Abstract

we establish a recurrence formula for 𝐷 numbers 𝐷(2π‘›βˆ’1)2𝑛. A generating function for 𝐷 numbers 𝐷(2π‘›βˆ’1)2𝑛 is also presented.

1. Introduction and Results

The Bernoulli polynomials 𝐡𝑛(π‘˜)(π‘₯) of order π‘˜, for any integer π‘˜, may be defined by (see [1–4])

ξ‚€π‘‘π‘’π‘‘ξ‚βˆ’1π‘˜π‘’π‘₯𝑑=βˆžξ“π‘›=0𝐡𝑛(π‘˜)(𝑑π‘₯)𝑛𝑛!,|𝑑|<2πœ‹.(1.1)

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

𝑑𝐡𝑑π‘₯𝑛(π‘˜)(π‘₯)=𝑛𝐡(π‘˜)π‘›βˆ’1(π‘₯),(1.2)𝐡𝑛(π‘˜+1)(π‘₯)=π‘˜βˆ’π‘›π‘˜π΅π‘›(π‘˜)𝑛(π‘₯)+(π‘₯βˆ’π‘˜)π‘˜π΅(π‘˜)π‘›βˆ’1(π‘₯),(1.3)𝐡𝑛(π‘˜+1)(π‘₯+1)=𝑛π‘₯π‘˜π΅(π‘˜)π‘›βˆ’1(π‘₯)βˆ’π‘›βˆ’π‘˜π‘˜π΅π‘›(π‘˜)(π‘₯),(1.4) where π‘›βˆˆβ„•, with β„• being the set of positive integers.

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

𝑑=(1+𝑑)log(1+𝑑)βˆžξ“π‘›=0𝐡𝑛(𝑛)𝑑𝑛.𝑛!(1.5)

The 𝐷 numbers 𝐷(π‘˜)2𝑛 may be defined by (see [4, 7, 8])

(𝑑csc𝑑)π‘˜=βˆžξ“π‘›=0(βˆ’1)𝑛𝐷(π‘˜)2𝑛𝑑2𝑛(2𝑛)!,|𝑑|<πœ‹.(1.6)

By (1.1), (1.6), and note that csc𝑑=2𝑖/(π‘’π‘–π‘‘βˆ’π‘’βˆ’π‘–π‘‘) (where 𝑖2=βˆ’1), we can get

𝐷(π‘˜)2𝑛=4𝑛𝐡(π‘˜)2π‘›ξ‚€π‘˜2.(1.7)

Taking π‘˜=1,2 in (1.7), and note that 𝐡(1)2𝑛(1/2)=(21βˆ’2π‘›βˆ’1)𝐡2𝑛, 𝐡(2)2𝑛(1)=(1βˆ’2𝑛)𝐡2𝑛 (see [4, page 22, page 145]), we have

𝐷(1)2𝑛=ξ€·2βˆ’22𝑛𝐡2𝑛,𝐷(2)2𝑛=4𝑛(1βˆ’2𝑛)𝐡2𝑛.(1.8)

The 𝐷 numbers 𝐷(π‘˜)2𝑛 satisfy the recurrence relation (see [7])

𝐷(π‘˜)2𝑛=(2π‘›βˆ’π‘˜+2)(2π‘›βˆ’π‘˜+1)(π·π‘˜βˆ’2)(π‘˜βˆ’1)(π‘˜βˆ’2)2π‘›βˆ’2𝑛(2π‘›βˆ’1)(π‘˜βˆ’2)π·π‘˜βˆ’1(π‘˜βˆ’2)2π‘›βˆ’2.(1.9)

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

𝐷(2𝑛+1)2𝑛=(βˆ’1)𝑛(2𝑛)!4𝑛𝑛ξƒͺ2𝑛,𝐷(2𝑛+2)2𝑛=(βˆ’1)𝑛4𝑛2𝑛+1(𝑛!)2,(1.10)𝐷(2𝑛+3)2𝑛=(βˆ’1)𝑛(2𝑛)!2β‹…42𝑛ξƒͺξ‚΅12𝑛+2𝑛+11+32+1521+β‹―+(2𝑛+1)2ξ‚Ά.(1.11)

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

𝑛𝑗=0(βˆ’1)𝑗4𝑗𝑗ξƒͺ𝐷(2𝑗+1)2𝑗(2π‘›βˆ’2𝑗)2π‘›βˆ’2𝑗=(2π‘›βˆ’2𝑗)!(βˆ’1)𝑛4𝑛𝑛ξƒͺ,2𝑛(1.12) so we find 𝐷0(0)=1,𝐷2(2)=βˆ’2/3,𝐷4(4)=88/15,𝐷6(6)=βˆ’3056/21,𝐷8(8)=319616/45,𝐷(10)10=βˆ’18940160/33,….

