Recurrence Relations of Poly-Cauchy Numbers by the Г-Stirling Transform

Home , Stirling number

arXiv:2103.15291v2 [math.NT] 22 Jun 2021 eeaiain ntrso trignmesadhroi n research. harmonic active and nu numbers an Bernoulli Stirling of expressing terms hand, in other generalizations the On polynomials. and l yusing numb by establi poly-Bernoulli als 17] of [16, formulae) Sasaki annihilation and (including Ohno lonesum Recently, example, partitions). (for set Combinatorics m Enumerative and and function zeta tion) generalized Num the in with relations applications and have example, sequences numbers new poly-Cauchy These and respectively. n of poly-Bernoulli family called a develop of been nomials properties have analytical authors and several combinatorial by the results of number rich A Introduction 1 eurnerltoso oyCuh ubr by numbers poly-Cauchy of relations Recurrence ubr ftescn id eas ics niiainfrua for of formulas case annihilation indices. Keywords: Stirling discuss our negative with also with In We are number kind. poly-Cauchy formulas kind. second the the first numbers, of the poly-Cauchy numbers numbers of or poly-Bernoulli numbers classical or Stirling the classical with the are of case formulas the In transform. 16,1B7 16,05A19 11B68, 11B37, 11B65, Classifications: Subject MR formulas annihilation bers, egv oefrua fpl-acynmesb the by numbers poly-Cauchy of formulas some give We eateto ahmtclSine,Sho fScience of School Sciences, Mathematical of Department r Siln ubr.W td hs fpl-acynumbers poly-Cauchy of those study We numbers. -Stirling trigtransform, Stirling the hjagSiTc University Sci-Tech Zhejiang r aghu301 China 310018 Hangzhou [email protected] Siln transform -Stirling aa Komatsu Takao Abstract r Siln ubr,pl-acynum- poly-Cauchy numbers, -Stirling rmr 17;Scnay11B73, Secondary 11B75; Primary 1 r n polynomi- and ers hsm formulae some sh mesadpoly- and umbers lil eafunc- zeta ultiple meshsbeen has umbers e hoy(for Theory ber r br rtheir or mbers dconcerning ed -Stirling arcsand matrices polynomials, the , In this paper, we mainly focus on the r-Stirling transform (analogous to the well-known binomial transform) of the poly-Cauchy numbers of both kinds. Expressing poly-Cauchy numbers in terms of Stirling numbers and harmonic numbers is also an active research. Some formulae are related to shifted poly-Cauchy numbers. We can find annihilation formulas for poly- Cauchy numbers with negative indices (analogous to the results given by Ohno and Sasaki for the poly-Bernoulli case). To obtain the results we use some combinatorial properties of the r-Stirling numbers of the second kind. The binomial transform takes the sequence {an}n≥0 to the sequence {bn}n≥0 via the transformation

n n b = a n k k Xk=0 with its inverse n n a = (−1)n−k b n k k Xk=0 (see, e.g., [6, 20]). For example, for Fibonacci numbers Fn, deﬁned by Fn = Fn−1 + Fn−2 (n ≥ 3) with F1 = F2 = 1, we have an = Fn → bn = F2n n and an = (−1) Fn → bn = −Fn. Similarly, the Stirling transform takes the sequence {an}n≥0 to the sequence {bn}n≥0 via the transformation

n n b = a (1) n k k k X=0 h i with its inverse n n a = (−1)n−k b (2) n k k Xk=0 n o n (see, e.g., [20]). Here, the (unsigned) Stirling numbers of the first kind k are defined by the coefficients of the rising factorial: n n (x)(n) = x(x + 1) · · · (x + n − 1) = xk . k k X=0 h i The Stirling numbers of the second kind are expressed as

