# Some Properties of a Sequence Similar to Generalized Euler Numbers

Authors: Haiqing Wang; Guodong Liu
Journal: ISRN Discrete Mathematics
 (2013)

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## Abstract

We introduce the sequence{Un(x)} given by generating function (1/(et+e-t-1))x=∑n=0∞Un(x)(tn/n!)(|t|<(1/3)π,1x:=1) and establish some explicit formulas for the sequence {Un(x)}. Several identities involving the sequence {Un(x)}, Stirling numbers, Euler polynomials, and the central factorial numbers are also presented.


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## Body

## 1. Introduction and Definitions

For a real or complex parameterα, the generalized Euler polynomials En(α)(x) are defined by the following generating function (see [1–4])
(1)(2et+1)αext=∑n=0∞‍En(α)(x)tnn!(|t|<π,1α:=1).
Obviously, we have
(2)En(1)(x)=En(x)(n∈ℕ0:=ℕ∪{0}),
in terms of the classical Euler polynomials En(x), ℕ being the set of positive integers. The classical Euler numbers En are given by the following:
(3)En=2nEn(12)(n∈ℕ0).The so-called the generalized Euler numbersE2n(x) are defined by (see [3, 5])
(4)(2et+e-t)x=∑n=0∞‍E2n(x)t2n(2n)!(|t|<π2,1x:=1).
In fact, E2n(k)(k∈ℤ) are the Euler numbers of order k, ℤ being the set of integers. The numbers E2n(1)=E2n are the ordinary Euler numbers.Zhi-Hong Sun introduces the sequence{Un} similar to Euler numbers as follows (see [6, 7]):
(5)U0=1,Un=-2∑k=1[n/2]‍(n2k)Un-2k,(n≥1),
where (and in what follows) [x] is the greatest integer not exceeding x.Clearly,U2n-1=0 for n≥1. The first few values of U2n are shown below
(6)U2=-2,U4=22,U6=-602,U8=30742,U10=-2523002,U12=303692662.The sequence{Un} is related to the classical  Bernoulli polynomials Bn(x) (see [8–11]) and the classical Euler polynomials En(x). Zhi-Hong Sun gets the generating function of {Un} and deduces many identities involving {Un}. As example, (see [6]),
(7)1et+e-t-1=∑n=0∞‍Untnn!=∑n=0∞‍U2nt2n(2n)!(|t|<13π),(8)12cost-1=∑n=0∞‍(-1)nU2nt2n(2n)!(|t|<13π),(9)U2n=32nE2n(13).Similarly, we can define the generalized sequence{Un(x)}. For a real or complex parameter x, the generalized sequence {Un(x)} is defined by the following generating function:
(10)(1et+e-t-1)x=∑n=0∞‍Un(x)tnn!(|t|<13π,1x:=1).
Obviously,
(11)U0(x)=1,Un(1)=Un(n∈ℕ).
By using (10), we can obtain
(12)Un(k)=n!∑v1,…,vk∈ℕ0(v1+⋯+vk=n)‍Uv1⋯Uvkv1!⋯vk!(k∈ℕ).We now return to the Stirling numberss(n,k) of the first kind, which are usually defined by (see [2, 5, 8, 11, 12])
(13)x(x-1)(x-2)⋯(x-n+1)=∑k=0n‍s(n,k)xk
or by the following generating function:
(14)(log(1+x))k=k!∑n=k∞‍s(n,k)xnn!.It follows from (13) or (14) that
(15)s(n,k)=s(n-1,k-1)-(n-1)s(n-1,k)
and that
(16)s(n,0)=0(n∈ℕ),s(n,n)=1(n∈ℕ0),s(n,1)=(-1)n-1(n-1)!(n∈ℕ),s(n,k)=0(k>nork<0).The central factorial numbersT(n,k) are given by the following expansion formula (see [3, 5, 13]):
(17)xn=∑k=0n‍T(n,k)x(x-12)×(x-22)⋯(x-(k-1)2)
or by means of the generating function
(18)(ex+e-x-2)k=(2k)!∑n=k∞‍T(n,k)x2n(2n)!.It follows from (17) or (18) that
(19)T(n,k)=T(n-1,k-1)+k2T(n-1,k),
with
(20)T(0,0)=1,T(n,0)=0(n∈ℕ),T(n,1)=1(n∈ℕ).
We also find from (18) that
(21)T(n,2)=14(4n-1-1),T(n,3)=9n360-4n60+124(n∈ℕ).The main purpose of this paper is to prove some formulas for the generalized sequence{Un(x)} and En(x). Some identities involving the sequence {Un(x)}, Stirling numbers s(n,k), and the central factorial numbers T(n,k) are deduced.


