Some Properties of a Sequence Similar to Generalized Euler Numbers

Abstract

We introduce the sequence {𝑈_𝑛^(𝑥)} given by generating function (1/(𝑒^𝑡+𝑒^−𝑡−1))^𝑥=∑^∞_𝑛=0𝑈_𝑛^(𝑥)(𝑡^𝑛/𝑛!)(|𝑡|<(1/3)𝜋,1^𝑥∶=1) and establish some explicit formulas for the sequence {𝑈_𝑛^(𝑥)}. Several identities involving the sequence {𝑈_𝑛^(𝑥)}, Stirling numbers, Euler polynomials, and the central factorial numbers are also presented.

1. Introduction and Definitions

For a real or complex parameter 𝛼, the generalized Euler polynomials 𝐸_𝑛^(𝛼)(𝑥) are defined by the following generating function (see [1–4]) 2𝑒^𝑡+1^𝛼𝑒^𝑥𝑡=^∞_𝑛=0𝐸_𝑛^(𝛼)(𝑥)𝑡^𝑛|𝑡|𝑛!<𝜋,1^𝛼∶=1. (1) Obviously, we have 𝐸_𝑛⁽¹⁾(𝑥)=𝐸_𝑛(𝑥)𝑛∈ℕ₀∶=ℕ∪{0}, (2) in terms of the classical Euler polynomials 𝐸_𝑛(𝑥), ℕ being the set of positive integers. The classical Euler numbers 𝐸_𝑛 are given by the following: 𝐸_𝑛=2^𝑛𝐸_𝑛12𝑛∈ℕ₀. (3)

The so-called the generalized Euler numbers 𝐸^(𝑥)_2𝑛 are defined by (see [3, 5]) 2𝑒^𝑡+𝑒^−𝑡𝑥=^∞_𝑛=0𝐸^(𝑥)_2𝑛𝑡^2𝑛()!|𝑡|<𝜋2𝑛2,1^𝑥∶=1. (4) In fact, 𝐸^(𝑘)_2𝑛(𝑘∈ℤ) are the Euler numbers of order 𝑘, ℤ being the set of integers. The numbers 𝐸⁽¹⁾_2𝑛=𝐸_2𝑛 are the ordinary Euler numbers.

Zhi-Hong Sun introduces the sequence {𝑈_𝑛} similar to Euler numbers as follows (see [6, 7]): 𝑈₀=1,𝑈_𝑛=−2^[𝑛/2]_𝑘=1𝑛𝑈2𝑘^𝑛−2𝑘,(𝑛≥1), (5) where (and in what follows) [𝑥] is the greatest integer not exceeding 𝑥.

Clearly, 𝑈_2𝑛−1=0 for 𝑛≥1. The first few values of 𝑈_2𝑛 are shown below 𝑈₂=−2,𝑈₄=22,𝑈₆=−602,𝑈₈𝑈=30742,₁₀=−2523002,𝑈₁₂=303692662. (6)

The sequence {𝑈_𝑛} is related to the classical Bernoulli polynomials 𝐵_𝑛(𝑥) (see [8–11]) and the classical Euler polynomials 𝐸_𝑛(𝑥). Zhi-Hong Sun gets the generating function of {𝑈_𝑛} and deduces many identities involving {𝑈_𝑛}. As example, (see [6]), 1𝑒^𝑡+𝑒^−𝑡=−1^∞_𝑛=0𝑈_𝑛𝑡^𝑛=𝑛!_∞𝑛=0𝑈_2𝑛𝑡^2𝑛()!|𝑡|<12𝑛3𝜋,1=2cos𝑡−1^∞_𝑛=0(−1)^𝑛𝑈_2𝑛𝑡^2𝑛()!|𝑡|<12𝑛3𝜋,𝑈_2𝑛=3^2𝑛𝐸_2𝑛13. (7)(8)(9)

