Some properties of higher-order Daehee polynomials of the second kind arising from umbral calculus

Abstract

In this paper, we study the higher-order Daehee polynomials of the second kind from the umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

1 Introduction

Let $k\in {\mathbb{Z}}_{\ge 0}$. The Daehee polynomials of the second kind of order k are defined by the generating function to be

$${\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_n^{(k)}(x)\frac{t^n}{n!}$$

(1)

(see [1]).

When $x=0$, ${\hat{D}}_n^{(k)}={\hat{D}}_n^{(k)}(0)$ are called the Daehee numbers of the second kind of order k.

The Stirling number of the first kind is defined by the falling factorial to be

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

(2)

Thus, by (2), we get

$${(log(1+t))}^m=m!\sum _{l=m}^{\mathrm{\infty }}{S}_1(l,m)\frac{t^l}{l!}$$

(3)

(see [2–4]), where $m\in {\mathbb{Z}}_{\ge 0}$.

For $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, the Frobenius-Euler polynomials of order s ($\in \mathbb{N}$) are given by

$${\left(\frac{1-\lambda }{e^t-\lambda }\right)}^s{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(s)}(x|\lambda )\frac{t^n}{n!}$$

(4)

(see [1–18]).

When $x=0$, $H_n^{(s)}(\lambda )=H_n^{(s)}(\lambda |0)$ are called the Frobenius-Euler numbers of order s.

As is well known, the Bernoulli polynomials of order k ($\in \mathbb{N}$) are defined by the generating function to be

$${\left(\frac{t}{e^t-1}\right)}^k{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(k)}(x)\frac{t^n}{n!}$$

(5)

(see [1–18]).

When $x=0$, $B_n^{(k)}=B_n^{(k)}(0)$ are called the Bernoulli numbers of order k.

In this paper, we study the higher-order Daehee polynomials of the second kind with umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

2 Umbral calculus

Let ℂ be the complex number field and let ℱ be the set of all formal power series

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

Let $\mathbb{P}=\mathbb{C}[x]$, and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on ℙ. $〈L|p(x)〉$ indicates the action of the linear functional L on the polynomial $p(x)$. Then the vector space operations on ${\mathbb{P}}^{\ast }$ are given by $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, and $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant in ℂ. For $f(t)\in \mathcal{F}$, the linear functional on ℙ is defined by $〈f(t)|x^n〉=a_n$. Then, in particular, we have

$$〈t^k|x^n〉=n!{\delta }_{n,k}(n,k\ge 0)$$

(6)

(see [3, 18]), where ${\delta }_{n,k}$ is the Kronecker symbol.

Let $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$. By (6), we get $〈f_L(t)|x^n〉=〈L|x^n〉$. That is, $L=f_L(t)$. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ denotes both the algebra of the formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of ℱ will be thought of as both a formal power series and a linear functional. We call ℱ the umbral algebra and the umbral calculus is the study of the umbral algebra. The order $o(f(t))$ of the power series $f(t)$ (≠0) is the smallest integer 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 $f(t),g(t)\in \mathcal{F}$ with $o(f(t))=1$ and $o(g(t))=0$. Then there exists a unique sequence $s_n(x)$ ($deg{s}_n(x)=n$) 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 $s_n(x)\sim (g(t),f(t))$. For $f(t),g(t)\in \mathcal{F}$, we have

$$〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉.$$

(7)

From (6), we note that

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}〈f(t)|x^k〉\frac{t^k}{k!},p(x)=\sum _{k=0}^{\mathrm{\infty }}〈t^k|p(x)〉\frac{x^k}{k!}$$

(8)

and, by (8), we get

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}\text{and}{e}^{yt}p(x)=p(x+y)$$

(9)

(see [3, 18]).

