Identities involving harmonic and hyperharmonic numbers

Abstract

In this paper, we give some new and interesting identities involving harmonic and hyperharmonic numbers which are derived from the transfer formula for the associated sequences.

1 Introduction

Let ℱ be the set of all formal power series in the variable t over C with

$$\mathcal{F}=\{f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k|a_k\in \mathbf{C}\}.$$

(1)

Suppose that ℙ is the algebra of polynomials in the variable x over C and that ${\mathbb{P}}^{\ast }$ is the vector space of all linear functionals on ℙ. The action of the linear functional L on a polynomial $p(x)$ is denoted by $〈L|p(x)〉$.

Let $f(t)\in \mathcal{F}$. Then we consider a linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n(n\ge 0)\text{ (see [1, 2])}.$$

(2)

From (1) and (2), we note that

$$〈t^k|x^n〉=n!{\delta }_{n,k}(n,k\ge 0)\text{ (see [1, 3–5])},$$

(3)

where ${\delta }_{n,k}$ is the Kronecker symbol.

Let $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^n〉}{k!}{t}^k$. Then we see that $〈f_L(t)|x^n〉=〈L|x^n〉$. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ is thought of as both a formal power series and a linear functional. We call ℱ the umbral algebra. The umbral calculus is the study of umbral algebra. The order $O(f(t))$ of the nonzero power series $f(t)$ is the smallest integer k for which the coefficient of $t^k$ does not vanish. If $O(f(t))=0$, then $f(t)$ is called an invertible series. If $O(f(t))=1$, then $f(t)$ is called a delta series. Let $O(f(t))=1$ and $O(g(t))=0$. Then there exists a unique sequence $s_n(x)$ of polynomials such that $〈g(t)f{(t)}^k|s_n(x)〉=n!{\delta }_{n,k}$ for $n,k\ge 0$. The sequence $s_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$ which is denoted by $s_n(x)\sim (g(t),f(t))$ (see [1, 3, 6]). If $s_n(x)\sim (1,f(t))$, then $s_n(x)$ is called the associated sequence for $f(t)$. By (3), we easily see that $〈e^{yt}|p(x)〉=p(y)$. Let $f(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$. Then we have

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f(t)|x^k〉}{k!}{t}^k,p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈t^k|p(x)〉}{k!}{x}^k\text{(see [1, 6, 7])}.$$

(4)

From (4), we note that

$$p^{(k)}(0)=〈t^k|p(x)〉,〈1|p^{(k)}(x)〉=p^{(k)}(0).$$

(5)

By (5), we easily see that

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}(k\ge 0)\text{ (see [2, 3, 6, 7])}.$$

(6)

Let ${\varphi }_n(x)$ be exponential polynomials which are given by

$$\sum _{k=0}^{\mathrm{\infty }}\frac{{\varphi }_k(x)}{k!}{t}^k=e^{x(e^t-1)}\text{(see [2, 6, 8])}.$$

(7)

Thus, by (7), we get

$${\varphi }_n(x)=\sum _{k=0}^n{S}_2(n,k)x^k\sim (1,log(1+t)),$$

(8)

where $S_2(n,k)$ is the Stirling number of the second kind.

The Stirling number of the first kind is defined by

$${(x)}_n=x(x-1)\cdots (x-n+1)=\sum _{k=0}^n{S}_1(n,k)x^k.$$

(9)

Thus, by (9), we get

$$S_1(n,k)=\frac{1}{k!}〈t^k|{(x)}_n〉\text{(see [2, 5])}.$$

(10)

Let $p_n(x)\sim (1,f(t))$, $q_n(x)\sim (1,g(t))$. Then the transfer formula for the associated sequences is given by

$$q_n(x)=x{\left(\frac{f(t)}{g(t)}\right)}^n{x}^{-1}{p}_n(x)\text{(see [2, 8])}.$$

(11)

The n th harmonic number is $H_n={\sum }_{i=1}^n\frac{1}{i}$ ($n\ge 1$) and $H_0=0$.

