Theory and Modern Applications

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

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}|{a}_{k}\in \mathbf{C}\right\}.$
(1)

Suppose that 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 . The action of the linear functional L on a polynomial $p\left(x\right)$ is denoted by $〈L|p\left(x\right)〉$.

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

(2)

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

(3)

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

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

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f\left(t\right)|{x}^{k}〉}{k!}{t}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈{t}^{k}|p\left(x\right)〉}{k!}{x}^{k}\phantom{\rule{1em}{0ex}}\text{(see [1, 6, 7])}.$
(4)

From (4), we note that

${p}^{\left(k\right)}\left(0\right)=〈{t}^{k}|p\left(x\right)〉,\phantom{\rule{2em}{0ex}}〈1|{p}^{\left(k\right)}\left(x\right)〉={p}^{\left(k\right)}\left(0\right).$
(5)

By (5), we easily see that

(6)

Let ${\varphi }_{n}\left(x\right)$ be exponential polynomials which are given by

$\sum _{k=0}^{\mathrm{\infty }}\frac{{\varphi }_{k}\left(x\right)}{k!}{t}^{k}={e}^{x\left({e}^{t}-1\right)}\phantom{\rule{1em}{0ex}}\text{(see [2, 6, 8])}.$
(7)

Thus, by (7), we get

${\varphi }_{n}\left(x\right)=\sum _{k=0}^{n}{S}_{2}\left(n,k\right){x}^{k}\sim \left(1,log\left(1+t\right)\right),$
(8)

where ${S}_{2}\left(n,k\right)$ is the Stirling number of the second kind.

The Stirling number of the first kind is defined by

${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{k=0}^{n}{S}_{1}\left(n,k\right){x}^{k}.$
(9)

Thus, by (9), we get

${S}_{1}\left(n,k\right)=\frac{1}{k!}〈{t}^{k}|{\left(x\right)}_{n}〉\phantom{\rule{1em}{0ex}}\text{(see [2, 5])}.$
(10)

Let ${p}_{n}\left(x\right)\sim \left(1,f\left(t\right)\right)$, ${q}_{n}\left(x\right)\sim \left(1,g\left(t\right)\right)$. Then the transfer formula for the associated sequences is given by

${q}_{n}\left(x\right)=x{\left(\frac{f\left(t\right)}{g\left(t\right)}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\text{(see [2, 8])}.$
(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 ${H}_{n}^{\left(r\right)}$ of order r is defined by

(12)

From (12), we note that ${H}_{n}^{\left(1\right)}$ is the ordinary harmonic number ${H}_{n}$. It is known that

${H}_{n}^{\left(r\right)}=\left(\genfrac{}{}{0}{}{n+r-1}{r-1}\right)\left({H}_{n+r-1}-{H}_{r-1}\right)\phantom{\rule{1em}{0ex}}\text{(see [9, 10])}.$
(13)

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

$\sum _{n=1}^{\mathrm{\infty }}{H}_{n}{t}^{n}=-\frac{log\left(1-t\right)}{1-t}$
(14)

and

$\sum _{n=1}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}{t}^{n}=-\frac{log\left(1-t\right)}{{\left(1-t\right)}^{r}},\phantom{\rule{1em}{0ex}}\text{respectively}.$
(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

${\varphi }_{n}\left(x\right)=\sum _{j=0}^{n}{S}_{2}\left(n,j\right){x}^{j}\sim \left(1,log\left(1+t\right)\right)$
(16)

and

${\left(-1\right)}^{n}{\varphi }_{n}\left(-x\right)\sim \left(1,-log\left(1-t\right)\right).$
(17)

