Abstract

The purpose of this paper is to give some identities on the Frobenius-Euler numbers and polynomials by using the fermionic 𝑝-adic 𝑞-integral equation on 𝑝.

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 algebraic closure of 𝑝, respectively. Let be the set of natural numbers and +={0}. The 𝑝-adic absolute value on 𝑝 is normalized so that |𝑝|𝑝=1/𝑝. Assume that 𝑞𝑝 with |1𝑞|𝑝<1.

Let 𝑓 be a continuous function on 𝑝. Then the fermionic 𝑝-adic 𝑞-integral on 𝑝 for 𝑓 is defined by Kim as follows:𝐼𝑞(𝑓)=𝑝𝑓(𝑥)𝑑𝜇𝑞(𝑥)=lim𝑁1+𝑞1+𝑞𝑝𝑁𝑝𝑁1𝑥=0𝑓(𝑥)(𝑞)𝑥[1]).,(see(1.1) From (1.1), we note that𝑞𝑛𝐼𝑞𝑓𝑛=(1)𝑛𝐼𝑞(𝑓)+(1+𝑞)𝑛1𝑙=0(1)𝑛1𝑙𝑓(𝑙)𝑞𝑙,(1.2) where 𝑛 and 𝑓𝑛(𝑥)=𝑓(𝑥+𝑛) (see [1]). The ordinary Euler polynomials 𝐸𝑛(𝑥) are defined by2𝑒𝑡e+1𝑥𝑡=𝑒𝐸(𝑥)𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛,𝑛!(1.3) with the usual convention about replacing 𝐸𝑛(𝑥) by 𝐸𝑛(𝑥) (see [110]). In the special case, 𝑥=0, 𝐸𝑛(0)=𝐸𝑛 is called the 𝑛th Euler number.

As the extension of (1.3), the Frobenius-Euler polynomials are defined by1𝑞𝑒𝑡𝑒𝑞𝑥𝑡=𝑛=0𝐻𝑛(𝑡𝑞,𝑥)𝑛[2]).𝑛!,(see(1.4) In the special case, 𝑥=0, 𝐻𝑛(𝑞,0)=𝐻𝑛(𝑞) is called the 𝑛th Frobenius-Euler number.

By (1.3) and (1.4), we easily get 𝐻𝑛(1,𝑥)=𝐸𝑛(𝑥).

From (1.4), we note that𝐻𝑛(𝑞,𝑥)=𝑛𝑙=0𝑛𝑙𝐻𝑙(𝑞)𝑥𝑛𝑙=(𝐻(𝑞)+𝑥)𝑛[2],(see),(1.5) with the usual convention about replacing 𝐻(𝑞)𝑛 by 𝐻𝑛(𝑞).

In this paper, we consider the fermionic 𝑝-adic 𝑞-integral on 𝑝 for the Frobenius-Euler numbers and polynomials. From these 𝑝-adic 𝑞-integrals on 𝑝, we derive some new and interesting identities on the Frobenius-Euler numbers and polynomials.

2. Identities on the Frobenius-Euler Numbers

From (1.2) and (1.4), we can derive the following:𝑝𝑒(𝑥+𝑦)𝑡𝑑𝜇𝑞(𝑦)=1+𝑞1𝑒𝑡+𝑞1𝑒𝑥𝑡=𝑛=0𝐻𝑛𝑞1𝑡,𝑥𝑛.𝑛!(2.1) Thus, by (2.1), we get Witt's formula for 𝐻𝑛(𝑞1,𝑥) as follows:𝑝(𝑥+𝑦)𝑛𝑑𝜇𝑞(𝑦)=𝐻𝑛𝑞1,𝑥,𝑛+.(2.2) By (1.5) and (2.1), we get𝐻𝑞1+1𝑛+𝑞1𝐻𝑛𝑞1=1+𝑞1,if𝑛=0,0,if𝑛>0,(2.3) with the usual convention about replacing 𝐻(𝑞1)𝑛 by 𝐻𝑛(𝑞1).

