Abstract

Kim et al. (2012) introduced an interesting p-adic analogue of the Eulerian polynomials. They studied some identities on the Eulerian polynomials in connection with the Genocchi, Euler, and tangent numbers. In this paper, by applying the symmetry of the fermionic p-adic q-integral on 𝑝, defined by Kim (2008), we show a symmetric relation between the q-extension of the alternating sum of integer powers and the Eulerian polynomials.

1. Introduction

The Eulerian polynomials 𝐴𝑛(𝑡),𝑛=0,1,, which can be defined by the generating function 1𝑡𝑒(𝑡1)𝑥=𝑡𝑛=0𝐴𝑛(𝑥𝑡)𝑛,𝑛!(1.1) have numerous important applications in number theory, combinatorics, and numerical analysis, among other areas. From (1.1), we note that (𝐴(𝑡)+(𝑡1))𝑛𝑡𝐴𝑛(𝑡)=(1𝑡)𝛿0,𝑛,(1.2) where 𝛿𝑛,𝑘 is the Kronecker symbol (see [1]). Thus far, few recurrences for the Eulerian polynomials other than (1.2) have been reported in the literature. Other recurrences are of importance as they might reveal new aspects and properties of the Eulerian polynomials, and they can help simplify the proofs of known properties. For more important properties, see, for instance, [1] or [2].

Let 𝑝 be a fixed odd prime number. Let 𝑝,𝑝, and 𝑝 be the ring of 𝑝-adic integers, the field of 𝑝-adic numbers, and the completion of the algebraic closure of 𝑝, respectively. Let ||𝑝 be the 𝑝-adic valuation on , where |𝑝|𝑝=𝑝1. The extended valuation on 𝑝 is denoted by the same symbol ||𝑝. Let 𝑞 be an indeterminate, where |1𝑞|𝑝<1. Then, the 𝑞-number is defined by [𝑥]𝑞=1𝑞𝑥,[𝑥]1𝑞𝑞=1(𝑞)𝑥.1+𝑞(1.3)

For a uniformly (or strictly) differentiable function 𝑓𝑝𝑝 (see [1, 36]), the fermionic 𝑝-adic 𝑞-integral on 𝑝 is defined by 𝐼𝑞(𝑓)=𝑝𝑓(𝑥)𝑑𝜇𝑞(𝑥)=lim𝑁1𝑝𝑁𝑝𝑞𝑁1𝑥=0𝑓(𝑥)(𝑞)𝑥.(1.4) Then, it is easy to see that 1𝑞𝐼1/𝑞𝑓1+𝐼1/𝑞[2](𝑓)=1/𝑞𝑓(0),(1.5) where 𝑓1(𝑥)=𝑓(𝑥+1).

By using the same method as that described in [1], and applying (1.5) to 𝑓, where 𝑓(𝑥)=𝑞(1𝜔)𝑥𝑒𝑥(1+𝑞)𝜔𝑡(1.6) for 𝜔>0, we consider the generalized Eulerian polynomials on 𝑝 by using the fermionic 𝑝-adic 𝑞-integral on 𝑝 as follows: 𝑝𝑞(1𝜔)𝑥𝑒𝑥(1+𝑞)𝜔𝑡𝑑𝜇1/𝑞(𝑥)=1+𝑞𝑞1𝜔𝑒(1+𝑞)𝜔𝑡=+𝑞𝑛=0𝐴𝑛𝑡(𝑞,𝜔)𝑛.𝑛!(1.7) By expanding the Taylor series on the left-hand side of (1.7) and comparing the coefficients of the terms 𝑡𝑛/𝑛!, we get 𝑝𝑞(1𝜔)𝑥𝑥𝑛𝑑𝜇1/𝑞(𝑥)=(1)𝑛𝜔𝑛(1+𝑞)𝑛𝐴𝑛(𝑞,𝜔).(1.8)

We note that, by substituting 𝜔=1 into (1.8), 𝐴𝑛(𝑞,1)=𝐴𝑛(𝑞)=(1)𝑛(1+𝑞)𝑛𝑝𝑥𝑛𝑑𝜇1/𝑞(𝑥)(1.9) is the Witt's formula for the Eulerian polynomials in [1, Theorem 1]. Recently, Kim et al. [1] investigated new properties of the Eulerian polynomials 𝐴𝑛(𝑞) at 𝑞=1 associated with the Genocchi, Euler, and tangent numbers.