k n 1 k = (−1)i (k − i)n . k k! i i n o X=0

2 n For example, by the Stirling transform we have an = ℓ → bn = (n + ℓ − (l) n+1 1)!/(ℓ − 1)! (ℓ ≥ 1) and an = (n) → bn = l! l+1 (l ≥ 0). Many more examples for binomial transforms and Stirlingh transformsi can be seen in [19]. There are many kinds of generalizations of the Stirling numbers. One of the most interesting ones is the r-Stirling number in [2]. The main purpose of this paper is to yield the relations associated with the inverse and extended transformation of (1) with (2). Namely, we consider the transformation by the r-Stirling numbers of the second kind: n n bn = ak (3) k r k X=0 n o with its inverse n n−k n an = (−1) bk , (4) k r k X=0 h i that is yielded from the orthogonal relation in (8) and (9). More details concerning the r-Stirling numbers, which are used in this paper, are men- tioned in the next section. There are some different definitions for the r- Stirling numbers of both kinds. They can defined combinatorially as follows. n The (unsigned) r-Stirling numbers of the first kind m r are defined as the number of permutations of the set {1,...,n} having m cycles, such that the numbers 1, . . . , r are in distinct cycles. The r-Stirling numbers of the n second kind m r are defined as the number of ways to partition the set {1,...,n} into m non-empty disjoint subsets such that the numbers 1, . . . , r are in different subsets. Bernoulli number Bn with its various generalizations is one of the most interesting and most popular numbers in history. For every integer k, the (k) poly-Bernoulli numbers Bn are defined by ∞ Li (1 − e−x) xn k = B(k) , 1 − e−x n n! n X=0 where ∞ zn Li (z)= k nk n=1 X (1) is the polylogarithm function. When k = 1, Bn = Bn is the Bernoulli number, defined by ∞ xex xn = B ex − 1 n n! n=0 X 3 with B1 = 1/2. In [16, 17], the annihilation formulas are obtained for positive integers µ and k with µ ≥ k: µ µ + 2 (−1)j B(n−j) = 0 j + 2 k j 2 X=0 or µ µ + 2 (−1)j B(−k) = 0 . j + 2 n+j j 2 X=0 On the other hand, similar formulas with positive index have been studied in [10, 18]. Surprisingly, relations include the harmonic numbers Hn := n ℓ=1 1/ℓ and their generalization. For example, we have n P n n! B(k) = (n ≥ 1) j j (n + 1)k j X=1 and n n + 1 B(k) = n!H(k) (n ≥ 0) , j + 1 j n+1 j X=0 (k) n k where Hn = i=1(1/i ) is the higher order harmonic number. Some other simpler cases are given by P n n + 1 (−1)j B = n!H (n ≥ 1) (5) j + 1 j n+1 j X=0 ([4]), and n n (n − 1)! B = − (n ≥ 1) j j n + 1 j X=1 ([21]), where the Bernoulli number Bn is defined by

∞ x xn = B ex − 1 n n! n=0 X with B1 = −1/2. In [3, Corollary 10] and [11, Theorem 5], Bernoulli polynomials are expressed in terms of Stirling numbers and hyperharmonic numbers, which is a generalization of harmonic numbers. Thus, the formula (5) can be obtained as a special case. Apart from that formula, other expres- sions can be found in [1].

4 x As the inverse function of e − 1 is log(1 + x), Cauchy number cn is also enough interesting in history. In [12, 13], for any integer k, the poly-Cauchy (k) numbers cn are defined by ∞ xn Lif log(1 + x) = c(k) , k n n! n=0 X where ∞ zn Lif (z)= k n!(n + 1)k n=0 X is the polylogarithm factorial (polyfactorial) function. The poly-Cauchy (k) numbers cn have an explicit expression in terms of the Stirling numbers of the first kind as n (−1)n−ℓ n c(k) = (n ≥ 1) . (6) n (ℓ + 1)k ℓ ℓ X=1 h i (1) When k = 1, cn = cn is the classical Cauchy number (e.g., see, [5]), defined by ∞ x xn = c . log(1 + x) n n! n X=0 Note that bn = cn/n! are call the Bernoulli numbers of the second kind (e.g., see, [7]). Cauchy numbers have many similar or corresponding relations to Bernoulli numbers. In this paper, we show that for integers n, r and k with n ≥ r ≥ 1, n r n (−1)r−ℓ r c(k) = j j (n − r + ℓ + 1)k ℓ j r r ℓ X= X=1 h i and for integers n, r and k with n ≥ r − 1 ≥ 0,

n r n−r+ℓ n + 1 r n − r + ℓ 1 c(k) = (−1)r−ℓ . j + 1 j ℓ i (i + 1)k j r r i =X−1 Xℓ=1 h i X=0 In the case of negative indices, we have for n ≥ k + 2,

k n − 1 c(−k) = 0 . n − l − 1 n−l l n−k−1 X=0 We also give their analogous results for the poly-Cauchy numbers of the (k) second kind cn , deﬁned in (10).

