙+1(1)=𝑙+1(1)𝑛𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐺2+𝑙+1𝑙+12𝛿1,𝑙+1=2(1)𝑛𝐵𝑛+𝛿1,𝑛+(1)𝑛𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐺𝑙+1𝑙+1+2(1)𝑛+1𝐵𝑛.(2.2) Equating (2.1) and (2.2), we obtain the following theorem.

Theorem 2.1. For 𝑛+, one has1+(1)𝑛+1𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐺𝑙+1𝑙+1=2(1)𝑛𝛿1,𝑛.(2.3)

By using the reflection symmetric property for the Euler polynomials, we can also obtain some interesting identities on the Euler numbers.

Now, we consider the fermionic 𝑝-adic integral on 𝑝 for the polynomials as follows: 𝐼2=𝑝𝐺𝑛(𝑥)𝑑𝜇1(=𝑥)𝑛𝑙=0𝑛𝑙𝐺𝑛𝑙𝑝𝑥𝑙𝑑𝜇1=(𝑥)𝑛𝑙=0𝑛𝑙𝐺𝑛𝑙𝐺𝑙+1𝑙+1,for𝑛+.(2.4) On the other hand, by (1.8), (1.10), and (1.11), we get 𝐼2=(1)𝑛1𝑝𝐺𝑛(1𝑥)𝑑𝜇1(𝑥)=(1)𝑛𝑛1𝑙=0𝑛𝑙𝐺𝑛𝑙𝑝(1𝑥)𝑙𝑑𝜇1(𝑥)=(1)𝑛𝑛1𝑙=0𝑛𝑙𝐺𝑛𝑙(1)𝑙𝐺𝑙+1(1)=𝑙+1(1)𝑛𝑛1𝑙=0𝑛𝑙𝐺𝑛𝑙𝐺2+𝑙+1𝑙+12𝛿1,𝑙+1=2(1)𝑛12𝛿1,𝑛𝐺𝑛+2(1)𝑛𝐺𝑛+(1)𝑛𝑛1𝑙=0𝑛𝑙𝐺𝑛𝑙𝐺𝑙+1.𝑙+1(2.5) Equating (2.4) and (2.5), we obtain the following theorem.

Theorem 2.2. For 𝑛+, one has(1+(1)𝑛)𝑛𝑙=0𝑛𝑙𝐺𝑛𝑙𝐺𝑙+1𝑙+1=4(1)𝑛𝐺𝑛+4(1)𝑛+1𝛿1,𝑛.(2.6)

Let us consider the fermionic 𝑝-adic integral on 𝑝 for the product of 𝐵𝑛(𝑥) and 𝐺𝑛(𝑥) as follows: 𝐼3=𝑝𝐵𝑚(𝑥)𝐺𝑛(𝑥)𝑑𝜇1(=𝑥)𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙𝑝𝑥𝑘+𝑙𝑑𝜇1=(𝑥)𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙𝐺𝑘+𝑙+1.𝑘+𝑙+1(2.7) On the other hand, by (1.10) and (1.14), we get 𝐼3=𝑝𝐵𝑚(𝑥)𝐺𝑛(𝑥)𝑑𝜇1(=𝑥)(1)𝑛+𝑚1𝑝𝐵𝑚(1𝑥)𝐺𝑛(1𝑥)𝑑𝜇1(𝑥)=(1)𝑚𝑛+𝑚1𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙𝑝(1𝑥)𝑘+𝑙𝑑𝜇1(𝑥)=2(1)𝑛+𝑚1𝐵𝑚(1)𝐺𝑛(1)+2(1)𝑚+𝑛𝐵𝑚𝐺𝑛+(1)𝑚𝑛+𝑚1𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙𝐺𝑘+𝑙+1.𝑘+𝑙+1(2.8) By (2.7) and (2.8), we easily see that 1+(1)𝑛+𝑚+1𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙𝐺𝑘+𝑙+1𝑘+𝑙+1=2(1)𝑚+𝑛1𝛿1,𝑚+𝐵𝑚2𝛿1,𝑛𝐺𝑛+2(1)𝑚+𝑛𝐵𝑚𝐺𝑛=4(1)𝑚+𝑛1𝐵𝑚𝛿1,𝑛+2(1)𝑚+𝑛𝐵𝑚𝐺𝑛+4(1)𝑚+𝑛1𝛿1,𝑚𝛿1,𝑛+2(1)𝑚+𝑛𝛿1,𝑚𝐺𝑛+2(1)𝑚+𝑛𝐵𝑚𝐺𝑛.(2.9) Therefore, by (2.9), we obtain the following theorem.

