# Some Computational Formulas forD-Nörlund Numbers

Authors: Guodong Liu
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

The author establishes some identities involving theD numbers, Bernoulli numbers, and central factorial numbers of the first kind. A generating function and several computational formulas for D-Nörlund numbers are 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 [1–5])
(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, 6, 7]). 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 [4, 8]). 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, 5])
(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, pages 22 and 145]), we have
(1.8)D2n(1)=(2-22n)B2n,D2n(2)=4n(1-2n)B2n.The numbersD2n(2n) are called the D-Nörlund numbers. These numbers D2n(2n) and D2n(2n-1) have many important applications. For example (see [4, page 246])
(1.9)∫0π/2sinttdt=∑n=0∞(-1)nD2n(2n)(2n+1)!,∫0π/2sinttdt=π2∑n=0∞(-1)n+1D2n(2n-1)22n(2n-1)(n!)2,(1.10)2π=∑n=0∞(-1)n+1D2n(2n-1)(2n-1)(2n)!.We now turn to the central factorial numberst(n,k) of the first kind, which are usually defined by (see [9–12])
(1.11)x(x+n2-1)(x+n2-2)⋯(x+n2-n+1)=∑k=0nt(n,k)xk,
or by means of the following generating function:
(1.12)(2log(x2+1+x24))k=k!∑n=k∞t(n,k)xnn!.It follows from (1.11) or (1.12) that
(1.13)t(n,k)=t(n-2,k-2)-14(n-2)2t(n-2,k),
and that
(1.14)t(n,0)=δn,0(n∈ℕ0:=ℕ∪{0}),t(n,n)=1(n∈ℕ),t(n,k)=0(n+kodd),t(n,k)=0(k>nork<0),
where δm,n denotes the Kronecker symbol.By (1.13), we have
(1.15)t(2n+1,1)=(-1)n(2n)!42n(2nn),t(2n+2,2)=(-1)n(n!)2(n∈ℕ0),(1.16)t(2n+2,4)=(-1)n+1(n!)2(1+122+132+⋯+1n2)(n∈ℕ),(1.17)t(2n+1,3)=(-1)n-1(2n)!42n-1(2nn)(1+132+152+⋯+1(2n-1)2)(n∈ℕ).The main purpose of this paper is to prove some identities involvingD numbers, Bernoulli numbers, and central factorial numbers of the first kind and obtain a generating function and several computational formulas for the D-Nörlund numbers. That is, we will prove the following main conclusion.Theorem 1.1.
Letn∈ℕ, k∈ℕ∖{1}. Then
(1.18)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).Remark 1.2.
By (1.18), we may immediately deduce the following (see [4, page 147]:
(1.19)D2n(2n+1)=(-1)n(2n)!4n(2nn),D2n(2n+2)=(-1)n4n2n+1(n!)2.Theorem 1.3.
Letn≥k(n,k∈ℕ0). Then
(1.20)D2n-2k(2n+1)=4n-k(2n2k)t(2n+1,2k+1),(1.21)D2n-2k(2n)=4n-k(2n-12k-1)t(2n,2k)(k≥1).Remark 1.4.
By (1.20) and (1.17), we may immediately deduce the following:
(1.22)D2n(2n+3)=(-1)n(2n)!2·42n(2n+2n+1)(1+132+152+⋯+1(2n+1)2).Theorem 1.5.
Letn∈ℕ0. Then
(1.23)∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!=(-1)n4n(2nn),
so one finds D0(0)=1,D2(2)=-2/3,D4(4)=88/15,D6(6)=-3056/21,D8(8)=319616/45,D10(10)=-18940160/33,….
By (1.23), and note that
(1.24)log(t+1+t2)=∑n=0∞(-1)n4n(2n+1)(2nn)t2n+1(|t|<1),11+t2=∑n=0∞(-1)n4n(2nn)t2n(|t|<1),
one may immediately deduce the following Corollary 1.6.Corollary 1.6.
Letn∈ℕ0. Then
(1.25)∑n=0∞D2n(2n)t2n(2n)!=t1+t2log(t+1+t2)(|t|<1).Theorem 1.7.
Letn∈ℕ. Then(i)
(1.26)D2n(2n)=(-1)n(2n)!4n(2nn)+n·4n∑j=1n(1-21-2j)jt(2n,2j)B2j(ii)
(1.27)D2n(2n)=(-1)n4n(n!)22n+1+n·4n∑j=1n2j-1jt(2n,2j)B2j(iii)
(1.28)D2n(2n)=∑j=0n4n-j2j+1t(2n+1,2j+1).Theorem 1.8.
Letn∈ℕ0. Then(i)
(1.29)∑j=0n2-4j4jt(2n+1,2j+1)B2j=(-1)n(n!)22n+1(ii)
(1.30)∑j=0n(1-2j)t(2n+1,2j+1)B2j=(-1)n(2n)!2·42n(2n+2n+1)(1+132+152+⋯+1(2n+1)2).