For $s_n(x)\sim (g(t),f(t))$, we have

$$\frac{d{s}_n(x)}{dx}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)〈\overline{f}(t)|x^{n-l}〉s_l(x),$$

(10)

where $\overline{f}(t)$ is the compositional inverse of $f(t)$ with $\overline{f}(f(t))=t$. We have

$$\frac{1}{g(\overline{f}(t))}{e}^{x\overline{f}(t)}=\sum _{n=0}^{\mathrm{\infty }}{s}_n(x)\frac{t^n}{n!},\text{for all }x\in \mathbb{C},$$

(11)

$$f(t)s_n(x)=n{s}_{n-1}(x)(n\ge 1),s_n(x)=\sum _{j=0}^n\frac{1}{j!}〈g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^j|x^n〉x^j,$$

(12)

$$s_n(x+y)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)s_j(x)p_{n-j}(y),$$

(13)

where $p_n(x)=g(t)s_n(x)$.

$$〈f(t)|xp(x)〉=〈{\partial }_{t}f(t)|p(x)〉,$$

(14)

with ${\partial }_{t}f(t)=\frac{df(t)}{dt}$, and

$$s_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})\frac{1}{f^{\prime }(t)}{s}_n(x)(n\ge 0)$$

(15)

(see [3, 18]).

Let us assume that $s_n(x)\sim (g(t),f(t))$ and $r_n(x)\sim (h(t),l(t))$. Then we see that

$$s_n(x)=\sum _{m=0}^n{C}_{n,m}{r}_m(x)(n\ge 0),$$

(16)

where

$$C_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉$$

(17)

(see [3, 18]).

3 Higher-order Daehee polynomials of the second kind

By (1), we see that

$${\hat{D}}_n^{(k)}(x)\sim ({\left(\frac{e^t-1}{t{e}^t}\right)}^k,e^t-1).$$

(18)

From (18), we have

$${\left(\frac{e^t-1}{t{e}^t}\right)}^k{\hat{D}}_n^{(k)}(x)\sim (1,e^t-1)\text{and}{(x)}_n\sim (1,e^t-1).$$

(19)

By (19), we get

$$\begin{array}{rl}{\hat{D}}_n^{(k)}(x)& ={\left(\frac{t{e}^t}{e^t-1}\right)}^k{(x)}_n\\ =\sum _{m=0}^n{S}_1(n,m){\left(\frac{t{e}^t}{e^t-1}\right)}^k{x}^m\\ =\sum _{m=0}^n{S}_1(n,m)e^{kt}{B}_n^{(k)}(x)\\ =\sum _{m=0}^n{S}_1(n,m)B_m^{(k)}(x+k).\end{array}$$

(20)

From (12) and (18), we have

$${\hat{D}}_n^{(k)}(x)=\sum _{j=0}^n\frac{1}{j!}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^j|x^n〉x^j,$$

(21)

where

$$\begin{array}{c}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^j|x^n〉\hfill \\ =〈{\left(\frac{log(1+t)}{t}\right)}^{k+j}{(1+t)}^k|t^j{x}^n〉\hfill \\ ={(n)}_j〈{\left(\frac{log(1+t)}{t}\right)}^{k+j}|\sum _{m=0}^{min\{k,n-j\}}\left(\genfrac{}{}{0ex}{}{k}{m}\right)t^m{x}^{n-j}〉\hfill \\ ={(n)}_j\sum _{m=0}^{n-j}\left(\genfrac{}{}{0ex}{}{k}{m}\right){(n-j)}_m\sum _{l=0}^{\mathrm{\infty }}\frac{(k+j)!}{(l+k+j)!}{S}_1(l+k+j,k+j)〈t^l|x^{n-j-m}〉\hfill \\ ={(n)}_j\sum _{m=0}^{n-j}\left(\genfrac{}{}{0ex}{}{k}{m}\right){(n-j)}_m\frac{(k+j)!}{(n+k-m)!}{S}_1(n+k-m,k+j)(n-j-m)!\hfill \\ ={(n)}_j\sum _{m=0}^{n-j}\left(\genfrac{}{}{0ex}{}{k}{m}\right){(n-j)}_m\frac{S_1(n+k-m,k+j)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}.\hfill \end{array}$$

(22)

Therefore, by (21) and (22), we obtain the following theorem.

