Korobov polynomials of the third kind and of the fourth kind

Abstract

The first degenerate version of the Bernoulli polynomials of the second kind appeared in the paper by Korobov (Math Notes 2:77–19, 1996; Proceedings of the IV international conference modern problems of number theory and its applications, pp 40–49, 2001). In this paper, we study two degenerate versions of the Bernoulli polynomials of the second kind which will be called Korobov polynomials of third kind and of the fourth kind. Some properties, identities, recurrence relations and connections with other polynomials are investigated by using umbral calculus.

Background

The Bernoulli polynomials of the second kind $b_n\left(x\right)$ are given by the generating function

$$\begin{array}{c}\hfill \frac{t}{log\left(1+t\right)}{\left(1+t\right)}^x=\sum _{n=0}^{\infty }{b}_n\left(x\right)\frac{t^n}{n!},\left(\text{see Kim et al. 2014, 2015; Roman 1984}\right).\end{array}$$ 1.1

When $x=0$, $b_n=b_n\left(0\right)$ are called Bernoulli numbers of the second kind. The degenerate version of the Bernoulli polynomials of the second kind are called Korobov polynomials of the first kind. We note here that the Carlitz degerate Bernoulli polynomials were rediscovered by Ustinov under the name of Korobov polynomial of the second kind (see Pylypiv and Maliarchuk 2014; Ustinov 2003).

The Daehee polynomials $D_n\left(x\right)$ are defined by the generating function

$$\begin{array}{c}\hfill \frac{log\left(1+t\right)}{t}{\left(1+t\right)}^x=\sum _{n=0}^{\infty }{D}_n\left(x\right)\frac{t^n}{n!},\left(\text{see Kim et al. 2014, 2015}\right).\end{array}$$ 1.2

For $x=0$, $D_n=D_n\left(0\right)$ are called Daehee numbers.

The Korobov polynomials $K_n\left(\mathit{\lambda },x\right)$ of the first kind are given by the generating function

$$\begin{array}{c}\hfill \frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^x=\sum _{n=0}^{\infty }{K}_n\left(x\mid \mathit{\lambda }\right)\frac{t^n}{n!},\left(\text{see Korobov 1996; Korobov 2001}\right).\end{array}$$ 1.3

When $x=0$, $K_n\left(\mathit{\lambda }\right)=K_n\left(0\mid \mathit{\lambda }\right)$ are called Korobov numbers of the first kind.

In the following, we will review very briefly some necessary things on umbral calculus. Our basic reference is Roman (1984). Also, one is asked to look at more recent papers on umbral calculus (Nisar et al. 2015; Srivastava et al. 2014).

Let $\mathbb{C}$ be the complex number field and let $\mathcal{F}$ be the set of all formal power series in the variable t over $\mathbb{C}$ with

$$\begin{array}{c}\hfill \mathcal{F}=\left\{\left.f\left(t\right)=\sum _{k=0}^{\infty }\frac{a_k}{k!}{t}^k\right|a_k\in \mathbb{C}\right\}.\end{array}$$ 1.4

Let $\mathbb{P}=\mathbb{C}\left[x\right]$ and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on $\mathbb{P}$. For $L\in {\mathbb{P}}^{\ast }$, the action of the linear functional L on a polynomial $p\left(x\right)$ is denoted by $〈\left.L\right|p\left(x\right)〉$ with

$$\begin{array}{c}\hfill 〈\left.L+M\right|p\left(x\right)〉=〈\left.L\right|p\left(x\right)〉+〈\left.M\right|p\left(x\right)〉,〈\left.cL\right|p\left(x\right)〉=c〈\left.L\right|p\left(x\right)〉,\end{array}$$

where c is a complex constant (see Kim 2014; Roman 1984).

For $f\left(t\right)={\sum }_{k=0}^{\infty }{a}_k\frac{t^k}{k!}\in \mathcal{F}$, we define a linear functional on $\mathbb{P}$ by setting

$$\begin{array}{c}\hfill 〈\left.f\left(t\right)\right|x^n〉=a_n,\text{for all}n\ge 0,\left(\text{see Kim et al. 2014; Roman 1984}\right).\end{array}$$ 1.5

Thus, by (1.5), we easily get

$$\begin{array}{c}\hfill 〈\left.t^k\right|x^n〉=n!{\mathit{\delta }}_{n,k},\left(n,k\ge 0\right),\end{array}$$ 1.6

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

Let $f_L\left(t\right)={\sum }_{k=0}^{\infty }〈\left.L\right|x^k〉\frac{t^n}{k!}$. Then, by (1.6), we get $〈\left.f_L\left(t\right)\right|x^n〉=〈\left.L\right|x^n〉$. Additionallly, the mapping $L\mapsto f_L\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto $\mathcal{F}$. Henceforth, $\mathcal{F}$ denotes both the algebra of formal power series in t and the vector space of all linear functionals on $\mathbb{P}$, and so an element $f\left(t\right)$ of $\mathcal{F}$ can be regarded as both a formal power series and a linear functional. We refer to $\mathcal{F}$ as the umbral algebra. The umbral calculus is the study of umbral algebra (see Kim 2014; Roman 1984). From (1.6), we can easily derive $〈\left.e^{yt}\right|x^n〉=y^n$. So $〈\left.e^{yt}\right|p\left(x\right)〉=p\left(y\right)$. The order $o\left(f\left(t\right)\right)$ of a power series $f\left(t\right)\left(\ne 0\right)$ is the smallest nonnegative integer k for which the coefficient at $t^k$ does not vanish. For $f\left(t\right)\in \mathcal{F}$ and $p\left(x\right)\in \mathbb{P}$, we have

$$\begin{array}{c}\hfill f\left(t\right)=\sum _{k=0}^{\infty }〈\left.f\left(t\right)\right|x^k〉\frac{t^k}{k!},p\left(x\right)=\sum _{k=0}^{\infty }〈\left.t^k\right|p\left(x\right)〉\frac{x^k}{k!}.\end{array}$$ 1.7

