Abstract

we establish a recurrence formula for 𝐷 numbers 𝐷(2𝑛−1)2𝑛. A generating function for 𝐷 numbers 𝐷(2𝑛−1)2𝑛 is also presented.

1. Introduction and Results

The Bernoulli polynomials 𝐵𝑛(𝑘)(𝑥) of order 𝑘, for any integer 𝑘, may be defined by (see [1–4])

𝑡𝑒𝑡−1𝑘𝑒𝑥𝑡=∞𝑛=0𝐵𝑛(𝑘)(𝑡𝑥)𝑛𝑛!,|𝑡|<2𝜋.(1.1)

The numbers 𝐵𝑛(𝑘)=𝐵𝑛(𝑘)(0) are the Bernoulli numbers of order 𝑘, 𝐵𝑛(1)=𝐵𝑛 are the ordinary Bernoulli numbers (see [2, 5]). By (1.1), we can get (see [4, page 145])

𝑑𝐵𝑑𝑥𝑛(𝑘)(𝑥)=𝑛𝐵(𝑘)𝑛−1(𝑥),(1.2)𝐵𝑛(𝑘+1)(𝑥)=𝑘−𝑛𝑘𝐵𝑛(𝑘)𝑛(𝑥)+(𝑥−𝑘)𝑘𝐵(𝑘)𝑛−1(𝑥),(1.3)𝐵𝑛(𝑘+1)(𝑥+1)=𝑛𝑥𝑘𝐵(𝑘)𝑛−1(𝑥)−𝑛−𝑘𝑘𝐵𝑛(𝑘)(𝑥),(1.4) where 𝑛∈ℕ, with ℕ being the set of positive integers.

The numbers 𝐵𝑛(𝑛) are called the Nörlund numbers (see [2, 4, 6]). A generating function for the Nörlund numbers 𝐵𝑛(𝑛) is (see [4, page 150])

𝑡=(1+𝑡)log(1+𝑡)∞𝑛=0𝐵𝑛(𝑛)𝑡𝑛.𝑛!(1.5)

The 𝐷 numbers 𝐷(𝑘)2𝑛 may be defined by (see [4, 7, 8])

(𝑡csc𝑡)𝑘=∞𝑛=0(−1)𝑛𝐷(𝑘)2𝑛𝑡2𝑛(2𝑛)!,|𝑡|<𝜋.(1.6)

By (1.1), (1.6), and note that csc𝑡=2𝑖/(𝑒𝑖𝑡−𝑒−𝑖𝑡) (where 𝑖2=−1), we can get

𝐷(𝑘)2𝑛=4𝑛𝐵(𝑘)2𝑛𝑘2.(1.7)

Taking 𝑘=1,2 in (1.7), and note that 𝐵(1)2𝑛(1/2)=(21−2𝑛−1)𝐵2𝑛, 𝐵(2)2𝑛(1)=(1−2𝑛)𝐵2𝑛 (see [4, page 22, page 145]), we have

𝐷(1)2𝑛=2−22𝑛𝐵2𝑛,𝐷(2)2𝑛=4𝑛(1−2𝑛)𝐵2𝑛.(1.8)

The 𝐷 numbers 𝐷(𝑘)2𝑛 satisfy the recurrence relation (see [7])

𝐷(𝑘)2𝑛=(2𝑛−𝑘+2)(2𝑛−𝑘+1)(𝐷𝑘−2)(𝑘−1)(𝑘−2)2𝑛−2𝑛(2𝑛−1)(𝑘−2)𝐷𝑘−1(𝑘−2)2𝑛−2.(1.9)

By (1.9), we may immediately deduce the following (see [4, page 147]):

𝐷(2𝑛+1)2𝑛=(−1)𝑛(2𝑛)!4𝑛𝑛2𝑛,𝐷(2𝑛+2)2𝑛=(−1)𝑛4𝑛2𝑛+1(𝑛!)2,(1.10)𝐷(2𝑛+3)2𝑛=(−1)𝑛(2𝑛)!2⋅42𝑛12𝑛+2𝑛+11+32+1521+⋯+(2𝑛+1)2.(1.11)