Let 𝑇𝑘,1/𝑞(𝑛) denote the 𝑞-extension of the alternating sum of integer powers, namely, 𝑇𝑘,1/𝑞(𝑛)=𝑛𝑖=0(1)𝑖𝑖𝑘𝑞𝑖=0𝑘𝑞01𝑘𝑞1++(1)𝑛𝑛𝑘𝑞𝑛,(1.10) where 00=1. If 𝑞1, 𝑇𝑘,𝑞(𝑛)𝑇𝑘(𝑛)=𝑛𝑖=0(1)𝑖𝑖𝑘 is the alternating sum of integer powers (see [4]). In particular, we have 𝑇𝑘,1/𝑞(0)=1,for𝑘=0,0,for𝑘>0.(1.11)

Let 𝜔1,𝜔2 be any positive odd integers. Our main result of symmetry between the 𝑞-extension of the alternating sum of integer powers and the Eulerian polynomials is given in the following theorem, which is symmetric in 𝜔1 and 𝜔2.

Theorem 1.1. Let 𝜔1,𝜔2 be any positive odd integers and 𝑛0. Then, one has 𝑛𝑖=0𝑛𝑖𝐴𝑖𝑞,𝜔1𝑇𝑛𝑖,𝑞2𝜔𝜔1𝜔12𝑛𝑖(1𝑞)𝑛𝑖=𝑛𝑖=0𝑛𝑖𝐴𝑖𝑞,𝜔2𝑇𝑛𝑖,𝑞1𝜔𝜔2𝜔11𝑛𝑖(1𝑞)𝑛𝑖.(1.12)

Observe that Theorem 1.1 can be obtained by the same method as that described in [4]. If 𝑞=1, Theorem 1.1 reduces to the form stated in the remark in [4, page 1275].

Using (1.11), if we take 𝜔2=1 in Theorem 1.1, we obtain the following corollary.

Corollary 1.2. Let 𝜔1 be any positive odd integer and 𝑛0. Then, one has 𝐴𝑛(𝑞)=𝑛𝑖=0𝑛𝑖𝐴𝑖𝑞,𝜔1𝑇𝑛𝑖,𝑞1𝜔11(1𝑞)𝑛𝑖.(1.13)

2. Proof of Theorem 1.1

For the proof of Theorem 1.1, we will need the following two identities (see (2.4) and (2.5)) related to the Eulerian polynomials and the 𝑞-extension of the alternating sum of integer powers.

Let 𝜔1,𝜔2 be any positive odd integers. From (1.7), we obtain 𝑝𝑞(1𝜔1)𝑥𝑒𝑥(1+𝑞)𝜔1𝑡𝑑𝜇1/𝑞(𝑥)𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=𝑞(𝑥)1+𝜔1𝑒(1+𝑞)𝜔1𝑡𝜔21+𝑞𝜔1𝑒(1+𝑞)𝜔1𝑡.(2.1) This has an interesting 𝑝-adic analytic interpretation, which we shall discuss below (see Remark 2.1). It is easy to see that the right-hand side of (2.1) can be written as 𝑞1+𝜔1𝑒(1+𝑞)𝜔1𝑡𝜔21+𝑞𝜔1𝑒(1+𝑞)𝜔1𝑡=𝜔21𝑖=0(1)𝑖𝑞𝜔1𝑖𝑒(1+𝑞)𝜔1𝑡𝑖=𝑘=0𝜔21𝑖=0(1)𝑖𝑖𝑘(𝑞𝜔1)𝑖𝜔𝑘1(1)𝑘(1+𝑞)𝑘𝑡𝑘.𝑘!(2.2) In (1.10), let 𝑞=𝑞𝜔1. The left-hand, right-hand side, by definition, becomes 𝑞1+𝜔1𝑒(1+𝑞)𝜔1𝑡𝜔21+𝑞𝜔1𝑒(1+𝑞)𝜔1𝑡=𝑘=0𝑇𝑘,𝑞1𝜔𝜔2𝜔1𝑘1(1)𝑘(1+𝑞)𝑘𝑡𝑘.𝑘!(2.3) A comparison of (2.1) and (2.3) yields the identity 𝑝𝑞(1𝜔1)𝑥𝑒𝑥(1+𝑞)𝜔1𝑡𝑑𝜇1/𝑞(𝑥)𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=(𝑥)𝑘=0𝑇𝑘,𝑞1𝜔𝜔2𝜔1𝑘1(1)𝑘(1+𝑞)𝑘𝑡𝑘.𝑘!(2.4) By slightly modifying the derivation of (2.4), we can obtain the following identity: 𝑝𝑞(1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔2𝑡𝑑𝜇1/𝑞(𝑥)𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=(𝑥)𝑘=0𝑇𝑘,𝑞2𝜔𝜔1𝜔1𝑘2(1)𝑘(1+𝑞)𝑘𝑡𝑘.𝑘!(2.5)

