# Sums of Products of Cauchy Numbers, Including Poly-Cauchy Numbers

Authors: Takao Komatsu
Journal: Journal of Discrete Mathematics
 (2013)

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## Abstract

We investigate sums of products of Cauchy numbers including poly-Cauchy numbers:Tm(k)(n)=∑i1+⋯+im=n,i1,…,im≥0(ni1,…,im)ci1⋯cim-1cim(k)(m≥1,n≥0). A relation among these sums Tmkn shown in the paper and explicit expressions of sums of two and three products (the case of m=2 and that of m=3 described in the paper) are given. We also study the other three types of sums of products related to the Cauchy numbers of both kinds and the poly-Cauchy numbers of both kinds.


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## Body

## 1. Introduction

The Cauchy numbers (of the first kind)cn are defined by the integral of the falling factorial:
(1)cn=∫01x(x-1)⋯(x-n+1)dx=n!∫01(xn)dx
(see [1, Chapter VII]). The numbers cn/n! are sometimes called the Bernoulli numbers of the second kind (see e.g., [2, 3]). Such numbers have been studied by several authors [4–8] because they are related to various special combinatorial numbers, including Stirling numbers of both kinds, Bernoulli numbers, and harmonic numbers. It is interesting to see that the Cauchy numbers of the first kind cn have the similar properties and expressions to the Bernoulli numbers Bn. For example, the generating function of the Cauchy numbers of the first kind cn is expressed in terms of the logarithmic function:
(2)xln(1+x)=∑n=0∞cnxnn!,
(see [1, 6]), and the generating function of Bernoulli numbers Bn is expressed in terms of the exponential function:
(3)xex-1=∑n=0∞Bnxnn!,
(see [1]) or
(4)x1-e-x=∑n=0∞Bnxnn!
(see [9]). In addition, Cauchy numbers of the first kind cn can be written explicitly as
(5)cn=(-1)n∑m=0n[nm](-1)mm+1
(see [1, Chapter VII], [6, page 1908]), where [nm] are the (unsigned) Stirling numbers of the first kind, arising as coefficients of the rising factorial
(6)x(x+1)⋯(x+n-1)=∑m=0n[nm]xm
(see e.g., [10]). Bernoulli numbers Bn (in the latter definition) can be also written explicitly as
(7)Bn=(-1)n∑m=0n{nm}(-1)mm!m+1,
where {nm} are the Stirling numbers of the second kind, determined by
(8){nm}=1m!∑j=0m(-1)j(mj)(m-j)n
(see, e.g., [10]). Recently, Liu et al. [5] established some recurrence relations about Cauchy numbers of the first kind as analogous results about Bernoulli numbers by Agoh and Dilcher [11].In 1997 Kaneko [9] introduced the poly-Bernoulli numbers Bn(k) (n≥0, k≥1) by the generating function
(9)Lik(1-e-x)1-e-x=∑n=0∞Bn(k)xnn!,
where
(10)Lik(z)=∑m=0∞zmmk
is the kth polylogarithm function. When k=1, Bn(1)=Bn is the classical Bernoulli number with B1(1)=1/2. On the other hand, the author [12] introduced the poly-Cauchy numbers (of the first kind) cn(k) as a generalization of the Cauchy numbers and an analogue of the poly-Bernoulli numbers by the following:
(11)cn(k)=∫01⋯∫01‍︸k(x1x2⋯xk)(x1x2⋯xk-1)⋯(x1x2⋯xk-n+1)dx1dx2⋯dxk.In addition, the generating function of poly-Cauchy numbers is given by(12)Lifk(ln(1+x))=∑n=0∞cn(k)xnn!,
where
(13)Lifk(z):=∑m=0∞zmm!(m+1)k
is the kth polylogarithm factorial function, which is also introduced by the author [12, 13]. If k=1, then cn(1)=cn is the classical Cauchy number.The following identity on sums of two products of Bernoulli numbers is known as Euler’s formula:(14)∑i=0n(ni)BiBn-i=-nBn-1-(n-1)Bn(n≥0).The corresponding formula for Cauchy numbers was discovered in [8]:
(15)∑i=0n(ni)cicn-i=-(n-1)cn-n(n-2)cn-1(n≥0).In this paper, we shall give more analogous results by investigating a general type of sums of products of Cauchy numbers including poly-Cauchy numbers:(16)∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)ci1⋯cim-1︸m-1cim(k)(m≥1,n≥0),
whose Bernoulli version is discussed in [14]. A relation among these sums and explicit expressions of sums of two and three products are also given.