Let us assume that

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{r}\right).$
(18)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t}{t{\left(1-t\right)}^{r}}\right)}^{n}{x}^{-1}{x}^{n}=x{\left(1-t\right)}^{-rn}{x}^{n-1}\\ =& x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{-rn}{k}\right){\left(-t\right)}^{k}{x}^{n-1}=x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{rn+k-1}{k}\right){t}^{k}{x}^{n-1}\\ =& x\sum _{k=0}^{n-1}\left(\genfrac{}{}{0}{}{rn+k-1}{k}\right){\left(n-1\right)}_{k}{x}^{n-1-k}=\sum _{k=1}^{n-1}\left(\genfrac{}{}{0}{}{rn+k-1}{k}\right){\left(n-1\right)}_{k}{x}^{n-k}\\ =& \sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{rn+n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.\end{array}$
(19)

Now, we use the following fact:

$\sum _{n=1}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}{t}^{n}=-\frac{log\left(1-t\right)}{\left(1-t\right)r}.$
(20)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{-log\left(1-t\right)}{t{\left(1-t\right)}^{r}}\right)}^{n}{x}^{-1}{\left(-1\right)}^{n}{\varphi }_{n}\left(-x\right)\\ =& x{\left(\sum _{l=0}^{\mathrm{\infty }}{H}_{l+1}^{\left(r\right)}{t}^{l}\right)}^{n}{x}^{-1}{\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-x\right)}^{j}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}x{\left(\sum _{l=0}^{\mathrm{\infty }}{H}_{l+1}^{\left(r\right)}{t}^{l}\right)}^{n}{x}^{j-1}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}x\left(\sum _{l=0}^{j-1}\left(\sum _{{l}_{1}+\cdots +{l}_{n}=l}{H}_{{l}_{1}+1}^{\left(r\right)}\cdots {H}_{{l}_{n}+1}^{\left(r\right)}\right){t}^{l}\right){x}^{j-1}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}\sum _{l=0}^{j-1}\sum _{{l}_{1}+\cdots +{l}_{n}=l}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}{H}_{{l}_{1}+1}^{\left(r\right)}\cdots {H}_{{l}_{n}+1}^{\left(r\right)}{\left(j-1\right)}_{l}{x}^{j-l}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}\sum _{k=1}^{j}\sum _{{l}_{1}+\cdots +{l}_{n}=j-k}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}{H}_{{l}_{1}+1}^{\left(r\right)}\cdots {H}_{{l}_{n}+1}^{\left(r\right)}{\left(j-1\right)}_{j-k}{x}^{k}\\ =& {\left(-1\right)}^{n}\sum _{k=1}^{n}\left\{\sum _{j=k}^{n}\sum _{{l}_{1}+\cdots +{l}_{n}=j-k}{\left(-1\right)}^{j}{S}_{2}\left(n,j\right){H}_{{l}_{1}+1}^{\left(r\right)}\cdots {H}_{{l}_{n}+1}^{\left(r\right)}{\left(j-1\right)}_{j-k}\right\}{x}^{k}.\end{array}$
(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

$\left(\genfrac{}{}{0}{}{rn+n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}={\left(-1\right)}^{n}\sum _{j=k}^{n}\sum _{{l}_{1}+\cdots +{l}_{n}=j-k}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}{H}_{{l}_{1}+1}^{\left(r\right)}\cdots {H}_{{l}_{n}+1}^{\left(r\right)}{\left(j-1\right)}_{j-k}.$

We recall the following equation:

${\left(\frac{log\left(1+t\right)}{t}\right)}^{n}=\sum _{l=0}^{\mathrm{\infty }}\frac{n!}{\left(l+n\right)!}{S}_{1}\left(l+n,n\right){t}^{l}.$
(22)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{-log\left(1-t\right)}{t{\left(1-t\right)}^{r}}\right)}^{n}{x}^{-1}{\left(-1\right)}^{n}{\varphi }_{n}\left(-x\right)\\ =& x{\left(\frac{log\left(1-t\right)}{-t}\right)}^{n}{\left(1-t\right)}^{-rn}{x}^{-1}{\left(-1\right)}^{n}{\varphi }_{n}\left(-x\right)\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}x{\left(\frac{log\left(1-t\right)}{-t}\right)}^{n}{\left(1-t\right)}^{-rn}{x}^{j-1}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}x{\left(\frac{log\left(1-t\right)}{-t}\right)}^{n}\sum _{l=0}^{j-1}\left(\genfrac{}{}{0}{}{rn+l-1}{l}\right){\left(j-1\right)}_{l}{x}^{j-1-l}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}{S}_{2}\left(n,j\right){\left(-1\right)}^{j}\sum _{l=0}^{j-1}\left(\genfrac{}{}{0}{}{rn+l-1}{l}\right){\left(j-1\right)}_{l}x\sum _{m=0}^{j-1-l}\frac{n!}{\left(m+n\right)!}\\ ×{S}_{1}\left(m+n,n\right){\left(-t\right)}^{m}{x}^{j-1-l}\\ =& {\left(-1\right)}^{n}\sum _{j=1}^{n}\sum _{l=0}^{j-1}\sum _{m=0}^{j-1-l}{\left(-1\right)}^{j+m}\left(\genfrac{}{}{0}{}{rn+l-1}{l}\right)\frac{n!}{\left(m+n\right)!}\frac{\left(j-1\right)!}{\left(j-1-l-m\right)!}\\ ×{S}_{1}\left(m+n,n\right){S}_{2}\left(n,j\right){x}^{j-l-m}\\ =& {\left(-1\right)}^{n}\sum _{k=1}^{n}\left\{\sum _{j=k}^{n}\sum _{l=0}^{j-k}{\left(-1\right)}^{k+l}\left(\genfrac{}{}{0}{}{rn+l-1}{l}\right)\frac{n!}{\left(j-l-k+n\right)!}\frac{\left(j-1\right)!}{\left(k-1\right)!}\\ ×{S}_{1}\left(j-l-k+n,n\right){S}_{2}\left(n,j\right)\right\}{x}^{k}.\end{array}$
(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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{rn+n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{j=k}^{n}\sum _{l=0}^{j-k}{\left(-1\right)}^{k+l}\left(\genfrac{}{}{0}{}{rn+l-1}{l}\right)\frac{n!}{\left(j-l-k+n\right)!}\frac{\left(j-1\right)!}{\left(k-1\right)!}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{S}_{1}\left(j-l-k+n,n\right){S}_{2}\left(n,j\right).\hfill \end{array}$

Here we invoke the following identity:

$\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{m=1}^{n}m{H}_{m}^{\left(r\right)}\right){t}^{n}=\frac{t\left(1-rlog\left(1-t\right)\right)}{{\left(1-t\right)}^{r+2}}.$
(24)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{r+2}\right).$
(25)