Similarly, we can define the generalized sequence {𝑈_𝑛^(𝑥)}. For a real or complex parameter 𝑥, the generalized sequence {𝑈_𝑛^(𝑥)} is defined by the following generating function: 1𝑒^𝑡+𝑒^−𝑡−1^𝑥=^∞_𝑛=0𝑈_𝑛^(𝑥)𝑡^𝑛|𝑡|<1𝑛!3𝜋,1^𝑥∶=1. (10) Obviously, 𝑈₀^(𝑥)=1,𝑈_𝑛⁽¹⁾=𝑈_𝑛(𝑛∈ℕ). (11) By using (10), we can obtain 𝑈_𝑛^(𝑘)=𝑛!^{(𝑣1+⋯+𝑣𝑘=𝑛)}_{𝑣1,…,𝑣𝑘∈ℕ0}𝑈_𝑣1⋯𝑈_𝑣𝑘𝑣₁!⋯𝑣_𝑘!(𝑘∈ℕ). (12)

We now return to the Stirling numbers 𝑠(𝑛,𝑘) of the first kind, which are usually defined by (see [2, 5, 8, 11, 12]) 𝑥(𝑥−1)(𝑥−2)⋯(𝑥−𝑛+1)=^𝑛_𝑘=0𝑠(𝑛,𝑘)𝑥^𝑘 (13) or by the following generating function: log(1+𝑥)^𝑘=𝑘!^∞_𝑛=𝑘𝑠()𝑥𝑛,𝑘^𝑛.𝑛! (14)

It follows from (13) or (14) that 𝑠(𝑛,𝑘)=𝑠(𝑛−1,𝑘−1)−(𝑛−1)𝑠(𝑛−1,𝑘) (15) and that 𝑠(𝑛,0)=0(𝑛∈ℕ),𝑠(𝑛,𝑛)=1𝑛∈ℕ₀,𝑠(𝑛,1)=(−1)^𝑛−1()!(),𝑠()().𝑛−1𝑛∈ℕ𝑛,𝑘=0𝑘>𝑛or𝑘<0 (16)

The central factorial numbers 𝑇(𝑛,𝑘) are given by the following expansion formula (see [3, 5, 13]): 𝑥^𝑛=^𝑛_𝑘=0𝑇(𝑛,𝑘)𝑥𝑥−1²×𝑥−2²⋯𝑥−(𝑘−1)² (17) or by means of the generating function 𝑒^𝑥+𝑒^−𝑥−2^𝑘=(2𝑘)!^∞_𝑛=𝑘𝑇()𝑥𝑛,𝑘^2𝑛()!.2𝑛 (18)

It follows from (17) or (18) that 𝑇(𝑛,𝑘)=𝑇(𝑛−1,𝑘−1)+𝑘²𝑇(𝑛−1,𝑘), (19) with 𝑇()()(),𝑇()().0,0=1,𝑇𝑛,0=0𝑛∈ℕ𝑛,1=1𝑛∈ℕ (20) We also find from (18) that 𝑇()=1𝑛,244^𝑛−1,𝑇()=9−1𝑛,3^𝑛−4360^𝑛+160().24𝑛∈ℕ (21)

The main purpose of this paper is to prove some formulas for the generalized sequence {𝑈_𝑛^(𝑥)} and 𝐸_𝑛(𝑥). Some identities involving the sequence {𝑈_𝑛^(𝑥)}, Stirling numbers 𝑠(𝑛,𝑘), and the central factorial numbers 𝑇(𝑛,𝑘) are deduced.

2. Main Results

Theorem 1. Let 𝑛≥𝑘(𝑛,𝑘∈ℕ) and 𝑞(𝑛,𝑘)=(−1)^𝑘𝑛_𝑗=𝑘2𝑗!𝑇𝑠.𝑗!𝑛,𝑗𝑗,𝑘 (22) Then, 𝑈^(𝑥)_2𝑛=^𝑛_𝑘=1𝑞(𝑛,𝑘)𝑥^𝑘. (23)

Remark 2. By (15), (19), (20), and Theorem 1, we know that 𝑈^(𝑥)_2𝑛 is a polynomial of 𝑥 with integral coefficients. For example, by setting 𝑛=1,2,3,4 in Theorem 1, we get 𝑈₂^(𝑥)=−2𝑥,𝑈₄^(𝑥)=10𝑥+12𝑥²,𝑈₆^(𝑥)=−182𝑥−300𝑥²−120𝑥³,𝑈₈^(𝑥)=6970𝑥+13692𝑥²+8400𝑥³+1680𝑥⁴. (24)

Taking 𝑥=1 in Theorem 1, we can obtain the following.