## 2. Proof of the Theorems

Proof of Theorem1.1.
By (1.4) and (1.3), we have
(2.1)B2n(k)(x+1)=2nxk-1B2n-1(k-1)(x)-2n-k+1k-1B2n(k-1)(x)=2nxk-1(k-2n-1k-2B2n-1(k-2)(x)+(x-k+2)2n-1k-2B2n-2(k-2)(x))-2n-k+1k-1(k-2n-2k-2B2n(k-2)(x)+(x-k+2)2nk-2B2n-1(k-2)(x))=(2n-k+1)(2n-k+2)(k-1)(k-2)B2n(k-2)(x)+2n(2n-1)(k-1)(k-2)x(x-k+2)B2n-2(k-2)(x)-2n(2n-k+1)(k-1)(k-2)(2x-k+2)B2n-1(k-2)(x).
Setting x=(k-2)/2 in (2.1), we get
(2.2)B2n(k)(k2)=(2n-k+1)(2n-k+2)(k-1)(k-2)B2n(k-2)(k-22)-2n(2n-1)(k-2)4(k-1)B2n-2(k-2)(k-22).
By (2.2) and (1.7), we immediately obtain (1.18). This completes the proof of Theorem 1.1.Proof of Theorem1.3.
By the usage of Theorem1.1 and (1.13).Proof of Theorem1.5.
Note the identity (see [4, page 203])
(2.3)B2n+1(k)(x+k2)=∑j=0n(2n+12j+1)D2n-2j(k-2j-1)22n-2jx(x2-(12)2)(x2-(32)2)⋯(x2-(2j-12)2),
we have
(2.4)limx→0B2n+1(2n+1)(x+(2n+1)/2)x=14n∑j=0n(2n+12j+1)D2n-2j(2n-2j)(-1)j12·32⋯(2j-1)2=(2n+1)!4n∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!.
By (2.4) and (1.2), we have
(2.5)limx→0(2n+1)B2n(2n+1)(x+2n+12)=(2n+1)!4n∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!,
that is,
(2.6)B2n(2n+1)(2n+12)=(2n)!4n∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!.
By (2.6) and (1.7), we have
(2.7)D2n(2n+1)=(2n)!∑j=0n(-1)j4j(2j+1)(2jj)D2n-2j(2n-2j)(2n-2j)!.
By (2.7) and (1.19), we immediately obtain (1.23). This completes the proof of Theorem 1.5.Proof of Theorem1.7.
By (1.6), we have
(2.8)D2n(k)=∑j=0n(2n2j)D2n-2j(k-l)D2j(l),
where l is an integer.
Settingk=2n+1,l=1 in (2.8), and note that D0(1)=1, we have
(2.9)D2n(2n+1)=∑j=0n(2n2j)D2n-2j(2n)D2j(1)=D2n(2n)+∑j=1n(2n2j)D2n-2j(2n)D2j(1).
By (2.9), (1.19), (1.8), and (1.21), we immediately obtain (1.26).
Settingk=2n+2,l=2 in (2.8), and note that D0(2)=1, we have
(2.10)D2n(2n+2)=∑j=0n(2n2j)D2n-2j(2n)D2j(2)=D2n(2n)+∑j=1n(2n2j)D2n-2j(2n)D2j(2).
By (2.10), (1.19), (1.8), and (1.21), we immediately obtain (1.27).
Settingk=2n,l=-1 in (2.8), and note that (1.20) and D2j(-1)=1/(2j+1), we immediately obtain (1.18). This completes the proof of Theorem 1.7.Proof of Theorem1.8.
Settingk=2n+2,l=1 in (2.8), and note (1.19), (1.20), and (1.8), we immediately obtain (1.29).
Settingk=2n+3,l=2 in (2.8), and note (1.22), (1.20), and (1.8), we immediately obtain (1.30). This completes the proof of Theorem 1.8.



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*Source: 430452-2009-11-18.xml*