Poly-Cauchy numbers and polynomials of the second kind with umbral calculus viewpoint

Abstract

In this paper, we consider the poly-Cauchy polynomials and numbers of the second kind which were studied by Komatsu. We note that the poly-Cauchy polynomials of the second kind are the special generalized Bernoulli polynomials of the second kind. The purpose of this paper is to give various identities of the poly-Cauchy polynomials of the second kind which are derived from umbral calculus.

1 Introduction

As is well known, the Bernoulli polynomials of the second kind are defined by the generating function to be

$$\frac{t}{log(1+t)}{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{b}_n(x)\frac{t^n}{n!}(\text{see [1, p.130]}).$$

(1)

When $x=0$, $b_n=b_n(0)$ are called the Bernoulli numbers of the second kind (see [[1], p.131]).

Let ${Lif}_k(x)$ be the polylogarithm factorial function, which is defined by

$${Lif}_k(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^m}{m!{(m+1)}^k}(\text{see [2–7]}).$$

(2)

The poly-Cauchy polynomials of the second kind ${\hat{c}}_n^{(k)}(x)$ ($k\in \mathbf{Z}$, $n\in {\mathbf{Z}}_{\ge 0}$) are defined by the generating function to be

$${Lif}_k(-log(1+t)){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{\hat{c}}_n^{(k)}(x)\frac{t^n}{n!}(\text{see [2, 3]}).$$

(3)

When $x=0$, ${\hat{c}}_n^{(k)}={\hat{c}}_n^{(k)}(0)$ are called the poly-Cauchy numbers of the second kind, defined by

$$\sum _{n=0}^{\mathrm{\infty }}{\hat{c}}_n^{(k)}\frac{t^n}{n!}={Lif}_k(-log(1+t)).$$

(4)

In particular, if we take $k=1$, then we have

$${Lif}_1(-log(1+t)){(1+t)}^x=\frac{t}{(1+t)log(1+t)}{(1+t)}^x=\frac{t{(1+t)}^{x-1}}{log(1+t)}.$$

(5)

Thus, we note that

$${\hat{c}}_n^{(1)}(x)=b_n(x-1)=B_n^{(n)}(x),$$

(6)

where $B_n^{(\alpha )}(x)$ are the Bernoulli polynomials of order α (see [8]) as their numbers [[9], p.257 and p.259].

When $x=0$, ${\hat{c}}_n^{(1)}={\hat{c}}_n^{(1)}(0)=b_n(-1)=B_n^{(n)}$, where $B_n^{(\alpha )}$ are the Bernoulli numbers of order α.

The falling factorial is defined by

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

(7)

where $S_1(n,l)$ is the signed Stirling number of the first kind.

For $m\in {\mathbf{Z}}_{\ge 0}$, it is well known that

$$\begin{array}{rl}{(log(1+t))}^m& =m!\sum _{l=m}^{\mathrm{\infty }}{S}_1(l,m)\frac{t^l}{l!}\\ =\sum _{l=0}^{\mathrm{\infty }}{S}_1(l+m,m)\frac{m!}{(l+m)!}{t}^{l+m}(\text{see [10, p.62]}).\end{array}$$

(8)

For $\lambda \in \mathbf{C}$ with $\lambda \ne 1$, the Frobenius-Euler polynomials of order r are defined by the generating function to be

$${\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|\lambda )\frac{t^n}{n!}(\text{see [11–13]}).$$

In this paper, we investigate the properties of the poly-Cauchy numbers and polynomials of the second kind with umbral calculus viewpoint. The purpose of this paper is to give various identities of the poly-Cauchy polynomials of the second kind which are derived from umbral calculus.

