# Arithmetic Identities Involving Genocchi and Stirling Numbers

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

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## Abstract

An explicit formula, the generalized Genocchi numbers, was established and some identities and congruences involving the Genocchi numbers, the Bernoulli numbers, and the Stirling numbers were obtained.


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## Body

## 1. Introduction

The Genocchi numbersGn and the Bernoulli numbers Bn (n∈ℕ0={0,1,2,…}) are defined by the following generating functions (see [1]):(1.1)2tet+1=∑n=0∞Gntnn!(|t|<π),(1.2)tet-1=∑n=0∞Bntnn!(|t|<2π),
respectively. By (1.1) and (1.2), we have(1.3)G2n+1=B2n+1=0,(n∈ℕ)Gn=2(1-2n)Bn,
with ℕ being the set of positive integers.The Genocchi numbersGn satisfy the recurrence relation(1.4)G0=0,G1=1,Gn=-12∑k=1n-1(nk)Gk(n≥2),
so we find G2=-1,G4=1,G6=-3,G8=17,G10=-155,G12=2073,G14=-38227,….The Stirling numbers of the first kinds(n,k)  can be defined by means of (see [2])(1.5)(x)n=x(x-1)…(x-n+1)=∑k=0ns(n,k)xk,
or by the generating function(1.6)(log(1+x))k=k!∑n=k∞s(n,k)xnn!.It follows from (1.5) or (1.6) that(1.7)s(n,k)=s(n-1,k-1)-(n-1)s(n-1,k),
with s(n,0)=0(n>0),  s(n,n)=1(n≥0),  s(n,1)=(-1)n-1(n-1)!(n>0), s(n,k)=0(k>n or k<0).Stirling numbers of the second kindS(n,k) can be defined by (see [2])(1.8)xn=∑k=0nS(n,k)(x)k
or by the generating function(1.9)(ex-1)k=k!∑n=k∞S(n,k)xnn!.
It follows from (1.8) or (1.9) that(1.10)S(n,k)=S(n-1,k-1)+kS(n-1,k),
with S(n,0)=0(n>0),  S(n,n)=1(n≥0),  S(n,1)=1(n>0),   S(n,k)=0(k>n or k<0).The study of Genocchi numbers and polynomials has received much attention; numerous interesting (and useful) properties for Genocchi numbers can be found in many books (see [1, 3–16]). The main purpose of this paper is to prove an explicit formula for the generalized Genocchi numbers (cf. Section 2). We also obtain some identities congruences involving the Genocchi numbers, the Bernoulli numbers, and the Stirling numbers. That is, we will prove the following main conclusion.Theorem 1.1.
Letn≥k(n,k∈ℕ), then
(1.11)∑v1,…,vk∈ℕv1+⋯+vk=nGv1⋯Gvk(v1⋯vk)v1!⋯vk!=(-1)n-k2kk!n!∑j=kn12jS(n,j)s(j,k).Remark 1.2.
Settingk=1 in (1.11), and noting that s(j,1)=(-1)j-1(j-1)!, we obtain
(1.12)Gn=2n∑j=1n(-1)n-j(j-1)!2jS(n,j)(n∈ℕ).Remark 1.3.
By (1.11) and (1.3), we have
(1.13)∑v1,…,vk∈ℕv1+⋯+vk=n(2v1-1)Bv1⋯(2vk-1)Bvk(v1⋯vk)v1!⋯vk!=(-1)nk!n!∑j=kn12jS(n,j)s(j,k).Theorem 1.4.
Letn,k∈ℕ, then
(1.14)∑j=0n(-1)j(k+j-1)!2jS(n,j)=2k-1∑j=0k-1(-1)js(k,k-j)Gn+k-jn+k-j.Remark 1.5.
Settingk=1,2,3,4  in (1.14), we get
(1.15)∑j=0n(-1)jj!2jS(n,j)=1n+1Gn+1,∑j=0n(-1)j(j+1)!2jS(n,j)=2n+1Gn+1+2n+2Gn+2,∑j=0n(-1)j(j+2)!2jS(n,j)=8n+1Gn+1+12n+2Gn+2+4n+3Gn+3,∑j=0n(-1)j(j+3)!2jS(n,j)=48n+1Gn+1+88n+2Gn+2+48n+3Gn+3+8n+4Gn+4.Theorem 1.6.
Letn∈ℕ, m∈ℕ0, then
(1.16)2nn+1Gn+1≡2n-m∑j=0m(mj)jn(modm+1).Remark 1.7.
Settingm=p-1 in (1.16),  we have
(1.17)1n+1Gn+1≡∑j=0p-1(-1)jjn(modp),
where p is any odd prime.