In general, the hyperharmonic number $H_n^{(r)}$ of order r is defined by

$$H_n^{(r)}=\{\begin{array}{cc}0\hfill & \text{if }n\le 0\text{ or }r<0,\hfill \\ \frac{1}{n}\hfill & \text{if }r=0,n\ge 1,\hfill \\ {\sum }_{i=1}^n{H}_i^{(r-1)}\hfill & \text{if }r,n\ge 1\hfill \end{array}\text{(see [9, 10])}.$$

(12)

From (12), we note that $H_n^{(1)}$ is the ordinary harmonic number $H_n$. It is known that

$$H_n^{(r)}=\left(\genfrac{}{}{0ex}{}{n+r-1}{r-1}\right)(H_{n+r-1}-H_{r-1})\text{(see [9, 10])}.$$

(13)

The generating functions of the harmonic and hyperharmonic numbers are given by

$$\sum _{n=1}^{\mathrm{\infty }}{H}_n{t}^n=-\frac{log(1-t)}{1-t}$$

(14)

and

$$\sum _{n=1}^{\mathrm{\infty }}{H}_n^{(r)}{t}^n=-\frac{log(1-t)}{{(1-t)}^r},\text{respectively}.$$

(15)

The purpose of this paper is to give some new and interesting identities involving harmonic and hyperharmonic numbers which are derived from the transfer formula for the associated sequences.

2 Identities involving harmonic and hyperharmonic numbers

From (7) and (8), we note that

$${\varphi }_n(x)=\sum _{j=0}^n{S}_2(n,j)x^j\sim (1,log(1+t))$$

(16)

and

$${(-1)}^n{\varphi }_n(-x)\sim (1,-log(1-t)).$$

(17)

Let us assume that

$$q_n(x)\sim (1,t{(1-t)}^r).$$

(18)

From (11), (18) and $x^n\sim (1,t)$, we note that

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t}{t{(1-t)}^r}\right)}^n{x}^{-1}{x}^n=x{(1-t)}^{-rn}{x}^{n-1}\\ =& x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{-rn}{k}\right){(-t)}^k{x}^{n-1}=x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{rn+k-1}{k}\right)t^k{x}^{n-1}\\ =& x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0ex}{}{rn+k-1}{k}\right){(n-1)}_k{x}^{n-1-k}=\sum _{k=1}^{n-1}\left(\genfrac{}{}{0ex}{}{rn+k-1}{k}\right){(n-1)}_k{x}^{n-k}\\ =& \sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{rn+n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.\end{array}$$

(19)

Now, we use the following fact:

$$\sum _{n=1}^{\mathrm{\infty }}{H}_n^{(r)}{t}^n=-\frac{log(1-t)}{(1-t)r}.$$

(20)

For $n\ge 1$, by (11), (17) and (18), we get

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{-log(1-t)}{t{(1-t)}^r}\right)}^n{x}^{-1}{(-1)}^n{\varphi }_n(-x)\\ =& x{\left(\sum _{l=0}^{\mathrm{\infty }}{H}_{l+1}^{(r)}{t}^l\right)}^n{x}^{-1}{(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-x)}^j\\ =& {(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-1)}^{j}x{\left(\sum _{l=0}^{\mathrm{\infty }}{H}_{l+1}^{(r)}{t}^l\right)}^n{x}^{j-1}\\ =& {(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-1)}^{j}x\left(\sum _{l=0}^{j-1}(\sum _{l_1+\cdots +l_n=l}{H}_{l_1+1}^{(r)}\cdots H_{l_n+1}^{(r)})t^l\right)x^{j-1}\\ =& {(-1)}^n\sum _{j=1}^n\sum _{l=0}^{j-1}\sum _{l_1+\cdots +l_n=l}{S}_2(n,j){(-1)}^j{H}_{l_1+1}^{(r)}\cdots H_{l_n+1}^{(r)}{(j-1)}_l{x}^{j-l}\\ =& {(-1)}^n\sum _{j=1}^n\sum _{k=1}^j\sum _{l_1+\cdots +l_n=j-k}{S}_2(n,j){(-1)}^j{H}_{l_1+1}^{(r)}\cdots H_{l_n+1}^{(r)}{(j-1)}_{j-k}{x}^k\\ =& {(-1)}^n\sum _{k=1}^n\{\sum _{j=k}^n\sum _{l_1+\cdots +l_n=j-k}{(-1)}^j{S}_2(n,j)H_{l_1+1}^{(r)}\cdots H_{l_n+1}^{(r)}{(j-1)}_{j-k}\}{x}^k.\end{array}$$

(21)

Therefore, by comparing coefficients on both sides of (19) and (20), we obtain the following theorem.