2 Umbral calculus

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

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

(9)

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

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(10)

Then, by (9) and (10), we get

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

(11)

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

For $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$, we have $〈f_L(t)|x^n〉=〈L|x^n〉$. 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 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 umbral algebra. The order $O(f(t))$ of a power series $f(t)$ (≠0) is the smallest integer k for which the coefficient of $t^k$ does not vanish. If $O(f(t))=1$, then $f(t)$ is called a delta series; if $O(f(t))=0$, then $f(t)$ is called an invertible series (see [10, 14, 15]). For $f(t),g(t)\in \mathcal{F}$ with $O(f(t))=1$ and $O(g(t))=0$, 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 $s_n(x)\sim (g(t),f(t))$ (see [10, 15]).

For $f(t),g(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$, we have

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

(12)

and

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

(13)

Thus, by (13), we get

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

(14)

Let us assume that $s_n(x)\sim (g(t),f(t))$. Then the generating function of $s_n(x)$ is given by

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

(15)

where $\overline{f}(t)$ is the compositional inverse of $f(t)$ with $\overline{f}(f(t))=t$ (see [10, 15]).

For $s_n(x)\sim (g(t),f(t))$, we have the following equation:

$$f(t)s_n(x)=n{s}_{n-1}(x)(n\ge 0),$$

(16)

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

(17)

and

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

(18)

where $p_n(x)=g(t)s_n(x)$ (see [[10], p.21]).

Let us assume that $p_n(x)\sim (1,f(t))$, $q_n(x)\sim (1,g(t))$. Then the transfer formula is given by

$$q_n(x)=x{\left(\frac{f(t)}{g(t)}\right)}^n{x}^{-1}{p}_n(x)(n\ge 0)(\text{see [10, p.51]}).$$

For $s_n(x)\sim (g(t),f(t))$, $r_n(x)\sim (h(t),l(t))$, let us assume that

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

(19)

Then we have

$$C_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉(\text{see [10, p.132]}).$$

(20)

3 Poly-Cauchy numbers and polynomials of the second kind

From (3), we note that ${\hat{c}}_n^{(k)}(x)$ is the Sheffer sequence for the pair

$$(g(t)=\frac{1}{{Lif}_k(-t)},f(t)=e^t-1),$$

that is,

$${\hat{c}}_n^{(k)}(x)\sim (\frac{1}{{Lif}_k(-t)},e^t-1).$$

(21)

Because for $\overline{f}(t)=log(1+t)$, using the formula (15), we get

$${Lif}_k(-log(1+t)){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{s}_n(x)\frac{t^n}{n!}$$

which is the generating function of ${\hat{c}}_n^{(k)}(x)$ in (3).

From (21), we have

$$\frac{1}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x)\sim (1,e^t-1),$$

(22)

and

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

(23)

By (22) and (23), we get

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)& ={Lif}_k(-t){(x)}_n=\sum _{m=0}^n{S}_1(n,m){Lif}_k(-t)x^m\\ =\sum _{m=0}^n{S}_1(n,m)\sum _{a=0}^m\frac{{(-1)}^a}{a!{(a+1)}^k}{t}^a{x}^m\\ =\sum _{m=0}^n\sum _{a=0}^m{S}_1(n,m)\frac{{(-1)}^a\left(\genfrac{}{}{0ex}{}{m}{a}\right)}{{(a+1)}^k}{x}^{m-a}\\ =\sum _{m=0}^n\sum _{j=0}^m{S}_1(n,m)\frac{{(-1)}^{m-j}\left(\genfrac{}{}{0ex}{}{m}{j}\right)}{{(m-j+1)}^k}{x}^j\\ =\sum _{j=0}^n\{\sum _{m=j}^n{S}_1(n,m)\frac{{(-1)}^{m-j}\left(\genfrac{}{}{0ex}{}{m}{j}\right)}{{(m-j+1)}^k}\}{x}^j.\end{array}$$

(24)

By (17) and (21), we get

$${\hat{c}}_n^{(k)}(x)=\sum _{j=0}^n\frac{1}{j!}〈{Lif}_k(-log(1+t)){(log(1+t))}^j|x^n〉x^j.$$

(25)