Theorem 2.3. For 𝑛,𝑚+, one has1+(1)𝑛+𝑚+1𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐺𝑛𝑙+1𝐺𝑛𝑙+1𝑘+𝑙+1𝑘+𝑙+1=4(1)𝑚+𝑛𝐵𝑚𝐺𝑛+4(1)𝑚+𝑛1𝐵𝑚𝛿1,𝑛+4(1)𝑚+𝑛1𝛿1,𝑚𝛿1,𝑛+2(1)𝑚+𝑛𝛿1,𝑚𝐺𝑛.(2.10)

Corollary 2.4. For 𝑛,𝑚, one has2𝑚𝑘=02𝑛𝑙=0𝑘𝑙𝐵2𝑚2𝑛2𝑚𝑘𝐺2𝑛𝑙𝐺𝑘+𝑙+1𝑘+𝑙+1=2𝐵2𝑚𝐺2𝑛.(2.11)

Let us consider the fermionic 𝑝-adic integral on 𝑝 for the product of the Bernoulli polynomials and the Bernstein polynomials. For 𝑛,𝑘+, with 0𝑘𝑛, 𝐵𝑘,𝑛((𝑥)=𝑛𝑘)𝑥𝑘(1𝑥)𝑛𝑘 are called the Bernstein polynomials of degree 𝑛, see [11]. It is easy to show that 𝐵𝑘,𝑛(𝑥)=𝐵𝑛𝑘,𝑛(1𝑥), 𝐼4=𝑝𝐵𝑚(𝑥)𝐵𝑘,𝑛(𝑥)𝑑𝜇1(=𝑛𝑘𝑥)𝑚𝑙=0𝑚𝑙𝐵𝑚𝑙𝑝𝑥𝑘+𝑙(1𝑥)𝑛𝑘𝑑𝜇1=𝑛𝑘(𝑥)𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐵𝑚𝑙𝑝𝑥𝑘+𝑙+𝑗𝑑𝜇1=𝑛𝑘(𝑥)𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐵𝑚𝑙𝐺𝑘+𝑙+𝑗+1.𝑘+𝑙+𝑗+1(2.12) On the other hand, by (1.14) and (2.12), we get 𝐼4=(1)𝑚𝑝𝐵𝑚(1𝑥)𝐵𝑛𝑘,𝑛(1𝑥)𝑑𝜇1(𝑥)=(1)𝑚𝑛𝑘𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐵𝑚𝑙𝑝(1𝑥)𝑛𝑘+𝑙+𝑗𝑑𝜇1(𝑥)=(1)𝑚𝑛𝑘𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐵𝑚𝑙×22𝛿1,𝑛𝑘+𝑙+𝑗+1+𝐺𝑛𝑘+𝑙+𝑗+1𝑛𝑘+𝑙+𝑗+1=2(1)𝑚𝑛𝑘𝐵𝑚(1)𝛿0,𝑘+2(1)𝑚+1𝑛𝑘𝐵𝑚𝛿𝑘,𝑛+(1)𝑚𝑛𝑘𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐵𝑚𝑙𝐺𝑛𝑘+𝑙+𝑗+1.𝑛𝑘+𝑙+𝑗+1(2.13) Equating (2.12) and (2.13), we see that 𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐵𝑚𝑙𝐺𝑘+𝑙+𝑗+1𝑘+𝑙+𝑗+1=2(1)𝑚𝐵𝑚(1)𝛿0,𝑘+2(1)𝑚+1𝐵𝑚𝛿𝑘,𝑛+(1)𝑚𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐵𝑚𝑙𝐺𝑛𝑘+𝑙+𝑗+1.𝑛𝑘+𝑙+𝑗+1(2.14) Thus, from (2.14), we obtain the following theorem.