b 5 2 The r-Stirling numbers

In this section, we mention some basic properties of the r-Stirling numbers, which will be used in this paper. By using the r-Stirling numbers, the original Stirling numbers can be expressed as n n n n = , = , m m 0 m m 0 h i h i n o h i and n n n n = , = (n> 0) , m m 1 m m 1 h i h i n o h i There exist recurrence relations as n n − 1 n − 1 = (n − 1) + (n > r) (7) m r m r m − 1 r h i with n n = 0 (n < r) and = δm,r (n = r) , m r m r h i h i and n n − 1 n − 1 = m + (n > r) m r m r m − 1 r n o with n n = 0 (n < r) and = δm,r (n = r) , m r m r n o n o where δm,r = 1 (m = r); 0 (m 6= r). Note that the orthogonal properties [2, Theorem 5, Theorem 6] hold between two kinds of r-Stirling numbers too. Namely, n k δ if n ≥ r; (−1)n−k = m,n (8) k r m k r (0 otherwise X h i and k m δ if n ≥ r; (−1)n−k = m,n (9) n r k r (0 otherwise . Xk n o The ordinary generating functions of both kinds of r-Stirling numbers [2, Corollary 9, Corollary 10] are given by

n zr(z + r)(z + r + 1) · · · (z + n − 1) if n ≥ r ≥ 0; zk = k r (0 otherwise Xk h i

6 and zm k if m ≥ r ≥ 0; zk = (1 − rz) 1 − (r + 1)z · · · 1 − mz m  k r X 0 otherwise , respectively. 

3 Main results

This is our ﬁrst main result.

Theorem 1. For integers n, r and k with n ≥ r ≥ 1, we have

n r n (−1)r−ℓ r c(k) = . j j (n − r + ℓ + 1)k ℓ j r r X= Xℓ=1 h i

Remark. When r = 1, 2, 3, 4, we have

n n 1 c(k) = ([12, Theorem 3]) , j j (n + 1)k j X=1 n n 1 1 c(k) = − + , j j nk (n + 1)k j 2 X=2 n n 2 3 1 c(k) = − + , j j (n − 1)k nk (n + 1)k j 3 X=3 n n 6 11 6 1 c(k) = − + − + . j j (n − 2)k (n − 1)k nk (n + 1)k j 4 X=4 r When n = r, by r r = 1, we have (6) with n = r. One variation is given as follows. Theorem 2. For integers n, r and k with n ≥ r − 1 ≥ 0, we have

n r n−r+ℓ n + 1 r n − r + ℓ 1 c(k) = (−1)r−ℓ . j + 1 j ℓ i (i + 1)k j r r ℓ i =X−1 X=1 h i X=0

7 Remark. When r = 1, we have

n n n + 1 n 1 c(k) = . j + 1 j i (i + 1)k j i X=0 X=0 When k = 1 and r = 1, 2, 3, we have

n n + 1 2n+1 − 1 c = , j + 1 j n + 1 j X=0 n n + 1 2n − 1 2n+1 − 1 c = − + , j + 1 j n n + 1 j 2 X=0 n n + 1 2(2n−1 − 1) 3(2n − 1) 2n+1 − 1 c = − + . j + 1 j n − 1 n n + 1 j 3 X=0 When n = r − 1, we have

r−1 ℓ−1 r ℓ − 1 1 c(k) = (−1)r−ℓ r−1 ℓ i (i + 1)k i Xℓ=0 h i X=0 r−1 r 1 r ℓ − 1 = (−1)r−ℓ (i + 1)k ℓ i i ℓ i X=0 =X+1 h i r−1 (−1)r−i−1 r − 1 = , (i + 1)k i i X=0 yielding the formula in (6) with n = r − 1. (k) Poly-Cauchy numbers of the second kind cn ([12]) are deﬁned by