Thus, by (1.7), we get

$$\begin{array}{c}\hfill p^{\left(k\right)}\left(0\right)={\left.{\left(\frac{d}{dx}\right)}^{k}p\left(x\right)\right|}_{x=0}=〈\left.t^k\right|p\left(x\right)〉=〈\left.1\right|p^{\left(k\right)}\left(x\right)〉.\end{array}$$ 1.8

From (1.8), we note that

$$\begin{array}{c}\hfill t^{k}p\left(x\right)=p^{\left(k\right)}\left(x\right)=\frac{d^k}{d{x}^k}p\left(x\right),e^{yt}p\left(x\right)=p\left(x+y\right),\left(\text{see Roman 1984}\right).\end{array}$$ 1.9

Let $f\left(t\right),g\left(t\right)\in \mathcal{F}$ such that $o\left(f\left(t\right)\right)=1$ and $o\left(g\left(t\right)\right)=0$. Then there exists a unique sequence $s_n\left(x\right)$$\left(deg{s}_n\left(x\right)=n\right)$ of polynomials such that $〈\left.g\left(t\right)f{\left(t\right)}^k\right|s_n\left(x\right)〉$$=n!{\mathit{\delta }}_{n,k}$, for $n,k\ge 0$. The sequence $s_n\left(x\right)$ is called the Sheffer sequence for the pair $\left(g\left(t\right),f\left(t\right)\right)$, which is denoted by $s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$. For $s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right),$ we have

$$\begin{array}{c}\hfill f\left(t\right)s_n\left(x\right)=n{s}_{n-1}\left(x\right),\left(n\in \mathbb{N}\cup \left\{0\right\}\right),\end{array}$$ 1.10

and

$$\begin{array}{c}\hfill \frac{1}{g\left(\overline{f}\left(t\right)\right)}{e}^{x\overline{f}\left(t\right)}=\sum _{k=0}^{\infty }\frac{s_k\left(x\right)}{k!}{t}^k,\text{for all}x\in \mathbb{C}.\end{array}$$ 1.11

Here $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$ (see Kim and Mansour 2014; Roman 1984).

The conjugation representation for $s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ is given by

$$\begin{array}{c}\hfill s_n\left(x\right)=\sum _{k=0}^n\frac{1}{k!}〈\left.g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^k\right|x^n〉x^k,\left(n\ge 0\right),\left(\text{see Carlitz 1979; Roman 1984}\right).\end{array}$$ 1.12

Let us consider the following two Sheffer sequences:

$$\begin{array}{c}\hfill s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right),r_n\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right).\end{array}$$ 1.13

Then, we have

$$\begin{array}{c}\hfill s_n\left(x\right)=\sum _{m=0}^n{C}_{n,m}{r}_m\left(x\right),\left(n\ge 0\right),\end{array}$$ 1.14

where

$$\begin{array}{c}\hfill C_{n,m}=\frac{1}{m!}〈\left.\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}{\left(l\left(\overline{f}\left(t\right)\right)\right)}^m\right|x^n〉,\left(\text{see Kim et al. 2014; Roman 1984}\right).\end{array}$$ 1.15

The first degenerate version of the Bernoulli polynomials of the second kind appeared in the paper by Korobov (2001; 1996). In this paper, we study two degenerate versions of the Bernoulli polynomials of the second kind which will be called Korobov polynomials of the third kind and of the fourth kind. Some properties, identities and recurrence relations for them are investigated by using umbral calculus. In addition, some connections with other polynomials are studied for which one refers to the related papers (Dattoli et al. 2006, 2004).

Korobov polynomials of the third kind and of the fourth kind

Now, we introduce Korobov polynomials of the third kind $K_{n,3}\left(x\mid \mathit{\lambda }\right)$ and of the fourth kind $K_{n,4}\left(x\mid \mathit{\lambda }\right)$, respectively, given by the generating functions

$$\begin{array}{c}\hfill \frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^x=\sum _{n=0}^{\infty }{K}_{n,3}\left(x\mid \mathit{\lambda }\right)\frac{t^n}{n!},\end{array}$$ 2.1

and

$$\begin{array}{c}\hfill \frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^x=\sum _{n=0}^{\infty }{K}_{n,4}\left(x\mid \mathit{\lambda }\right)\frac{t^n}{n!}.\end{array}$$ 2.2

When $x=0$, $K_{n,3}\left(\mathit{\lambda }\right)=K_{n,3}\left(0,\mathit{\lambda }\right)$ and $K_{n,4}\left(\mathit{\lambda }\right)=K_{n,4}\left(0\mid \mathit{\lambda }\right)$ are called Korobov numbers of the third kind and of the fourth kind, respectively.