Corollary 3. Let 𝑛∈ℕ. Then, 𝑈_2𝑛=^𝑛_𝑗=0(−1)^𝑗2𝑗!𝑇𝑛,𝑗. (25)

From Corollary 3, we may immediately deduce the following results.

Corollary 4. Let 𝑛∈ℕ. Then, 𝑈_2𝑛(),𝑈≡−2mod24_2𝑛(),𝑈≡−2+24𝑇𝑛,2)(mod720_2𝑛≡−2+24𝑇(𝑛,2)−720𝑇(𝑛,3)(mod40320). (26)

Theorem 5. Let 𝑛≥𝑘(𝑛,𝑘∈ℕ). Then, 𝑈_2𝑛=^𝑛_𝑘=1𝑞(),𝑈𝑛,𝑘_2𝑛=2^[𝑛/2]_𝑘=1𝑞(𝑛,2𝑘)−2=2^{[(𝑛−1)/2]}_𝑘=1𝑞(𝑛,2𝑘+1)+2. (27)

Theorem 6. Let 𝑛≥𝑘(𝑛,𝑘∈ℕ). Suppose also that 𝑞(𝑛,𝑘) is defined by (22). Then, 𝑘!𝑞(𝑛,𝑘)=(2𝑛)!3^2𝑛−𝑘×^{(𝑣1+⋯+𝑣𝑘=𝑛)}_{𝑣1,…,𝑣𝑘∈ℕ}𝐸_2𝑣1−1(0)−𝐸_2𝑣1−123⋯𝐸_{2𝑣𝑘−1}(0)−𝐸_{2𝑣𝑘−1}23×(2𝑣₁)!⋯(2𝑣_𝑘)!⁻¹. (28)

Theorem 7. Let 𝑛∈ℕ. Then, −2^𝑛−1_𝑘=0𝑈2𝑛−12𝑘^2𝑘=3^2𝑛−1𝐸_2𝑛−1(0)−𝐸_2𝑛−123. (29)

Theorem 8. Let 𝑛∈ℕ. Then, 𝑈_𝑛+1=^𝑛−1_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1−2^𝑛−𝑘𝑈_𝑘. (30)

Theorem 9. Let 𝑛∈ℕ₀. Then, ^∞_𝑛=01()!𝑈𝑛+1^𝑛=1√3√log2𝑒−1−322−√3𝑒−5+3√3. (31)

3. Proofs of Theorems

Proof of Theorem 1. By (10), (13), and (18), we have ^∞_𝑛=0𝑈^(𝑥)_2𝑛𝑡^2𝑛()!=12𝑛𝑒^𝑡+𝑒^−𝑡−1^𝑥=11+(𝑒^𝑡+𝑒^−𝑡−2)^𝑥=^∞_𝑗=0(−1)^𝑗𝑗𝑒𝑥+𝑗−1_𝑡+𝑒^−𝑡−2^𝑗=^∞_𝑗=0(−1)^𝑗𝑗!𝑥+𝑗−12𝑗^∞_𝑛=𝑗𝑇𝑡𝑛,𝑗^2𝑛()!=2𝑛_∞𝑛=0𝑡^2𝑛(2𝑛)!^𝑛_𝑗=0(−1)^𝑗!𝑗2𝑗𝑇,𝑥+𝑗−1𝑛,𝑗 (32) which readily yields 𝑈^(𝑥)_2𝑛=^𝑛_𝑗=0(−1)^𝑗!𝑗𝑇=2𝑗𝑥+𝑗−1𝑛,𝑗_𝑛𝑗=0(−1)^𝑗12𝑗!𝑇𝑛,𝑗𝑥()⋯=𝑗!𝑥+1𝑥+𝑗−1_𝑛𝑗=02𝑗!𝑇𝑗!𝑛,𝑗^𝑗_𝑘=1(−1)^𝑘𝑠𝑗,𝑘𝑥^𝑘=^𝑛_𝑘=1(−1)^𝑘𝑛_𝑗=𝑘2𝑗!𝑇𝑠𝑥𝑗!𝑛,𝑗𝑗,𝑘^𝑘=^𝑛_𝑘=1𝑞(𝑛,𝑘)𝑥^𝑘. (33) This completes the proof of Theorem 1.