Now, we observe that

$$\begin{array}{r}〈{Lif}_k(-log(1+t)){(log(1+t))}^j|x^n〉\\ =\sum _{m=0}^{\mathrm{\infty }}\frac{{(-1)}^m}{m!{(m+1)}^k}〈{(log(1+t))}^{m+j}|x^n〉\\ =\sum _{m=0}^{n-j}\frac{{(-1)}^m}{m!{(m+1)}^k}\sum _{l=0}^{n-j-m}\frac{S_1(l+m+j,m+j)}{(l+m+j)!}(m+j)!〈t^{m+j+l}|x^n〉\\ =\sum _{m=0}^{n-j}\frac{{(-1)}^m}{m!{(m+1)}^k}\sum _{l=0}^{n-m-j}\frac{S_1(l+m+j,m+j)}{(l+m+j)!}(m+j)!n!{\delta }_{n,l+m+j}\\ =\sum _{m=0}^{n-j}\frac{{(-1)}^m(m+j)!}{m!{(m+1)}^k}{S}_1(n,m+j).\end{array}$$

(26)

From (25) and (26), we have

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)& =\sum _{j=0}^n\frac{1}{j!}\sum _{m=0}^{n-j}\frac{{(-1)}^m(m+j)!}{m!{(m+1)}^k}{S}_1(n,m+j)x^j=\sum _{j=0}^n\{\sum _{m=0}^{n-j}\frac{{(-1)}^m\left(\genfrac{}{}{0ex}{}{m+j}{m}\right)}{{(m+1)}^k}{S}_1(n,m+j)\}{x}^j\\ =\sum _{j=0}^n\{\sum _{m=j}^n\frac{{(-1)}^{m-j}\left(\genfrac{}{}{0ex}{}{m}{j}\right)}{{(m-j+1)}^k}{S}_1(n,m)\}{x}^j,\end{array}$$

(27)

which is the same as the expression in (24). From (1), we note that

$$\frac{1}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x)\sim (1,e^t-1),x^n\sim (1,t).$$

(28)

For $n\ge 1$, by (19) and (28), we get

$$\begin{array}{rl}\frac{1}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x)& =x{\left(\frac{t}{e^t-1}\right)}^n{x}^{-1}{x}^n=x{\left(\frac{t}{e^t-1}\right)}^n{x}^{n-1}\\ =x{B}_{n-1}^{(n)}(x)=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)B_{n-1-l}^{(n)}{x}^{l+1}.\end{array}$$

(29)

Thus, by (29), we see that

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)=& \sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)B_{n-1-l}^{(n)}{Lif}_k(-t)x^{l+1}\\ =& \sum _{l=0}^{n-1}\sum _{m=0}^{l+1}{(-1)}^m\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l+1}{m}\right)\frac{B_{n-1-l}^{(n)}}{{(m+1)}^k}{x}^{l+1-m}\\ =& \sum _{l=0}^{n-1}\sum _{j=0}^{l+1}{(-1)}^{l+1-j}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l+1}{j}\right)\frac{B_{n-1-l}^{(n)}}{{(l+2-j)}^k}{x}^j\\ =& \sum _{l=0}^{n-1}{(-1)}^{l+1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\frac{B_{n-1-l}^{(n)}}{{(l+2)}^k}\\ +\sum _{j=1}^n\left\{\sum _{l=j-1}^{n-1}{(-1)}^{l+1-j}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l+1}{j}\right)\frac{B_{n-1-l}^{(n)}}{{(l+2-j)}^k}\right\}{x}^j.\end{array}$$

(30)

Therefore, by (27) and (30), we obtain the following theorem.