Theorem 1 For $n\ge 1$, $r\ge 1$, $1\le k\le n$, we have

$$\left(\genfrac{}{}{0ex}{}{rn+n-k-1}{n-k}\right){(n-1)}_{n-k}={(-1)}^n\sum _{j=k}^n\sum _{l_1+\cdots +l_n=j-k}{S}_2(n,j){(-1)}^j{H}_{l_1+1}^{(r)}\cdots H_{l_n+1}^{(r)}{(j-1)}_{j-k}.$$

We recall the following equation:

$${\left(\frac{log(1+t)}{t}\right)}^n=\sum _{l=0}^{\mathrm{\infty }}\frac{n!}{(l+n)!}{S}_1(l+n,n)t^l.$$

(22)

For $n\ge 1$, from (11), (17) and (18), we have

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{-log(1-t)}{t{(1-t)}^r}\right)}^n{x}^{-1}{(-1)}^n{\varphi }_n(-x)\\ =& x{\left(\frac{log(1-t)}{-t}\right)}^n{(1-t)}^{-rn}{x}^{-1}{(-1)}^n{\varphi }_n(-x)\\ =& {(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-1)}^{j}x{\left(\frac{log(1-t)}{-t}\right)}^n{(1-t)}^{-rn}{x}^{j-1}\\ =& {(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-1)}^{j}x{\left(\frac{log(1-t)}{-t}\right)}^n\sum _{l=0}^{j-1}\left(\genfrac{}{}{0ex}{}{rn+l-1}{l}\right){(j-1)}_l{x}^{j-1-l}\\ =& {(-1)}^n\sum _{j=1}^n{S}_2(n,j){(-1)}^j\sum _{l=0}^{j-1}\left(\genfrac{}{}{0ex}{}{rn+l-1}{l}\right){(j-1)}_{l}x\sum _{m=0}^{j-1-l}\frac{n!}{(m+n)!}\\ \times S_1(m+n,n){(-t)}^m{x}^{j-1-l}\\ =& {(-1)}^n\sum _{j=1}^n\sum _{l=0}^{j-1}\sum _{m=0}^{j-1-l}{(-1)}^{j+m}\left(\genfrac{}{}{0ex}{}{rn+l-1}{l}\right)\frac{n!}{(m+n)!}\frac{(j-1)!}{(j-1-l-m)!}\\ \times S_1(m+n,n)S_2(n,j)x^{j-l-m}\\ =& {(-1)}^n\sum _{k=1}^n\{\sum _{j=k}^n\sum _{l=0}^{j-k}{(-1)}^{k+l}\left(\genfrac{}{}{0ex}{}{rn+l-1}{l}\right)\frac{n!}{(j-l-k+n)!}\frac{(j-1)!}{(k-1)!}\\ \times S_1(j-l-k+n,n)S_2(n,j)\}{x}^k.\end{array}$$

(23)

Therefore, by (19) and (23), we obtain the following theorem.

Theorem 2 For $r,n\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{rn+n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ ={(-1)}^n\sum _{j=k}^n\sum _{l=0}^{j-k}{(-1)}^{k+l}\left(\genfrac{}{}{0ex}{}{rn+l-1}{l}\right)\frac{n!}{(j-l-k+n)!}\frac{(j-1)!}{(k-1)!}\hfill \\ \times S_1(j-l-k+n,n)S_2(n,j).\hfill \end{array}$$

Here we invoke the following identity:

$$\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{m=1}^{n}m{H}_m^{(r)}\right)t^n=\frac{t(1-rlog(1-t))}{{(1-t)}^{r+2}}.$$

(24)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^{r+2}).$$

(25)

For $n\ge 1$, by (19) and (25), we get

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{(r+3)n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.$$

(26)

Let us assume that

$$p_n(x)\sim (1,t(1-rlog(1-t))).$$

(27)

For $n\ge 1$, by (11), (27) and $x^n\sim (1,t)$, we get

$$\begin{array}{rcl}{p}_n(x)& =& 7x{\left(\frac{t}{t(1-rlog(1-t))}\right)}^n{x}^{-1}{x}^n\\ =& x{(1-rlog(1-t))}^{-n}{x}^{n-1}\\ =& x\sum _{l=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)r^l{(log(1-t))}^l{x}^{n-1}\\ =& x\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)r^l\sum _{j=0}^{n-1-l}\frac{l!}{(j+l)!}{S}_1(j+l,l)t^{j+l}{x}^{n-1}\\ =& \sum _{l=0}^{n-1}\sum _{j=0}^{n-1-l}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{j+l}\right)S_1(j+l,l)x^{n-j-l}\\ =& \sum _{k=1}^n\{\sum _{l=0}^{n-k}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)S_1(n-k,l)\}{x}^k.\end{array}$$