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

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{\left(r+3\right)n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.$
(26)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t\left(1-rlog\left(1-t\right)\right)\right).$
(27)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& 7x{\left(\frac{t}{t\left(1-rlog\left(1-t\right)\right)}\right)}^{n}{x}^{-1}{x}^{n}\\ =& x{\left(1-rlog\left(1-t\right)\right)}^{-n}{x}^{n-1}\\ =& x\sum _{l=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right){r}^{l}{\left(log\left(1-t\right)\right)}^{l}{x}^{n-1}\\ =& x\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right){r}^{l}\sum _{j=0}^{n-1-l}\frac{l!}{\left(j+l\right)!}{S}_{1}\left(j+l,l\right){t}^{j+l}{x}^{n-1}\\ =& \sum _{l=0}^{n-1}\sum _{j=0}^{n-1-l}l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{j+l}\right){S}_{1}\left(j+l,l\right){x}^{n-j-l}\\ =& \sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{k-1}\right){S}_{1}\left(n-k,l\right)\right\}{x}^{k}.\end{array}$
(28)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t\left(1-rlog\left(1-t\right)\right)}{t{\left(1-t\right)}^{r+2}}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& x{\left(\sum _{j=1}^{\mathrm{\infty }}\left(\sum _{m=1}^{j}m{H}_{m}^{\left(r\right)}\right){t}^{j-1}\right)}^{n}\sum _{a=1}^{n}\left\{\sum _{l=0}^{n-a}l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right)\\ ×{S}_{1}\left(n-a,l\right)\right\}{x}^{a-1}\\ =& \sum _{a=1}^{n}\left\{\sum _{l=0}^{n-a}l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right){S}_{1}\left(n-a,l\right)\right\}\\ ×x\left[\sum _{j=0}^{\mathrm{\infty }}\left\{\sum _{{j}_{1}+\cdots +{j}_{n}=j}\left(\sum _{{m}_{1}=1}^{{j}_{1}+1}\cdots \sum _{{m}_{n}=1}^{{j}_{n}+1}{m}_{1}\cdots {m}_{n}{H}_{{m}_{1}}^{\left(r\right)}\cdots {H}_{{m}_{n}}^{\left(r\right)}\right)\right\}{t}^{j}\right]{x}^{a-1}\\ =& \sum _{a=1}^{n}\sum _{l=0}^{n-a}\sum _{k=1}^{a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}\left(\sum _{{m}_{1}=1}^{{j}_{1}+1}\cdots \sum _{{m}_{n}=1}^{{j}_{n}+1}{m}_{1}\cdots {m}_{n}{H}_{{m}_{1}}^{\left(r\right)}\cdots {H}_{{m}_{n}}^{\left(r\right)}\right)\\ ×l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right){S}_{1}\left(n-a,l\right){\left(a-1\right)}_{a-k}{x}^{k}\\ =& \sum _{k=1}^{n}\left\{\sum _{a=k}^{n}\sum _{l=0}^{n-a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}\left(\sum _{{m}_{1}=1}^{{j}_{1}+1}\cdots \sum _{{m}_{n}=1}^{{j}_{n}+1}{m}_{1}\cdots {m}_{n}{H}_{{m}_{1}}^{\left(r\right)}\cdots {H}_{{m}_{n}}^{\left(r\right)}\right)\\ ×l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right){S}_{1}\left(n-a,l\right){\left(a-1\right)}_{a-k}\right\}{x}^{k}.\end{array}$
(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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{\left(r+3\right)n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=k}^{n}\sum _{l=0}^{n-a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}\left(\sum _{{m}_{1}=1}^{{j}_{1}+1}\cdots \sum _{{m}_{n}=1}^{{j}_{n}+1}{m}_{1}\cdots {m}_{n}{H}_{{m}_{1}}^{\left(r\right)}\cdots {H}_{{m}_{n}}^{\left(r\right)}\right)l!{r}^{l}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right){S}_{1}\left(n-a,l\right){\left(a-1\right)}_{n-k}.\hfill \end{array}$

Here we use the following identity:

$\sum _{n=1}^{\mathrm{\infty }}n{H}_{n}^{\left(r\right)}{t}^{n}=\frac{t\left(1-rlog\left(1-t\right)\right)}{{\left(1-t\right)}^{r+1}}.$
(30)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{r+1}\right).$
(31)

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

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{\left(r+2\right)n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.$
(32)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t\left(1-rlog\left(1-t\right)\right)\right).$
(33)

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

${p}_{n}\left(x\right)=\sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}l!{r}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{k-1}\right){S}_{1}\left(n-k,l\right)\right\}{x}^{k}.$
(34)

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

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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{\left(r+2\right)n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=k}^{n}\sum _{l=0}^{n-a}\left(\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}\left({j}_{1}+1\right)\cdots \left({j}_{n}+1\right){H}_{{j}_{1}+1}^{\left(r\right)}\cdots {H}_{{j}_{n}+1}^{\left(r\right)}\right)l!{r}^{l}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{n-1}{a-1}\right){S}_{1}\left(n-a,l\right){\left(a-1\right)}_{a-k}.\hfill \end{array}$

Now, we utilize the following identity:

$\sum _{n=1}^{\mathrm{\infty }}\left(n+1\right){H}_{n}{t}^{n}=\frac{t-log\left(1-t\right)}{{\left(1-t\right)}^{2}}.$
(36)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{2}\right).$
(37)