Theorem 1 For $n\ge 1$, $1\le j\le n$, we have

$$\sum _{m=j}^n\frac{{(-1)}^{m-j}\left(\genfrac{}{}{0ex}{}{m}{j}\right)}{{(m-j+1)}^k}{S}_1(n,m)=\sum _{l=j-1}^{n-1}{(-1)}^{l+1-j}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\left(\genfrac{}{}{0ex}{}{l+1}{j}\right)\frac{B_{n-1-l}^{(n)}}{{(l+2-j)}^k}.$$

In addition, for $n\ge 1$, we have

$${\hat{c}}_n^{(k)}=\sum _{m=0}^n{S}_1(n,m)\frac{{(-1)}^m}{{(m+1)}^k}=\sum _{l=0}^{n-1}{(-1)}^{l+1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\frac{B_{n-1-l}^{(n)}}{{(l+2)}^k}.$$

From (18), we note that

$${\hat{c}}_n^{(k)}(x+y)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right){\hat{c}}_j^{(k)}(x)p_{n-j}(y),$$

(31)

where $p_n(y)=\frac{1}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(y)\sim (1,e^t-1)$.

By (22) and (23), we get

$${(y)}_n=p_n(y)\sim (1,e^t-1).$$

(32)

Thus, from (31) and (32), we have

$${\hat{c}}_n^{(k)}(x+y)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right){\hat{c}}_j^{(k)}(x){(y)}_{n-j}.$$

(33)

By (14), (16), and (21), we get

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

For $s_n(x)\sim (g(t),f(t))$, the recurrence formula for $s_n(x)$ is given by

$$s_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})\frac{1}{f^{\prime }(t)}{s}_n(x)(\text{see [10]}).$$

(34)

By (21) and (34), we get

$$\begin{array}{rl}{\hat{c}}_{n+1}^{(k)}(x)& =(x-\frac{{Lif}_k^{\prime }(-t)}{{Lif}_k(-t)})e^{-t}{\hat{c}}_n^{(k)}(x)\\ =x{\hat{c}}_n^{(k)}(x-1)-e^{-t}\frac{{Lif}_k^{\prime }(-t)}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x).\end{array}$$

(35)

We observe that

$$\begin{array}{rl}\frac{{Lif}_k^{\prime }(-t){Lif}_k^{\prime }(-t)}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x)& ={Lif}_k^{\prime }(-t)\frac{1}{{Lif}_k(-t)}{\hat{c}}_n^{(k)}(x)={Lif}_k^{\prime }(-t){(x)}_n\\ =\sum _{l=0}^n{S}_1(n,l){Lif}_k^{\prime }(-t)x^l\\ =\sum _{l=0}^n{S}_1(n,l)\sum _{m=0}^l\frac{{(-1)}^m\left(\genfrac{}{}{0ex}{}{l}{m}\right)}{{(m+2)}^k}{x}^{l-m}\\ =\sum _{j=0}^n\{\sum _{l=j}^n\frac{{(-1)}^{l-j}\left(\genfrac{}{}{0ex}{}{l}{j}\right)}{{(l-j+2)}^k}{S}_1(n,l)\}{x}^j.\end{array}$$

(36)

Therefore, by (35) and (36), we obtain the following theorem.

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

$${\hat{c}}_{n+1}^{(k)}(x)=x{\hat{c}}_n^{(k)}(x-1)-\sum _{j=0}^n\{\sum _{l=j}^n{S}_1(n,l)\frac{{(-1)}^{l-j}}{{(l-j+2)}^k}\left(\genfrac{}{}{0ex}{}{l}{j}\right)\}{(x-1)}^j.$$

From (11), we note that

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(y)=& 〈\sum _{l=0}^{\mathrm{\infty }}{\hat{c}}_l^{(k)}(y)\frac{t^l}{l!}|x^n〉=〈{Lif}_k(-log(1+t)){(1+t)}^y|x^n〉\\ =& 〈{Lif}_k(-log(1+t)){(1+t)}^y|x{x}^{n-1}〉\\ =& 〈{\partial }_t\left({Lif}_k(-log(1+t)){(1+t)}^y\right)|x^{n-1}〉\\ =& 〈{\partial }_t\left({Lif}_k(-log(1+t))\right){(1+t)}^y|x^{n-1}〉\\ +〈{Lif}_k(-log(1+t)){\partial }_t{(1+t)}^y|x^{n-1}〉\\ =& 〈{\partial }_t\left({Lif}_k(-log(1+t))\right){(1+t)}^y|x^{n-1}〉+y{\hat{c}}_{n-1}^{(k)}(y-1),\end{array}$$

(37)

where ${\partial }_{t}f(t)=\frac{df(t)}{dt}$.