(28)

For $n\ge 1$, from (11), (25) and (27), we can derive the following equation:

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t(1-rlog(1-t))}{t{(1-t)}^{r+2}}\right)}^n{x}^{-1}{p}_n(x)\\ =& x{\left(\sum _{j=1}^{\mathrm{\infty }}\left(\sum _{m=1}^{j}m{H}_m^{(r)}\right)t^{j-1}\right)}^n\sum _{a=1}^n\{\sum _{l=0}^{n-a}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)\\ \times S_1(n-a,l)\}{x}^{a-1}\\ =& \sum _{a=1}^n\{\sum _{l=0}^{n-a}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l)\}\\ \times x\left[\sum _{j=0}^{\mathrm{\infty }}\left\{\sum _{j_1+\cdots +j_n=j}(\sum _{m_1=1}^{j_1+1}\cdots \sum _{m_n=1}^{j_n+1}{m}_1\cdots m_n{H}_{m_1}^{(r)}\cdots H_{m_n}^{(r)})\right\}{t}^j\right]x^{a-1}\\ =& \sum _{a=1}^n\sum _{l=0}^{n-a}\sum _{k=1}^a\sum _{j_1+\cdots +j_n=a-k}(\sum _{m_1=1}^{j_1+1}\cdots \sum _{m_n=1}^{j_n+1}{m}_1\cdots m_n{H}_{m_1}^{(r)}\cdots H_{m_n}^{(r)})\\ \times l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_{a-k}{x}^k\\ =& \sum _{k=1}^n\{\sum _{a=k}^n\sum _{l=0}^{n-a}\sum _{j_1+\cdots +j_n=a-k}(\sum _{m_1=1}^{j_1+1}\cdots \sum _{m_n=1}^{j_n+1}{m}_1\cdots m_n{H}_{m_1}^{(r)}\cdots H_{m_n}^{(r)})\\ \times l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_{a-k}\}{x}^k.\end{array}$$

(29)

Therefore, by (26) and (29), we obtain the following theorem.

Theorem 3 For $n,r\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{(r+3)n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =\sum _{a=k}^n\sum _{l=0}^{n-a}\sum _{j_1+\cdots +j_n=a-k}(\sum _{m_1=1}^{j_1+1}\cdots \sum _{m_n=1}^{j_n+1}{m}_1\cdots m_n{H}_{m_1}^{(r)}\cdots H_{m_n}^{(r)})l!r^l\hfill \\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_{n-k}.\hfill \end{array}$$

Here we use the following identity:

$$\sum _{n=1}^{\mathrm{\infty }}n{H}_n^{(r)}{t}^n=\frac{t(1-rlog(1-t))}{{(1-t)}^{r+1}}.$$

(30)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^{r+1}).$$

(31)

For $n\ge 1$, from (19) and (31), we have

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{(r+2)n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.$$

(32)

Let us assume that

$$p_n(x)\sim (1,t(1-rlog(1-t))).$$

(33)

Then, from (28) and (33), we note that, for $n\ge 1$,

$$p_n(x)=\sum _{k=1}^n\{\sum _{l=0}^{n-k}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)S_1(n-k,l)\}{x}^k.$$

(34)

For $n\ge 1$, by (11), (32) and (33), we get

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t(1-rlog(1-t))}{t{(1-t)}^{r+1}}\right)}^n{x}^{-1}{p}_n(x)\\ =& x{\left(\sum _{j=1}^{\mathrm{\infty }}j{H}_j^{(r)}{t}^{j-1}\right)}^n{x}^{-1}\sum _{a=1}^n\{\sum _{l=0}^{n-a}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l)\}{x}^a\\ =& \sum _{a=1}^n\{\sum _{l=0}^{n-a}l!r^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l)\}\\ \times x\sum _{j=0}^{a-1}(\sum _{j_1+\cdots +j_n=j}(j_1+1)\cdots (j_n+1)H_{j_1+1}^{(r)}\cdots H_{j_n+1}^{(r)})t^j{x}^{a-1}\\ =& \sum _{a=1}^n\sum _{l=0}^{n-a}\sum _{j=0}^{a-1}(\sum _{j_1+\cdots +j_n=j}(j_1+1)\cdots (j_n+1)H_{j_1+1}^{(r)}\cdots H_{j_n+1}^{(r)})l!r^l\\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_j{x}^{a-j}\\ =& \sum _{k=1}^n\{\sum _{a=k}^n\sum _{l=0}^{n-a}(\sum _{j_1+\cdots +j_n=a-k}(j_1+1)\cdots (j_n+1)H_{j_1+1}^{(r)}\cdots H_{j_n+1}^{(r)})l!r^l\\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_{a-k}\}{x}^k.\end{array}$$

(35)

Therefore, by (32) and (35), we obtain the following theorem.