As all $\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}=\frac{t}{\frac{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\mathit{\lambda }}}$, $\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}=\frac{log{\left(1+\mathit{\lambda }t\right)}^{\frac{1}{\mathit{\lambda }}}}{log\left(1+t\right)}$, $\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}=\frac{log{\left(1+\mathit{\lambda }t\right)}^{\frac{1}{\mathit{\lambda }}}}{\frac{{\left(1+\mathit{\lambda }t\right)}^{\mathit{\lambda }}-1}{\mathit{\lambda }}}$ tend to $\frac{t}{log\left(1+t\right)}$ as $\mathit{\lambda }\to 0$, ${lim}_{\mathit{\lambda }\to 0}{K}_n\left(x\mid \mathit{\lambda }\right)={lim}_{\mathit{\lambda }\to 0}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)={lim}_{\mathit{\lambda }\to 0}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)=b_n\left(x\right)$, $\left(n\ge 0\right)$. We observe first that $K_{n,3}\left(x\mid \mathit{\lambda }\right)$ and $K_{n,4}\left(x\mid \mathit{\lambda }\right)$ are Sheffer sequences for the respective pairs $\left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)$ and $\left(\frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)$. That is,

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right),\end{array}$$

and

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right).\end{array}$$ 2.3

From (1.12) and (2.2), we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{k=0}^n\frac{1}{k!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(log\left(1+t\right)\right)}^k\right|x^n〉x^k.\end{array}$$ 2.4

We observe that

$$\begin{array}{cc}& 〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(log\left(1+x\right)\right)}^k\right|x^n〉\hfill \\ \hfill & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|{\left(log\left(1+x\right)\right)}^k{x}^n〉\hfill \\ \hfill & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|k!\sum _{l=k}^{\infty }{S}_1\left(l,k\right)\frac{t^l}{l!}{x}^n〉\hfill \\ \hfill & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\frac{t}{log\left(1+t\right)}{x}^{n-l}〉\hfill \\ \hfill & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\sum _{m=0}^{\infty }{b}_m\frac{t^m}{m!}{x}^{n-l}〉\hfill \\ \hfill & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)b_m〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-l-m}〉\hfill \\ \hfill & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)b_m〈\left.\sum _{j=0}^{\infty }{D}_j{\mathit{\lambda }}^j\frac{t^j}{j!}\right|x^{n-l-m}〉\hfill \\ \hfill & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)b_m{D}_{n-l-m}{\mathit{\lambda }}^{n-l-m}\hfill \\ \hfill & =k!\sum _{l=k}^n\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)S_1\left(l,k\right)b_m{D}_{n-l-m}{\mathit{\lambda }}^{n-l-m},\hfill \end{array}$$ 2.5

where $S_1\left(n,m\right)$ is the Stirling number of the first kind defined by

$$\begin{array}{c}\hfill x\left(x-1\right)\dots \left(x-n+1\right)={\left(x\right)}_n=\sum _{l=0}^n{S}_1\left(n,l\right)x^l,\left(n\ge 0\right).\end{array}$$

Therefore, by (2.4) and (2.5), we have

Theorem 1

For$n\ge 0$, we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{k=0}^n\left(\sum _{l=k}^n\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)S_1\left(l,k\right)b_{n-l-m}{D}_m{\mathit{\lambda }}^m\right)x^k.\end{array}$$

From (1.12) and (2.3), we have

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{k=0}^n\frac{1}{k!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(log\left(1+t\right)\right)}^k\right|x^n〉x^k.\end{array}$$ 2.6

We observe that

$$\begin{array}{cc}& 〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(log\left(1+t\right)\right)}^k\right|x^n〉\hfill \\ & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right|x^{n-l}〉\hfill \\ & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\sum _{m=0}^{\infty }{K}_m\left(\mathit{\lambda }\right)\frac{t^m}{m!}{x}^{n-l}〉\hfill \\ & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)K_m\left(\mathit{\lambda }\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-l-m}〉\hfill \\ & =k!\sum _{l=k}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,k\right)\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)K_m\left(\mathit{\lambda }\right)D_{n-l-m}{\mathit{\lambda }}^{n-l-m}\hfill \\ & =k!\sum _{l=k}^n\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)S_1\left(l,k\right)K_{n-l-m}\left(\mathit{\lambda }\right)D_m{\mathit{\lambda }}^m.\hfill \end{array}$$ 2.7

Therefore, by (2.6) and (2.7), we obtain the following theorem.

Theorem 2

For$n\ge 0$, we have

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{k=0}^n\left(\sum _{l=k}^n\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)S_1\left(l,k\right)K_{n-l-m}\left(\mathit{\lambda }\right)D_m{\mathit{\lambda }}^m\right)x^k.\end{array}$$

By (1.6) and (2.1), we easily get

$$\begin{array}{cc}\hfill K_{n,3}\left(y\mid \mathit{\lambda }\right)& =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^y\right|x^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|{\left(1+t\right)}^y{x}^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\sum _{l=0}^{\infty }{\left(y\right)}_l\frac{t^l}{l!}{x}^n〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)b_m{D}_{n-l-m}{\mathit{\lambda }}^{n-l-m}\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)b_{n-l-m}{D}_m{\mathit{\lambda }}^m\hfill \\ & =\sum _{l=0}^n\left(\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)b_{n-l-m}{D}_m{\mathit{\lambda }}^m\right){\left(y\right)}_l.\hfill \end{array}$$ 2.8

Thus, by (2.8), we get

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)b_{n-l-m}{D}_m{\mathit{\lambda }}^m\right){\left(x\right)}_l,\left(n\ge 0\right).\end{array}$$ 2.9

From (1.6) and (2.2), we note that

$$\begin{array}{cc}\hfill K_{n,4}\left(y\mid \mathit{\lambda }\right)& =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^y\right|x^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right|{\left(1+t\right)}^y{x}^n〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)K_m\left(\mathit{\lambda }\right)D_{n-l-m}{\mathit{\lambda }}^{n-l-m}\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l\sum _{m=0}^{n-l}\left(\begin{array}{c}n-l\\ m\end{array}\right)K_{n-l-m}\left(\mathit{\lambda }\right)D_m{\mathit{\lambda }}^m\hfill \\ & =\sum _{l=0}^n\left(\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)K_{n-l-m}\left(\mathit{\lambda }\right)D_m{\mathit{\lambda }}^m\right){\left(y\right)}_l.\hfill \end{array}$$ 2.10