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

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{3n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.$
(38)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t-log\left(1-t\right)\right).$
(39)

We observe that

$t-log\left(1-t\right)=t+\sum _{n=1}^{\mathrm{\infty }}\frac{{t}^{n}}{n}=2t+\sum _{n=2}^{\mathrm{\infty }}\frac{{t}^{n}}{n}.$
(40)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& x{\left(\frac{t}{2\left(t+{\sum }_{n=2}^{\mathrm{\infty }}\frac{{t}^{n}}{n}\right)}\right)}^{n}{x}^{-1}{x}^{n}\\ =& {2}^{-n}x{\left(1+\sum _{n=2}^{\mathrm{\infty }}\frac{{t}^{n-1}}{2n}\right)}^{-n}{x}^{n-1}\\ =& {2}^{-n}x\sum _{l=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{-n}{l}\right){\left(\sum _{n=2}^{\mathrm{\infty }}\frac{{t}^{n-1}}{2n}\right)}^{l}{x}^{n-1}\\ =& {2}^{-n}x\sum _{l=0}^{n-1}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\\ ×\sum _{m=0}^{n-1-l}\sum _{{m}_{1}+\cdots +{m}_{l}=m}\frac{1}{{2}^{l}\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}{t}^{m+l}{x}^{n-1}\\ =& {2}^{-n}\sum _{l=0}^{n-1}\sum _{m=0}^{n-1-l}\sum _{{m}_{1}+\cdots +{m}_{l}=m}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\frac{{\left(n-1\right)}_{m+l}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}{x}^{n-l-m}\\ =& {2}^{-n}\sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-k}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\frac{{\left(n-1\right)}_{n-k}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\right\}{x}^{k}.\end{array}$
(41)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t-log\left(1-t\right)}{t-{\left(1-t\right)}^{2}}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& x{\left(\sum _{j=0}^{\mathrm{\infty }}\left(j+2\right){H}_{j+1}{t}^{j}\right)}^{n}{x}^{-1}{2}^{-n}\sum _{a=1}^{n}\left\{\sum _{l=0}^{n-a}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-a}{\left(-\frac{1}{2}\right)}^{l}\\ ×\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\frac{{\left(n-1\right)}_{n-a}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\right\}{x}^{a}\\ =& {2}^{-n}\sum _{a=1}^{n}\left\{\sum _{l=0}^{n-a}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-a}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\\ ×\frac{{\left(n-1\right)}_{n-a}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\right\}x\sum _{j=0}^{a-1}\left(\sum _{{j}_{1}+\cdots +{j}_{n}=j}\left({j}_{1}+2\right)\cdots \left({j}_{n}+2\right)\\ ×{H}_{{j}_{1}+1}\cdots {H}_{{j}_{n}+1}\right){\left(a-1\right)}_{j}{x}^{a-1-j}\\ =& {2}^{-n}\sum _{a=1}^{n}\sum _{l=0}^{n-a}\sum _{k=1}^{a}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\\ ×\frac{{\left(n-1\right)}_{n-a}{\left(a-1\right)}_{a-k}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\left({j}_{1}+2\right)\cdots \left({j}_{n}+2\right){H}_{{j}_{1}+1}\cdots {H}_{{j}_{n}+1}{x}^{k}\\ =& {2}^{-n}\sum _{k=1}^{n}\left\{\sum _{a=k}^{n}\sum _{l=0}^{n-a}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\\ ×\frac{{\left(n-1\right)}_{n-a}{\left(a-1\right)}_{a-k}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\left({j}_{1}+2\right)\cdots \left({j}_{n}+2\right){H}_{{j}_{1}+1}\cdots {H}_{{j}_{n}+1}\right\}{x}^{k}.\end{array}$
(42)