Theorem 1 For $n\in {\mathbb{Z}}_{\ge 0}$ and $k\ge 1$, we have

$${\hat{D}}_n^{(k)}(x)=\sum _{j=0}^n\left\{\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{m=0}^{n-j}\left(\genfrac{}{}{0ex}{}{k}{m}\right){(n-j)}_m\frac{S_1(n+k-m,k+j)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\right\}{x}^j.$$

By (1) and (6), we get

$$\begin{array}{rcl}{\hat{D}}_n^{(k)}(y)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\hat{D}}_l^{(k)}(y)\frac{t^l}{l!}|x^n〉\\ =& 〈{\left(\frac{log(1+t)}{t}\right)}^k{(1+t)}^y|{(1+t)}^k{x}^n〉\\ =& \sum _{0\le r\le min\{k,n\}}\left(\genfrac{}{}{0ex}{}{k}{r}\right){(n)}_r〈{\left(\frac{log(1+t)}{t}\right)}^k{(1+t)}^y|x^{n-r}〉\\ =& \sum _{0\le r\le min\{k,n\}}\left(\genfrac{}{}{0ex}{}{k}{r}\right){(n)}_r\sum _{0\le m\le n-r}\left(\genfrac{}{}{0ex}{}{y}{m}\right){(n-r)}_m\\ \times \sum _{0\le l\le n-r-m}\frac{k!S_1(l+k,k)}{(l+k)!}〈t^l|x^{n-r-m}〉\\ =& \sum _{0\le r\le n}\sum _{0\le m\le n-r}\frac{{(n)}_r\left(\genfrac{}{}{0ex}{}{k}{r}\right)\left(\genfrac{}{}{0ex}{}{n-r}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n-r-m+k}{k}\right)}{S}_1(n-r-m+k,k){(y)}_m.\end{array}$$

(23)

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

Theorem 2 For $n\ge 0$, we have

$$\begin{array}{c}{\hat{D}}_n^{(k)}(x)\hfill \\ =\sum _{0\le m\le n}\{\sum _{0\le r\le n-m}\frac{{(n)}_r\left(\genfrac{}{}{0ex}{}{k}{r}\right)\left(\genfrac{}{}{0ex}{}{n-r}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n-r-m+k}{k}\right)}{S}_1(n-r-m+k,k)\}{(x)}_m\hfill \\ =\sum _{0\le m\le n}\{\sum _{0\le r\le n-m}\frac{{(n)}_r\left(\genfrac{}{}{0ex}{}{k}{r}\right)\left(\genfrac{}{}{0ex}{}{n-r}{n-m}\right)}{\left(\genfrac{}{}{0ex}{}{m-r+k}{k}\right)}{S}_1(m-r+k,k)\}{(x)}_{n-m}.\hfill \end{array}$$

From (12) and (18), we have

$$(e^t-1){\hat{D}}_n^{(k)}(x)=n{\hat{D}}_{n-1}^{(k)}(x)$$

(24)

and

$$(e^t-1){\hat{D}}_n^{(k)}(x)={\hat{D}}_n^{(k)}(x+1)-{\hat{D}}_n^{(k)}(x).$$

Thus, by (24), we get

$${\hat{D}}_n^{(k)}(x+1)-{\hat{D}}_n^{(k)}(x)=n{\hat{D}}_{n-1}^{(k)}(x)(n\ge 1).$$

(25)

From (15) and (18), we derive the following equation:

$$\begin{array}{rl}{\hat{D}}_{n+1}^{(k)}(x)& =(x+k\frac{e^t-1-t}{t(e^t-1)})e^{-t}{\hat{D}}_n^{(k)}(x)\\ =x{\hat{D}}_n^{(k)}(x-1)+k{e}^{-t}\frac{e^t-1-t}{t(e^t-1)}{\hat{D}}_n^{(k)}(x),\end{array}$$