Thus, by (2.10), we get

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\sum _{m=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ m\end{array}\right)K_{n-l-m}\left(\mathit{\lambda }\right)D_m{\mathit{\lambda }}^m\right){\left(x\right)}_l.\end{array}$$ 2.11

From (2.2), we note that

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)\hfill \\ \hfill ⟺& \frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)={\left(x\right)}_n\sim \left(1,e^t-1\right).\hfill \end{array}$$ 2.12

By (2.12), we get

$$\begin{array}{cc}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)& =\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{\mathit{\lambda }t}{\left(x\right)}_n\hfill \\ \hfill & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{\mathit{\lambda }t}{x}^k\hfill \\ \hfill & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{e^t-1}{t}\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{\mathit{\lambda }\left(e^t-1\right)}{x}^k\hfill \\ \hfill & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{e^t-1}{t}\sum _{l=0}^k{D}_l{\mathit{\lambda }}^l\frac{{\left(e^t-1\right)}^l}{l!}{x}^k\hfill \\ \hfill & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{e^t-1}{t}\sum _{l=0}^k{D}_l{\mathit{\lambda }}^l\sum _{m=l}^{\infty }{S}_2\left(m,l\right)\frac{t^m}{m!}{x}^k\hfill \\ \hfill & =\sum _{k=0}^n{S}_1\left(n,k\right)\sum _{l=0}^k{D}_l{\mathit{\lambda }}^l\sum _{m=l}^k\left(\begin{array}{c}k\\ m\end{array}\right)S_2\left(m,l\right)\frac{e^t-1}{t}{x}^{k-m}.\hfill \end{array}$$ 2.13

We observe that

$$\begin{array}{cc}\hfill \frac{e^t-1}{t}{x}^{k-m}& =\sum _{j=1}^{\infty }\frac{t^{j-1}}{j!}{x}^{k-m}\hfill \\ \hfill & =\sum _{j=0}^{\infty }\frac{1}{\left(j+1\right)!}{t}^j{x}^{k-m}\hfill \\ \hfill & =\sum _{j=0}^{k-m}\frac{1}{\left(j+1\right)!}{t}^j{x}^{k-m}\hfill \\ \hfill & =\sum _{j=0}^{k-m}\frac{1}{j+1}\left(\begin{array}{c}k-m\\ j\end{array}\right)x^{k-m-j}\hfill \\ \hfill & =\sum _{j=0}^{k-m}\frac{1}{k-m-j+1}\left(\begin{array}{c}k-m\\ j\end{array}\right)x^j.\hfill \end{array}$$ 2.14

Thus, by (2.13) and (2.14), we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{k=0}^n\sum _{l=0}^k\sum _{m=l}^k\sum _{j=0}^{k-m}\frac{1}{k-m-j+1}\left(\begin{array}{c}k\\ m\end{array}\right)\left(\begin{array}{c}k-m\\ j\end{array}\right)S_1\left(n,k\right)S_2\left(m,l\right)D_l{\mathit{\lambda }}^l{x}^j\hfill \\ & =\sum _{k=0}^n\sum _{l=0}^k\sum _{m=0}^{k-l}\sum _{j=0}^m\frac{1}{m-j+1}\left(\begin{array}{c}k\\ m\end{array}\right)\left(\begin{array}{c}m\\ j\end{array}\right)S_1\left(n,k\right)S_2\left(k-m,l\right)D_l{\mathit{\lambda }}^l{x}^j\hfill \\ & =\sum _{j=0}^n\left(\sum _{k=j}^n\sum _{l=0}^{k-j}\sum _{m=j}^{k-l}\frac{1}{k+1}\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_1\left(n,k\right)S_2\left(k-m,l\right)D_l{\mathit{\lambda }}^l\right)x^j,\hfill \end{array}$$ 2.15

where $S_2\left(n,k\right)$ is the Stirling number of the second kind given by

$$\begin{array}{c}\hfill x^n=\sum _{l=0}^n{S}_2\left(n,l\right){\left(x\right)}_l,\left(n\ge 0\right).\end{array}$$

Therefore, by (2.15), we obtain the following theorem expressing $K_{n,3}\left(x\mid \mathit{\lambda }\right)$ in terms of the Stirling numbers of the first kind and of the second and Daehee numbers.

Theorem 3

For$n\ge 0$, we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ \hfill & =\sum _{j=0}^n\left(\sum _{k=j}^n\sum _{l=0}^{k-j}\sum _{m=j}^{k-l}\frac{1}{k+1}\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_1\left(n,k\right)S_2\left(k-m,l\right)D_l{\mathit{\lambda }}^l\right)x^j.\hfill \end{array}$$

From (2.3), we have

$$\begin{array}{c}\hfill \frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)={\left(x\right)}_n\sim \left(1,e^t-1\right),\left(n\ge 0\right).\end{array}$$ 2.16

Thus, by (2.16), we get

$$\begin{array}{cc}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)& =\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{e^{\mathit{\lambda }t}-1}{\left(x\right)}_n\hfill \\ & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{e^{\mathit{\lambda }t}-1}{x}^k\hfill \\ & =\sum _{k=0}^n{S}_1\left(n,k\right)\frac{\mathit{\lambda }\left(e^t-1\right)}{e^{\mathit{\lambda }t}-1}\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{\mathit{\lambda }\left(e^t-1\right)}{x}^k\hfill \\ & =\sum _{k=0}^n{S}_1\left(n,k\right)\sum _{l=0}^k{D}_l{\mathit{\lambda }}^l\sum _{m=l}^k\left(\begin{array}{c}k\\ m\end{array}\right)S_2\left(m,l\right)\frac{e^t-1}{t}\frac{\mathit{\lambda }t}{e^{\mathit{\lambda }t}-1}{x}^{k-m},\hfill \end{array}$$ 2.17