∞ xn Lif − log(1 + x) = bc(k) . (10) k n n! n=0 X (1) b When k = 1, cn = cn are the classical Cauchy numbers of the second kind (see, e.g., [5]), deﬁned by

b b ∞ x xn = c(k) . (1 + x) log(1 + x) n n! n=0 X b (k) Concerning the poly-Cauchy numbers of the second kind cn , we have the following corresponding results. b 8 Theorem 3. For integers n, r and k with n ≥ r ≥ 1, we have n r n (−1)n r c(k) = . j j (n − r + ℓ + 1)k ℓ j r r ℓ X= X=1 h i b Theorem 4. For integers n, r and k with n ≥ r − 1 ≥ 0, we have

n r n−r+ℓ n + 1 r n − r + ℓ (−1)i c(k) = (−1)r−ℓ . j + 1 j ℓ i (i + 1)k j r r ℓ i =X−1 X=1 h i X=0 b

4 Proof of the main results

We need the following relations in order to prove Theorem 1. Lemma 1. Let r be any integer with 1 ≤ r ≤ n. Then for 1 ≤ m ≤ n−r+1, m r n n = , ℓ r − ℓ + m r m Xℓ=1 h i h i and for n − r + 2 ≤ m ≤ n,

n+1−max{m,r} r n n = . m − n + r − 1+ ℓ n − ℓ + 1 m ℓ r X=1 h i

Remark. There are some different ways to express the Stirling numbers of the first kind in terms of the r-Stirling numbers of the first kind by choosing different r. For example, 6 2 6 2 6 = 225 = + 3 1 4 2 3 2 2 = 1 · 71 + 1 · 154 3 6 3 6 3 6 = + + 1 5 2 4 3 3 3 3 3 = 2 · 12 + 3 · 47 + 1 · 60 4 6 4 6 4 6 = + + 1 6 2 5 3 4 4 4 4 = 6 · 1 + 11 · 9 + 6 · 20 .

9 Proof. According to the definition of the r-Stirling numbers of the first kind, the relation 6 4 6 4 6 4 6 = + + 3 1 6 2 5 3 4 4 4 4 can be explained as follows. The permutations of the set {1, 2, 3, 4, 5, 6} with 3 cycles has three different ways, by fixing the first numbers 1, 2, 3, 4.

(1) 6 numbers are divided into 6 diﬀerent cycles (so that 1, 2, 3, 4 are in distinct cycles). Then, the cycles including 1, 2, 3, 4 are collected into the same group (permutation).

(1)(2)(3)(4)(. . .)(. . .)=⇒ (1)(2)(3)(4) ...... (2) 6 numbers are divided into 5 diﬀerent cycles so that 1, 2, 3, 4 are in distinct cycles. Then, 4 cycles including 1, 2, 3, 4 are collected into 2 diﬀerent groups (permutation).

(1 . . .)(2 . . .)(3 . . .)(4 . . .)(. . .)=⇒ (a...)(b...) (c...)(d...) . . . or =⇒ (a...)(b...)(c...) d... . . . , where {a, b, c, d} = {1, 2, 3, 4}

(3) 6 numbers are divided into 4 diﬀerent cycles so that 1, 2, 3, 4 are in distinct cycles. Then, 4 cycles including 1, 2, 3, 4 are collected into 3 diﬀerent groups (permutation).

(1 . . .)(2 . . .)(3 . . .)(4 . . .)=⇒ (a...)(b...) c... d...

where {a, b, c, d} = {1, 2, 3, 4}

In general, the permutations of the set {1, 2,...,n} with m cycles are done by the following way. When 1 ≤ m ≤ n − r + 1, n numbers are divided into r − ℓ + m diﬀerent cycles so that 1, 2, . . . , r are in distinct cycles. Then, the r cycles including 1, 2, . . . , r are collected into ℓ diﬀerent groups. Here, 1 ≤ ℓ ≤ m.