(26)

where

$$\begin{array}{c}{e}^{-t}\frac{e^t-1-t}{t(e^t-1)}{\hat{D}}_n^{(k)}(x)\hfill \\ =e^{-t}\frac{e^t-1-t}{t(e^t-1)}\sum _{0\le j\le n}\{\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times S_1(n+k-m,k+j)\}{x}^j\hfill \\ =\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times S_1(n+k-m,k+j)e^{-t}\frac{e^t-1-t}{t(e^t-1)}{x}^j\hfill \\ =\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times S_1(n+k-m,k+j)e^{-t}\left(\frac{e^t-1-t}{e^t-1}\right)\frac{x^{j+1}}{j+1}\hfill \\ =\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times \frac{S_1(n+k-m,k+j)}{j+1}{e}^{-t}(x^{j+1}-B_{j+1}(x))\hfill \\ =\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times \frac{S_1(n+k-m,k+j)}{j+1}{e}^{-t}({(x-1)}^{j+1}-B_{j+1}(x-1)).\hfill \end{array}$$

(27)

Therefore, from (26) and (27), we obtain the following theorem.

Theorem 3 For $n\ge 0$, $k\ge 1$, we have

$$\begin{array}{c}{\hat{D}}_{n+1}^{(k)}(x)\hfill \\ =x{\hat{D}}_n^{(k)}(x-1)+k\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)\left(\genfrac{}{}{0ex}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0ex}{}{n+k-m}{k+j}\right)}\hfill \\ \times \frac{S_1(n+k-m,k+j)}{j+1}\{{(x-1)}^{j+1}-B_{j+1}(x-1)\}.\hfill \end{array}$$

Now, we observe that

$$\begin{array}{c}{e}^{-t}\frac{e^t-1-t}{t(e^t-1)}{\hat{D}}_n^{(k)}(x)\hfill \\ =\sum _{j=0}^n\frac{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}{\left(\genfrac{}{}{0ex}{}{n+k}{j+k}\right)}{S}_1(n+k,j+k)e^{-t}\frac{e^t-1-t}{t(e^t-1)}{(x+k)}^j\hfill \\ =\sum _{j=0}^n\frac{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}{\left(\genfrac{}{}{0ex}{}{n+k}{j+k}\right)}{S}_1(n+k,j+k)e^{(k-1)t}\frac{e^t-1-t}{t(e^t-1)}{x}^j\hfill \\ =\sum _{j=0}^n\frac{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}{\left(\genfrac{}{}{0ex}{}{n+k}{j+k}\right)}\frac{S_1(n+k,j+k)}{j+1}{e}^{(k-1)t}(x^{j+1}-B_{j+1}(x))\hfill \\ =\sum _{j=0}^n\frac{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}{\left(\genfrac{}{}{0ex}{}{n+k}{j+k}\right)}\frac{S_1(n+k,j+k)}{j+1}({(x+k-1)}^{j+1}-B_{j+1}(x+k-1)).\hfill \end{array}$$

(28)

Thus, by (28), we get

$${\hat{D}}_{n+1}^{(k)}(x)=x{\hat{D}}_n^{(k)}(x-1)+k\sum _{j=0}^n\frac{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}{\left(\genfrac{}{}{0ex}{}{n+k}{j+k}\right)}\frac{S_1(n+k,j+k)}{j+1}({(x+k-1)}^{j+1}-B_{j+1}(x+k-1)).$$

From (10) and (18), we note that

$$\frac{d}{dx}{\hat{D}}_n^{(k)}(x)=n!\sum _{l=0}^{n-1}\frac{{(-1)}^{n-l-1}}{l!(n-l)}{\hat{D}}_l^{(k)}(x).$$

(29)