Proof of Theorem 5. By (10), we have ^∞_𝑛=0𝑈⁽⁻¹⁾_2𝑛𝑡^2𝑛(2𝑛)!=𝑒^𝑡+𝑒^−𝑡−1=2^∞_𝑛=0𝑡^2𝑛(2𝑛)!−1, (34) and 𝑈₀^(𝑥)=1, thus ^∞_𝑛=1𝑈⁽⁻¹⁾_2𝑛𝑡^2𝑛(2𝑛)!=𝑒^𝑡+𝑒^−𝑡−1=2^∞_𝑛=1𝑡^2𝑛()!.2𝑛 (35) By Theorem 1 and comparing the coefficient of 𝑡^2𝑛/(2𝑛)! on both sides of (35), we get ^𝑛_𝑘=1𝑞(𝑛,𝑘)(−1)^𝑘=𝑈⁽⁻¹⁾_2𝑛=2. (36) Again, by taking 𝑥=1 in Theorem 1, we have ^𝑛_𝑘=1𝑞(𝑛,𝑘)=𝑈_2𝑛. (37) By (36) and (37), we immediately obtain (27). This completes the proof of Theorem 5.

Proof of Theorem 6. By applying Theorem 1, we have ()=𝑑𝑘!𝑞𝑛,𝑘^𝑘𝑑𝑥^𝑘𝑈_𝑛^(𝑥)_𝑥=0. (38) On the other hand, it follows from (10) that ^∞_𝑛=𝑘𝑑^𝑘𝑑𝑥^𝑘𝑈_𝑛^(𝑥)_𝑥=0𝑡^2𝑛()!=12𝑛log𝑒^𝑡+𝑒^−𝑡−1^𝑘. (39) By using (38) and (39), we find that 𝑘!^∞_𝑛=𝑘𝑞()𝑡𝑛,𝑘^2𝑛()!=12𝑛log𝑒^𝑡+𝑒^−𝑡−1^𝑘. (40) We now note that 𝑑1𝑑𝑡log𝑒^𝑡+𝑒^−𝑡=𝑒−1_−𝑡−𝑒^𝑡𝑒^𝑡+𝑒^−𝑡=𝑒−1_−𝑡−𝑒^𝑡22𝑒^𝑡𝑒^3𝑡++12𝑒^−𝑡𝑒^−3𝑡=1+122𝑒^3𝑡−2+1𝑒^−3𝑡−+12𝑒^2𝑡𝑒^3𝑡−+12𝑒^−2𝑡𝑒^−3𝑡=1+12^∞_𝑛=0𝐸_𝑛(0)(3𝑡)^𝑛−𝑛!^∞_𝑛=0𝐸_𝑛(0)(−3𝑡)^𝑛−1𝑛!2^∞_𝑛=0𝐸_𝑛23(3𝑡)^𝑛−𝑛!^∞_𝑛=0𝐸_𝑛23(−3𝑡)^𝑛=𝑛!_∞𝑛=03^2𝑛+1𝐸_2𝑛+1(0)−𝐸_2𝑛+123𝑡^2𝑛+1()!.2𝑛+1 (41) Hence, 1log𝑒^𝑡+𝑒^−𝑡=−1^∞_𝑛=03^2𝑛+1𝐸_2𝑛+1(0)−𝐸_2𝑛+123𝑡^2𝑛+2()!=2𝑛+2_∞𝑛=13^2𝑛−1𝐸_2𝑛−1(0)−𝐸_2𝑛−123𝑡^2𝑛(2𝑛)! (42) yields 𝑘!^∞_𝑛=𝑘𝑞()𝑡𝑛,𝑘^2𝑛()!=2𝑛_∞𝑛=13^2𝑛−1𝐸_2𝑛−1(0)−𝐸_2𝑛−123𝑡^2𝑛(2𝑛)!^𝑘=^∞_𝑛=𝑘𝑡^2𝑛()!()2𝑛2𝑛!3^2𝑛−𝑘×^{(𝑣1+⋯+𝑣𝑘=𝑛)}_{𝑣1,…,𝑣𝑘∈ℕ}𝐸_2𝑣1−1(0)−𝐸_2𝑣1−123⋯𝐸_{2𝑣𝑘−1}(0)−𝐸_{2𝑣𝑘−1}23×(2𝑣₁)!⋯(2𝑣_𝑘)!⁻¹. (43) Comparing the coefficient of 𝑡^2𝑛/(2𝑛)! on both sides of (43), we immediately get (28). This completes the proof of Theorem 6.

