DISCRETE.MATHEMATICS ISRN Discrete Mathematics 2090-7788 Hindawi Publishing Corporation 810245 10.1155/2013/810245 810245 Research Article Some Properties of a Sequence Similar to Generalized Euler Numbers Wang Haiqing Liu Guodong Klostermeyer W. F. Prellberg T. Rim S. Smyth W. F. 1 Department of Mathematics Huizhou University Huizhou Guangdong 516007 China hzu.edu.cn 2013 11 3 2013 2013 27 12 2012 07 02 2013 2013 Copyright © 2013 Haiqing Wang and Guodong Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce the sequence {Un(x)} given by generating function (1/(et+e-t-1))x=n=0Un(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.

1. Introduction and Definitions

For a real or complex parameter α, the generalized Euler polynomials En(α)(x) are defined by the following generating function (see ) (1)(2et+1)αext=n=0En(α)(x)tnn!(|t|<π,1α:=1). Obviously, we have (2)En(1)(x)=En(x)(n0:={0}), in terms of the classical Euler polynomials En(x), being the set of positive integers. The classical Euler numbers En are given by the following: (3)En=2nEn(12)(n0).

The so-called the generalized Euler numbers E2n(x) are defined by (see [3, 5]) (4)(2et+e-t)x=n=0E2n(x)t2n(2n)!(|t|<π2,1x:=1). In fact, E2n(k)(k) are the Euler numbers of order k, 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=-2k=1[n/2](n2k)Un-2k,(n1), where (and in what follows) [x] is the greatest integer not exceeding x.

Clearly, U2n-1=0 for n1. 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 ) 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 ), (7)1et+e-t-1=n=0Untnn!=n=0U2nt2n(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=0Un(x)tnn!(|t|<13π,1x:=1). Obviously, (11)U0(x)=1,Un(1)=Un(n). By using (10), we can obtain (12)Un(k)=n!v1,,vk0(v1++vk=n)Uv1Uvkv1!vk!(k).

We now return to the Stirling numbers s(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=ks(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),s(n,n)=1(n0),s(n,1)=(-1)n-1(n-1)!(n),s(n,k)=0(k>nork<0).

The central factorial numbers T(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=kT(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),T(n,1)=1(n). We also find from (18) that (21)T(n,2)=14(4n-1-1),T(n,3)=9n360-4n60+124(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.

Let nk(n,k) and (22)q(n,k)=(-1)kj=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.

Taking x=1 in Theorem 1, we can obtain the following.

Corollary 3.

Let n. Then, (25)U2n=j=0n(-1)j(2j)!T(n,j).

From Corollary 3, we may immediately deduce the following results.

Corollary 4.

Let n. Then, (26)U2n-2(mod24),U2n-2+24T(n,2)(mod720),U2n-2+24T(n,2)-720T(n,3)(mod40320).

Theorem 5.

Let nk(n,k). Then, (27)U2n=k=1nq(n,k),U2n=2k=1[n/2]q(n,2k)-2=2k=1[(n-1)/2]q(n,2k+1)+2.

Theorem 6.

Let nk(n,k). Suppose also that q(n,k) is defined by (22). Then, (28)k!q(n,k)=(2n)!32n-k×v1,,vk(v1++vk=n)(E2v1-1(0)-E2v1-1(23))(E2vk-1(0)-E2vk-1(23))×((2v1)!(2vk)!)-1.

Theorem 7.

Let n. Then, (29)-2k=0n-1(2n-12k)U2k=32n-1(E2n-1(0)-E2n-1(23)).

Theorem 8.

Let n. Then, (30)Un+1=k=0n-1(nk)((1-2n-k)Uk+1-2n-kUk).

Theorem 9.

Let n0. Then, (31)n=01(n+1)!Un=13log2e-1-32(2-3)e-5+33.

3. Proofs of Theorems Proof of Theorem <xref ref-type="statement" rid="thm2.1">1</xref>.

By (10), (13), and (18), we have (32)n=0U2n(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=jT(n,j)t2n(2n)!=n=0t2n(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)kj=kn(2j)!j!T(n,j)s(j,k)xk=k=1nq(n,k)xk. This completes the proof of Theorem 1.

Proof of Theorem <xref ref-type="statement" rid="thm2.2">5</xref>.

By (10), we have (34)n=0U2n(-1)t2n(2n)!=et+e-t-1=2n=0t2n(2n)!-1, and U0(x)=1, thus (35)n=1U2n(-1)t2n(2n)!=et+e-t-1=2n=1t2n(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 Theorem <xref ref-type="statement" rid="thm2.3">6</xref>.