Theorem 4 For $n,r\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{(r+2)n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =\sum _{a=k}^n\sum _{l=0}^{n-a}(\sum _{j_1+\cdots +j_n=a-k}(j_1+1)\cdots (j_n+1)H_{j_1+1}^{(r)}\cdots H_{j_n+1}^{(r)})l!r^l\hfill \\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{n-1}{a-1}\right)S_1(n-a,l){(a-1)}_{a-k}.\hfill \end{array}$$

Now, we utilize the following identity:

$$\sum _{n=1}^{\mathrm{\infty }}(n+1)H_n{t}^n=\frac{t-log(1-t)}{{(1-t)}^2}.$$

(36)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^2).$$

(37)

For $n\ge 1$, from (19) and (37), we have

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{3n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.$$

(38)

Let us assume that

$$p_n(x)\sim (1,t-log(1-t)).$$

(39)

We observe that

$$t-log(1-t)=t+\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n}=2t+\sum _{n=2}^{\mathrm{\infty }}\frac{t^n}{n}.$$

(40)

From (11), (39), (40) and $x^n\sim (1,t)$, we can derive the following equation:

$$\begin{array}{rcl}{p}_n(x)& =& x{\left(\frac{t}{2(t+{\sum }_{n=2}^{\mathrm{\infty }}\frac{t^n}{n})}\right)}^n{x}^{-1}{x}^n\\ =& 2^{-n}x{(1+\sum _{n=2}^{\mathrm{\infty }}\frac{t^{n-1}}{2n})}^{-n}{x}^{n-1}\\ =& 2^{-n}x\sum _{l=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{-n}{l}\right){\left(\sum _{n=2}^{\mathrm{\infty }}\frac{t^{n-1}}{2n}\right)}^l{x}^{n-1}\\ =& 2^{-n}x\sum _{l=0}^{n-1}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\\ \times \sum _{m=0}^{n-1-l}\sum _{m_1+\cdots +m_l=m}\frac{1}{2^l(m_1+2)\cdots (m_l+2)}{t}^{m+l}{x}^{n-1}\\ =& 2^{-n}\sum _{l=0}^{n-1}\sum _{m=0}^{n-1-l}\sum _{m_1+\cdots +m_l=m}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\frac{{(n-1)}_{m+l}}{(m_1+2)\cdots (m_l+2)}{x}^{n-l-m}\\ =& 2^{-n}\sum _{k=1}^n\left\{\sum _{l=0}^{n-k}\sum _{m_1+\cdots +m_l=n-l-k}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\frac{{(n-1)}_{n-k}}{(m_1+2)\cdots (m_l+2)}\right\}{x}^k.\end{array}$$

(41)

For $n\ge 1$, by (11), (37), (39) and (41), we get

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t-log(1-t)}{t-{(1-t)}^2}\right)}^n{x}^{-1}{p}_n(x)\\ =& x{(\sum _{j=0}^{\mathrm{\infty }}(j+2)H_{j+1}{t}^j)}^n{x}^{-1}{2}^{-n}\sum _{a=1}^n\{\sum _{l=0}^{n-a}\sum _{m_1+\cdots +m_l=n-l-a}{(-\frac{1}{2})}^l\\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\frac{{(n-1)}_{n-a}}{(m_1+2)\cdots (m_l+2)}\}{x}^a\\ =& 2^{-n}\sum _{a=1}^n\{\sum _{l=0}^{n-a}\sum _{m_1+\cdots +m_l=n-l-a}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\\ \times \frac{{(n-1)}_{n-a}}{(m_1+2)\cdots (m_l+2)}\left\}x\sum _{j=0}^{a-1}\right(\sum _{j_1+\cdots +j_n=j}(j_1+2)\cdots (j_n+2)\\ \times H_{j_1+1}\cdots H_{j_n+1}){(a-1)}_j{x}^{a-1-j}\\ =& 2^{-n}\sum _{a=1}^n\sum _{l=0}^{n-a}\sum _{k=1}^a\sum _{m_1+\cdots +m_l=n-l-a}\sum _{j_1+\cdots +j_n=a-k}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\\ \times \frac{{(n-1)}_{n-a}{(a-1)}_{a-k}}{(m_1+2)\cdots (m_l+2)}(j_1+2)\cdots (j_n+2)H_{j_1+1}\cdots H_{j_n+1}{x}^k\\ =& 2^{-n}\sum _{k=1}^n\{\sum _{a=k}^n\sum _{l=0}^{n-a}\sum _{m_1+\cdots +m_l=n-l-a}\sum _{j_1+\cdots +j_n=a-k}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\\ \times \frac{{(n-1)}_{n-a}{(a-1)}_{a-k}}{(m_1+2)\cdots (m_l+2)}(j_1+2)\cdots (j_n+2)H_{j_1+1}\cdots H_{j_n+1}\}{x}^k.\end{array}$$