Since $t{Lif}_k^{\prime }(t)={Lif}_{k-1}(t)-{Lif}_k(t)$, we get

$${Lif}_k^{\prime }(t)=\frac{{Lif}_{k-1}(t)-{Lif}_k(t)}{t}.$$

(38)

By (37) and (38), we see that

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(y)=& y{\hat{c}}_{n-1}^{(k)}(y-1)\\ +〈\frac{{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))}{(1+t)log(1+t)}{(1+t)}^y|x^{n-1}〉\\ =& y{\hat{c}}_{n-1}^{(k)}(y-1)\\ +〈\frac{{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))}{t(1+t)}{(1+t)}^y|\frac{t}{log(1+t)}{x}^{n-1}〉.\end{array}$$

(39)

From (1), (6), and (38), we note that

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(y)=& y{\hat{c}}_{n-1}^{(k)}(y-1)+\sum _{l=0}^{n-1}\frac{B_l^{(l)}(1)}{l!}{(n-1)}_l\\ \times 〈\frac{{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))}{t}{(1+t)}^{y-1}|x^{n-l-1}〉\\ =& y{\hat{c}}_{n-1}^{(k)}(y-1)+\sum _{l=0}^{n-1}\frac{B_l^{(l)}(1)}{l!}{(n-1)}_l\\ \times 〈\frac{{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))}{t}{(1+t)}^{y-1}|t\frac{x^{n-l}}{n-l}〉\\ =& y{\hat{c}}_{n-1}^{(k)}(y-1)+\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\frac{B_l^{(l)}(1)}{n-l}\{{\hat{c}}_{n-l}^{(k-1)}(y-1)-{\hat{c}}_{n-l}^{(k)}(y-1)\}\\ =& y{\hat{c}}_{n-1}^{(k)}(y-1)+\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(l)}(1)\{{\hat{c}}_{n-l}^{(k-1)}(y-1)-{\hat{c}}_{n-l}^{(k)}(y-1)\}.\end{array}$$

(40)

It is not difficult to show that ${\hat{c}}_0^{(k)}(y-1)={\hat{c}}_0^{(k-1)}(y-1)$. Since ${\hat{c}}_0^{(k)}(y-1)={\hat{c}}_0^{(k-1)}(y-1)$, by (40), we obtain the following theorem.

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

$${\hat{c}}_n^{(k)}(x)=x{\hat{c}}_{n-1}^{(k)}(x-1)+\frac{1}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(l)}(1)\{{\hat{c}}_{n-l}^{(k-1)}(x-1)-{\hat{c}}_{n-l}^{(k)}(x-1)\}.$$

For $n\ge m\ge 1$, we compute

$$〈{(log(1+t))}^m{Lif}_k(-log(1+t))|x^n〉$$

in two different ways.

On the one hand,

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

(41)

On the other hand, we get

$$\begin{array}{r}〈{(log(1+t))}^m{Lif}_k(-log(1+t))|x^n〉\\ =〈{(log(1+t))}^m{Lif}_k(-log(1+t))|x{x}^{n-1}〉\\ =〈{\partial }_t\left({(log(1+t))}^m{Lif}_k(-log(1+t))\right)|x^{n-1}〉.\end{array}$$

(42)

Now, we observe that

$$\begin{array}{r}{\partial }_t\left({(log(1+t))}^m{Lif}_k(-log(1+t))\right)\\ =m{(log(1+t))}^{m-1}\frac{1}{1+t}{Lif}_k(-log(1+t))\\ +{(log(1+t))}^m\frac{{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))}{(1+t)log(1+t)}\\ ={(log(1+t))}^{m-1}{(1+t)}^{-1}\{m{Lif}_k(-log(1+t))\\ +{Lif}_{k-1}(-log(1+t))-{Lif}_k(-log(1+t))\}.\end{array}$$