## 2. Main Results

Theorem 1.
Letn≥k(n,k∈ℕ) and
(22)q(n,k)=(-1)k∑j=kn‍(2j)!j!T(n,j)s(j,k).
Then,
(23)U2n(x)=∑k=1n‍q(n,k)xk.Remark 2.
By (15), (19), (20), and Theorem 1, we know that U2n(x) is a polynomial of x with integral coefficients. For example, by setting n=1,2,3,4 in Theorem 1, we get
(24)U2(x)=-2x,U4(x)=10x+12x2,U6(x)=-182x-300x2-120x3,U8(x)=6970x+13692x2+8400x3+1680x4.Takingx=1 in Theorem 1, we can obtain the following.Corollary 3.
Letn∈ℕ. Then,
(25)U2n=∑j=0n‍(-1)j(2j)!T(n,j).From Corollary3, we may immediately deduce the following results.Corollary 4.
Letn∈ℕ. Then,
(26)U2n≡-2(mod24),U2n≡-2+24T(n,2)(mod720),U2n≡-2+24T(n,2)-720T(n,3)(mod40320).Theorem 5.
Letn≥k(n,k∈ℕ). Then,
(27)U2n=∑k=1n‍q(n,k),U2n=2∑k=1[n/2]‍q(n,2k)-2=2∑k=1[(n-1)/2]‍q(n,2k+1)+2.Theorem 6.
Letn≥k(n,k∈ℕ). Suppose also that q(n,k) is defined by (22). Then,
(28)k!q(n,k)=(2n)!32n-k×∑v1,…,vk∈ℕ(v1+⋯+vk=n)(E2v1-1(0)-E2v1-1(23))‍⋯(E2vk-1(0)-E2vk-1(23))×((2v1)!⋯(2vk)!)-1.Theorem 7.
Letn∈ℕ. Then,
(29)-2∑k=0n-1(2n-12k)U2k=32n-1(E2n-1(0)-E2n-1(23)).Theorem 8.
Letn∈ℕ. Then,
(30)Un+1=∑k=0n-1‍(nk)((1-2n-k)Uk+1-2n-kUk).Theorem 9.
Letn∈ℕ0. Then,
(31)∑n=0∞‍1(n+1)!Un=13log2e-1-32(2-3)e-5+33.