Now, we observe that

$$\begin{array}{cc}\hfill \frac{e^t-1}{t}\frac{\mathit{\lambda }t}{e^{\mathit{\lambda }t}-1}{x}^{k-m}& =\frac{e^t-1}{t}\sum _{j=0}^{k-m}{B}_j{\mathit{\lambda }}^j\frac{t^j}{j!}{x}^{k-m}\hfill \\ & =\sum _{j=0}^{k-m}\left(\begin{array}{c}k-m\\ j\end{array}\right)B_j{\mathit{\lambda }}^j\frac{e^t-1}{t}{x}^{k-m-j}\hfill \\ & =\sum _{j=0}^{k-m}\left(\begin{array}{c}k-m\\ j\end{array}\right)B_j{\mathit{\lambda }}^j\sum _{i=0}^{k-m-j}\frac{1}{i+1}\left(\begin{array}{c}k-m-j\\ i\end{array}\right)x^{k-m-j-i},\hfill \end{array}$$ 2.18

where $B_n$ is the n-th Bernoulli number given by the generating function

$$\begin{array}{c}\hfill \frac{t}{e^t-1}=\sum _{n=0}^{\infty }{B}_n\frac{t^n}{n!},\left(\text{see Bayad and kim 2010; Kim et al. 2012, 2015}\right).\end{array}$$

Thus, by (2.17) and (2.18), we get

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{k=0}^n{S}_1\left(n,k\right)\sum _{l=0}^k{D}_l{\mathit{\lambda }}^l\sum _{m=0}^{k-l}\left(\begin{array}{c}k\\ m\end{array}\right)S_2\left(k-m,l\right)\hfill \\ & \times \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)B_{m-j}{\mathit{\lambda }}^{m-j}\sum _{i=0}^j\frac{1}{j-i+1}\left(\begin{array}{c}j\\ i\end{array}\right)x^i\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{j-i+1}\left(\begin{array}{c}k\\ m\end{array}\right)\right.\hfill \\ & \times \left.\left(\begin{array}{c}m\\ j\end{array}\right)\left(\begin{array}{c}j\\ i\end{array}\right){\mathit{\lambda }}^{l+m-j}{S}_1\left(n,k\right)S_2\left(k-m,l\right)D_l{B}_{m-j}\right)x^i\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{k+1}\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j+1\end{array}\right)\left(\begin{array}{c}j+1\\ i\end{array}\right){\mathit{\lambda }}^{l+m-j}\right.\hfill \\ & \times \left.S_1\left(n,k\right)S_2\left(k-m,l\right)D_l{B}_{m-j}\right)x^i.\hfill \end{array}$$ 2.19

Therefore, by (2.19), we obtain the following theorem expressing $K_{n,4}\left(x\mid \mathit{\lambda }\right)$ in terms of the Stirling numbers of the first kind and of the second kind, Daehee numbers and Bernoulli numbers.

Theorem 4

For $n\ge 0$, we have

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{k+1}\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j+1\end{array}\right)\left(\begin{array}{c}j+1\\ i\end{array}\right)\right.\hfill \\ & \times \left.{\mathit{\lambda }}^{l+m-j}{S}_1\left(n,k\right)S_2\left(k-m,l\right)D_l{B}_{m-j}\right)x^i.\hfill \end{array}$$

From (2.8), we have

$$\begin{array}{cc}\hfill K_{n,3}\left(y\mid \mathit{\lambda }\right)& =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\sum _{i=0}^{\infty }{K}_{i,3}\left(\mathit{\lambda }\right)\frac{t^i}{i!}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l{K}_{n-l,3}\left(\mathit{\lambda }\right).\hfill \end{array}$$ 2.20

Thus, by (2.20), we get

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_{n-l,3}\left(\mathit{\lambda }\right){\left(x\right)}_l.\end{array}$$ 2.21

From (2.10), we have

$$\begin{array}{cc}\hfill K_{n,4}\left(y\mid \mathit{\lambda }\right)& =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l〈\left.\sum _{i=0}^{\infty }{K}_{i,4}\left(\mathit{\lambda }\right)\frac{t^i}{i!}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){\left(y\right)}_l{K}_{n-l,4}\left(\mathit{\lambda }\right).\hfill \end{array}$$ 2.22

By (2.22), we get

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_{n-l,4}\left(\mathit{\lambda }\right){\left(x\right)}_l,\left(n\ge 0\right).\end{array}$$ 2.23

From (2.19), we have

$$\begin{array}{cc}\hfill K_{n,3}\left(y\mid \mathit{\lambda }\right)& =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^y\right|x^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\frac{t}{log\left(1+t\right)}{\left(1+t\right)}^y{x}^n〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left(y\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left(y\right)〈\left.\sum _{m=0}^{\infty }{D}_m{\mathit{\lambda }}^m\frac{t^m}{m!}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left(y\right)D_{n-l}{\mathit{\lambda }}^{n-l}.\hfill \end{array}$$ 2.24

Thus, by (2.24), we get

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)D_{n-l}{\mathit{\lambda }}^{n-l}{b}_l\left(x\right),\left(n\ge 0\right).\end{array}$$ 2.25

From (2.10), we can also derive the following equation:

$$\begin{array}{cc}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)& =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^y\right|x^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^y{x}^n〉\hfill \\ & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\sum _{l=0}^{\infty }{K}_l\left(y\mid \mathit{\lambda }\right)\frac{t^l}{l!}{x}^n〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(y\mid \mathit{\lambda }\right)〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(y\mid \mathit{\lambda }\right)〈\left.\sum _{m=0}^{\infty }{D}_m{\mathit{\lambda }}^m\frac{t^m}{m!}\right|x^{n-l}〉\hfill \\ & =\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(y\mid \mathit{\lambda }\right)D_{n-l}{\mathit{\lambda }}^{n-l}.\hfill \end{array}$$ 2.26