(43)

By (42) and (43), we get

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

(44)

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

Theorem 4 For $n\ge m\ge 1$, we have

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

In particular, if we take $m=1$, then we get

$${\hat{c}}_n^{(k-1)}(-1)=\sum _{l=0}^{n-1}{(-1)}^{l}l!\left(\genfrac{}{}{0ex}{}{n}{l+1}\right){\hat{c}}_{n-l-1}^{(k)}.$$

Remark For $s_n(x)\sim (g(t),f(t))$, it is known that

$$\frac{d}{dx}{s}_n(x)=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)〈\overline{f}(t)|x^{n-l}〉s_l(x)(\text{see [10, p.108]}).$$

(45)

By (21) and (45), we easily show that

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

which is a special case of Proposition 2 in [4].

Let us consider the following two Sheffer sequences:

$${\hat{c}}_n^{(k)}(x)\sim (\frac{1}{{Lif}_k(-t)},e^t-1),$$

(46)

and

$$B_n^{(r)}(x)\sim ({\left(\frac{e^t-1}{t}\right)}^r,t).$$

Suppose that

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

(47)

By (20), we see that

$$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!}〈\frac{{(\frac{t}{log(1+t)})}^r}{\frac{1}{{Lif}_k(-log(1+t))}}{(log(1+t))}^m|x^n〉\\ =& \frac{1}{m!}〈{Lif}_k(-log(1+t)){\left(\frac{t}{log(1+t)}\right)}^r{(log(1+t))}^m|x^n〉\\ =& \frac{1}{m!}\sum _{l=0}^{n-m}\frac{m!}{(l+m)!}{S}_1(l+m,m){(n)}_{l+m}\\ \times 〈{Lif}_k(-log(1+t)){\left(\frac{t}{log(1+t)}\right)}^r|x^{n-l-m}〉\\ =& \frac{1}{m!}\sum _{l=0}^{n-m}\frac{m!}{(l+m)!}{S}_1(l+m,m){(n)}_{l+m}\sum _{a=0}^{n-l-m}{B}_a^{(a-r+1)}\frac{1}{a!}\\ \times 〈{Lif}_k(-log(1+t))|t^a{x}^{n-l-m}〉\\ =& \sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_1(l+m,m)\sum _{a=0}^{n-l-m}{B}_a^{(a-r+1)}\frac{{(n-l-m)}_a}{a!}\\ \times 〈{Lif}_k(-log(1+t))|x^{n-l-m-a}〉\\ =& \sum _{l=0}^{n-m}\sum _{a=0}^{n-l-m}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-m-l}{a}\right)S_1(l+m,m)B_a^{(a-r+1)}(1){\hat{c}}_{n-l-m-a}^{(k)}.\end{array}$$

(48)

Therefore, by (47) and (48), we obtain the following theorem.

Theorem 5 For $n\ge 0$, we have

$${\hat{c}}_n^{(k)}(x)=\sum _{m=0}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-m-l}{a}\right)S_1(l+m,m)B_a^{(a-r+1)}(1){\hat{c}}_{n-m-l-a}^{(k)}\}{B}_m^{(r)}(x).$$

Remark The Narumi polynomials of order a are defined by the generating function to be

$$\sum _{k=0}^{\mathrm{\infty }}\frac{N_k^{(a)}(x)}{k!}{t}^k={\left(\frac{t}{log(1+t)}\right)}^{-a}{(1+t)}^x(\text{see [10, p.127])}.$$

(49)

Indeed, $N_a^{(k)}(x)=B_k^{(k+a+1)}(x+1)$, $N_k^{(a)}(x)\sim ({(\frac{e^t-1}{t})}^a,e^t-1)$.