(42)

Therefore, by (38) and (42), we obtain the following theorem.

Theorem 5 For $n\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{3n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =2^{-n}\sum _{a=k}^n\sum _{l=0}^{n-a}\sum _{m_1+\cdots +m_l=n-l-a}\sum _{j_1+\cdots +j_n=a-k}{(-\frac{1}{2})}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\hfill \\ \times \frac{{(n-1)}_{n-a}{(a-1)}_{a-k}}{(m_1+2)\cdots (m_l+2)}(j_1+2)\cdots (j_n+2)H_{j_1+1}\cdots H_{j_n+1}.\hfill \end{array}$$

Now, we recall the following identity:

$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{H}_n{t}^n=\frac{t\{1+2t-(1+t)log(1-t)\}}{{(1-t)}^3}.$$

(43)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^3).$$

(44)

For $n\ge 1$, from (19) and (44), we can derive the following equation:

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{4n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.$$

(45)

Let us assume that

$$p_n(x)\sim (1,t\{1+2t-(1+t)log(1-t)\}).$$

(46)

We observe that

$$\begin{array}{rcl}1+2t-(1+t)log(1-t)& =& 1+2t+(1+t)\sum _{j=1}^{\mathrm{\infty }}\frac{t^j}{j}\\ =& 1+2t+t+\sum _{j=2}^{\mathrm{\infty }}\frac{t^j}{j}+\sum _{j=1}^{\mathrm{\infty }}\frac{t^{j+1}}{j}\\ =& 1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{t^{j+2}}{j+2}+\sum _{j=0}^{\mathrm{\infty }}\frac{t^{j+2}}{j+1}\\ =& 1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{(j+2)(j+1)}{t}^{j+2}.\end{array}$$

(47)

For $n\ge 1$, by (11), (46), (47) and $x^n\sim (1,t)$, we get

$$\begin{array}{rcl}{p}_n(x)& =& x{\left(\frac{t}{t\{1+2t-(1+t)log(1-t)\}}\right)}^n{x}^{-1}{x}^n\\ =& x{(1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{(j+1)(j+2)}{t}^{j+2})}^{-n}{x}^{n-1}\\ =& x\sum _{l=0}^{n-1}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right){(3+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{(j+1)(j+2)}{t}^{j+1})}^l{t}^l{x}^{n-1}\\ =& \sum _{l=0}^{n-1}\sum _{a=0}^{n-1-l}\sum _{k=1}^{n-a-l}\sum _{j_1+\cdots +j_a=n-a-k-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-k}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)x^k\\ =& \sum _{k=1}^n\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{j_1+\cdots +j_a=n-a-k-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-a)}_{n-k}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\}{x}^k.\end{array}$$

(48)

For $n\ge 1$, from (11), (44), (46) and (48), we have

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t(1+2t-(1+t)log(1-t))}{t{(1-t)}^3}\right)}^n{x}^{-1}{p}_n(x)\\ =& \sum _{m=1}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-m}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)x\sum _{b=0}^{m-1}\sum _{b_1+\cdots +b_n=b}\left(\prod _{i=1}^n{(b_i+1)}^2{H}_{b_i+1}\right)t^b{x}^{m-1}\\ =& \sum _{m=1}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-m}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\sum _{b=0}^{m-1}\sum _{b_1+\cdots +b_n=b}\prod _{i=1}^n{(b_i+1)}^2{H}_{b_i+1}{(m-1)}_b{x}^{m-b}\\ =& \sum _{k=1}^n\{\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)\\ \times 3^{l-a}{(n-1)}_{n-m}{(m-1)}_{m-k}\left(\frac{{\prod }_{i=1}^a(2j_i+3){\prod }_{i=1}^n{(b_i+1)}^2{H}_{b_i+1}}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\}{x}^k.\end{array}$$

(49)

Therefore, by (45) and (49), we obtain the following theorem.

