# A Recurrence Formula forD

Authors: Guodong Liu
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

we establish a recurrence formula forD numbers D2n(2n−1). A generating function for D numbers D2n(2n−1) is also presented.


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## Body

## 1. Introduction and Results

The Bernoulli polynomialsBn(k)(x) of order k, for any integer k, may be defined by (see [1–4])(1.1)(tet-1)kext=∑n=0∞Bn(k)(x)tnn!,|t|<2π.The numbersBn(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])(1.2)ddxBn(k)(x)=nBn-1(k)(x),(1.3)Bn(k+1)(x)=k-nkBn(k)(x)+(x-k)nkBn-1(k)(x),(1.4)Bn(k+1)(x+1)=nxkBn-1(k)(x)-n-kkBn(k)(x),
where n∈ℕ, with ℕ being the set of positive integers.The numbersBn(n) are called the Nörlund numbers (see [2, 4, 6]). A generating function for the Nörlund numbers Bn(n) is (see [4, page 150])(1.5)t(1+t)log(1+t)=∑n=0∞Bn(n)tnn!.TheD numbers D2n(k) may be defined by (see [4, 7, 8])(1.6)(tcsct)k=∑n=0∞(-1)nD2n(k)t2n(2n)!,|t|<π.By (1.1), (1.6), and note that csct=2i/(eit-e-it) (where i2=-1), we can get(1.7)D2n(k)=4nB2n(k)(k2).Takingk=1,2 in (1.7), and note that B2n(1)(1/2)=(21-2n-1)B2n, B2n(2)(1)=(1-2n)B2n (see [4, page 22, page 145]), we have(1.8)D2n(1)=(2-22n)B2n,D2n(2)=4n(1-2n)B2n.TheD numbers D2n(k) satisfy the recurrence relation (see [7])(1.9)D2n(k)=(2n-k+2)(2n-k+1)(k-2)(k-1)D2n(k-2)-2n(2n-1)(k-2)k-1D2n-2(k-2).By (1.9), we may immediately deduce the following (see [4, page 147]):(1.10)D2n(2n+1)=(-1)n(2n)!4n(2nn),D2n(2n+2)=(-1)n4n2n+1(n!)2,(1.11)D2n(2n+3)=(-1)n(2n)!2·42n(2n+2n+1)(1+132+152+⋯+1(2n+1)2).The numbersD2n(2n) are called the D-Nörlund numbers that satisfy the recurrence relation (see [7])(1.12)∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!=(-1)n4n(2nn),
so we find D0(0)=1,D2(2)=-2/3,D4(4)=88/15,D6(6)=-3056/21,D8(8)=319616/45,D10(10)=-18940160/33,….A generating function for theD-Nörlund numbers D2n(2n) is (see [7])(1.13)t1+t2log(t+1+t2)=∑n=0∞D2n(2n)t2n(2n)!,|t|<1.These numbersD2n(2n) and D2n(2n-1) have many important applications. For example (see [4, page 246])(1.14)∫0π/2sinttdt=∑n=0∞(-1)nD2n(2n)(2n+1)!,∫0π/2sinttdt=π2∑n=0∞(-1)n+1D2n(2n-1)22n(2n-1)(n!)2,2π=∑n=0∞(-1)n+1D2n(2n-1)(2n-1)(2n)!.The main purpose of this paper is to prove a recurrence formula forD numbers D2n(2n-1) and to obtain a generating function for D numbers D2n(2n-1). That is, we will prove the following main conclusion.Theorem 1.1.
Letn∈ℕ. Then
(1.15)∑j=1n(2n2j)(-1)j-14j-1((j-1)!)2D2n-2j(2n-1-2j)=(-1)n-12(2n)!4n(2n-2n-1),
so one finds D2(1)=-1/3,D4(3)=17/5,D6(5)=-1835/21,D8(7)=195013/45,D10(9)=-3887409/11,….Theorem 1.2.
Lett be a complex number with |t|<1. Then
(1.16)∑n=0∞D2n(2n-1)t2n(2n)!=11+t2(tlog(t+1+t2))2.


## 2. Proof of the Theorems

Proof of Theorem1.1.
Note the identity (see [4, page 203])
(2.1)B2n(k)(x+k2)=∑j=0n(2n2j)D2n-2j(k-2j)22n-2jx2(x2-12)(x2-22)⋯(x2-(j-1)2),
we have
(2.2)B2n(k)(x+k/2)-B2n(k)(k/2)x2=∑j=1n(2n2j)D2n-2j(k-2j)22n-2j(x2-12)(x2-22)⋯(x2-(j-1)2).
Therefore,
(2.3)limx→0B2n(k)(x+k/2)-B2n(k)(k/2)x2=∑j=1n(2n2j)D2n-2j(k-2j)22n-2j(-1)j-1((j-1)!)2.
By (2.3) and (1.2), we have
(2.4)limx→02n(2n-1)B2n-2(k)(x+k/2)2=∑j=1n(2n2j)D2n-2j(k-2j)22n-2j(-1)j-1((j-1)!)2.
That is,
(2.5)n(2n-1)B2n-2(k)(k2)=∑j=1n(2n2j)D2n-2j(k-2j)22n-2j(-1)j-1((j-1)!)2.
By (2.5) and (1.7), we have
(2.6)D2n-2(k)=1n(2n-1)∑j=1n(2n2j)(-1)j-14j-1((j-1)!)2D2n-2j(k-2j).
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.
Settingk=2n in (2.6), and note (1.10), we may immediately deduce the following recurrence formula for D-Nörlund numbers D2n(2n):
(2.7)∑j=1n(2n2j)(-1)j4j((j-1)!)2D2n-2j(2n-2j)=(-1)nn4n((n-1)!)2(n∈ℕ).Proof of Theorem1.2.
Note the identity (see [9])
(2.8)∑n=0∞(-1)n4n(n!)2t2n(2n)!=11+t2(1-t1+t2log(t+1+t2)),
where |t|<1. We have
(2.9)∑n=0∞(-1)n4n(n!)2t2n+2(2n+2)!=12(log(t+1+t2))2,
That is,
(2.10)∑n=1∞(-1)n-14n-1((n-1)!)2t2n(2n)!=12(log(t+1+t2))2.
On the other hand,(2.11)∑n=1∞(-1)n-12(2n)!4n(2n-2n-1)t2n(2n)!=12∑n=0∞(-1)n4n(2nn)t2n+2=t221+t2.
Thus, by (2.10), (2.11), and Theorem 1.1, we have
(2.12)∑n=1∞(-1)n-14n-1((n-1)!)2t2n(2n)!∑n=1∞D2n(2n-1)t2n(2n)!=∑n=1∞(-1)n-12(2n)!4n(2n-2n-1)t2n(2n)!.
That is,
(2.13)12(log(t+1+t2))2∑n=1∞D2n(2n-1)t2n(2n)!=t221+t2.
By (2.13), and note that
(2.14)limt→0tlog(t+1+t2)=1,D0(-1)=1,
we immediately obtain Theorem 1.2. This completes the proof of Theorem 1.2.



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*Source: 605313-2009-11-19.xml*