Remark 2.1. The derivations of identities are based on the fermionic 𝑝-adic 𝑞-integral expression of the generating function for the Eulerian polynomials in (1.7) and the quotient of integrals in (2.4), (2.5) that can be expressed as the exponential generating function for the 𝑞-extension of the alternating sum of integer powers.
Observe that similar identities related to the Eulerian polynomials and the 𝑞-extension of the alternating sum of integer powers in (2.4) and (2.5) can be found, for instance, in [3, (1.8)], [4, (21)], and [6, Theorem 4].

Proof of Theorem 1.1. Let 𝜔1,𝜔2 be any positive odd integers. Using the iterated fermionic 𝑝-adic 𝑞-integral on 𝑝 and (1.7), we have 𝑝𝑞(1𝜔1)𝑥1+(1𝜔2)𝑥2𝑒(1+𝑞)(𝜔1𝑥1+𝜔2𝑥2)𝑡𝑑𝜇1/𝑞𝑥1𝑑𝜇1/𝑞𝑥2𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=[2](𝑥)1/𝑞𝑞𝜔1𝜔2𝑒(1+𝑞)𝜔1𝜔2𝑡+1𝑞𝜔1𝑒(1+𝑞)𝜔1𝑡𝑞+1𝜔2𝑒(1+𝑞)𝜔2𝑡.+1(2.6) Now, we put 𝐼=𝑝𝑞(1𝜔1)𝑥1+(1𝜔2)𝑥2𝑒(1+𝑞)(𝜔1𝑥1+𝜔2𝑥2)𝑡𝑑𝜇1/𝑞𝑥1𝑑𝜇1/𝑞𝑥2𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞.(𝑥)(2.7) From (1.7) and (2.5), we see that 𝐼=𝑝𝑞(1𝜔1)𝑥1𝑒(1+𝑞)(𝜔1𝑥1)𝑡𝑑𝜇1/𝑞𝑥1×𝑝𝑞(1𝜔2)𝑥2𝑒(1+𝑞)(𝜔2𝑥2)𝑡𝑑𝜇1/𝑞𝑥2𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=(𝑥)𝑘=0𝐴𝑘𝑞,𝜔1𝑡𝑘×𝑘!𝑙=0𝑇𝑙,𝑞2𝜔𝜔1𝜔1𝑙2(1)𝑙(1+𝑞)𝑙𝑡𝑙=𝑙!𝑛=0𝑛𝑖=0(1)𝑛𝑖𝑛𝑖𝐴𝑖𝑞,𝜔1𝑇𝑛𝑖,𝑞2𝜔𝜔1𝜔12𝑛𝑖(1+𝑞)𝑛𝑖𝑡𝑛.𝑛!(2.8) On the other hand, from (1.7) and (2.4), we have 𝐼=𝑝𝑞(1𝜔2)𝑥2𝑒(1+𝑞)(𝜔2𝑥2)𝑡𝑑𝜇1/𝑞𝑥2×𝑝𝑞(1𝜔1)𝑥1𝑒(1+𝑞)(𝜔1𝑥1)𝑡𝑑𝜇1/𝑞𝑥1𝑝𝑞(1𝜔1𝜔2)𝑥𝑒𝑥(1+𝑞)𝜔1𝜔2𝑡𝑑𝜇1/𝑞=(𝑥)𝑘=0𝐴𝑘𝑞,𝜔2𝑡𝑘×𝑘!𝑙=0𝑇𝑙,𝑞1𝜔𝜔2𝜔1𝑙1(1)𝑙(1+𝑞)𝑙𝑡𝑙=𝑙!𝑛=0𝑛𝑖=0(1)𝑛𝑖𝑛𝑖𝐴𝑖𝑞,𝜔2𝑇𝑛𝑖,𝑞1𝜔𝜔2𝜔11𝑛𝑖(1+𝑞)𝑛𝑖𝑡𝑛.𝑛!(2.9) By comparing the coefficients on both sides of (2.8) and (2.9), we obtain the result in Theorem 1.1.