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

π‘‘βˆš1+𝑑2ξ‚€βˆšlog𝑑+1+𝑑2=βˆžξ“π‘›=0𝐷(2𝑛)2𝑛𝑑2𝑛(2𝑛)!,|𝑑|<1.(1.13)

These numbers 𝐷(2𝑛)2𝑛 and 𝐷(2π‘›βˆ’1)2𝑛 have many important applications. For example (see [4, page 246])

ξ€œ0πœ‹/2sin𝑑𝑑𝑑𝑑=βˆžξ“π‘›=0(βˆ’1)𝑛𝐷(2𝑛)2𝑛,ξ€œ(2𝑛+1)!0πœ‹/2sinπ‘‘π‘‘πœ‹π‘‘π‘‘=2βˆžξ“π‘›=0(βˆ’1)𝑛+1𝐷(2π‘›βˆ’1)2𝑛22𝑛(2π‘›βˆ’1)(𝑛!)2,2πœ‹=βˆžβˆ‘π‘›=0(βˆ’1)𝑛+1𝐷(2π‘›βˆ’1)2𝑛(.2π‘›βˆ’1)(2𝑛)!(1.14)

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

Theorem 1.1. Let π‘›βˆˆβ„•. Then 𝑛𝑗=1ξƒͺ2𝑛2𝑗(βˆ’1)π‘—βˆ’14π‘—βˆ’1((π‘—βˆ’1)!)2𝐷(2π‘›βˆ’1βˆ’2𝑗)2π‘›βˆ’2𝑗=(βˆ’1)π‘›βˆ’12(2𝑛)!4𝑛ξƒͺ,2π‘›βˆ’2π‘›βˆ’1(1.15) so one finds 𝐷2(1)=βˆ’1/3,𝐷4(3)=17/5,𝐷6(5)=βˆ’1835/21,𝐷8(7)=195013/45,𝐷(9)10=βˆ’3887409/11,….

Theorem 1.2. Let 𝑑 be a complex number with |𝑑|<1. Then βˆžξ“π‘›=0𝐷(2π‘›βˆ’1)2𝑛𝑑2𝑛=1(2𝑛)!√1+𝑑2ξƒ©π‘‘βˆšlog(𝑑+1+𝑑2)ξƒͺ2.(1.16)