## 2. Main Results

We shall consider the sums of products of Cauchy numbers including poly-Cauchy numbers. Kamano [14] investigated the following types of sums of products:
(17)Sm(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)Bi1⋯Bim-1︸m-1Bim(k)(m≥1,n≥0),
where Bernoulli numbers Bn are defined by the generating function (3) and poly-Bernoulli numbers Bn(k) are defined by the generating function (9) and Lik(z) is the kth polylogarithm function defined in (10). It is shown [14] that
(18)∑l=0m(-1)m-l[m+1l+1]Sm+1(k-l)(n)={n!(n-m)!∑l=0m[ml]Bn-m+l(k),(n≥m),0,(0≤n≤m-1).Consider an analogous type of sums of products of Cauchy numbers including poly-Cauchy numbers:(19)Tm(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)ci1⋯cim-1︸m-1cim(k)(m≥1,n≥0).Then we show the following result.Theorem 1.
For an integerk and a nonnegative integer m, one has
(20)∑l=0m(-1)m-l[m+1l+1]Tm+1(k-l)(n)={∑l=0m∑i=0n-mn!i!(ln-m-i){ml}cl+i(k)(n≥m),0(0≤n≤m-1).Note that the generating function ofTm(k) is given by
(21)(xln(1+x))m-1Lifk(ln(1+x))=∑n=0∞Tm(k)(n)xnn!.Put(22)Gk(x):=Lifk(ln(1+x))=∑n=0∞cn(k)xnn!.Since(23)Lif1(z)=ez-1z,Lif0(z)=ez,Lif-1(z)=(z+1)ez,
we have
(24)G1(x)=xln(1+x),G0(x)=1+x,G-1(x)=(1+x)(ln(1+x)+1).Since(25)xmdldxlGk(x)=∑i=0∞cl+i(k)xm+ii!(m,l≥0,k≥1),
the coefficient of xn in
(26)xmdldxlGk(x)
is equal to
(27)cn-m+l(k)(n-m)!(n≥m),0(0≤n≤m-1).We need the following lemma in order to prove Theorem1.Lemma 2.
For an integerk and a positive integer m, one has
(28)(∑i=1m{mi}1(1+x)m-ididxi)Gk(x)=1(1+x)m(ln(1+x))m∑l=0m(-1)m-l[m+1l+1]Gk-l(x).Proof of Lemma2.
Since(29)ddxLifk(x)=1x∑m=0∞mxmm!(m+1)k=1x∑m=0∞(xmm!(m+1)k-1-xmm!(m+1)k)=Lifk-1(x)-Lifk(x)x,
we have
(30)ddxGk(x)=Gk-1(x)-Gk(x)(1+x)ln(1+x).
By induction, we can show that form≥1(31)dmdxmGk(x)=∑ν=1m(-1)m-ν(1+x)m(log(1+x))ν[mν]gν+1(x),
where
(32)gν+1(x):=∑l=0ν(-1)ν-l[ν+1l+1]Gk-l(x)(ν=1,2,…,m).
Thus, by using the inversion relationship(33)∑j=νm(-1)j-ν{mj}[jν]={1(ν=m);0(ν=1,2,…,m-1)
(see e.g., [10, Chapter 6]), the left-hand side of the identity in the previous lemma is equal to
(34)∑j=1m{mj}1(1+x)m-j∑ν=1j(-1)j-ν(1+x)j(ln(1+x))ν[jν]gν+1(x)=1(1+x)m∑ν=1mgν+1(x)(ln(1+x))ν∑j=νm(-1)j-ν{mj}[jν]=gm+1(x)(1+x)m(ln(1+x))m,
which is the right-hand side of the desired identity.Now, by the generating function (21), the identity (25), and Lemma 2,
(35)∑n=0∞(∑l=0m(-1)m-l[m+1l+1]Tm+1(k-l)(n))xnn!=(xln(1+x))m∑l=0m(-1)m-l[m+1l+1]Gk-l(x)=xm(∑i=1m(1+x)i{mi}didxi)Gk(x)=∑l=0m(1+x)l{ml}∑i=0∞cl+i(k)xm+ii!=∑l=0m∑κ=0l(lκ)xκ{ml}∑i=0∞cl+i(k)xm+ii!=∑l=0m∑n=0∞∑i=0n(ln-i){ml}cl+i(k)i!xm+n=∑n=m∞(∑l=0m∑i=0n-mn!i!(ln-m-i){ml}cl+i(k))xnn!.Note that(lν)=0 (l<ν) and {m0}=0 (m≥1). Therefore,
(36)∑l=0m(-1)m-l[m+1l+1]Tm+1(k-l)(n)={∑l=0m∑i=0n-mn!i!(ln-m-i){ml}cl+i(k)(n≥m);0(0≤n≤m-1).If we putm=1 in Theorem 1, we get an analogous formula to Euler's formula (14) for sums of products of Cauchy number and a poly-Cauchy number.Corollary 3.
One has(37)∑i=0n(ni)ci(cn-i(k-1)-cn-i(k))=n(n-1)cn-1(k)+ncn(k)(n≥0).