Thus, by (2.26), we get

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)D_{n-l}{\mathit{\lambda }}^{n-l}{K}_l\left(x\mid \mathit{\lambda }\right).\end{array}$$ 2.27

Therefore, by (2.21), (2.23), (2.25) and (2.27), we obtain the following theorem expressing $K_{n,3}\left(x\mid \mathit{\lambda }\right)$ and $K_{n,4}\left(x\mid \mathit{\lambda }\right)$ both in terms of falling factorial polynomials. Also, we express $K_{n,3}\left(x\mid \mathit{\lambda }\right)$ and $K_{n,4}\left(x\mid \mathit{\lambda }\right)$ respectively by Bernoulli polynomials of the second kind and Korobov polynomials of the first kind.

Theorem 5

For$n\ge 0$, we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_{n-l,3}\left(\mathit{\lambda }\right){\left(x\right)}_l=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)D_{n-l}{\mathit{\lambda }}^{n-l}{b}_l\left(x\right),\end{array}$$

and

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_{n-l,4}\left(\mathit{\lambda }\right){\left(x\right)}_l=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)D_{n-l}{\mathit{\lambda }}^{n-l}{K}_l\left(x\mid \mathit{\lambda }\right).\end{array}$$

It is easy to see that

$$\begin{array}{c}\hfill x^n\sim \left(1,t\right),\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(1,e^t-1\right).\end{array}$$ 2.28

For $n\ge 1$, we have

$$\begin{array}{cc}\hfill \frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)& =x{\left(\frac{t}{e^t-1}\right)}^n{x}^{-1}{x}^n\hfill \\ \hfill & =x{B}_{n-1}^{\left(n\right)}\left(x\right)\hfill \\ \hfill & =\sum _{k=0}^{n-1}\left(\begin{array}{c}n-1\\ k\end{array}\right)B_k^{\left(n\right)}{x}^{n-k}\hfill \\ \hfill & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}{x}^k.\hfill \end{array}$$ 2.29

Thus, by (2.29), we get

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{\mathit{\lambda }t}{x}^k\hfill \\ & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}\sum _{l=0}^k\sum _{m=0}^{k-l}\sum _{j=0}^m\frac{1}{k+1}\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_2\left(k-m,l\right)D_l{\mathit{\lambda }}^l{x}^j\hfill \\ & =\sum _{k=0}^n\sum _{l=0}^k\sum _{m=0}^{k-l}\sum _{j=0}^m\frac{1}{k+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_2\left(k-m,l\right){\mathit{\lambda }}^l{D}_l{B}_{n-k}^{\left(n\right)}{x}^j\hfill \\ & =\sum _{j=0}^n\left(\sum _{k=j}^n\sum _{l=0}^{k-j}\sum _{m=j}^{k-l}\frac{1}{k+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_2\left(k-m,l\right){\mathit{\lambda }}^l{D}_l{B}_{n-k}^{\left(n\right)}\right)x^j.\hfill \end{array}$$ 2.30

From (2.3), we note that

$$\begin{array}{c}\hfill x^n\sim \left(1,t\right),\frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)\sim \left(1,e^t-1\right).\end{array}$$ 2.31

For $n\ge 1$, by (2.31), we get

$$\begin{array}{cc}\hfill \frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)& =x{\left(\frac{t}{e^t-1}\right)}^n{x}^{-1}{x}^n=x{B}_{n-1}^{\left(n\right)}\left(x\right)\hfill \\ & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}{x}^k.\hfill \end{array}$$ 2.32

Thus, by (2.32), we have

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}\frac{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{e^{\mathit{\lambda }t}-1}{x}^k\hfill \\ & =\sum _{k=1}^n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)B_{n-k}^{\left(n\right)}\sum _{l=0}^k\sum _{m=0}^{k-l}\sum _{j=0}^m\sum _{i=0}^j\frac{1}{j-i+1}\left(\begin{array}{c}k\\ m\end{array}\right)\hfill \\ & \times \left(\begin{array}{c}m\\ j\end{array}\right)\left(\begin{array}{c}j\\ i\end{array}\right){\mathit{\lambda }}^{l+m-j}{S}_2\left(k-m,l\right)D_l{B}_{m-j}{x}^i\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{j-i+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k\\ m\end{array}\right)\left(\begin{array}{c}m\\ j\end{array}\right)\left(\begin{array}{c}j\\ i\end{array}\right)\right.\hfill \\ & \times \left.{\mathit{\lambda }}^{l+m-j}{S}_2\left(k-m,l\right)D_l{B}_{m-j}{B}_{n-k}^{\left(n\right)}\right)x^i\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{k+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j+1\end{array}\right)\right.\hfill \\ & \times \left.\left(\begin{array}{c}j+1\\ i\end{array}\right){\mathit{\lambda }}^{l+m-j}{S}_2\left(k-m,l\right)D_l{B}_{m-j}{B}_{n-k}^{\left(n\right)}\right)x^i.\hfill \end{array}$$ 2.33

Therefore, by (2.30) and (2.33), we obtain the following theorem.