2. Proof of the Theorems

Proof of Theorem 1.1. Note the identity (see [4, page 203]) 𝐡(π‘˜)2π‘›ξ‚€π‘˜π‘₯+2=𝑛𝑗=0ξƒͺ𝐷2𝑛2𝑗(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗22π‘›βˆ’2𝑗π‘₯2ξ€·π‘₯2βˆ’12π‘₯ξ€Έξ€·2βˆ’22ξ€Έβ‹―ξ€·π‘₯2βˆ’(π‘—βˆ’1)2ξ€Έ,(2.1) we have 𝐡(π‘˜)2𝑛(π‘₯+π‘˜/2)βˆ’π΅(π‘˜)2𝑛(π‘˜/2)π‘₯2=𝑛𝑗=1ξƒͺ𝐷2𝑛2𝑗(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗22π‘›βˆ’2𝑗π‘₯2βˆ’12π‘₯ξ€Έξ€·2βˆ’22ξ€Έβ‹―ξ€·π‘₯2βˆ’(π‘—βˆ’1)2ξ€Έ.(2.2) Therefore, limπ‘₯β†’0𝐡(π‘˜)2𝑛(π‘₯+π‘˜/2)βˆ’π΅(π‘˜)2𝑛(π‘˜/2)π‘₯2=𝑛𝑗=1ξƒͺ𝐷2𝑛2𝑗(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗22π‘›βˆ’2𝑗(βˆ’1)π‘—βˆ’1((π‘—βˆ’1)!)2.(2.3) By (2.3) and (1.2), we have limπ‘₯β†’02𝑛(2π‘›βˆ’1)𝐡(π‘˜)2π‘›βˆ’2(π‘₯+π‘˜/2)2=𝑛𝑗=1ξƒͺ𝐷2𝑛2𝑗(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗22π‘›βˆ’2𝑗(βˆ’1)π‘—βˆ’1((π‘—βˆ’1)!)2.(2.4) That is, 𝑛(2π‘›βˆ’1)𝐡(π‘˜)2π‘›βˆ’2ξ‚€π‘˜2=𝑛𝑗=1ξƒͺ𝐷2𝑛2𝑗(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗22π‘›βˆ’2𝑗(βˆ’1)π‘—βˆ’1((π‘—βˆ’1)!)2.(2.5) By (2.5) and (1.7), we have 𝐷(π‘˜)2π‘›βˆ’2=1𝑛(2π‘›βˆ’1)𝑛𝑗=1ξƒͺ2𝑛2𝑗(βˆ’1)π‘—βˆ’14π‘—βˆ’1((π‘—βˆ’1)!)2𝐷(π‘˜βˆ’2𝑗)2π‘›βˆ’2𝑗.(2.6) Setting π‘˜=2π‘›βˆ’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 π‘˜=2𝑛 in (2.6), and note (1.10), we may immediately deduce the following recurrence formula for 𝐷-NΓΆrlund numbers 𝐷(2𝑛)2𝑛: 𝑛𝑗=1ξƒͺ2𝑛2𝑗(βˆ’1)𝑗4𝑗((π‘—βˆ’1)!)2𝐷(2π‘›βˆ’2𝑗)2π‘›βˆ’2𝑗=(βˆ’1)𝑛𝑛4𝑛((π‘›βˆ’1)!)2(π‘›βˆˆβ„•).(2.7)

Proof of Theorem 1.2. Note the identity (see [9]) βˆžξ“π‘›=0(βˆ’1)𝑛4𝑛(𝑛!)2𝑑2𝑛=1(2𝑛)!1+𝑑2𝑑1βˆ’βˆš1+𝑑2ξ‚€βˆšlog𝑑+1+𝑑2ξƒͺ,(2.8) where |𝑑|<1. We have βˆžξ“π‘›=0(βˆ’1)𝑛4𝑛(𝑛!)2𝑑2𝑛+2=1(2𝑛+2)!2ξ‚€ξ‚€βˆšlog𝑑+1+𝑑22,(2.9) That is, βˆžξ“π‘›=1(βˆ’1)π‘›βˆ’14π‘›βˆ’1((π‘›βˆ’1)!)2𝑑2𝑛=1(2𝑛)!2ξ‚€ξ‚€βˆšlog𝑑+1+𝑑22.(2.10)
On the other hand, βˆžξ“π‘›=1(βˆ’1)π‘›βˆ’12(2𝑛)!4𝑛ξƒͺ𝑑2π‘›βˆ’2π‘›βˆ’12𝑛=1(2𝑛)!2βˆžξ“π‘›=0(βˆ’1)𝑛4𝑛𝑛ξƒͺ𝑑2𝑛2𝑛+2=𝑑22√1+𝑑2.(2.11) Thus, by (2.10), (2.11), and Theorem 1.1, we have βˆžξ“π‘›=1(βˆ’1)π‘›βˆ’14π‘›βˆ’1((π‘›βˆ’1)!)2𝑑2𝑛(2𝑛)!βˆžξ“π‘›=1𝐷(2π‘›βˆ’1)2𝑛𝑑2𝑛=(2𝑛)!βˆžξ“π‘›=1(βˆ’1)π‘›βˆ’12(2𝑛)!4𝑛ξƒͺ𝑑2π‘›βˆ’2π‘›βˆ’12𝑛.(2𝑛)!(2.12) That is, 12ξ‚€ξ‚€βˆšlog𝑑+1+𝑑22βˆžξ“π‘›=1𝐷(2π‘›βˆ’1)2𝑛𝑑2𝑛=𝑑(2𝑛)!22√1+𝑑2.(2.13) By (2.13), and note that lim𝑑→0π‘‘ξ‚€βˆšlog𝑑+1+𝑑2=1,𝐷0(βˆ’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).