## 3. Proofs of Theorems

Proof of Theorem1.
By (10), (13), and (18), we have
(32)∑n=0∞‍U2n(x)t2n(2n)!=(1et+e-t-1)x=(11+(et+e-t-2))x=∑j=0∞‍(-1)j(x+j-1j)(et+e-t-2)j=∑j=0∞‍(-1)j(x+j-1j)(2j)!∑n=j∞‍T(n,j)t2n(2n)!=∑n=0∞‍t2n(2n)!∑j=0n‍(-1)j(2j)!(x+j-1j)T(n,j),
which readily yields
(33)U2n(x)=∑j=0n‍(-1)j(2j)!(x+j-1j)T(n,j)=∑j=0n‍(-1)j(2j)!T(n,j)1j!x(x+1)⋯(x+j-1)=∑j=0n‍(2j)!j!T(n,j)∑k=1j‍(-1)ks(j,k)xk=∑k=1n‍(-1)k∑j=kn‍(2j)!j!T(n,j)s(j,k)xk=∑k=1n‍q(n,k)xk.
This completes the proof of Theorem 1.Proof of Theorem5.
By (10), we have
(34)∑n=0∞‍U2n(-1)t2n(2n)!=et+e-t-1=2∑n=0∞‍t2n(2n)!-1,
and U0(x)=1, thus
(35)∑n=1∞‍U2n(-1)t2n(2n)!=et+e-t-1=2∑n=1∞‍t2n(2n)!.
By Theorem 1 and comparing the coefficient of t2n/(2n)! on both sides of (35), we get
(36)∑k=1n‍q(n,k)(-1)k=U2n(-1)=2.
Again, by taking x=1 in Theorem 1, we have
(37)∑k=1n‍q(n,k)=U2n.
By (36) and (37), we immediately obtain (27). This completes the proof of Theorem 5.Proof of Theorem6.
By applying Theorem1, we have
(38)k!q(n,k)=dkdxk{Un(x)}|x=0.
On the other hand, it follows from (10) that
(39)∑n=k∞‍dkdxk{Un(x)}|x=0t2n(2n)!=(log(1et+e-t-1))k.
By using (38) and (39), we find that
(40)k!∑n=k∞‍q(n,k)t2n(2n)!=(log(1et+e-t-1))k.
We now note that
(41)ddt{log(1et+e-t-1)}=e-t-etet+e-t-1=e-t-et2(2ete3t+1+2e-te-3t+1)=12((2e3t+1-2e-3t+1)-(2e2te3t+1-2e-2te-3t+1))=12(∑n=0∞‍En(0)(3t)nn!-∑n=0∞‍En(0)(-3t)nn!)-12(∑n=0∞‍En(23)(3t)nn!-∑n=0∞‍En(23)(-3t)nn!)=∑n=0∞‍32n+1(E2n+1(0)-E2n+1(23))t2n+1(2n+1)!.
Hence,
(42)log1et+e-t-1=∑n=0∞‍32n+1(E2n+1(0)-E2n+1(23))t2n+2(2n+2)!=∑n=1∞‍32n-1(E2n-1(0)-E2n-1(23))t2n(2n)!
yields
(43)k!∑n=k∞‍q(n,k)t2n(2n)!=(∑n=1∞‍32n-1(E2n-1(0)-E2n-1(23))t2n(2n)!)k=∑n=k∞‍t2n(2n)!(2n)!32n-k×∑v1,…,vk∈ℕ(v1+⋯+vk=n)(E2v1-1(0)-E2v1-1(23))‍⋯(E2vk-1(0)-E2vk-1(23))×((2v1)!⋯(2vk)!)-1.
Comparing the coefficient of t2n/(2n)! on both sides of (43), we immediately get (28). This completes the proof of Theorem 6.Proof of Theorem7.
Consider(44)ddt{log(1et+e-t-1)}=e-t-etet+e-t-1=∑n=0∞‍U2nt2n(2n)!(-2∑n=0∞‍t2n+1(2n+1)!)=-2∑n=0∞‍∑k=0n‍(2n+12k)U2kt2n+1(2n+1)!.
Thus,
(45)log1et+e-t-1=-2∑n=1∞‍∑k=0n-1‍(2n-12k)U2kt2n(2n)!.
By (42) and (45) we obtain (29). This completes the proof of Theorem 7.Proof of Theorem8.
By using (7), we have
(46)∑n=1∞‍Untn-1(n-1)!=e-t-et(et+e-t-1)2.
Thus
(47)(e2t-et+1)∑n=1∞‍Untn-1(n-1)!=(1-e2t)∑n=0∞‍Untnn!,∑n=0∞‍(2n-1)tnn!∑n=0∞‍Un+1tnn!+∑n=0∞‍Un+1tnn!=∑n=0∞‍Untnn!-∑n=0∞‍2ntnn!∑n=0∞‍Untnn!.
That is,
(48)∑n=0∞‍∑k=0n‍(nk)(2n-k-1)Uk+1tnn!+∑n=0∞‍Un+1tnn!=∑n=0∞‍Untnn!-∑n=0∞‍∑k=0n‍(nk)2n-kUktnn!.
Comparing the coefficient of tn/n! on both sides of (48), we get the following:
(49)Un+1-Un=∑k=0n‍(nk)((1-2n-k)Uk+1-2n-kUk).
By (49) we immediately obtain (30). This completes the proof of Theorem 8.Proof of Theorem9.
By integrating (7) with respect to t from 0 to 1, we have
(50)∑n=0∞‍1(n+1)!Un=∫01‍1et+e-t-1dt=∫01‍1e2t-et+1det=∫1e‍1x2-x+1dx.
By (50) and ∫‍(1/(ax2+bx+c))dx=(1/b2-4ac)log|(2ax+b-b2-4ac)/(2ax+b+b2-4ac)|+c (c is constant), we have (31). This completes the proof of Theorem 9.



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