Theorem 6

For$n\ge 0$, we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{j=0}^n\left(\sum _{k=j}^n\sum _{l=0}^{k-j}\sum _{m=j}^{k-l}\frac{1}{k+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j\end{array}\right)S_2\left(k-m,l\right){\mathit{\lambda }}^l{D}_l{B}_{n-k}^{\left(n\right)}\right)x^j\hfill \end{array}$$

and

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{i=0}^n\left(\sum _{k=i}^n\sum _{l=0}^{k-i}\sum _{m=i}^{k-l}\sum _{j=i}^m\frac{1}{k+1}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}k+1\\ m+1\end{array}\right)\left(\begin{array}{c}m+1\\ j+1\end{array}\right)\right.\hfill \\ & \times \left.\left(\begin{array}{c}j+1\\ i\end{array}\right){\mathit{\lambda }}^{l+m-j}{S}_2\left(k-m,l\right)D_l{B}_{m-j}{B}_{n-k}^{\left(n\right)}\right)x^i\hfill \end{array}$$

For $s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right),$ we note that Sheffer identity is given by

$$\begin{array}{c}\hfill s_n\left(x+y\right)=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)s_j\left(x\right)p_{n-j}\left(y\right),\text{where}{p}_n\left(x\right)=g\left(t\right)s_n\left(x\right).\end{array}$$ 2.34

By (2.2) and (2.34), we get

$$\begin{array}{c}\hfill K_{n,3}\left(x+y\mid \mathit{\lambda }\right)=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)K_{j,3}\left(x\mid \mathit{\lambda }\right){\left(y\right)}_{n-j},\end{array}$$ 2.35

where $p_n\left(x\right)=\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)={\left(x\right)}_n,\left(n\ge 0\right)$.

From (2.3) and (2.34), we have

$$\begin{array}{c}\hfill K_{n,4}\left(x+y\mid \mathit{\lambda }\right)=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)K_{j,4}\left(x\mid \mathit{\lambda }\right){\left(y\right)}_{n-j},\end{array}$$ 2.36

where $p_n\left(x\right)=\frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)={\left(x\right)}_n$.

By (1.10), we see that

$$\begin{array}{c}\hfill \left(e^t-1\right)K_{n,3}\left(x\mid \mathit{\lambda }\right)=n{K}_{n-1,3}\left(x\mid \mathit{\lambda }\right),\end{array}$$ 2.37

and

$$\begin{array}{cc}\hfill \left(e^t-1\right)K_{n,3}\left(x\mid \mathit{\lambda }\right)& =e^t{K}_{n,3}\left(x\mid \mathit{\lambda }\right)-K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ \hfill & =K_{n,3}\left(x+1\mid \mathit{\lambda }\right)-K_{n,3}\left(x\mid \mathit{\lambda }\right).\hfill \end{array}$$ 2.38

From (2.37) and (2.38), we have

$$\begin{array}{c}\hfill n{K}_{n-1,3}\left(x\mid \mathit{\lambda }\right)=K_{n,3}\left(x+1\mid \mathit{\lambda }\right)-K_{n,3}\left(x\mid \mathit{\lambda }\right).\end{array}$$ 2.39

By (1.10) and (2.3), we get

$$\begin{array}{c}\hfill \left(e^t-1\right)K_{n,4}\left(x\mid \mathit{\lambda }\right)=n{K}_{n-1,4}\left(x\mid \mathit{\lambda }\right).\end{array}$$ 2.40

Thus, by (2.40), we have

$$\begin{array}{c}\hfill K_{n,4}\left(x+1\mid \mathit{\lambda }\right)-K_{n,4}\left(x\mid \mathit{\lambda }\right)=n{K}_{n-1,4}\left(x\mid \mathit{\lambda }\right).\end{array}$$ 2.41

Therefore, by (2.35), (2.36), (2.39) and (2.41), we obtain the following theorem.

Theorem 7

For$n\ge 0$, we have

$$\begin{array}{cc}\hfill K_{n,3}\left(x+y\mid \mathit{\lambda }\right)& =\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)K_{j,3}\left(x\mid \mathit{\lambda }\right){\left(y\right)}_{n-j},\hfill \\ \hfill K_{n,4}\left(x+y\mid \mathit{\lambda }\right)& =\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)K_{j,4}\left(x\mid \mathit{\lambda }\right){\left(y\right)}_{n-j},\hfill \\ \hfill n{K}_{n-1,3}\left(x\mid \mathit{\lambda }\right)& =K_{n,3}\left(x+1\mid \mathit{\lambda }\right)-K_{n,3}\left(x\mid \mathit{\lambda }\right),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill n{K}_{n-1,4}\left(x\mid \mathit{\lambda }\right)=K_{n,4}\left(x+1\mid \mathit{\lambda }\right)-K_{n,4}\left(x\mid \mathit{\lambda }\right).\end{array}$$

For $s_n\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we note that

$$\begin{array}{c}\hfill \frac{d}{dx}{s}_n\left(x\right)=\sum _{l=0}^{n-1}\left(\begin{array}{c}n\\ l\end{array}\right)〈\left.\overline{f}\left(t\right)\right|x^{n-l}〉s_l\left(x\right).\end{array}$$ 2.42

For $K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)$, by (2.42), we get

$$\begin{array}{cc}\hfill 〈\left.\overline{f}\left(t\right)\right|x^{n-l}〉& =〈\left.log\left(1+t\right)\right|x^{n-l}〉\hfill \\ \hfill & =〈\left.\sum _{m=1}^{\infty }{\left(-1\right)}^{m-1}\left(m-1\right)!\frac{t^m}{m!}\right|x^{n-l}〉\hfill \\ \hfill & ={\left(-1\right)}^{n-l-1}\left(n-l-1\right)!.\hfill \end{array}$$ 2.43

Thus, by (2.42) and (2.43), we have

$$\begin{array}{cc}\hfill \frac{d}{dx}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)& =\sum _{l=0}^{n-1}\left(\begin{array}{c}n\\ l\end{array}\right){\left(-1\right)}^{n-l-1}\left(n-l-1\right)!K_{l,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{K}_{l,3}\left(x\mid \mathit{\lambda }\right).\hfill \end{array}$$ 2.44

By the same method as (2.44), we get

$$\begin{array}{c}\hfill \frac{d}{dx}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{K}_{l,4}\left(x\mid \mathit{\lambda }\right).\end{array}$$ 2.45

Therefore, by (2.44) and (2.45), we obtain the following theorem.