By (6) and (18), we see that

$$\begin{array}{rcl}{\hat{D}}_n^{(k)}(y)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\hat{D}}_l^{(k)}(y)\frac{t^l}{l!}|x^n〉(n\ge 1)\\ =& 〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^y|x^n〉\\ =& 〈{\partial }_t\left({\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^y\right)|x^{n-1}〉\\ =& 〈\left({\partial }_t{\left(\frac{(1+t)log(1+t)}{t}\right)}^k\right){(1+t)}^y|x^{n-1}〉\\ +y〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^{y-1}|x^{n-1}〉\\ =& y{\hat{D}}_{n-1}^{(k)}(y-1)\\ +k〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(1+t)}^y|(log(1+t)+1-\frac{(1+t)log(1+t)}{t})\frac{x^n}{n}〉\\ =& y{\hat{D}}_{n-1}^{(k)}(y-1)+\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(1+t)}^y|log(1+t)x^n〉\\ +\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(1+t)}^y|x^n〉\\ -\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^y|x^n〉\\ =& y{\hat{D}}_{n-1}^{(k)}(y-1)+\frac{k}{n}{\hat{D}}_n^{(k-1)}(y)-\frac{k}{n}{\hat{D}}_n^{(k)}(y)\\ +\frac{k}{n}\sum _{1\le l\le n}\frac{{(-1)}^{l-1}{(n)}_l}{l}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(1+t)}^y|x^{n-l}〉.\end{array}$$

(30)

Thus, by (30), we get

$$\begin{array}{rcl}{\hat{D}}_n^{(k)}(x)& =& \frac{n}{n+k}x{\hat{D}}_{n-1}^{(k)}(x-1)+\frac{k}{n+k}{\hat{D}}_n^{(k-1)}(x)\\ +\frac{k}{n+k}\sum _{1\le l\le n}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)(l-1)!{\hat{D}}_{n-l}^{(k-1)}(x).\end{array}$$

(31)

Therefore, by (31), we obtain the following theorem.

Theorem 4 For $n\ge 0$, $k\ge 1$, we have

$$\begin{array}{rcl}{\hat{D}}_n^{(k)}(x)& =& \frac{n}{n+k}x{\hat{D}}_{n-1}^{(k)}(x-1)+\frac{k}{n+k}{\hat{D}}_n^{(k-1)}(x)\\ +\frac{k}{n+k}\sum _{1\le l\le n}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)(l-1)!{\hat{D}}_{n-l}^{(k-1)}(x).\end{array}$$

Now, we compute $〈{(\frac{(1+t)log(1+t)}{t})}^k{(log(1+t))}^m|x^n〉$ in two different ways:

$$\begin{array}{c}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m|x^n〉\hfill \\ =〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k|{(log(1+t))}^m{x}^n〉\hfill \\ =\sum _{0\le l\le n-m}\frac{m!}{(l+m)!}{S}_1(l+m,m){(n)}_{l+m}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k|x^{n-l-m}〉\hfill \\ =\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_1(l+m,m){\hat{D}}_{n-l-m}^{(k)}\hfill \\ =\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l^{(k)}.\hfill \end{array}$$

(32)

On the other hand,

$$\begin{array}{c}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m|x^n〉\hfill \\ =〈{\partial }_t\left({\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m\right)|x^{n-1}〉\hfill \\ =k〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}\left(\frac{log(1+t)+1-\frac{(1+t)log(1+t)}{t}}{t}\right){(log(1+t))}^m|x^{n-1}〉\hfill \\ +m〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^{-1}{(log(1+t))}^{m-1}|x^{n-1}〉\hfill \\ =\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(log(1+t))}^{m+1}|x^n〉\hfill \\ +\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(log(1+t))}^m|x^n〉\hfill \\ -\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m|x^n〉\hfill \\ +m〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^{-1}{(log(1+t))}^{m-1}|x^{n-1}〉.\hfill \end{array}$$

(33)