## 2. Definition and Lemma

Definition 2.1.
For a real or complex parameterx, we have the generalized Genocchi numbers Gn(x), which are defined by
(2.1)(2e2t+1)x=∑n=0∞Gn(x)tnn!(|t|<π2;1x:=1).
By (1.1) and (2.1), we have
(2.2)nGn-1(1)=2n-1Gn.Remark 2.2.
For an integerx, the higher-order Euler numbers  E2n(x)  are defined by the following generating functions (see [17]):
(2.3)(2et+e-t)x=∑n=0∞E2n(x)t2n(2n)!(|t|<π2).
Then we have
(2.4)Gn(x)=(-1)n∑k=0[n/2](n2k)E2k(x)xn-2k,
where [n/2] denotes the greatest integer not exceeding n/2.Lemma 2.3.
Letn≥k(n,k∈ℕ), then
(2.5)Gn(x)=∑k=1nω(n,k)xk,
where
(2.6)ω(n,k)=(-1)k∑j=kn2n-jS(n,j)s(j,k).Proof.
By (2.1), (1.5), and (1.9) we have
(2.7)∑n=0∞Gn(x)tnn!=(2e2t+1)x=(11+(1/2)(e2t-1))x=∑j=0∞(-1)j2j(x+j-1j)(e2t-1)j=∑j=0∞(-1)j2j(x+j-1j)j!∑n=j∞2nS(n,j)tnn!=∑n=0∞‍∑j=0n(-1)jj!2n-j(x+j-1j)S(n,j)tnn!,
which readily yields
(2.8)Gn(x)=∑j=0n(-1)jj!2n-j(x+j-1j)S(n,j)=∑j=0n(-1)j2n-jS(n,j)(x+j-1)(x+j-2)⋯(x+1)x=∑j=0n(-1)j2n-jS(n,j)∑k=1j(-1)j-ks(j,k)xk=∑k=1n(-1)k∑j=kn2n-jS(n,j)s(j,k)xk=∑k=1nω(n,k)xk.
This completes the  proof of Lemma 2.3.Remark 2.4.
From (1.7), (1.10), and Lemma 2.3 we know that Gn(x) is a polynomial of x with integral coefficients. For example, setting  n=1,2,3,4  in Lemma 2.3, we get
(2.9)G1(x)=-x,G2(x)=-x+x2,G3(x)=3x2-x3,G4(x)=2x+3x2-6x3+x4.Remark 2.5.
Letn,m∈ℕ, then by (2.5), we have
(2.10)∑k=1nω(n,k)=2nn+1Gn+1.
Therefore, if q∈ℕ is odd, then by (2.10) we get
(2.11)G2kq≡0(modq),
where k∈ℕ.


## 3. Proof of the Theorems

Proof of Theorem1.1.
By applying Lemma2.3, we have
(3.1)k!ω(n,k)=dkdxkGn(x)|x=0.
On the other hand, it follows from (2.1) that
(3.2)∑n=k∞dkdxkGn(x)|x=0tnn!=(log2e2t+1)k,
where  log(2/(e2t+1))  is the principal branch of logarithm of 2/(e2t+1).
Thus, by (3.1) and (3.2), we have
(3.3)k!∑n=k∞ω(n,k)tnn!=(log2e2t+1)k.
Now note that(3.4)ddtlog2e2t+1=2e2t+1-2=∑n=0∞Gn(1)tnn!-2=∑n=0∞2nGn+1n+1tnn!-2,
whence by integrating from  0  to t, we deduce that
(3.5)log2e2t+1=∑n=1∞2n-1Gnntnn!-2t=∑n=1∞(-1)n2n-1Gnntnn!.
Since  G2n+1=0   (n∈ℕ). Substituting (3.5) in (3.3) we get
(3.6)ω(n,k)=(-1)nn!2n-kk!∑v1,…,vk∈ℕv1+⋯+vk=nGv1⋯Gvk(v1⋯vk)v1!⋯vk!.
By (3.6) and (2.6), we may immediately obtain Theorem 1.1. This completes the proof of Theorem 1.1.Proof of Theorem1.4.
By (2.1) and note the identity
(3.7)2(2e2t+1)x+1xddt(2e2t+1)x=(2e2t+1)x+1,
we have
(3.8)Gn(x+1)=2Gn(x)+1xGn+1(x).
By (3.8), (1.7), and note that Gn(1)=2n/(n+1)Gn+1, we obtain
(3.9)Gn(k)=1(k-1)!∑j=0k-1(-1)j2js(k,k-j)Gn+k-1-j(1)=2n+k-1(k-1)!∑j=0k-1(-1)js(k,k-j)Gn+k-jn+k-j.
Comparing (3.9) and (2.8), we immediately obtain Theorem 1.4. This completes the proof of Theorem 1.4.Proof of Theorem1.6.
By Lemma2.3, we have
(3.10)Gn(m+x)=∑j=1nω(n,j)(m+x)j≡∑j=1nω(n,j)xj=Gn(x)(modm).
Therefore
(3.11)Gn(k)=Gn(m+k-m)≡Gn(-m)=2n-m∑j=0m(mj)jn(modm+k).
Taking k=1 in (3.11) and note that Gn(1)=2n/(n+1)Gn+1, we immediately obtain Theorem 1.6. This completes the proof of Theorem 1.6.



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*Source: 621068-2009-10-27.xml*