### 2.1. Explicit Formula forT2(k)(n)

Theorem1 gives only relations among sums of products Tm(k)(n). For m=2, an explicit formula for
(38)T2(k)(n)=∑i=0n(ni)cicn-i(k)
is given.Theorem 4.
Forn≥0 and k≥1 one has
(39)T2(0)(n)=cn(1)(-1)=(-1)n∑i=0n∑m=in(-1)m[nm](mi)1m-i+1,(40)T2(k)(n)=cn(1)(-1)-n∑j=1k(cn(j)+(n-1)cn-1(j)),(41)T2(-k)(n)=cn(1)(-1)+n∑j=0k-1(cn(-j)+(n-1)cn-1(-j)),
where cn(1)(-1)=cn+ncn-1.Proof.
Consider(42)∑n=0∞T2(0)(n)xnn!=xln(1+x)Lif0(ln(1+x))=(1+x)xln(1+x)=∑n=0∞cn(1)(-1)xnn!,
where cn(k)(z) are poly-Cauchy polynomials of the first kind, defined by the generating function
(43)Lifk(ln(1+x))(1+x)z=∑n=0∞cn(k)(z)xnn!.cn(k)(z) are expressed explicitly in terms of the Stirling numbers of the first kind [13, Theorem 1]:
(44)cn(k)(z)=∑m=0n[nm](-1)n-m∑i=0m(mi)(-z)i(m-i+1)k.
Hence, the identity (39) holds because
(45)T2(0)(n)=cn(1)(-1)=∑m=0n[nm](-1)n-m∑i=0m(mi)1m-i+1=(-1)n∑i=0n∑m=in(-1)m[nm](mi)1m-i+1.
Next, by (30) and G0(x)=1+x we have
(46)∑j=1kddxGj(x)=∑j=1kGj-1(x)-Gj(x)(1+x)ln(1+x)=1ln(1+x)-Gk(x)(1+x)ln(1+x).
Hence,(47)∑n=0∞T2(k)(n)xnn!=xln(1+x)Gk(x)=x(1+x)ln(1+x)-x(1+x)∑j=1kddxGj(x)=∑n=0∞cn(1)(-1)xnn!-x(1+x)∑j=1k∑n=0∞cn+1(j)xnn!=∑n=0∞cn(1)(-1)xnn!-∑n=0∞∑j=1k(ncn(j)+n(n-1)cn-1(j))xnn!.
Therefore, we get the identity (40).
Finally, by(48)ddxLif-k(x)=Lif-k-1(x)-Lif-k(x)x,
we have
(49)∑j=0k-1ddxG-j(x)=∑j=0k-1G-j-1(x)-G-j(x)(1+x)ln(1+x)=G-k(x)(1+x)ln(1+x)-1ln(1+x).
Hence,(50)∑n=0∞T2(-k)(n)xnn!=xln(1+x)G-k(x)=x(1+x)ln(1+x)+x(1+x)∑j=0k-1ddxG-j(x)=∑n=0∞cn(1)(-1)xnn!+x(1+x)∑j=0k-1∑n=0∞cn+1(-j)xnn!=∑n=0∞cn(1)(-1)xnn!+∑n=0∞∑j=0k-1(ncn(-j)+n(n-1)cn-1(-j))xnn!.
Therefore, we get the identity (41).Puttingk=1 in (40), we have the following identity, which is also found in [8]. This is also an analogous formula to Euler’s formula (14).Corollary 5.
One has (see also Table1)
(51)∑i=0n(ni)cicn-i=-(n-1)cn-n(n-2)cn-1(n≥0).Table 1
n
0
1
2
3
4
5
6
7
8
9
c
n
1
1
/
2
-
1
/
6
1
/
4
-
19
/
30
9
/
4
-
863
/
84
1375
/
24
-
33953
/
90
57281
/
20
c
n
(
1
)
(
-
1
)
1
3
/
2
5
/
6
-
1
/
4
11
/
30
-
11
/
12
271
/
84
-
117
/
8
7297
/
90
-
2125
/
4Proof.
Since(52)∑n=0∞T2(0)(n)xnn!=(1+x)xln(1+x)=(1+x)∑n=0∞cnxnn!=∑n=0∞(cn+ncn-1)xnn!,
we have cn(1)(-1)=cn+ncn-1. Hence,
(53)∑i=0n(ni)cicn-i=T2(1)(n)=cn(1)(-1)-n(cn+(n-1)cn-1)=-(n-1)cn-n(n-2)cn-1.


### 2.2. Explicit Formulae forT3(k)(n)

Form=3, an explicit formula for
(54)T3(k)(n)=∑i=0n∑j=0n-in!i!j!(n-i-j)!cicjcn-i-j(k)
is also given.Theorem 6.
Forn≥0 and k≥1 one has
(55)T3(0)(n)=-(n-1)cn-2n(n-2)cn-1-n(n-1)(n-3)cn-2,(56)T3(k)(n)=T3(0)(n)-(1-2-k)n(cn-1+(n-1)cn-2)+n(n-1)∑j=1k(1-2j-k-1)×((n-2)2cn-2(j)+(2n-3)cn-1(j)+cn(j)),(57)T3(-k)(n)=T3(0)(n)+(2k-1)n(cn-1+(n-1)cn-2)+n(n-1)∑j=0k-2(2k-j-1-1)×((n-2)2cn-2(-j)+(2n-3)cn-1(-j)+cn(-j)).Proof.
Consider(58)∑n=0∞T3(0)(n)xnn!=x2(1+x)(ln(1+x))2=(1+x)∑n=0∞T2(1)(n)xnn!=∑n=0∞(T2(1)(n)+nT2(1)(n-1))xnn!.
Hence, by Corollary5(59)T3(0)(n)=T2(1)(n)+nT2(1)(n-1)=-(n-1)cn-n(n-2)cn-1+n(-(n-2)cn-1-(n-1)(n-3)cn-2)=-(n-1)cn-2n(n-2)cn-1-n(n-1)(n-3)cn-2,
which is the identity (55).
By Lemma2 with m=2(60)(d2dx2+11+xddx)Gk(x)=(2Gk(x)-Gk-1(x))-(2Gk-1(x)-Gk-2(x))(1+x)2(ln(1+x))2.
Thus,(61)∑j=1l(d2dx2+11+xddx)Gj(x)=2Gl(x)-Gl-1(x)(1+x)2(ln(1+x))2-2G0(x)-G-1(x)(1+x)2(ln(1+x))2.
By multiplying both sides by2l-1 and summing over l from 1 to k, we obtain
(62)2kGk(x)(1+x)2(ln(1+x))2=G0(x)(1+x)2(ln(1+x))2+(∑l=1k2l-1)2G0(x)-G-1(x)(1+x)2(ln(1+x))2+∑l=1k2l-1∑j=1l(d2dx2+11+xddx)Gj(x)=(2k+1-1)G0(x)-(2k-1)G-1(x)(1+x)2(ln(1+x))2+∑j=1k(∑l=jk2l-1)(d2dx2+11+xddx)Gj(x).
Hence, we have(63)2k(xln(1+x))2Gk(x)=(2k+1-1)(1+x)x2(ln(1+x))2-(2k-1)(1+x)x2(ln(1+x)+1)(ln(1+x))2+x2(1+x)2∑j=1k(2k-2j-1)(d2dx2+11+xddx)Gj(x).
By comparing the coefficients ofxn/n! in both sides,
(64)2kT3(k)(n)=(2k+1-1)T3(0)(n)-(2k-1)nT2(0)(n-1)-(2k-1)T3(0)(n)+∑j=1k(2k-2j-1)×(n(n-1)cn(j)+2n(n-1)(n-2)cn-1(j)+n(n-1)(n-2)(n-3)cn-2(j)+n(n-1)cn-1(j)+n(n-1)(n-2)cn-2(j))=2kT3(0)(n)-(2k-1)nT2(0)(n-1)+∑j=1k(2k-2j-1)n(n-1)×((n-2)2cn-2(j)+(2n-3)cn-1(j)+cn(j)).
Dividing both sides by2k, we have the identity (56).
Finally, since(65)∑j=0l-2(d2dx2+11+xddx)G-j(x)=∑j=0l-2(2G-j(x)-G-j-1)-(2G-j-1(x)-G-j-2)(1+x)2(ln(1+x))2=∑j=0l-2(2G0(x)-G-1(1+x)2(ln(1+x))2-2G-l+1(x)-G-l(1+x)2(ln(1+x))2),
we obtain
(66)2-kG-k(x)(1+x)2(ln(1+x))2=(1-2-k)G-1(x)-(1-2-k+1)G0(x)(1+x)2(ln(1+x))2+∑j=0k-2(∑l=j+2k2-l)(d2dx2+11+xddx)G-j(x).
Hence, we have(67)2-k(xln(1+x))2G-k(x)=(1-2-k)(1+x)x2ln(1+x)+2-k(1+x)x2(ln(1+x))2+x2(1+x)2∑j=0k-2(2-j-1-2-k)×(d2dx2+11+xddx)G-j(x),
yielding the identity (57).


