On generalized hyperharmonic numbers of order r, H rn,m ( σ )

: In this paper, we define generalized hyperharmonic numbers of order r, H rn,m ( σ ) , for m ∈ Z + and give some applications by using generating functions of these numbers. For example, for n, r, s ∈ Z + such that 1 ≤ s ≤ r,


Introduction
The harmonic numbers, denoted by H n , are defined by and their generating function is In [10], it is known that n k=0 where H n,2 = n k=1 1 k 2 .Harmonic numbers are interesting research objects.Recently, these numbers have been generalized by several authors.There are a lot of works involving harmonic numbers and their generalizations ( [3,[5][6][7][8][9]).
The exponential generating function is The derangement numbers d n are given by the closed form formula These numbers satisfy the recursive formula given by In [11], for 0 ≤ r ≤ n, D r (n) denotes the number of derangements on n + r elements under the restriction that the first r elements are in disjoint cycles.A closed form formula for D r (n) is also given by The r-derangement numbers D r (n) satisfy the recursive formula with initial conditions The generating function of the r-derangement numbers D r (n) is given by (1.3) Note that for r = 0, we have (1.4) In [1,2], for m ∈ Z, the polylogarithm is defined by The Stirling numbers of the second kind S 2 (n, k) are defined by where x n stands for the falling factorial defined by x 0 = 1 and x n = x(x − 1)...(x − n + 1).
The generating function of the Stirling numbers of the second kind S 2 (n, k) is given by

Some results
In this section, we will define generalized harmonic numbers, H n,m (σ) and then give some applications of them.
Definition 2.1.For n, m ∈ Z + , the generalized harmonic numbers, H n,m (σ) , are defined by where σ is an appropriate parameter.
It is clearly seen that for m > 0, we have Definition 2.2.For r < 0 or n ≤ 0, H r 0,m (σ) = 0 and for n ≥ 1, the generalized hyperharmonic numbers of order r, H r n,m (σ) , are defined by ) Proof.By (2.2) and (2.3), we have as claimed.
From Theorem 2.1, it is clearly seen that Proof.By (1.4) and (2.4), we have Thus, by comparing the coefficients on both sides, the proof is complete.
Theorem 2.3.Let r, s be positive integers such that 1 ≤ s ≤ r.For n, m ∈ Z + , we have and Proof.By (1.4) and (2.4), we have and by (2.4), Thus, by comparing the coefficients on both sides, we have the proof.
For example, when r = s in (2.5), we obtained Theorem 2.2.
Theorem 2.4.For n, m, r ∈ Z + , we have Proof.Inserting 1 − e −x in the place of x in (2.4), by (1.6), we have and from (1.1) and (1.6), Thus, by comparing the coefficients on right sides of (2.7) and (2.8), we have the proof.
For the second equality, by (1.1) and (2.4), we have and by (1.2) and (2.4) (2.10) Thus, by comparing the coefficients on right sides of (2.9) and (2.10), we have the second equality.
Theorem 2.7.For n, m, r ∈ Z + , we have Proof.Inserting 1−e −x σ in the place of x in (2.6), by (1.6), then and inserting 1 − e −x in the place of x in (1.5), by (1.6), we have (2.12) By comparing the coefficients on right sides of (2.11) and (2.12), we obtain the first equality.For the second equality, we have By comparing the coefficients on right sides of (2.13) and (2.14), the second equality is given.Thus, we have the proof.