The numbers 𝐷(2𝑛)2𝑛 are called the 𝐷-Nörlund numbers that satisfy the recurrence relation (see [7])

𝑛𝑗=0(−1)𝑗4𝑗𝑗𝐷(2𝑗+1)2𝑗(2𝑛−2𝑗)2𝑛−2𝑗=(2𝑛−2𝑗)!(−1)𝑛4𝑛𝑛,2𝑛(1.12) so we find 𝐷0(0)=1,𝐷2(2)=−2/3,𝐷4(4)=88/15,𝐷6(6)=−3056/21,𝐷8(8)=319616/45,𝐷(10)10=−18940160/33,….

A generating function for the 𝐷-Nörlund numbers 𝐷(2𝑛)2𝑛 is (see [7])

𝑡√1+𝑡2√log𝑡+1+𝑡2=∞𝑛=0𝐷(2𝑛)2𝑛𝑡2𝑛(2𝑛)!,|𝑡|<1.(1.13)

These numbers 𝐷(2𝑛)2𝑛 and 𝐷(2𝑛−1)2𝑛 have many important applications. For example (see [4, page 246])

0𝜋/2sin𝑡𝑡𝑑𝑡=∞𝑛=0(−1)𝑛𝐷(2𝑛)2𝑛,(2𝑛+1)!0𝜋/2sin𝑡𝑡𝜋𝑑𝑡=2∞𝑛=0(−1)𝑛+1𝐷(2𝑛−1)2𝑛22𝑛(2𝑛−1)(𝑛!)2,2𝜋=∞∑𝑛=0(−1)𝑛+1𝐷(2𝑛−1)2𝑛(.2𝑛−1)(2𝑛)!(1.14)

The main purpose of this paper is to prove a recurrence formula for 𝐷 numbers 𝐷(2𝑛−1)2𝑛 and to obtain a generating function for 𝐷 numbers 𝐷(2𝑛−1)2𝑛. That is, we will prove the following main conclusion.

Theorem 1.1. Let 𝑛∈ℕ. Then 𝑛𝑗=12𝑛2𝑗(−1)𝑗−14𝑗−1((𝑗−1)!)2𝐷(2𝑛−1−2𝑗)2𝑛−2𝑗=(−1)𝑛−12(2𝑛)!4𝑛,2𝑛−2𝑛−1(1.15) so one finds 𝐷2(1)=−1/3,𝐷4(3)=17/5,𝐷6(5)=−1835/21,𝐷8(7)=195013/45,𝐷(9)10=−3887409/11,….

Theorem 1.2. Let 𝑡 be a complex number with |𝑡|<1. Then ∞𝑛=0𝐷(2𝑛−1)2𝑛𝑡2𝑛=1(2𝑛)!√1+𝑡2𝑡√log(𝑡+1+𝑡2)2.(1.16)