(1 . . .)(2 . . .) · · · (r...) (. . .) · · · (. . .)

m−ℓ | =⇒{z (a1}. . .) . . . · · · . . . (ar . . .) . . . · · · . . . , ℓ m−ℓ | {z } | {z } 10 where {a1, . . . , ar} = {1, . . . , r}. When n − r + 2 ≤ m ≤ n, by putting m′ = m − n + r − 1, n numbers are divided into n − ℓ + m = r + (m − m′ − ℓ) diﬀerent cycles so that 1, 2, . . . , r are in distinct cycles. Then, the r cycles including 1, 2, . . . , r are collected into m′ + ℓ diﬀerent groups. Here, 1 ≤ ℓ ≤ n + 1 − max{m, r}.

(1 . . .)(2 . . .) · · · (r...) (. . .) · · · (. . .)

m−m′−ℓ | =⇒{z (a1}. . .) . . . · · · . . . (ar . . .) . . . · · · . . . , m′+ℓ m−m′−ℓ where {a1, . . . , ar} = {1, . . . , r}.| {z } | {z } Now, it is ready to prove Theorem 1.

Proof of Theorem 1. By the transformation of the r-Stirling numbers in (3) with (4), equivalently, we shall prove that

n r r−ℓ (k) n−i n (−1) r cn = (−1) i r (i − r + ℓ + 1)k ℓ i=r h i ℓ=1 h i Xr nX r (−1)n−i n = (−1)r−ℓ (n ≥ r) . (11) ℓ (i − r + ℓ + 1)k i r ℓ i r X=1 h i X= h i By Lemma 1, setting m = i − r + ℓ, the right-hand side of (11) is equal to

n−r+1 m (−1)n−m r n (m + 1)k ℓ r − ℓ + m m=1 ℓ r X X=1 h i n n+1−max{m,r} (−1)n−m r n + (m + 1)k m − n + r − 1+ ℓ n − ℓ + 1 m=n−r+2 ℓ=1 r n X X (−1)n−m n = = c(k) (6) . (m + 1)k m n m=1 X h i

Proof of Theorem 2. As similar orthogonal relations to (8) and (9), we have

n + 1 k + 1 δ if n ≥ r − 1; (−1)n−k = m,n k + 1 m + 1 k r r (0 otherwise X 11 and k + 1 m + 1 δ if n ≥ r − 1; (−1)n−k = m,n n + 1 r k + 1 r (0 otherwise . Xk Thus, equivalently, we shall prove that

n r j−r+ℓ n + 1 r j − r + ℓ 1 c(k) = (−1)n−j (−1)r−ℓ n j + 1 ℓ i (i + 1)k j r r ℓ i =X−1 X=1 h i X=0 r n+ℓ−r m r n + 1 m 1 = (−1)n−m . ℓ m − ℓ + r + 1 i (i + 1)m r i Xℓ=1 h i mX=ℓ−1 X=0 Because of the expression in (6), it is suﬃcient to prove that for n ≥ r

r n−r+2 n r ℓ + j − 2 n + 1 = (−1)ℓ+j−i . (12) i ℓ i r + j − 1 j r h i Xℓ=1 h i X=1 Put for 0 ≤ i ≤ n and n ≥ 1

r n−r+2 r ℓ + j − 2 n + 1 a := (−1)ℓ+j−i . n,i ℓ i r + j − 1 ℓ j r X=1 h i X=1 First, for i = 0 and r ≥ 2, by

r r (−1)ℓ = 0 , ℓ ℓ X=1 h i we have

r n−r+2 r ℓ + j − 2 n + 1 a = (−1)ℓ+j n,0 ℓ 0 r + j − 1 j r Xℓ=1 h i X=1 n−r+2 r n + 1 r = (−1)j (−1)ℓ r + j − 1 ℓ j r X=1 Xℓ=1 h i = 0 .