By (48) and (49), we get

$$C_{n,m}=\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-l-m}{a}\right)S_1(l+m,m)N_a^{(-r)}{\hat{c}}_{n-l-m-a}^{(k)}.$$

(50)

From (47) and (50), we have

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)=& \sum _{m=0}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-l-m}{a}\right)\\ \times S_1(l+m,m)N_a^{(-r)}{\hat{c}}_{n-l-m-a}^{(k)}\}{B}_m^{(r)}(x).\end{array}$$

(51)

By (1), we easily show that

$$\begin{array}{rl}{C}_{n,m}=& \sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{a_1+\cdots +a_r=a}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-l-m}{a}\right)\left(\genfrac{}{}{0ex}{}{a}{a_1,\dots ,a_r}\right)\\ \times S_1(l+m,m)b_{a_1}\cdots b_{a_r}{\hat{c}}_{n-m-l-a}^{(k)}.\end{array}$$

(52)

From (47) and (52), we can derive the following equation:

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)=& \sum _{m=0}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^{n-m-l}\sum _{a_1+\cdots +a_r=a}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{n-l-m}{a}\right)\left(\genfrac{}{}{0ex}{}{a}{a_1,\dots ,a_r}\right)\\ \times S_1(l+m,m)\left(\prod _{i=1}^r{b}_{a_i}\right){\hat{c}}_{n-m-l-a}^{(k)}\}{B}_m^{(r)}(x).\end{array}$$

(53)

For (20) and (24), let

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

(54)

where, by (20), we get

$$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!{(1-\lambda )}^r}〈{Lif}_k(-log(1+t)){(1+t-\lambda )}^r|{(log(1+t))}^m{x}^n〉\\ =& \frac{1}{m!{(1-\lambda )}^r}\sum _{l=0}^{n-m}\frac{m!}{(l+m)!}{S}_1(l+m,m){(n)}_{l+m}\\ \times 〈{Lif}_k(-log(1+t)){(1+t-\lambda )}^r|x^{n-l-m}〉.\end{array}$$

(55)

We observe that

$$\begin{array}{r}〈{Lif}_k(-log(1+t)){(1+t-\lambda )}^r|x^{n-l-m}〉\\ =\sum _{a=0}^r\left(\genfrac{}{}{0ex}{}{r}{a}\right){(1-\lambda )}^{r-a}〈{Lif}_k(-log(1+t))|t^a{x}^{n-l-m}〉\\ =\sum _{a=0}^r\left(\genfrac{}{}{0ex}{}{r}{a}\right){(1-\lambda )}^{r-a}{(n-m-l)}_a〈{Lif}_k(-log(1+t))|x^{n-l-m-a}〉\\ =\sum _{a=0}^r\left(\genfrac{}{}{0ex}{}{r}{a}\right){(1-\lambda )}^{r-a}{(n-m-l)}_a{\hat{c}}_{n-l-m-a}^{(k)}.\end{array}$$

(56)

Thus, by (55) and (56), we get

$$C_{n,m}=\sum _{l=0}^{n-m}\sum _{a=0}^r\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{r}{a}\right){(n-m-l)}_a{(1-\lambda )}^{-a}{S}_1(l+m,m){\hat{c}}_{n-m-l-a}^{(k)}.$$

(57)

Therefore, by (54) and (57), we obtain the following theorem.

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

$$\begin{array}{rl}{\hat{c}}_n^{(k)}(x)=& \sum _{m=0}^n\{\sum _{l=0}^{n-m}\sum _{a=0}^r\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)\left(\genfrac{}{}{0ex}{}{r}{a}\right){(n-m-l)}_a{(1-\lambda )}^{-a}{S}_1(l+m,m)\\ \times {\hat{c}}_{n-m-l-a}^{(k)}\}{H}_m^{(r)}(x|\lambda ).\end{array}$$

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

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

(58)

From (20), we note that

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

(59)

Therefore, by (58) and (59), we obtain the following theorem.

Theorem 7 For $n\ge 0$, we have

$${\hat{c}}_n^{(k)}(x)=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{c}}_{n-m}^{(k)}{(x)}_m.$$