By applying Theorem 1, we have (38)k!q(n,k)=dkdxk{Un(x)}|x=0. On the other hand, it follows from (10) that (39)n=kdkdxk{Un(x)}|x=0t2n(2n)!=(log(1et+e-t-1))k. By using (38) and (39), we find that (40)k!n=kq(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=0En(0)(3t)nn!-n=0En(0)(-3t)nn!)-12(n=0En(23)(3t)nn!-n=0En(23)(-3t)nn!)=n=032n+1(E2n+1(0)-E2n+1(23))t2n+1(2n+1)!. Hence, (42)log1et+e-t-1=n=032n+1(E2n+1(0)-E2n+1(23))t2n+2(2n+2)!=n=132n-1(E2n-1(0)-E2n-1(23))t2n(2n)! yields (43)k!n=kq(n,k)t2n(2n)!=(n=132n-1(E2n-1(0)-E2n-1(23))t2n(2n)!)k=n=kt2n(2n)!(2n)!32n-k×v1,,vk(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 Theorem <xref ref-type="statement" rid="thm2.4">7</xref>.

Consider (44)ddt{log(1et+e-t-1)}=e-t-etet+e-t-1=n=0U2nt2n(2n)!(-2n=0t2n+1(2n+1)!)=-2n=0k=0n(2n+12k)U2kt2n+1(2n+1)!. Thus, (45)log1et+e-t-1=-2n=1k=0n-1(2n-12k)U2kt2n(2n)!. By (42) and (45) we obtain (29). This completes the proof of Theorem 7.

Proof of Theorem <xref ref-type="statement" rid="thm2.5">8</xref>.

By using (7), we have (46)n=1Untn-1(n-1)!=e-t-et(et+e-t-1)2. Thus (47)(e2t-et+1)n=1Untn-1(n-1)!=(1-e2t)n=0Untnn!,n=0(2n-1)tnn!n=0Un+1tnn!+n=0Un+1tnn!=n=0Untnn!-n=02ntnn!n=0Untnn!. That is, (48)n=0k=0n(nk)(2n-k-1)Uk+1tnn!+n=0Un+1tnn!=n=0Untnn!-n=0k=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 Theorem <xref ref-type="statement" rid="thm2.6">9</xref>.

By integrating (7) with respect to t from 0 to 1, we have (50)n=01(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.

Acknowledgments

This work is partly supported by the Social Science Foundation (no. 2012YB03) of Huizhou University and the Key Discipline Foundation (no. JG2011019) of Huizhou University.

Liu G. D. On congruences of Euler numbers modulo powers of two Journal of Number Theory 2008 128 12 3063 3071 10.1016/j.jnt.2008.04.003 MR2464854 ZBL1204.11051 Liu G. D. Srivastava H. M. Explicit formulas for the Nörlund polynomials Bn(x) and bn(x) Computers & Mathematics with Applications 2006 51 9-10 1377 1384 10.1016/j.camwa.2006.02.003 MR2237635 ZBL1161.11314 Liu G. D. Zhang W. P. Applications of an explicit formula for the generalized Euler numbers Acta Mathematica Sinica 2008 24 2 343 352 10.1007/s10114-007-1013-x MR2383361 ZBL1145.11020 Luke Y. L. The Special Functions and Their Approximations, Vol. I 1969 New York, NY, USA Academic Press MR0241700 Srivastava H. M. Liu G. D. Some identities and congruences involving a certain family of numbers Russian Journal of Mathematical Physics 2009 16 4 536 542 10.1134/S1061920809040086 MR2587810 ZBL1192.05014 Sun Z. H. Identities and congruences for a new sequence International Journal of Number Theory 2012 8 1 207 225 10.1142/S1793042112500121 MR2887891 Sun Z. H. Some properties of a sequence analogous to Euler numbers Bulletin of the Australian Mathematical Society 2012 10.1017/S0004972712000433 Jordan C. Calculus of Finite Differences 1965 New York, NY, USA Chelsea MR0183987 ZBL0156.04904 Liu G. D. The D numbers and the central factorial numbers Publicationes Mathematicae Debrecen 2011 79 1-2 41 53 10.5486/PMD.2011.4917 MR2850033 ZBL1249.11032 Liu G. D. A recurrence formula for D numbers D2n(2n-1) Discrete Dynamics in Nature and Society 2009 6 605313 10.1155/2009/605313 MR2563607 Nörlund N. E. Vorlesungen über Differenzenrechnung 1924 Berlin, Germany Springer Reprinted by Chelsea, Bronx, New York, NY, USA, 1954 Comtet L. Advanced Combinatorics 1974 Dordrecht, The Netherlands D. Reidel (Translated from the French by J. W. Nienhuys) MR0460128 ZBL0307.26011 Riordan J. Combinatorial Identities 1968 New York, NY, YSA John Wiley & Sons MR0231725