For i = 0 and r = 1,

n+1 1 n + 1 a = (−1)j+1 = 0 (n ≥ 1) . n,0 1 j j X=1 12 Next, let i ≥ 1. For convenience, put r r a = b n,i ℓ n,i,ℓ Xℓ=1 h i with for ﬁxed ℓ n−r+2 ℓ + j − 2 n + 1 b = b := (−1)ℓ+j−i . n,i n,i,ℓ i r + j − 1 j r X=1 Then, using the recurrence relation (7), by n = 0 (j = 0, j = n − r + 2) r + j − 1 r and ℓ + j − 1 ℓ + j − 2 ℓ + j − 2 = + , i i i − 1 we have n−r+2 ℓ + j − 2 n b = n (−1)ℓ+j−i n,i i r + j − 1 j r X=1 n−r+2 ℓ + j − 2 n + (−1)ℓ+j−i i r + j − 2 j r X=1 n−r+1 ℓ + j − 2 n = n (−1)ℓ+j−i i r + j − 1 j r X=1 n−r+1 ℓ + j − 1 n − (−1)ℓ+j−i i r + j − 1 j r X=0 n−r+1 ℓ + j − 2 n = (n − 1) (−1)ℓ+j−i i r + j − 1 j r X=1 n−r+1 ℓ + j − 2 n − (−1)ℓ+j−i i − 1 r + j − 1 j r X=0 = (n − 1)bn−1,i + bn−1,i−1 . n Hence, an,i = (n − 1)an−1,i + an−1,i−1 (i ≥ 1). Since an,i and i satisfy the same recurrence relation as (7) with a = n (n ≥ 1), the relation n,0 0 (12) holds.

13 The proofs of Theorem 3 and Theorem 4 are similar to those of Theorem 1 and Theorem 2, respectively, and omitted.

5 Poly-Cauchy polynomials

(k) Poly-Cauchy polynomials cn (z) are deﬁned by ∞ xn (1 + x)zLif log(1 + x) = c(k)(z) (13) k n n! n=0 X ([8, Theorem 2]), and an explicit expression is given by n m n m zm−i c(k)(z)= (−1)n−m . (14) n m i (i + 1)k m i X=0 h i X=0 (k) (k) ([8, Theorem 1]). When z = 0, cn = cn (0) are the poly-Cauchy numbers. The deﬁnition in [13] is an alternative way, simply by replacing z by −z. As an extension of Theorem 1 and Theorem 2, we have the following. Theorem 5. For integers n, r and k with n ≥ r ≥ 1 and a real number q, we have n r n−r+ℓ n r n − r + ℓ qn−i c(k)(q)= (−1)r−ℓ . j j ℓ i (i + 1)k j r r ℓ i X= X=1 h i X=0

Remark. When k = 1, we have the relation of the Cauchy polynomials (1) cj(z)= cn (z): n r n r (q + 1)n−r+ℓ+1 − qn−r+ℓ+1 c (q)= (−1)r−ℓ . j j ℓ n − r + ℓ + 1 j r r X= Xℓ=1 h i When q = 0 in Theorem 5, Theorem 1 is reduced. When q = 1 in Theorem 5, we ﬁnd n n n n + 1 c(k)(1) = c(k) . j j j + 1 j j r r j r r X= =X−1 Proof of Theorem 5. The proof is similar to that of Theorem 2. We prove that n r j−r+ℓ n + 1 r j − r + ℓ qj−r+ℓ−i c(k)(q)= (−1)n−j (−1)r−ℓ n j + 1 ℓ i (i + 1)k j r r ℓ i =X−1 X=1 h i X=0 14 r n+ℓ−r m r n + 1 m qm−i = (−1)n−m . ℓ m − ℓ + r + 1 i (i + 1)m r i Xℓ=1 h i mX=ℓ−1 X=0 Then, it is also suﬃcient to prove (12).

(k) Poly-Cauchy polynomials of the second kind cn (z) are deﬁned by

∞ n Lifk − log(1 + x) x = c(k)b(z) (15) (1 + x)z n n! n=0 X ([8, Theorem 5]), and an explit expression is givenb by n m n m zm−i c(k)(z)= (−1)n . (16) n m i (i + 1)k m=0 i X h i X=0 b (k) (k) ([8, Theorem 4]). When z = 0, cn = cn (0) are the poly-Cauchy numbers of the second kind. As an extension of Theorem 3, we have the following. The proof is similar and omitted.b b Theorem 6. For integers n, r and k with n ≥ r ≥ 1 and a real number q, we have n r−1 n−r+ℓ n r n − r + ℓ (−q)n−r−ℓq−i c(k)(q)= (−1)r−ℓ . j j ℓ i (i + 1)k j r r ℓ i X= X=0 h i X=0 b Notice that when q = 1, n n n n + 1 c(k)(1) 6= c(k) . j j j + 1 j j r r j r r X= =X−1 b b 6 Shifted poly-Cauchy numbers and harmonic numbers