Theorem 6 For $n\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{4n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}\hfill \\ \times {(n-1)}_{n-m}{(m-1)}_{m-k}\left(\frac{{\prod }_{i=1}^a(2j_i+3){\prod }_{i=1}^n{(b_i+1)}^2{H}_{b_i+1}}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right).\hfill \end{array}$$

Here we invoke the following identity:

$$\sum _{b=1}^{\mathrm{\infty }}\left(\sum _{c=1}^b{c}^2{H}_c\right)t^b=\frac{t\{1+2t-(1+t)log(1-t)\}}{{(1-t)}^4}.$$

(50)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^4).$$

(51)

From (19) and (51), we note that

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{5n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k.$$

(52)

Let us assume that

$$p_n(x)\sim (1,t(1+2t-(1+t)log(1-t))).$$

(53)

For $n\ge 1$, from (48) and (49), we have

$$\begin{array}{rcl}{p}_n(x)& =& \sum _{k=1}^n\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{j_1+\cdots +j_a=n-a-k-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-k}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\}{x}^k.\end{array}$$

(54)

For $n\ge 1$, from (11), (51), (53) and (50), we can derive the following identity:

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t\{1+2t-(1+t)log(1-t)\}}{t{(1-t)}^4}\right)}^n{x}^{-1}{p}_n(x)\\ =& x{\left(\sum _{b=0}^{\mathrm{\infty }}\left(\sum _{c=1}^{b+1}{c}^2{H}_c\right)t^b\right)}^n{x}^{-1}{p}_n(x)\\ =& x\sum _{b=0}^{\mathrm{\infty }}\sum _{b_1+\cdots +b_n=b}\{\sum _{c_1=1}^{b_1+1}\cdots \sum _{c_n=1}^{b_n+1}{c}_1^2\cdots c_n^2{H}_{c_1}\cdots H_{c_n}\}{t}^b\\ \times \sum _{m=1}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}\\ \times {(n-1)}_{n-m}\left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\}{x}^{m-1}\\ =& \sum _{m=1}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-m}\\ \times \left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\sum _{b=0}^{m-1}\sum _{b_1+\cdots +b_n=b}\{\sum _{c_1=1}^{b_1+1}\cdots \sum _{c_n=1}^{b_n+1}{c}_1^2\cdots c_n^2{H}_{c_1}\cdots H_{c_n}\}\\ \times {(m-1)}_b{x}^{m-b}\\ =& \sum _{k=1}^n\{\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\\ \times \left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}{(n-1)}_{n-m}{(m-1)}_{m-k}\left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\\ \times \sum _{c_1=1}^{b_1+1}\cdots \sum _{c_n=1}^{b_n+1}\prod _{i=1}^n{c}_i^2{H}_{c_i}\}{x}^k.\end{array}$$

(55)

Therefore, by (52) and (55), we obtain the following theorem.

Theorem 7 For $n\ge 1$, $1\le k\le n$, we have

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{5n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-m-l}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)3^{l-a}\hfill \\ \times {(n-1)}_{n-m}{(m-1)}_{m-k}\left(\frac{{\prod }_{i=1}^a(2j_i+3)}{{\prod }_{i=1}^a(j_i+1)(j_i+2)}\right)\sum _{c_1=1}^{b_1+1}\cdots \sum _{c_n=1}^{b_n+1}\prod _{i=1}^n{c}_i^2{H}_{c_i}.\hfill \end{array}$$

Here we use the following identity:

$$\sum _{n=1}^{\mathrm{\infty }}n(2n+1)H_n{t}^n=\frac{t\{3(1+t)-(t+3)log(1-t)\}}{{(1-t)}^3}.$$

(56)

Let us consider the following associated sequence:

$$q_n(x)\sim (1,t{(1-t)}^3).$$

(57)

By (19) and (57), we get

$$q_n(x)=\sum _{k=1}^n\left(\genfrac{}{}{0ex}{}{4n-k-1}{n-k}\right){(n-1)}_{n-k}{x}^k(n\ge 1).$$