Thus, by (33), we get

$$\begin{array}{c}\frac{n+k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m|x^n〉\hfill \\ =\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(log(1+t))}^{m+1}|x^n〉\hfill \\ +\frac{k}{n}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-1}{(log(1+t))}^m|x^n〉\hfill \\ +m〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(1+t)}^{-1}{(log(1+t))}^{m-1}|x^{n-1}〉.\hfill \end{array}$$

(34)

From (34), we derive the following equation:

$$\begin{array}{c}\frac{n+k}{k}\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l^{(k)}\hfill \\ =\frac{k}{n}\sum _{0\le l\le n-m-1}(m+1)!\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m+1){\hat{D}}_l^{(k-1)}\hfill \\ +\frac{k}{n}\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l^{(k-1)}\hfill \\ +m\sum _{0\le l\le n-m}(m-1)!\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1){\hat{D}}_l^{(k)}(-1).\hfill \end{array}$$

(35)

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

Theorem 5 For $n-1\ge m\ge 1$, we have

$$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l^{(k)}\hfill \\ =\frac{k(m+1)}{n+k}\sum _{0\le l\le n-m-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m+1){\hat{D}}_l^{(k-1)}\hfill \\ +\frac{k}{n+k}\sum _{0\le l\le n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l^{(k-1)}\hfill \\ +\frac{n}{n+k}\sum _{0\le l\le n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1){\hat{D}}_l^{(k)}(-1).\hfill \end{array}$$

For ${\hat{D}}_n^{(k)}(x)\sim ({(\frac{e^t-1}{t{e}^t})}^k,e^t-1)$, and ${(x)}_n\sim (1,e^t-1)$, let us assume that

$${\hat{D}}_n^{(k)}(x)=\sum _{m=0}^n{C}_{n,m}{(x)}_m.$$

(36)

Then, by (16) and (17), we get

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k|t^m{x}^n〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k|x^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{D}}_{n-m}^{(k)}.\end{array}$$

(37)

Therefore, by (36) and (37), we obtain the following theorem.

Theorem 6 For $n\ge 0$, we have

$$\begin{array}{rl}{\hat{D}}_n^{(k)}(x)& =\sum _{0\le m\le n}\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{D}}_{n-m}^{(k)}{(x)}_m\\ =\sum _{0\le m\le n}m!\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{D}}_{n-m}^{(k)}\left(\genfrac{}{}{0ex}{}{x}{m}\right).\end{array}$$

Now, we consider the following two Sheffer sequences:

$${\hat{D}}_n^{(k)}(x)\sim ({\left(\frac{e^t-1}{t{e}^t}\right)}^k,e^t-1)$$

(38)

and

$$H_n^{(s)}(x|\lambda )\sim ({\left(\frac{e^t-\lambda }{1-\lambda }\right)}^s,t),s\in \mathbb{N},\lambda \in \mathbb{C}\text{ with }\lambda \ne 1.$$

Let

$${\hat{D}}_n^{(k)}(x)=\sum _{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda ).$$

(39)

Here

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!{(1-\lambda )}^s}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m{(1-\lambda +t)}^s|x^n〉\\ =& \frac{1}{m!{(1-\lambda )}^s}\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\lambda )}^{s-j}{(n)}_j\\ \times 〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^k{(log(1+t))}^m|x^{n-j}〉\\ =& \sum _{j=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\lambda )}^{-j}{(n)}_j\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n-j}{l+m}\right)S_1(l+m,m){\hat{D}}_{n-j-l-m}^{(k)}\\ =& \sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j{(1-\lambda )}^{-j}{S}_1(n-j-l,m){\hat{D}}_l^{(k)}.\end{array}$$

(40)

Therefore, by (39) and (40), we obtain the following theorem.