The right-hand sides of Theorem 1 and Theorem 3 can be written in terms of shifted poly-Cauchy numbers of both kinds, defined in [15]. For integers n ≥ 0 and k ≥ 1 and a real number α > 0, the shifted (k) poly-Cauchy numbers of the first kind cn,α are defined by 1 1 (k) α cn,α = · · · (x1 · · · xk) (x1 · · · xk − 1) · · · (x1 · · · xk − n + 1)dx1 ...dxk Z0 Z0 k | {z } 15 and can be expressed in terms of the Stirling numbers of the first kind as n n (−1)n−m c(k) = (n ≥ 0, k ≥ 1) n,α m (m + α)k m=0 X h i (k) (k) ([15, Theorem 1]). When α = 1, cn = cn,1 are the original poly-Cauchy numbers of the first kind. Similarly, for integers n ≥ 0 and k ≥ 1 and a real (k) number α> 0, the shifted poly-Cauchy numbers of the second kind cn,α are defined by 1 1 b (k) α cn,α = · · · (−x1 · · · xk) (−x1 · · · xk−1) · · · (−x1 · · · xk−n+1)dx1 ...dxk Z0 Z0 b k and can| be{z expressed} in terms of the Stirling numbers of the first kind as n n 1 c(k) = (−1)n (n ≥ 0, k ≥ 1) n,α m (m + α)k m X=0 h i b (k) (k) ([15, Proposition 2]). When α = 1, cn = cn,1 are the original poly-Cauchy numbers of the second kind. By using the shifted poly-Cauchyb numbersb of the both kinds, the identities in Theorem 1 and Theorem 3 can be written in terms of shifted poly- Cauchy numbers of both kinds. Corollary 1. For integers n, r and k with n ≥ r ≥ 1, we have n n c(k) = c(k) , j j r,n−r+1 j r r X= n n c(k) = c(k) . j j r,n−r+1 j r r X= b b In addition, it would be noticed that a similar formula in Theorem 1 is given as a special case in [9, Corollary 8].

Latter identities in Theorem 4, Theorem 5 and Theorem 6 are relate to the higher order harmonic numbers. For a sequence t = (t1,t2,... ), the Bell polynomials Ωi(t):=Ωi(t1,t2,...,ti) are deﬁned by ∞ ∞ xi xk Ω (t) = exp t i i! k k! i ! X=0 Xk=1 16 and expressed as

a1 a1 ai i! t1 t2 ti Ωi(t)= · · · . a1!a2! · · · ai! 1 2 i a1+2a2+···+iai=i a1,a2X,...,ai≥0

It is known and has been studied by several people that

n m + n n 1 i!m (−1)k n k (m + k)i+1 Xk=0 (2) (2) (i) (i) =Ωi(Hm+n − Hm−1,Hm+n − Hm−1,...,Hm+n − Hm−1) (17)

(see, e.g., [22, (3.56)] and references therein). By using this general harmonic number identity (17), Theorem 4, the case q = −1 of Theorem 5, and the cases q = −1 of Theorem 6 can be written in terms of the higher order harmonic number numbers.