### 2.3. Poly-Cauchy Numbers of the Second Kind

Similarly, defineT^m(k)(n) by
(68)T^m(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)c^i1⋯c^im-1︸m-1c^im(k)(m≥1,n≥0),
where c^n(k) is poly-Cauchy number of the second kind [12], whose generating function is given by
(69)Lifk(-ln(1+x))=∑n=0∞c^n(k)xnn!.c^n(k) can be also defined by
(70)c^n(k)=∫01‍⋯∫01‍︸k(-x1x2⋯xk)(-x1x2⋯xk-1)⋯(-x1x2⋯xk-n+1)dx1dx2⋯dxk
(see [12]). In this sense, cn(k) is called poly-Cauchy number of the first kind. When k=1, c^n=c^n(1) is the classical Cauchy number of the second kind, whose generating function is given by
(71)x(1+x)ln(1+x)=∑n=0∞c^nxnn!.By using the corresponding lemma to Lemma2, where Gk(x) is replaced by G^k(x)=Lifk(-ln(1+x)), we can obtain the following result.Theorem 7.
For an integerk and a nonnegative integer m, one has
(72)∑l=0m(-1)m-l[m+1l+1]T^m+1(k-1)(n)={(-1)n-m∑l=0m-1∑i=0n-m(-1)in!i!(n-l-i-1n-m-i){ml}c^l+i(k)+n!(n-m)!c^n(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 7, one has the following.Corollary 8.
(73)
∑
i
=
0
n
(
n
i
)
c
^

i
(
c
^

n
-
i
(
k
-
1
)
-
c
^

n
-
i
(
k
)
)
=
n
c
^

n
(
k
)
(
n
≥
0
)
.Consider the casem=2. Note that the generating function of poly-Cauchy polynomial of the second kind c^n(k)(z) [13] is given by
(74)(1+x)zLifk(-ln(1+x))=∑n=0∞c^n(k)(z)xnn!.c^n(k)(z) are expressed explicitly in terms of the Stirling numbers of the first kind [13, Theorem 4]
(75)c^n(k)(z)=∑m=0n[nm](-1)n∑i=0m(mi)(-z)i(m-i+1)k.Hence,(76)c^n(1)(-1)=∑m=0n[nm](-1)n∑i=0m(mi)1m-i+1=(-1)n∑i=0n∑m=in[nm](mi)1m-i+1.On the other hand,(77)∑n=0∞c^n(1)(-1)xnn!=x(1+x)2ln(1+x)=11+x∑n=0∞c^nxnn!=(∑μ=0∞(-x)μ)(∑ν=0∞c^νxνν!)=∑n=0∞(-1)n∑i=0nn!i!(-1)ic^ixnn!.Thus,(78)c^n(1)(-1)=(-1)n∑i=0nn!i!(-1)ic^i
(see Table 2).Table 2
n
0
1
2
3
4
5
6
7
c
^
n
1
-
1
/
2
5
/
6
-
9
/
4
251
/
30
-
475
/
12
19087
/
84
-
36799
/
24
c
^
n
(
1
)
(
-
1
)
1
-
3
/
2
23
/
6
-
55
/
4
1901
/
30
-
4277
/
12
198721
/
84
-
144747
/
8Theorem 9.
Forn≥0 and k≥1 one has
(79)T^2(0)(n)=c^n(1)(-1),(80)T^2(k)(n)=c^n(1)(-1)-n∑j=1kc^n(j),(81)T^2(-k)(n)=c^n(1)(-1)+n∑j=0k-1c^n(-j).Puttingk=1 in (80), we have the following identity. This is also an analogous formula to Euler’s formula (14).Corollary 10.
One has(82)∑i=0n(ni)c^ic^n-i=c^n(1)(-1)-nc^n(n≥0).Ifm=3, then T^3(k) can be expressed as follows.Theorem 11.
Forn≥0 and k≥1 one has
(83)T^3(0)(n)=(-1)n∑l=0n(-1)ln!l!T^2(1)(l)=(-1)n∑l=0n(-1)ln!l!(c^l(1)(-1)-lc^l),T^3(k)(n)=T^3(0)(n)+(1-2-k)nc^n-1(1)(-2)+n(n-1)∑j=1k(1-2j-k-1)×(c^n(j)+(-1)n(n-2)!∑l=0n-2(-1)lc^l+1(j)l!),T^3(-k)(n)=T^3(0)(n)-(2k-1)nc^n-1(1)(-2)+n(n-1)∑j=0k-2(2k-j-1-1)×(c^n(-j)+(-1)n(n-2)!∑l=0n-2(-1)lc^l+1(-j)l!).Remark 12.
Note that(84)nc^n-1(1)(-2)=(-1)n-1n!∑l=0n-1(n-l)(-1)lc^ll!.