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

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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{3n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}={2}^{-n}\sum _{a=k}^{n}\sum _{l=0}^{n-a}\sum _{{m}_{1}+\cdots +{m}_{l}=n-l-a}\sum _{{j}_{1}+\cdots +{j}_{n}=a-k}{\left(-\frac{1}{2}\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\frac{{\left(n-1\right)}_{n-a}{\left(a-1\right)}_{a-k}}{\left({m}_{1}+2\right)\cdots \left({m}_{l}+2\right)}\left({j}_{1}+2\right)\cdots \left({j}_{n}+2\right){H}_{{j}_{1}+1}\cdots {H}_{{j}_{n}+1}.\hfill \end{array}$

Now, we recall the following identity:

$\sum _{n=1}^{\mathrm{\infty }}{n}^{2}{H}_{n}{t}^{n}=\frac{t\left\{1+2t-\left(1+t\right)log\left(1-t\right)\right\}}{{\left(1-t\right)}^{3}}.$
(43)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{3}\right).$
(44)

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

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{4n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.$
(45)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t\left\{1+2t-\left(1+t\right)log\left(1-t\right)\right\}\right).$
(46)

We observe that

$\begin{array}{rcl}1+2t-\left(1+t\right)log\left(1-t\right)& =& 1+2t+\left(1+t\right)\sum _{j=1}^{\mathrm{\infty }}\frac{{t}^{j}}{j}\\ =& 1+2t+t+\sum _{j=2}^{\mathrm{\infty }}\frac{{t}^{j}}{j}+\sum _{j=1}^{\mathrm{\infty }}\frac{{t}^{j+1}}{j}\\ =& 1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{{t}^{j+2}}{j+2}+\sum _{j=0}^{\mathrm{\infty }}\frac{{t}^{j+2}}{j+1}\\ =& 1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{\left(j+2\right)\left(j+1\right)}{t}^{j+2}.\end{array}$
(47)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& x{\left(\frac{t}{t\left\{1+2t-\left(1+t\right)log\left(1-t\right)\right\}}\right)}^{n}{x}^{-1}{x}^{n}\\ =& x{\left(1+3t+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{\left(j+1\right)\left(j+2\right)}{t}^{j+2}\right)}^{-n}{x}^{n-1}\\ =& x\sum _{l=0}^{n-1}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right){\left(3+\sum _{j=0}^{\mathrm{\infty }}\frac{2j+3}{\left(j+1\right)\left(j+2\right)}{t}^{j+1}\right)}^{l}{t}^{l}{x}^{n-1}\\ =& \sum _{l=0}^{n-1}\sum _{a=0}^{n-1-l}\sum _{k=1}^{n-a-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-k-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-k}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right){x}^{k}\\ =& \sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-k-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-a\right)}_{n-k}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\right\}{x}^{k}.\end{array}$
(48)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t\left(1+2t-\left(1+t\right)log\left(1-t\right)\right)}{t{\left(1-t\right)}^{3}}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& \sum _{m=1}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-m}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)x\sum _{b=0}^{m-1}\sum _{{b}_{1}+\cdots +{b}_{n}=b}\left(\prod _{i=1}^{n}{\left({b}_{i}+1\right)}^{2}{H}_{{b}_{i}+1}\right){t}^{b}{x}^{m-1}\\ =& \sum _{m=1}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-m}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\sum _{b=0}^{m-1}\sum _{{b}_{1}+\cdots +{b}_{n}=b}\prod _{i=1}^{n}{\left({b}_{i}+1\right)}^{2}{H}_{{b}_{i}+1}{\left(m-1\right)}_{b}{x}^{m-b}\\ =& \sum _{k=1}^{n}\left\{\sum _{m=k}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}\sum _{{b}_{1}+\cdots +{b}_{n}=m-k}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right)\\ ×{3}^{l-a}{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right){\prod }_{i=1}^{n}{\left({b}_{i}+1\right)}^{2}{H}_{{b}_{i}+1}}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\right\}{x}^{k}.\end{array}$
(49)