Proof of Theorem 7. Consider 𝑑1𝑑𝑡log𝑒^𝑡+𝑒^−𝑡=𝑒−1^−𝑡−𝑒^𝑡𝑒^𝑡+𝑒^−𝑡=−1_∞𝑛=0𝑈_2𝑛𝑡^2𝑛(2𝑛)!−2^∞_𝑛=0𝑡^2𝑛+1(2𝑛+1)!=−2^∞𝑛_𝑛=0𝑘=0𝑈2𝑛+12𝑘^2𝑘𝑡^2𝑛+1()!.2𝑛+1 (44) Thus, 1log𝑒^𝑡+𝑒^−𝑡−1=−2^∞_𝑛=1^𝑛−1_𝑘=0𝑈2𝑛−12𝑘^2𝑘𝑡^2𝑛(2𝑛)!. (45) By (42) and (45) we obtain (29). This completes the proof of Theorem 7.

Proof of Theorem 8. By using (7), we have ^∞_𝑛=1𝑈_𝑛𝑡^𝑛−1()!=𝑒𝑛−1^−𝑡−𝑒^𝑡(𝑒^𝑡+𝑒^−𝑡−1)². (46) Thus 𝑒^2𝑡−𝑒^𝑡+1^∞_𝑛=1𝑈_𝑛𝑡^𝑛−1()!=𝑛−11−𝑒^2𝑡∞_𝑛=0𝑈_𝑛𝑡^𝑛,𝑛!_∞𝑛=02^𝑛𝑡−1^𝑛𝑛!^∞_𝑛=0𝑈_𝑛+1𝑡^𝑛+𝑛!^∞_𝑛=0𝑈_𝑛+1𝑡^𝑛=𝑛!_∞𝑛=0𝑈_𝑛𝑡^𝑛−𝑛!^∞_𝑛=02^𝑛𝑡^𝑛𝑛!^∞_𝑛=0𝑈_𝑛𝑡^𝑛.𝑛! (47) That is, ^∞𝑛_𝑛=0𝑘=0𝑛𝑘2^𝑛−𝑘−1𝑈_𝑘+1𝑡^𝑛+𝑛!^∞_𝑛=0𝑈_𝑛+1𝑡^𝑛=𝑛!_∞𝑛=0𝑈_𝑛𝑡^𝑛−𝑛!^∞𝑛_𝑛=0𝑘=0𝑛𝑘2^𝑛−𝑘𝑈_𝑘𝑡^𝑛.𝑛! (48) Comparing the coefficient of 𝑡^𝑛/𝑛! on both sides of (48), we get the following: 𝑈_𝑛+1−𝑈_𝑛=^𝑛_𝑘=0𝑛𝑘1−2^𝑛−𝑘𝑈_𝑘+1−2^𝑛−𝑘𝑈_𝑘. (49) By (49) we immediately obtain (30). This completes the proof of Theorem 8.

Proof of Theorem 9. By integrating (7) with respect to 𝑡 from 0 to 1, we have ^∞_𝑛=01()!𝑈𝑛+1^𝑛=¹₀1𝑒^𝑡+𝑒^−𝑡=−1𝑑𝑡₁₀1𝑒^2𝑡−𝑒^𝑡+1𝑑𝑒^𝑡=^𝑒₁1𝑥²−𝑥+1𝑑𝑥. (50) By (50) and ∫(1/(𝑎𝑥²+𝑏𝑥+𝑐))𝑑𝑥=(1/𝑏²−4𝑎𝑐)log|(2𝑎𝑥+𝑏−𝑏²−4𝑎𝑐)/(2𝑎𝑥+𝑏+𝑏²−4𝑎𝑐)|+𝑐 (𝑐 is constant), we have (31). This completes the proof of Theorem 9.

Acknowledgments

This work is partly supported by the Social Science Foundation (no. 2012YB03) of Huizhou University and the Key Discipline Foundation (no. JG2011019) of Huizhou University.