### 2.4. Two Kinds of Poly-Cauchy Numbers

DefineUm(k)(n) by
(85)Um(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)ci1⋯cim-1︸m-1c^im(k)(m≥1,n≥0).Then we obtain the following.Theorem 13.
For an integerk and a nonnegative integer m, one has
(86)∑l=0m(-1)m-l[m+1l+1]Um+1(k-l)(n)={∑l=0m∑i=0n-mn!i!(ln-m-i){ml}c^l+i(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 13, we have the following.Corollary 14.
(87)
∑
i
=
0
n
(
n
i
)
c
i
(
c
^

n
-
i
(
k
-
1
)
-
c
^

n
-
i
(
k
)
)
=
n
(
(
n
-
1
)
c
^

n
-
1
(
k
)
+
c
^

n
(
k
)
)
(
n
≥
0
)
.Ifm=2, then U2(k) can be expressed explicitly.Theorem 15.
Forn≥0 and k≥1 one has
(88)U2(0)(n)=c^n,(89)U2(k)(n)=c^n-n∑j=1k(c^n(j)+(n-1)c^n-1(j)),(90)U2(-k)(n)=c^n+n∑j=0k-1(c^n(-j)+(n-1)c^n-1(-j)).Puttingk=1 in (89), we have the alternative identity (2.3) in [6, Theorem 2.4] because cn=c^n+nc^n-1 by (2.2) in [6, Theorem 2.4].Corollary 16.
One has(91)∑i=0n(ni)cic^n-i=-(n-1)(c^n+nc^n-1)(n≥1).Ifm=3, then U3(k) can be expressed as follows.Theorem 17.
Forn≥0 and k≥1 one has
(92)U3(0)(n)=-(n-1)(c^n+nc^n-1)=-(n-1)cn,(93)U3(k)(n)=U3(0)(n)+(1-2-k)nc^n-1+n(n-1)∑j=1k(1-2j-k-1)×((n-2)2c^n-2(j)+(2n-3)c^n-1(j)+c^n(j)),(94)U3(-k)(n)=U3(0)(n)-(2k-1)nc^n-1+n(n-1)∑j=0k-2(2k-j-1-1)×((n-2)2c^n-2(-j)+(2n-3)c^n-1(-j)+c^n(-j)).DefineVm(k)(n) by
(95)Vm(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)c^i1⋯c^im-1︸m-1cim(k)(m≥1,n≥0).Then we obtain the following.Theorem 18.
For an integerk and a nonnegative integer m, one has
(96)∑l=0m(-1)m-l[m+1l+1]Vm+1(k-l)(n)={(-1)n-m∑l=0m-1∑i=0n-m(-1)in!i!(n-l-i-1n-m-i){ml}cl+i(k)+n!(n-m)!cn(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 18, we have the following.Corollary 19.
(97)
∑
i
=
0
n
(
n
i
)
c
^

i
(
c
n
-
i
(
k
-
1
)
-
c
n
-
i
(
k
)
)
=
n
c
n
(
k
)
(
n
≥
0
)
.Ifm=2, then V2(k) can be expressed explicitly.Theorem 20.
Forn≥0 and k≥1 one has
(98)V2(0)(n)=cn,(99)V2(k)(n)=cn-n∑j=1kcn(j),(100)V2(-k)(n)=cn+n∑j=0k-1cn(-j).Puttingk=1 in (99), we have the identity (2.3) in [6, Theorem 2.4].Corollary 21.
One has(101)∑i=0n(ni)c^icn-i=-(n-1)cn(n≥1).Ifm=3, then V3(k) can be expressed as follows.Theorem 22.
Forn≥0 and k≥1 one has
(102)V3(0)(n)=-(n-1)cn,V3(k)(n)=V3(0)(n)-(1-2-k)nc^n-1+n(n-1)∑j=1k(1-2j-k-1)×(cn(j)+(-1)n(n-2)!∑l=0n-2(-1)lcl+1(j)l!),V3(-k)(n)=V3(0)(n)+(2k-1)nc^n-1+n(n-1)∑j=0k-2(2k-j-1-1)×(cn(-j)+(-1)n(n-2)!∑l=0n-2(-1)lcl+1(-j)l!).