From (1.5) and (2.3), we note that𝐻0𝑞1=1,𝐻𝑛𝑞1,1+𝑞1𝐻𝑛𝑞1=0,if𝑛>0.(2.4) By (2.1) and (2.2), we get𝑝(𝑦+1𝑥)𝑛𝑑𝜇𝑞(𝑦)=(1)𝑛𝑝(𝑦+𝑥)𝑛𝑑𝜇𝑞1(𝑦).(2.5) Therefore, by (2.5), we obtain the following lemma.

Lemma 2.1. For 𝑛+, one has 𝐻𝑛𝑞1,1𝑥=(1)𝑛𝐻𝑛(𝑞,𝑥).(2.6)

From (2.3), we can derive the following:𝑞2𝐻𝑛𝑞1,2𝑞2𝑞=𝑞2𝑛𝑙=0𝑛𝑙𝐻𝑞1+1𝑙𝑞2𝑞=𝑞𝑛𝑙=1𝑛𝑙𝑞𝐻𝑞1+1𝑙𝑞=𝑞𝑛𝑙=0𝑛𝑙𝐻𝑙𝑞1=(1+𝑞)𝛿0,𝑛+𝐻𝑛𝑞1,(2.7) where 𝛿𝑘,𝑛 is the Kronecker symbol.

Therefore, by (2.7), we obtain the following theorem.

Theorem 2.2. For 𝑛+, one has 𝐻𝑛𝑞1,2=1+𝑞1𝑞2(1+𝑞)𝛿0,𝑛+𝑞2𝐻𝑛𝑞1.(2.8)

First we consider the fermionic 𝑝-adic 𝑞-integral on 𝑝 for the 𝑛th Frobenius-Euler polynomials as follows:𝐼1=𝑝𝐻𝑛𝑞1,𝑥𝑑𝜇𝑞(=𝑥)𝑛𝑙=0𝑛𝑙𝐻𝑙𝑞1𝑝𝑥𝑛𝑙𝑑𝜇𝑞=(𝑥)𝑛𝑙=0𝑛𝑙𝐻𝑙𝑞1𝐻𝑛𝑙𝑞1,where𝑛+.(2.9) On the other hand, by (2.5) and Lemma 2.1, we get𝐼1=𝑝𝐻𝑛𝑞1,𝑥𝑑𝜇𝑞(𝑥)=(1)𝑛𝑝𝐻𝑛(𝑞,1𝑥)𝑑𝜇𝑞(𝑥)=(1)𝑛𝑛𝑙=0𝑛𝑙𝐻𝑛𝑙(𝑞)𝑝(1𝑥)𝑙𝑑𝜇𝑞=(𝑥)𝑛𝑙=0𝑛𝑙(1)𝑛𝑙𝐻𝑛𝑙(𝑞)𝑝(𝑥1)𝑙𝑑𝜇𝑞=(𝑥)𝑛𝑙=0𝑛𝑙(1)𝑛𝑙𝐻𝑛𝑙(𝑞)𝐻𝑙𝑞1.,1(2.10) From Lemma 2.1, Theorem 2.2, and (2.10), we note that𝐼1=𝑛𝑙=0𝑛𝑙(1)𝑛𝑙𝐻𝑛𝑙(𝑞)𝐻𝑙𝑞1=,1𝑛𝑙=0𝑛𝑙(1)𝑛𝐻𝑛𝑙(𝑞)𝐻𝑙(=𝑞,2)𝑛𝑙=0𝑛𝑙(1)𝑛𝐻𝑛𝑙(𝑞)1+𝑞𝑞21+𝑞1𝛿0,𝑙+𝑞2𝐻𝑙(𝑞)=(1)𝑛(1+𝑞)(1+𝑞)𝛿0,𝑛𝑞𝐻𝑛(𝑞)𝐻𝑛(𝑞)𝑞+𝑞2(1)𝑛+(1)𝑛𝑞2𝑛𝑙=0𝑛𝑙𝐻𝑛𝑙(𝑞)𝐻𝑙(𝑞).(2.11) Therefore, by (2.10) and (2.11), we obtain the following theorem.