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

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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{4n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=k}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}\sum _{{b}_{1}+\cdots +{b}_{n}=m-k}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right){\prod }_{i=1}^{n}{\left({b}_{i}+1\right)}^{2}{H}_{{b}_{i}+1}}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right).\hfill \end{array}$

Here we invoke the following identity:

$\sum _{b=1}^{\mathrm{\infty }}\left(\sum _{c=1}^{b}{c}^{2}{H}_{c}\right){t}^{b}=\frac{t\left\{1+2t-\left(1+t\right)log\left(1-t\right)\right\}}{{\left(1-t\right)}^{4}}.$
(50)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{4}\right).$
(51)

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

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{5n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}.$
(52)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t\left(1+2t-\left(1+t\right)log\left(1-t\right)\right)\right).$
(53)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& \sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-k-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-k}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\right\}{x}^{k}.\end{array}$
(54)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t\left\{1+2t-\left(1+t\right)log\left(1-t\right)\right\}}{t{\left(1-t\right)}^{4}}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& 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}\left(x\right)\\ =& x\sum _{b=0}^{\mathrm{\infty }}\sum _{{b}_{1}+\cdots +{b}_{n}=b}\left\{\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}}\right\}{t}^{b}\\ ×\sum _{m=1}^{n}\left\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}\\ ×{\left(n-1\right)}_{n-m}\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\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}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-m}\\ ×\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\sum _{b=0}^{m-1}\sum _{{b}_{1}+\cdots +{b}_{n}=b}\left\{\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}}\right\}\\ ×{\left(m-1\right)}_{b}{x}^{m-b}\\ =& \sum _{k=1}^{n}\left\{\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}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\\ ×\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\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}}\right\}{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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{5n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=k}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-m-l}\sum _{{b}_{1}+\cdots +{b}_{n}=m-k}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){3}^{l-a}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\frac{{\prod }_{i=1}^{a}\left(2{j}_{i}+3\right)}{{\prod }_{i=1}^{a}\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\sum _{{c}_{1}=1}^{{b}_{1}+1}\cdots \sum _{{c}_{n}=1}^{{b}_{n}+1}\prod _{i=1}^{n}{c}_{i}^{2}{H}_{{c}_{i}}.\hfill \end{array}$

Here we use the following identity:

$\sum _{n=1}^{\mathrm{\infty }}n\left(2n+1\right){H}_{n}{t}^{n}=\frac{t\left\{3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)\right\}}{{\left(1-t\right)}^{3}}.$
(56)

Let us consider the following associated sequence:

${q}_{n}\left(x\right)\sim \left(1,t{\left(1-t\right)}^{3}\right).$
(57)

By (19) and (57), we get

${q}_{n}\left(x\right)=\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{4n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}{x}^{k}\phantom{\rule{1em}{0ex}}\left(n\ge 1\right).$
(58)

Let us assume that

${p}_{n}\left(x\right)\sim \left(1,t\left\{3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)\right\}\right).$
(59)

We see that

$3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)=3+6t+\sum _{n=1}^{\mathrm{\infty }}\frac{4n+1}{n\left(n+1\right)}{t}^{n+1}.$
(60)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& x{\left(\frac{t}{t\left\{3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)\right\}}\right)}^{n}{x}^{-1}{x}^{n}\\ =& x{\left(3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)\right)}^{-n}{x}^{n-1}\\ =& x{\left(3+6t+\sum _{j=1}^{\mathrm{\infty }}\frac{4j+1}{j\left(j+1\right)}{t}^{j+1}\right)}^{-n}{x}^{n-1}.\end{array}$
(61)