## 2.1. Explicit Formula forT2(k)(n)

Theorem1 gives only relations among sums of products Tm(k)(n). For m=2, an explicit formula for
(38)T2(k)(n)=∑i=0n(ni)cicn-i(k)
is given.Theorem 4.
Forn≥0 and k≥1 one has
(39)T2(0)(n)=cn(1)(-1)=(-1)n∑i=0n∑m=in(-1)m[nm](mi)1m-i+1,(40)T2(k)(n)=cn(1)(-1)-n∑j=1k(cn(j)+(n-1)cn-1(j)),(41)T2(-k)(n)=cn(1)(-1)+n∑j=0k-1(cn(-j)+(n-1)cn-1(-j)),
where cn(1)(-1)=cn+ncn-1.Proof.
Consider(42)∑n=0∞T2(0)(n)xnn!=xln(1+x)Lif0(ln(1+x))=(1+x)xln(1+x)=∑n=0∞cn(1)(-1)xnn!,
where cn(k)(z) are poly-Cauchy polynomials of the first kind, defined by the generating function
(43)Lifk(ln(1+x))(1+x)z=∑n=0∞cn(k)(z)xnn!.cn(k)(z) are expressed explicitly in terms of the Stirling numbers of the first kind [13, Theorem 1]:
(44)cn(k)(z)=∑m=0n[nm](-1)n-m∑i=0m(mi)(-z)i(m-i+1)k.
Hence, the identity (39) holds because
(45)T2(0)(n)=cn(1)(-1)=∑m=0n[nm](-1)n-m∑i=0m(mi)1m-i+1=(-1)n∑i=0n∑m=in(-1)m[nm](mi)1m-i+1.
Next, by (30) and G0(x)=1+x we have
(46)∑j=1kddxGj(x)=∑j=1kGj-1(x)-Gj(x)(1+x)ln(1+x)=1ln(1+x)-Gk(x)(1+x)ln(1+x).
Hence,(47)∑n=0∞T2(k)(n)xnn!=xln(1+x)Gk(x)=x(1+x)ln(1+x)-x(1+x)∑j=1kddxGj(x)=∑n=0∞cn(1)(-1)xnn!-x(1+x)∑j=1k∑n=0∞cn+1(j)xnn!=∑n=0∞cn(1)(-1)xnn!-∑n=0∞∑j=1k(ncn(j)+n(n-1)cn-1(j))xnn!.
Therefore, we get the identity (40).
Finally, by(48)ddxLif-k(x)=Lif-k-1(x)-Lif-k(x)x,
we have
(49)∑j=0k-1ddxG-j(x)=∑j=0k-1G-j-1(x)-G-j(x)(1+x)ln(1+x)=G-k(x)(1+x)ln(1+x)-1ln(1+x).
Hence,(50)∑n=0∞T2(-k)(n)xnn!=xln(1+x)G-k(x)=x(1+x)ln(1+x)+x(1+x)∑j=0k-1ddxG-j(x)=∑n=0∞cn(1)(-1)xnn!+x(1+x)∑j=0k-1∑n=0∞cn+1(-j)xnn!=∑n=0∞cn(1)(-1)xnn!+∑n=0∞∑j=0k-1(ncn(-j)+n(n-1)cn-1(-j))xnn!.
Therefore, we get the identity (41).Puttingk=1 in (40), we have the following identity, which is also found in [8]. This is also an analogous formula to Euler’s formula (14).Corollary 5.
One has (see also Table1)
(51)∑i=0n(ni)cicn-i=-(n-1)cn-n(n-2)cn-1(n≥0).Table 1
n
0
1
2
3
4
5
6
7
8
9
c
n
1
1
/
2
-
1
/
6
1
/
4
-
19
/
30
9
/
4
-
863
/
84
1375
/
24
-
33953
/
90
57281
/
20
c
n
(
1
)
(
-
1
)
1
3
/
2
5
/
6
-
1
/
4
11
/
30
-
11
/
12
271
/
84
-
117
/
8
7297
/
90
-
2125
/
4Proof.
Since(52)∑n=0∞T2(0)(n)xnn!=(1+x)xln(1+x)=(1+x)∑n=0∞cnxnn!=∑n=0∞(cn+ncn-1)xnn!,
we have cn(1)(-1)=cn+ncn-1. Hence,
(53)∑i=0n(ni)cicn-i=T2(1)(n)=cn(1)(-1)-n(cn+(n-1)cn-1)=-(n-1)cn-n(n-2)cn-1.


## 2.2. Explicit Formulae forT3(k)(n)

Form=3, an explicit formula for
(54)T3(k)(n)=∑i=0n∑j=0n-in!i!j!(n-i-j)!cicjcn-i-j(k)
is also given.Theorem 6.
Forn≥0 and k≥1 one has
(55)T3(0)(n)=-(n-1)cn-2n(n-2)cn-1-n(n-1)(n-3)cn-2,(56)T3(k)(n)=T3(0)(n)-(1-2-k)n(cn-1+(n-1)cn-2)+n(n-1)∑j=1k(1-2j-k-1)×((n-2)2cn-2(j)+(2n-3)cn-1(j)+cn(j)),(57)T3(-k)(n)=T3(0)(n)+(2k-1)n(cn-1+(n-1)cn-2)+n(n-1)∑j=0k-2(2k-j-1-1)×((n-2)2cn-2(-j)+(2n-3)cn-1(-j)+cn(-j)).Proof.
Consider(58)∑n=0∞T3(0)(n)xnn!=x2(1+x)(ln(1+x))2=(1+x)∑n=0∞T2(1)(n)xnn!=∑n=0∞(T2(1)(n)+nT2(1)(n-1))xnn!.
Hence, by Corollary5(59)T3(0)(n)=T2(1)(n)+nT2(1)(n-1)=-(n-1)cn-n(n-2)cn-1+n(-(n-2)cn-1-(n-1)(n-3)cn-2)=-(n-1)cn-2n(n-2)cn-1-n(n-1)(n-3)cn-2,
which is the identity (55).
By Lemma2 with m=2(60)(d2dx2+11+xddx)Gk(x)=(2Gk(x)-Gk-1(x))-(2Gk-1(x)-Gk-2(x))(1+x)2(ln(1+x))2.
Thus,(61)∑j=1l(d2dx2+11+xddx)Gj(x)=2Gl(x)-Gl-1(x)(1+x)2(ln(1+x))2-2G0(x)-G-1(x)(1+x)2(ln(1+x))2.
By multiplying both sides by2l-1 and summing over l from 1 to k, we obtain
(62)2kGk(x)(1+x)2(ln(1+x))2=G0(x)(1+x)2(ln(1+x))2+(∑l=1k2l-1)2G0(x)-G-1(x)(1+x)2(ln(1+x))2+∑l=1k2l-1∑j=1l(d2dx2+11+xddx)Gj(x)=(2k+1-1)G0(x)-(2k-1)G-1(x)(1+x)2(ln(1+x))2+∑j=1k(∑l=jk2l-1)(d2dx2+11+xddx)Gj(x).
Hence, we have(63)2k(xln(1+x))2Gk(x)=(2k+1-1)(1+x)x2(ln(1+x))2-(2k-1)(1+x)x2(ln(1+x)+1)(ln(1+x))2+x2(1+x)2∑j=1k(2k-2j-1)(d2dx2+11+xddx)Gj(x).
By comparing the coefficients ofxn/n! in both sides,
(64)2kT3(k)(n)=(2k+1-1)T3(0)(n)-(2k-1)nT2(0)(n-1)-(2k-1)T3(0)(n)+∑j=1k(2k-2j-1)×(n(n-1)cn(j)+2n(n-1)(n-2)cn-1(j)+n(n-1)(n-2)(n-3)cn-2(j)+n(n-1)cn-1(j)+n(n-1)(n-2)cn-2(j))=2kT3(0)(n)-(2k-1)nT2(0)(n-1)+∑j=1k(2k-2j-1)n(n-1)×((n-2)2cn-2(j)+(2n-3)cn-1(j)+cn(j)).
Dividing both sides by2k, we have the identity (56).
Finally, since(65)∑j=0l-2(d2dx2+11+xddx)G-j(x)=∑j=0l-2(2G-j(x)-G-j-1)-(2G-j-1(x)-G-j-2)(1+x)2(ln(1+x))2=∑j=0l-2(2G0(x)-G-1(1+x)2(ln(1+x))2-2G-l+1(x)-G-l(1+x)2(ln(1+x))2),
we obtain
(66)2-kG-k(x)(1+x)2(ln(1+x))2=(1-2-k)G-1(x)-(1-2-k+1)G0(x)(1+x)2(ln(1+x))2+∑j=0k-2(∑l=j+2k2-l)(d2dx2+11+xddx)G-j(x).
Hence, we have(67)2-k(xln(1+x))2G-k(x)=(1-2-k)(1+x)x2ln(1+x)+2-k(1+x)x2(ln(1+x))2+x2(1+x)2∑j=0k-2(2-j-1-2-k)×(d2dx2+11+xddx)G-j(x),
yielding the identity (57).