2. Proof of the Theorems

Proof of Theorem 1.1. Note the identity (see [4, page 203]) 𝐵(𝑘)2𝑛𝑘𝑥+2=𝑛𝑗=0𝐷2𝑛2𝑗(𝑘−2𝑗)2𝑛−2𝑗22𝑛−2𝑗𝑥2𝑥2−12𝑥2−22⋯𝑥2−(𝑗−1)2,(2.1) we have 𝐵(𝑘)2𝑛(𝑥+𝑘/2)−𝐵(𝑘)2𝑛(𝑘/2)𝑥2=𝑛𝑗=1𝐷2𝑛2𝑗(𝑘−2𝑗)2𝑛−2𝑗22𝑛−2𝑗𝑥2−12𝑥2−22⋯𝑥2−(𝑗−1)2.(2.2) Therefore, lim𝑥→0𝐵(𝑘)2𝑛(𝑥+𝑘/2)−𝐵(𝑘)2𝑛(𝑘/2)𝑥2=𝑛𝑗=1𝐷2𝑛2𝑗(𝑘−2𝑗)2𝑛−2𝑗22𝑛−2𝑗(−1)𝑗−1((𝑗−1)!)2.(2.3) By (2.3) and (1.2), we have lim𝑥→02𝑛(2𝑛−1)𝐵(𝑘)2𝑛−2(𝑥+𝑘/2)2=𝑛𝑗=1𝐷2𝑛2𝑗(𝑘−2𝑗)2𝑛−2𝑗22𝑛−2𝑗(−1)𝑗−1((𝑗−1)!)2.(2.4) That is, 𝑛(2𝑛−1)𝐵(𝑘)2𝑛−2𝑘2=𝑛𝑗=1𝐷2𝑛2𝑗(𝑘−2𝑗)2𝑛−2𝑗22𝑛−2𝑗(−1)𝑗−1((𝑗−1)!)2.(2.5) By (2.5) and (1.7), we have 𝐷(𝑘)2𝑛−2=1𝑛(2𝑛−1)𝑛𝑗=12𝑛2𝑗(−1)𝑗−14𝑗−1((𝑗−1)!)2𝐷(𝑘−2𝑗)2𝑛−2𝑗.(2.6) Setting 𝑘=2𝑛−1 in (2.6), and note (1.10), we immediately obtain Theorem 1.1. This completes the proof of Theorem 1.1.

Remark 2.1. Setting 𝑘=2𝑛 in (2.6), and note (1.10), we may immediately deduce the following recurrence formula for 𝐷-Nörlund numbers 𝐷(2𝑛)2𝑛: 𝑛𝑗=12𝑛2𝑗(−1)𝑗4𝑗((𝑗−1)!)2𝐷(2𝑛−2𝑗)2𝑛−2𝑗=(−1)𝑛𝑛4𝑛((𝑛−1)!)2(𝑛∈ℕ).(2.7)

Proof of Theorem 1.2. Note the identity (see [9]) ∞𝑛=0(−1)𝑛4𝑛(𝑛!)2𝑡2𝑛=1(2𝑛)!1+𝑡2𝑡1−√1+𝑡2√log𝑡+1+𝑡2,(2.8) where |𝑡|<1. We have ∞𝑛=0(−1)𝑛4𝑛(𝑛!)2𝑡2𝑛+2=1(2𝑛+2)!2√log𝑡+1+𝑡22,(2.9) That is, ∞𝑛=1(−1)𝑛−14𝑛−1((𝑛−1)!)2𝑡2𝑛=1(2𝑛)!2√log𝑡+1+𝑡22.(2.10)
On the other hand, ∞𝑛=1(−1)𝑛−12(2𝑛)!4𝑛𝑡2𝑛−2𝑛−12𝑛=1(2𝑛)!2∞𝑛=0(−1)𝑛4𝑛𝑛𝑡2𝑛2𝑛+2=𝑡22√1+𝑡2.(2.11) Thus, by (2.10), (2.11), and Theorem 1.1, we have ∞𝑛=1(−1)𝑛−14𝑛−1((𝑛−1)!)2𝑡2𝑛(2𝑛)!∞𝑛=1𝐷(2𝑛−1)2𝑛𝑡2𝑛=(2𝑛)!∞𝑛=1(−1)𝑛−12(2𝑛)!4𝑛𝑡2𝑛−2𝑛−12𝑛.(2𝑛)!(2.12) That is, 12√log𝑡+1+𝑡22∞𝑛=1𝐷(2𝑛−1)2𝑛𝑡2𝑛=𝑡(2𝑛)!22√1+𝑡2.(2.13) By (2.13), and note that lim𝑡→0𝑡√log𝑡+1+𝑡2=1,𝐷0(−1)=1,(2.14) we immediately obtain Theorem 1.2. This completes the proof of Theorem 1.2.

Acknowledgment

This work was Supported by the Guangdong Provincial Natural Science Foundation (no. 8151601501000002).