Corollary 2. For integers n, r and k with n ≥ r − 1 ≥ 0, we have

n r (2) (k−1) n + 1 r Ωk−1(Hn−r+ℓ+1,H ,...,H ) c(k) = (−1)r−ℓ n−r+ℓ+1 n−r+ℓ+1 . j + 1 j ℓ (n − r + ℓ + 1)(k − 1)! j r r ℓ =X−1 X=1 h i b For integers n, r and k with n ≥ r ≥ 1, we have

n r (2) (k−1) n r Ωk−1(Hn−r+ℓ+1,H ,...,H ) c(k)(−1) = (−1)n−r+ℓ n−r+ℓ+1 n−r+ℓ+1 , j j ℓ (n − r + ℓ + 1)(k − 1)! j r r X= Xℓ=1 h i n r−1 (2) (k−1) n r Ωk−1(Hn−r+ℓ+1,H ,...,H ) c(k)(−1) = (−1)r−ℓ n−r+ℓ+1 n−r+ℓ+1 . j j ℓ (n − r + ℓ + 1)(k − 1)! j r r X= Xℓ=0 h i b

7 Annihilation formulas for poly-Cauchy numbers

Some annihilation formulas for poly-Bernoulli numbers with negative indices have been established in [16, 17]. In this section, we show annihilation formulas for poly-Cauchy numbers with negative indices. In [14], some annihilation formulas for poly-Cauchy numbers have been done, but the ex- pressions are not so elegant. With the aid of the r-Stirling numbers, we can give more elegant forms.

17 Theorem 7. For n ≥ k + 2,

k n − 1 c(−k) = 0 . n − l − 1 n−l l n−k−1 X=0

Remark. For k = 1, 2, 3 we have

n − 1 n − 1 0= c(−1) + c(−1) n − 1 n n − 2 n−1 n−2 n−2 (−1) (−1) = cn + (n − 2)cn−1 (n ≥ 3) , n − 1 n − 1 n − 1 0= c(−2) + c(−2) + c(−2) n − 1 n n − 2 n−1 n − 3 n−2 n−3 n−3 n−3 (−2) (−2) 2 (−2) = cn + (2n − 5)cn−1 + (n − 3) cn−2 (n ≥ 4) , n − 1 n − 1 0= c(−3) + c(−3) n − 1 n n − 2 n−1 n−4 n−4 n − 1 n − 1 + c(−3) + c(−3) n − 3 n−2 n − 4 n−3 n−4 n−4 (−3) (−3) 2 (−3) 3 (−3) = cn + (3n − 9)cn−1 + (3n − 21n + 37)cn−2 + (n − 4) cn−3 (n ≥ 5) .

Proof of Theorem 7. Since the r-Stirling numbers of the second kind can be expressed as n + m = i i · · · i n 1 2 m r r i i n ≤ 1≤···≤X m≤ ([2, Theorem 8]), together with [14, Theorem 2.1], we have

k 0= (n − i ) · · · (n − i ) c(−k)  1 l  n−l l l i i k X=0 +1≤ 1≤···≤X l≤ +1 k   n − 1 = c(−k) . n − l − 1 n−l l n−k−1 X=0

(−k) Similarly, concerning the poly-Cauchy numbers of the second kind cn , we have the following annihilation formula. b

18 Theorem 8. For n ≥ k + 2,

k n + 1 c(−k) = 0 . n − l + 1 n−l l n−k X=0 b Remark. For k = 0, 1, 2 we have

n + 1 n + 1 0= c(0) + c(0) n + 1 n n n−1 n n (0) (0) = cn + ncn−b1 (n ≥ 1) , b n + 1 n + 1 n + 1 0= c(−1) + c(−1) + c(−1) b n + 1b n n n−1 n − 1 n−2 n−1 n−1 n−1 (−1) (−1) 2 (−1) = cn + (2n −b1)cn−1 + (n − 1) cn−b2 (n ≥ 2) , b n + 1 n + 1 0= c(−2) + c(−2) b n + 1 n b n b n−1 n−2 n−2 n + 1 n + 1 + b c(−2) + b c(−2) n − 1 n−2 n − 2 n−3 n−2 n−2 (−2) (−2) 2 (−2) 3 (−2) = cn + (3n − 3)cnb−1 + (3n − 9n + 7)cbn−2 + (n − 2) cn−3 (n ≥ 3) . Proof of Theorem 8. b bBy using [2, Theorem 8])b again, togetherb with [14, The- orem 3.1], we have

k+1 0= (n − i ) · · · (n − i ) c(−k)  1 l  n−l l l i i k X=0 −1≤ 1X≤···≤ l≤ k   n + 1 b = c(−k) . n − l + 1 n−l l n−k X=0 b

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