Theorem 2.3. For 𝑛+, one has 𝑛𝑙=0𝑛𝑙𝐻𝑙𝑞1𝐻𝑛𝑙𝑞1=(1)𝑛(1+𝑞)(1+𝑞)𝛿0,𝑛2𝑞𝐻𝑛(𝑞)+(1)𝑛𝑞2𝑛𝑙=0𝑛𝑙𝐻𝑛𝑙(𝑞)𝐻𝑙(𝑞).(2.12) In particular, for 𝑛, one has 𝑛𝑙=0𝑛𝑙𝐻𝑙𝑞1𝐻𝑛𝑙𝑞1=2(1)𝑛+1𝑞(1+𝑞)𝐻𝑛(𝑞)+(1)𝑛𝑞2𝑛𝑙=0𝑛𝑙𝐻𝑛𝑙(𝑞)𝐻𝑙(𝑞).(2.13)

Let us consider the following fermionic 𝑝-adic 𝑞-integral on 𝑝 for the product of Bernoulli and Frobenius-Euler polynomials as follows:𝐼2=𝑝𝐵𝑚(𝑥)𝐻𝑛𝑞1,𝑥𝑑𝜇𝑞(=𝑥)𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙𝑞1𝑝𝑥𝑘+𝑙𝑑𝜇𝑞=(𝑥)𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙𝑞1𝐻𝑘+𝑙𝑞1.(2.14) It is known that 𝐵𝑛(𝑥)=(1)𝑛𝐵𝑛(1𝑥).

On the other hand, by Lemma 2.1, we get𝐼2=(1)𝑚+𝑛𝑝𝐵𝑚(1𝑥)𝐻𝑛(𝑞,1𝑥)𝑑𝜇𝑞(𝑥)=(1)𝑚𝑚+𝑛𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙(𝑞)𝑝(1𝑥)𝑘+𝑙𝑑𝜇𝑞(𝑥)=(1)𝑚𝑚+𝑛𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙(𝑞)(1+𝑞)𝑞21+𝑞1𝛿0,𝑘+𝑙+𝑞2𝐻𝑘+𝑙(𝑞)=(1)𝑚+𝑛(1+𝑞)𝐵𝑚(1)𝐻𝑛(𝑞𝑞,1)2(+𝑞1)𝑚+𝑛𝐵𝑚𝐻𝑛(𝑞)+𝑞2(1)𝑚+𝑛m𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙(𝑞)𝐻𝑘+𝑙(𝑞).(2.15) Therefore, by (2.14) and (2.15), we obtain the following theorem.

Theorem 2.4. For 𝑚,𝑛+, one has 𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙𝑞1𝐻𝑘+𝑙𝑞1=(1)𝑚+𝑛(1+𝑞)𝐵𝑚(1)𝐻𝑛𝑞(𝑞,1)2+𝑞(1)𝑚+𝑛𝐵𝑚𝐻𝑛(𝑞)+𝑞2(1)𝑚𝑚+𝑛𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙(𝑞)𝐻𝑘+𝑙(𝑞).(2.16) In particular, for 𝑚{1}, 𝑛, one has 𝑚𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙𝑞1𝐻𝑘+𝑙𝑞1=2(1)𝑚+𝑛+1𝑞2𝐵+𝑞𝑚𝐻𝑛(𝑞)+𝑞2(1)𝑚𝑚+𝑛𝑛𝑘=0𝑙=0𝑚𝑘𝑛𝑙𝐵𝑚𝑘𝐻𝑛𝑙(𝑞)𝐻𝑘+𝑙(𝑞).(2.17)