Theorem 7 For $n\ge 0$, $k\ge 1$ and $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, we have

$$\begin{array}{rcl}{\hat{D}}_n^{(k)}(x)& =& \sum _{m=0}^n\{\sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j\\ \times {(1-\lambda )}^{-j}{S}_1(n-j-l,m){\hat{D}}_l^{(k)}\}{H}_m^{(s)}(x|\lambda ).\end{array}$$

We consider the following two Sheffer sequences:

$${\hat{D}}_n^{(k)}(x)\sim ({\left(\frac{e^t-1}{t{e}^t}\right)}^k,e^t-1),B_n^{(s)}(x)\sim ({\left(\frac{e^t-1}{t}\right)}^s,t).$$

Let

$${\hat{D}}_n^{(k)}(x)=\sum _{m=0}^n{C}_{n,m}{B}_m^{(s)}(x).$$

(41)

Here

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{t}{log(1+t)})}^s}{{(\frac{t}{(1+t)log(1+t)})}^k}{(log(1+t))}^m|x^n〉\\ =\frac{1}{m!}〈{(1+t)}^s\frac{{(\frac{t}{(1+t)log(1+t)})}^s}{{(\frac{t}{(1+t)log(1+t)})}^k}{(log(1+t))}^m|x^n〉.\end{array}$$

(42)

Case 1. For $s>k$, we have

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{t}{(1+t)log(1+t)}\right)}^{s-k}{(log(1+t))}^m|{(1+t)}^s{x}^n〉\\ =& \frac{1}{m!}\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(n)}_j〈{\left(\frac{t}{(1+t)log(1+t)}\right)}^{s-k}|{(log(1+t))}^m{x}^{n-j}〉\\ =& \sum _{0\le j\le n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(n)}_j\sum _{m\le l\le n-j}{S}_1(l,m)\\ \times \left(\genfrac{}{}{0ex}{}{n-j}{l}\right)〈{\left(\frac{t}{(1+t)log(1+t)}\right)}^{s-k}|x^{n-j-l}〉\\ =& \sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j{S}_1(l,m){\hat{C}}_{n-j-l}^{(s-k)},\end{array}$$

(43)

where ${\hat{C}}_i^{(s-k)}$ is the i th Cauchy number of the second kind of order $s-k$ (see [14]).

Case 2. For $s=k$, we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{(log(1+t))}^m|{(1+t)}^s{x}^n〉\\ =\frac{1}{m!}〈{(log(1+t))}^m|\sum _{j=0}^s\left(\genfrac{}{}{0ex}{}{s}{j}\right)t^j{x}^n〉\\ =\sum _{0\le j\le n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(n)}_j\sum _{m\le l<\mathrm{\infty }}\frac{S_1(l,m)}{l!}〈t^l|x^{n-j}〉\\ =\sum _{0\le j\le n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(n)}_j{S}_1(n-j,m).\end{array}$$

(44)

Case 3. For $s<k$, we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{(1+t)log(1+t)}{t}\right)}^{k-s}{(log(1+t))}^m|{(1+t)}^s{x}^n〉\\ =\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j{S}_1(l,m){\hat{D}}_{n-j-l}^{(k-s)}.\end{array}$$

(45)

Therefore, by (41), (42), (43), (44), and (45), we obtain the following theorem.

Theorem 8 Let $n\ge 0$, we have:

• (I) For $s>k$, we have

  $$\begin{array}{rcl}{\hat{D}}_n^{(k)}(x)& =& \sum _{0\le m\le n}\{\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right)\\ \times {(n)}_j{S}_1(l,m){\hat{C}}_{n-j-l}^{(s-k)}\}{B}_m^{(s)}(x).\end{array}$$

• (II) For $s=k$, we have

  $${\hat{D}}_n^{(k)}(x)=\sum _{0\le m\le n}\{\sum _{0\le j\le n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(n)}_j{S}_1(n-j,m)\}{B}_m^{(s)}(x).$$

• (III) For $s<k$, we have

  $$\begin{array}{rcl}{\hat{D}}_n^{(k)}(x)& =& \sum _{0\le m\le n}\{\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right)\\ \times {(n)}_j{S}_1(l,m){\hat{D}}_{n-j-l}^{(k-s)}\}{B}_m^{(s)}(x).\end{array}$$