Theorem 2.5. For 𝑛,𝑚, one has2𝑚𝑛𝑙=0𝑗=0𝑙𝑛𝑗2𝑚(1)𝑗𝐵2𝑚𝑙𝐺𝑙+𝑗+1𝑙+𝑗+1=2𝐵2𝑚(1)+2𝑚𝑙=0𝑙𝐵2𝑚2𝑚𝑙𝐺𝑛+𝑙+1𝑛+𝑙+1.(2.15)

Finally, we consider the fermionic 𝑝-adic integral on 𝑝 for the product of the Euler polynomials and the Bernstein polynomials as follows: 𝐼5=𝑝𝐺𝑚(𝑥)𝐵𝑘,𝑛(𝑥)𝑑𝜇1(=𝑛𝑘𝑥)𝑚𝑙=0𝑚𝑙𝐺𝑚𝑙𝑝𝑥𝑘+𝑙(1𝑥)𝑛𝑘𝑑𝜇1=𝑛𝑘(𝑥)𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐺𝑚𝑙𝑝𝑥𝑘+𝑙+𝑗𝑑𝜇1=𝑛𝑘(𝑥)𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐺𝑚𝑙𝐺𝑘+𝑙+𝑗+1.𝑘+𝑙+𝑗+1(2.16) On the other hand, by (1.10) and (2.12), we get 𝐼5=(1)𝑚1𝑝𝐺𝑚(1𝑥)𝐵𝑛𝑘,𝑛(1𝑥)𝑑𝜇1(𝑥)=(1)𝑚1𝑛𝑘𝑚𝑙=0𝑚𝑙𝐺𝑘𝑚𝑙𝑗=0𝑘𝑗(1)𝑗𝑝(1𝑥)𝑛𝑘+𝑙+𝑗𝑑𝜇1(𝑥)=(1)𝑚1𝑛𝑘𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐺𝑚𝑙×𝐺2+𝑛𝑘+𝑙+𝑗+1𝑛𝑘+𝑙+𝑗+12𝛿1,𝑛𝑘+𝑙+𝑗+1=2(1)𝑚1𝑛𝑘𝐺𝑚(1)𝛿0,𝑘+2(1)𝑚𝑛𝑘𝐺𝑚𝛿𝑘,𝑛+(1)𝑚1𝑛𝑘𝑚𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐺𝑚𝑙𝐺𝑛𝑘+𝑙+𝑗+1.𝑛𝑘+𝑙+𝑗+1(2.17) Equating (2.16) and (2.17), we obtain 𝑚𝑙=0𝑛𝑘𝑗=0𝑚𝑙𝑗𝑛𝑘(1)𝑗𝐺𝑚𝑙𝐺𝑘+𝑙+𝑗+1𝑘+𝑙+𝑗+1=2(1)𝑚1𝐺𝑚(1)𝛿0,𝑘+2(1)𝑚𝐺𝑚𝛿𝑘,𝑛+(1)𝑚𝑚1𝑘𝑙=0𝑗=0𝑚𝑙𝑘𝑗(1)𝑗𝐺𝑚𝑙𝐺𝑛𝑘+𝑙+𝑗+1.𝑛𝑘+𝑙+𝑗+1(2.18) Therefore, by (2.18), we obtain the following theorem.

Theorem 2.6. For 𝑛,𝑚, one has2𝑚𝑛𝑙=0𝑗=0𝑙𝑛𝑗2𝑚(1)𝑗𝐺2𝑚𝑙𝐺𝑙+𝑗+1𝑙+𝑗+1=2𝐺2𝑚(1)2𝑚𝑙=0𝑙𝐺2𝑚2𝑚𝑙𝐺𝑛+𝑙+1𝑛+𝑙+1.(2.19)

Acknowledgment

This paper was supported by Kynugpook National University Research Fund, 2012.