It is known that 𝑥𝑛=(1/(𝑛+1))𝑛𝑙=0𝑙𝑛+1𝐵𝑙(𝑥). Let us consider the following fermionic 𝑝-adic 𝑞-integral on 𝑝:𝑝𝑥𝑛𝑑𝜇𝑞1(𝑥)=𝑛+1𝑛𝑙=0𝑙𝑛+1𝑝𝐵𝑙(𝑥)𝑑𝜇𝑞=1(𝑥).𝑛+1𝑛𝑙=0𝑙𝑛+1𝑙𝑘=0𝑙𝑘𝐵𝑙𝑘𝑝𝑥𝑘𝑑𝜇𝑞(=1𝑥)𝑛+1𝑛𝑙=0𝑙𝑛+1𝑙𝑘=0𝑙𝑘𝐵𝑙𝑘𝐻𝑘𝑞1.(2.18) Therefore by (2.18), we obtain the following theorem.

Theorem 2.5. For 𝑛+, one has 𝐻𝑛𝑞1=1𝑛+1𝑛𝑙=0𝑙𝑛+1𝑙𝑘=0𝑙𝑘𝐵𝑙𝑘𝐻𝑘𝑞1.(2.19)

From (1.3), we can derive the following:𝑥𝑛=𝐸𝑛1(𝑥)+2𝑛1𝑙=0𝑛𝑙𝐸𝑙(𝑥).(2.20) Let us take the fermionic 𝑝-adic 𝑞-integral on 𝑝 in (2.20) as follows:𝑝𝑥𝑛𝑑𝜇𝑞(𝑥)=𝑝𝐸𝑛(𝑥)𝑑𝜇𝑞1(𝑥)+2𝑛1𝑙=0𝑛𝑙𝑝𝐸𝑙(𝑥)𝑑𝜇𝑞=(𝑥)𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐻𝑙𝑞1+12𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘𝐻𝑘𝑞1.(2.21) Therefore, by (2.21), we obtain the following theorem.

Theorem 2.6. For 𝑛, one has 𝐻𝑛𝑞1=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐻𝑙𝑞1+12𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘𝐻𝑘𝑞1.(2.22)

By Theorems 2.5 and 2.6, we obtain the following corollary.

Corollary 2.7. For 𝑛, one has 1𝑛+1𝑛𝑙=0𝑙𝑛+1𝑙𝑘=0𝑙𝑘𝐵𝑙𝑘𝐻𝑘𝑞1=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐻𝑙𝑞1+12𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘𝐻𝑘𝑞1.(2.23)

By (1.3), we easily get 𝐸𝑛(𝑥)=(1)𝑛𝐸𝑛(1𝑥).

Thus, we have𝑝𝑥𝑛𝑑𝜇𝑞(𝑥)=(1)𝑛𝑝𝐸𝑛(1𝑥)𝑑𝜇𝑞1(𝑥)+2𝑛1𝑙=0𝑛𝑙(1)𝑙𝑝𝐸𝑙(1𝑥)𝑑𝜇𝑞(𝑥)=(1)𝑛𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝑝(1𝑥)𝑙𝑑𝜇𝑞+1(𝑥)2𝑛1𝑙=0𝑛𝑙(1)𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘𝑝(1𝑥)𝑘𝑑𝜇𝑞=(𝑥)𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙(1)𝑛𝑙𝐻𝑙𝑞1+1,12𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘(1)𝑙𝑘𝐻𝑘𝑞1=,1𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙(1)𝑛𝐻𝑙1(𝑞,2)+2𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘(1)𝑙𝐻𝑘(𝑞,2).(2.24) Therefore, by (2.24), we obtain the following theorem.

Theorem 2.8. For 𝑛, one has 𝐻𝑛𝑞1=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙(1)𝑛𝐻𝑙+1(𝑞,2)2𝑛1𝑙=0𝑛𝑙𝑙𝑘=0𝑙𝑘𝐸𝑙𝑘(1)𝑙𝐻𝑘(𝑞,2).(2.25)

Acknowledgment

The second author was supported by National Research Foundation of Korea Grant funded by the Korean Government 2009-0072514.