(58)

Let us assume that

$$p_n(x)\sim (1,t\{3(1+t)-(t+3)log(1-t)\}).$$

(59)

We see that

$$3(1+t)-(t+3)log(1-t)=3+6t+\sum _{n=1}^{\mathrm{\infty }}\frac{4n+1}{n(n+1)}{t}^{n+1}.$$

(60)

For $n\ge 1$, from (11), (59), (60) and $x^n\sim (1,t)$, we have

$$\begin{array}{rcl}{p}_n(x)& =& x{\left(\frac{t}{t\{3(1+t)-(t+3)log(1-t)\}}\right)}^n{x}^{-1}{x}^n\\ =& x{(3(1+t)-(t+3)log(1-t))}^{-n}{x}^{n-1}\\ =& x{(3+6t+\sum _{j=1}^{\mathrm{\infty }}\frac{4j+1}{j(j+1)}{t}^{j+1})}^{-n}{x}^{n-1}.\end{array}$$

(61)

From (61), by the same method of (48), we get

$$\begin{array}{rcl}{p}_n(x)& =& 3^{-n}\sum _{k=1}^n\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{j_1+\cdots +j_a=n-a-l-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)2^{l-a}\\ \times {(n-1)}_{n-k}\left(\prod _{i=1}^a\frac{(4j_i+5)}{3(j_i+1)(j_i+2)}\right)\}{x}^k.\end{array}$$

(62)

For $n\ge 1$, by (11), (56), (57), (59) and (62), we get

$$\begin{array}{rcl}{q}_n(x)& =& x{\left(\frac{t\{3(1+t)-(t+3)log(1-t)\}}{t{(1-t)}^3}\right)}^n{x}^{-1}{p}_n(x)\\ =& x{(\sum _{b=0}^{\mathrm{\infty }}(b+1)(2b+3)H_{b+1}{t}^b)}^n{x}^{-1}{p}_n(x)\\ =& x\sum _{b=0}^{\mathrm{\infty }}\left(\sum _{b_1+\cdots +b_n=b}(\prod _{i=1}^b(b_i+1)(2b_i+3)H_{b_i+1})t^b\right)\\ \times 3^{-n}\sum _{m=1}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-l-m}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)2^{l-a}\\ \times {(n-1)}_{n-m}\prod _{i=1}^a\frac{(4j_i+5)}{3(j_i+1)(j_i+2)}\}{x}^{m-1}\\ =& 3^{-n}\sum _{m=1}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-l-m}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)2^{l-a}{(n-1)}_{n-m}\\ \times \left(\prod _{i=1}^a\frac{(4j_i+5)}{3(j_i+1)(j_i+2)}\right)\sum _{b=0}^{m-1}\sum _{b_1+\cdots +b_n=b}(\prod _{i=1}^n(b_i+1)(2b_i+3)H_{b_i+1})\\ \times {(m-1)}_b{x}^{m-b}.\end{array}$$

(63)

By the same method, we can derive the following identity from (63):

$$\begin{array}{rcl}{q}_n(x)& =& 3^{-n}\sum _{k=1}^n\{\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-l-m}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\\ \times \left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)2^{l-a}{(n-1)}_{n-m}{(m-1)}_{m-k}\left(\prod _{i=1}^a\frac{(4j_i+5)}{3(j_i+1)(j_i+2)}\right)\\ \times \prod _{i=1}^n(b_i+1)(2b_i+3)H_{b_i+1}\}{x}^k.\end{array}$$

(64)

By comparing coefficients on both sides of (58) and (64), we get

$$\begin{array}{c}\left(\genfrac{}{}{0ex}{}{4n-k-1}{n-k}\right){(n-1)}_{n-k}\hfill \\ =3^{-n}\sum _{m=k}^n\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{j_1+\cdots +j_a=n-a-l-m}\sum _{b_1+\cdots +b_n=m-k}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{a}\right)\hfill \\ \times 2^{l-a}{(n-1)}_{n-m}{(m-1)}_{m-k}\left(\prod _{i=1}^a\frac{(4j_i+5)}{3(j_i+1)(j_i+2)}\right)\hfill \\ \times (\prod _{i=1}^n(b_i+1)(2b_i+3)H_{b_i+1}).\hfill \end{array}$$

(65)

Remark Recently, several authors have studied the q-extension of harmonic and hyperharmonic numbers (see [11–13]).