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

$\begin{array}{rcl}{p}_{n}\left(x\right)& =& {3}^{-n}\sum _{k=1}^{n}\left\{\sum _{l=0}^{n-k}\sum _{a=0}^{n-k-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-l-k}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){2}^{l-a}\\ ×{\left(n-1\right)}_{n-k}\left(\prod _{i=1}^{a}\frac{\left(4{j}_{i}+5\right)}{3\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\right\}{x}^{k}.\end{array}$
(62)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& x{\left(\frac{t\left\{3\left(1+t\right)-\left(t+3\right)log\left(1-t\right)\right\}}{t{\left(1-t\right)}^{3}}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& x{\left(\sum _{b=0}^{\mathrm{\infty }}\left(b+1\right)\left(2b+3\right){H}_{b+1}{t}^{b}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\\ =& x\sum _{b=0}^{\mathrm{\infty }}\left(\sum _{{b}_{1}+\cdots +{b}_{n}=b}\left(\prod _{i=1}^{b}\left({b}_{i}+1\right)\left(2{b}_{i}+3\right){H}_{{b}_{i}+1}\right){t}^{b}\right)\\ ×{3}^{-n}\sum _{m=1}^{n}\left\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-l-m}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){2}^{l-a}\\ ×{\left(n-1\right)}_{n-m}\prod _{i=1}^{a}\frac{\left(4{j}_{i}+5\right)}{3\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right\}{x}^{m-1}\\ =& {3}^{-n}\sum _{m=1}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-l-m}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){2}^{l-a}{\left(n-1\right)}_{n-m}\\ ×\left(\prod _{i=1}^{a}\frac{\left(4{j}_{i}+5\right)}{3\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\sum _{b=0}^{m-1}\sum _{{b}_{1}+\cdots +{b}_{n}=b}\left(\prod _{i=1}^{n}\left({b}_{i}+1\right)\left(2{b}_{i}+3\right){H}_{{b}_{i}+1}\right)\\ ×{\left(m-1\right)}_{b}{x}^{m-b}.\end{array}$
(63)

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

$\begin{array}{rcl}{q}_{n}\left(x\right)& =& {3}^{-n}\sum _{k=1}^{n}\left\{\sum _{m=k}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-l-m}\sum _{{b}_{1}+\cdots +{b}_{n}=m-k}{\left(-1\right)}^{l}\\ ×\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right){2}^{l-a}{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\prod _{i=1}^{a}\frac{\left(4{j}_{i}+5\right)}{3\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\\ ×\prod _{i=1}^{n}\left({b}_{i}+1\right)\left(2{b}_{i}+3\right){H}_{{b}_{i}+1}\right\}{x}^{k}.\end{array}$
(64)

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

$\begin{array}{c}\left(\genfrac{}{}{0}{}{4n-k-1}{n-k}\right){\left(n-1\right)}_{n-k}\hfill \\ \phantom{\rule{1em}{0ex}}={3}^{-n}\sum _{m=k}^{n}\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{{j}_{1}+\cdots +{j}_{a}=n-a-l-m}\sum _{{b}_{1}+\cdots +{b}_{n}=m-k}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n+l-1}{l}\right)\left(\genfrac{}{}{0}{}{l}{a}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{2}^{l-a}{\left(n-1\right)}_{n-m}{\left(m-1\right)}_{m-k}\left(\prod _{i=1}^{a}\frac{\left(4{j}_{i}+5\right)}{3\left({j}_{i}+1\right)\left({j}_{i}+2\right)}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left(\prod _{i=1}^{n}\left({b}_{i}+1\right)\left(2{b}_{i}+3\right){H}_{{b}_{i}+1}\right).\hfill \end{array}$
(65)

Remark Recently, several authors have studied the q-extension of harmonic and hyperharmonic numbers (see [1113]).

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## Acknowledgements

The authors express their sincere gratitude to the referees for their valuable suggestions and comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786).

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read and approved the final manuscript.

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Kim, D.S., Kim, T. Identities involving harmonic and hyperharmonic numbers. Adv Differ Equ 2013, 235 (2013). https://doi.org/10.1186/1687-1847-2013-235

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• DOI: https://doi.org/10.1186/1687-1847-2013-235

### Keywords

• Vector Space
• Formal Power Series
• Linear Functional
• Small Integer
• Kronecker Symbol