3. Concluding Remarks

Note that many other interesting symmetric properties for the Euler, Genocchi, and tangent numbers are derivable as corollaries of the results presented herein. For instance, considering [1, (5)], 𝐴𝑛(1,𝜔)=(2𝜔)𝑛𝐸𝑛(𝑛0),(3.1) where 𝐸𝑛 denotes the 𝑛th Euler number defined by 𝐸𝑛=𝐸𝑛(0), and the Euler polynomials are defined by the generating function 2𝑒𝑡𝑒+1𝑥𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛,𝑛!(3.2) and on putting 𝑞=1 in Theorem 1.1 and Corollary 1.2, we obtain 𝑛𝑖=0𝑛𝑖𝜔𝑖1𝐸𝑖𝑇𝑛𝑖𝜔1𝜔12𝑛𝑖=𝑛𝑖=0𝑛𝑖𝜔𝑖2𝐸𝑖𝑇𝑛𝑖𝜔2𝜔11𝑛𝑖,(3.3)𝐸𝑛=𝑛𝑖=0𝑛𝑖𝜔𝑖1𝐸𝑖𝑇𝑛𝑖𝜔1.1(3.4) These formulae are valid for any positive odd integers 𝜔1,𝜔2. The Genocchi numbers 𝐺𝑛 may be defined by the generating function 2𝑡𝑒𝑡=+1𝑛=0𝐺𝑛𝑡𝑛,𝑛!(3.5) which have several combinatorial interpretations in terms of certain surjective maps on finite sets. The well-known identity 𝐺𝑛=2(12𝑛)𝐵𝑛(3.6) shows the relation between the Genocchi and the Bernoulli numbers. It follows from (3.6) and the Staudt-Clausen theorem that the Genocchi numbers are integers. It is easy to see that 𝐺𝑛=2𝑛𝐸2𝑛1(𝑛1),(3.7) and from (3.2), (3.5) we deduce that 𝐸𝑛(𝑥)=𝑛𝑘=0𝑛𝑘𝐺𝑘+1𝑥𝑘+1𝑛𝑘.(3.8) It is well known that the tangent coefficients (or numbers) 𝑇𝑛, defined by tan𝑡=𝑛=1(1)𝑛1𝑇2𝑛𝑡2𝑛1,(2𝑛1)!(3.9) are closely related to the Bernoulli numbers, that is, (see [1]) 𝑇𝑛=2𝑛(2𝑛𝐵1)𝑛𝑛.(3.10) Ramanujan ([7, page 5]) observed that 2𝑛(2𝑛1)𝐵𝑛/𝑛 and, therefore, the tangent coefficients, are integers for 𝑛1. From (3.3), (3.6), (3.7), and (3.10), the obtained symmetric formulae involve the Bernoulli, Genocchi, and tangent numbers (see [1]).

Acknowledgment

This work was supported by the Kyungnam University Foundation grant, 2012.