## 2.3. Poly-Cauchy Numbers of the Second Kind

Similarly, defineT^m(k)(n) by
(68)T^m(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)c^i1⋯c^im-1︸m-1c^im(k)(m≥1,n≥0),
where c^n(k) is poly-Cauchy number of the second kind [12], whose generating function is given by
(69)Lifk(-ln(1+x))=∑n=0∞c^n(k)xnn!.c^n(k) can be also defined by
(70)c^n(k)=∫01‍⋯∫01‍︸k(-x1x2⋯xk)(-x1x2⋯xk-1)⋯(-x1x2⋯xk-n+1)dx1dx2⋯dxk
(see [12]). In this sense, cn(k) is called poly-Cauchy number of the first kind. When k=1, c^n=c^n(1) is the classical Cauchy number of the second kind, whose generating function is given by
(71)x(1+x)ln(1+x)=∑n=0∞c^nxnn!.By using the corresponding lemma to Lemma2, where Gk(x) is replaced by G^k(x)=Lifk(-ln(1+x)), we can obtain the following result.Theorem 7.
For an integerk and a nonnegative integer m, one has
(72)∑l=0m(-1)m-l[m+1l+1]T^m+1(k-1)(n)={(-1)n-m∑l=0m-1∑i=0n-m(-1)in!i!(n-l-i-1n-m-i){ml}c^l+i(k)+n!(n-m)!c^n(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 7, one has the following.Corollary 8.
(73)
∑
i
=
0
n
(
n
i
)
c
^

i
(
c
^

n
-
i
(
k
-
1
)
-
c
^

n
-
i
(
k
)
)
=
n
c
^

n
(
k
)
(
n
≥
0
)
.Consider the casem=2. Note that the generating function of poly-Cauchy polynomial of the second kind c^n(k)(z) [13] is given by
(74)(1+x)zLifk(-ln(1+x))=∑n=0∞c^n(k)(z)xnn!.c^n(k)(z) are expressed explicitly in terms of the Stirling numbers of the first kind [13, Theorem 4]
(75)c^n(k)(z)=∑m=0n[nm](-1)n∑i=0m(mi)(-z)i(m-i+1)k.Hence,(76)c^n(1)(-1)=∑m=0n[nm](-1)n∑i=0m(mi)1m-i+1=(-1)n∑i=0n∑m=in[nm](mi)1m-i+1.On the other hand,(77)∑n=0∞c^n(1)(-1)xnn!=x(1+x)2ln(1+x)=11+x∑n=0∞c^nxnn!=(∑μ=0∞(-x)μ)(∑ν=0∞c^νxνν!)=∑n=0∞(-1)n∑i=0nn!i!(-1)ic^ixnn!.Thus,(78)c^n(1)(-1)=(-1)n∑i=0nn!i!(-1)ic^i
(see Table 2).Table 2
n
0
1
2
3
4
5
6
7
c
^
n
1
-
1
/
2
5
/
6
-
9
/
4
251
/
30
-
475
/
12
19087
/
84
-
36799
/
24
c
^
n
(
1
)
(
-
1
)
1
-
3
/
2
23
/
6
-
55
/
4
1901
/
30
-
4277
/
12
198721
/
84
-
144747
/
8Theorem 9.
Forn≥0 and k≥1 one has
(79)T^2(0)(n)=c^n(1)(-1),(80)T^2(k)(n)=c^n(1)(-1)-n∑j=1kc^n(j),(81)T^2(-k)(n)=c^n(1)(-1)+n∑j=0k-1c^n(-j).Puttingk=1 in (80), we have the following identity. This is also an analogous formula to Euler’s formula (14).Corollary 10.
One has(82)∑i=0n(ni)c^ic^n-i=c^n(1)(-1)-nc^n(n≥0).Ifm=3, then T^3(k) can be expressed as follows.Theorem 11.
Forn≥0 and k≥1 one has
(83)T^3(0)(n)=(-1)n∑l=0n(-1)ln!l!T^2(1)(l)=(-1)n∑l=0n(-1)ln!l!(c^l(1)(-1)-lc^l),T^3(k)(n)=T^3(0)(n)+(1-2-k)nc^n-1(1)(-2)+n(n-1)∑j=1k(1-2j-k-1)×(c^n(j)+(-1)n(n-2)!∑l=0n-2(-1)lc^l+1(j)l!),T^3(-k)(n)=T^3(0)(n)-(2k-1)nc^n-1(1)(-2)+n(n-1)∑j=0k-2(2k-j-1-1)×(c^n(-j)+(-1)n(n-2)!∑l=0n-2(-1)lc^l+1(-j)l!).Remark 12.
Note that(84)nc^n-1(1)(-2)=(-1)n-1n!∑l=0n-1(n-l)(-1)lc^ll!.


## 2.4. Two Kinds of Poly-Cauchy Numbers

DefineUm(k)(n) by
(85)Um(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)ci1⋯cim-1︸m-1c^im(k)(m≥1,n≥0).Then we obtain the following.Theorem 13.
For an integerk and a nonnegative integer m, one has
(86)∑l=0m(-1)m-l[m+1l+1]Um+1(k-l)(n)={∑l=0m∑i=0n-mn!i!(ln-m-i){ml}c^l+i(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 13, we have the following.Corollary 14.
(87)
∑
i
=
0
n
(
n
i
)
c
i
(
c
^

n
-
i
(
k
-
1
)
-
c
^

n
-
i
(
k
)
)
=
n
(
(
n
-
1
)
c
^

n
-
1
(
k
)
+
c
^

n
(
k
)
)
(
n
≥
0
)
.Ifm=2, then U2(k) can be expressed explicitly.Theorem 15.
Forn≥0 and k≥1 one has
(88)U2(0)(n)=c^n,(89)U2(k)(n)=c^n-n∑j=1k(c^n(j)+(n-1)c^n-1(j)),(90)U2(-k)(n)=c^n+n∑j=0k-1(c^n(-j)+(n-1)c^n-1(-j)).Puttingk=1 in (89), we have the alternative identity (2.3) in [6, Theorem 2.4] because cn=c^n+nc^n-1 by (2.2) in [6, Theorem 2.4].Corollary 16.
One has(91)∑i=0n(ni)cic^n-i=-(n-1)(c^n+nc^n-1)(n≥1).Ifm=3, then U3(k) can be expressed as follows.Theorem 17.
Forn≥0 and k≥1 one has
(92)U3(0)(n)=-(n-1)(c^n+nc^n-1)=-(n-1)cn,(93)U3(k)(n)=U3(0)(n)+(1-2-k)nc^n-1+n(n-1)∑j=1k(1-2j-k-1)×((n-2)2c^n-2(j)+(2n-3)c^n-1(j)+c^n(j)),(94)U3(-k)(n)=U3(0)(n)-(2k-1)nc^n-1+n(n-1)∑j=0k-2(2k-j-1-1)×((n-2)2c^n-2(-j)+(2n-3)c^n-1(-j)+c^n(-j)).DefineVm(k)(n) by
(95)Vm(k)(n):=∑i1+⋯+im=ni1,…,im≥0(ni1,…,im)c^i1⋯c^im-1︸m-1cim(k)(m≥1,n≥0).Then we obtain the following.Theorem 18.
For an integerk and a nonnegative integer m, one has
(96)∑l=0m(-1)m-l[m+1l+1]Vm+1(k-l)(n)={(-1)n-m∑l=0m-1∑i=0n-m(-1)in!i!(n-l-i-1n-m-i){ml}cl+i(k)+n!(n-m)!cn(k)(n≥m);0(0≤n≤m-1).Puttingm=1 in Theorem 18, we have the following.Corollary 19.
(97)
∑
i
=
0
n
(
n
i
)
c
^

i
(
c
n
-
i
(
k
-
1
)
-
c
n
-
i
(
k
)
)
=
n
c
n
(
k
)
(
n
≥
0
)
.Ifm=2, then V2(k) can be expressed explicitly.Theorem 20.
Forn≥0 and k≥1 one has
(98)V2(0)(n)=cn,(99)V2(k)(n)=cn-n∑j=1kcn(j),(100)V2(-k)(n)=cn+n∑j=0k-1cn(-j).Puttingk=1 in (99), we have the identity (2.3) in [6, Theorem 2.4].Corollary 21.
One has(101)∑i=0n(ni)c^icn-i=-(n-1)cn(n≥1).Ifm=3, then V3(k) can be expressed as follows.Theorem 22.
Forn≥0 and k≥1 one has
(102)V3(0)(n)=-(n-1)cn,V3(k)(n)=V3(0)(n)-(1-2-k)nc^n-1+n(n-1)∑j=1k(1-2j-k-1)×(cn(j)+(-1)n(n-2)!∑l=0n-2(-1)lcl+1(j)l!),V3(-k)(n)=V3(0)(n)+(2k-1)nc^n-1+n(n-1)∑j=0k-2(2k-j-1-1)×(cn(-j)+(-1)n(n-2)!∑l=0n-2(-1)lcl+1(-j)l!).


## 3. Further Study

Kamano [14] mentioned that explicit formulae of Sm(k)(n) for m≥4 seemed to be complicated to describe. We will give explicit formulae of Tm(k)(n) for any m≥2 later anywhere else. In addition, one may consider the sums of products of (m-k) Cauchy numbers and k poly-Cauchy numbers. It would be an interesting work to establish the explicit expressions of such summations.



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*Source: 373927-2013-01-30.xml*