Theorem 8

For$n\ge 1$, we have

$$\begin{array}{c}\hfill \frac{d}{dx}{K}_{n,3}\left(x\mid \mathit{\lambda }\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{K}_{l,3}\left(x\mid \mathit{\lambda }\right),\end{array}$$

and

$$\begin{array}{c}\hfill \frac{d}{dx}{K}_{n,4}\left(x\mid \mathit{\lambda }\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{K}_{l,4}\left(x\mid \mathit{\lambda }\right).\end{array}$$

Let $n\ge 1$. Then, by (1.6), (2.1) and (2.3), we get

$$\begin{array}{cc}& K_{n,3}\left(y\mid \mathit{\lambda }\right)\hfill \\ \hfill & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^y\right|x^n〉\hfill \\ \hfill & =〈\left.{\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^y\right)\right|x^{n-1}〉\hfill \\ \hfill & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\mathit{\partial }}_t{\left(1+t\right)}^y\right|x^{n-1}〉+〈\left.\left({\mathit{\partial }}_t\frac{log\left(1+\mathit{\lambda }t\right)}{log\left(1+t\right)}\right){\left(1+t\right)}^y\right|x^{n-1}〉.\hfill \end{array}$$ 2.46

We observe that

$$\begin{array}{cc}\hfill 〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\mathit{\partial }}_t{\left(1+t\right)}^y\right|x^{n-1}〉& =y〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{y-1}\right|x^{n-1}〉\hfill \\ \hfill & =y{K}_{n-1,3}\left(y-1\mid \mathit{\lambda }\right),\hfill \end{array}$$ 2.47

and

$$\begin{array}{cc}\hfill {\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right)& =\frac{\frac{\mathit{\lambda }}{1+\mathit{\lambda }t}·\mathit{\lambda }log\left(1+t\right)-log\left(1+\mathit{\lambda }t\right)\frac{\mathit{\lambda }}{1+t}}{{\left(\mathit{\lambda }log\left(1+t\right)\right)}^2}\hfill \\ \hfill & =\frac{t}{log\left(1+t\right)}\frac{1}{t}\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{-1}\right\}.\hfill \end{array}$$ 2.48

By (2.48), we get

$$\begin{array}{cc}& 〈\left.\left({\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{log\left(1+t\right)}\right)\right){\left(1+t\right)}^y\right|x^{n-1}〉\hfill \\ & =〈\left.\frac{t}{log\left(1+t\right)}\frac{1}{t}\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{-1}\right\}{\left(1+t\right)}^y\right|x^{n-1}〉\hfill \\ & =\frac{1}{n}〈\left.\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{-1}\right\}{\left(1+t\right)}^y\right|\frac{t}{log\left(1+t\right)}{x}^n〉\hfill \\ & =\frac{1}{n}〈\left.\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{-1}\right\}{\left(1+t\right)}^y\right|\sum _{l=0}^{\infty }{b}_l\frac{t^l}{l!}{x}^n〉\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{〈\left.\frac{1}{1+\mathit{\lambda }t}{\left(1+t\right)}^y\right|x^{n-l}〉-〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1+t\right)}^{y-1}\right|x^{n-l}〉\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{〈{\left(1+t\right)}^y\left|\sum _{m=0}^{\infty }{\left(-\mathit{\lambda }t\right)}^m{x}^{n-l}\right.〉-K_{n-l,3}\left(y-1\mid \mathit{\lambda }\right)\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m〈\left.{\left(1+t\right)}^y\right|x^{n-l-m}〉-K_{n-l,3}\left(y-1\mid \mathit{\lambda }\right)\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(y\right)}_{n-l-m}-K_{n-l,3}\left(y-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$ 2.49

For $n\ge 1$, from (2.46), (2.47) and (2.49), we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)-x{K}_{n-1,3}\left(x-1\mid \mathit{\lambda }\right)\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(x\right)}_{n-l-m}-K_{n-l,3}\left(x-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$ 2.50

Therefore, by (2.50), we obtain the following theorem.

Theorem 9

For$n\ge 1$, we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)-x{K}_{n-1,3}\left(x-1\mid \mathit{\lambda }\right)\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(x\right)}_{n-l-m}-K_{n-l,3}\left(x-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$

Remark

We note that

$$\begin{array}{cc}\hfill b_n\left(x\right)-x{b}_{n-1}\left(x-1\right)& =\underset{\mathit{\lambda }\to 0}{lim}\left(K_{n,3}\left(x\mid \mathit{\lambda }\right)-x{K}_{n-1,3}\left(x-1\mid \mathit{\lambda }\right)\right)\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)b_l\left({\left(x\right)}_{n-l}-b_{n-l}\left(x-1\right)\right).\hfill \end{array}$$

From (1.6) and (2.2), we have

$$\begin{array}{cc}\hfill K_{n,4}\left(y\mid \mathit{\lambda }\right)& =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^y\right|x^n〉\hfill \\ \hfill & =〈\left.{\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^y\right)\right|x^{n-1}〉\hfill \\ \hfill & =〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\mathit{\partial }}_t{\left(1+t\right)}^y\right|x^{n-1}〉\hfill \\ \hfill & +〈\left.\left({\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right)\right){\left(1+t\right)}^y\right|x^{n-1}〉,\hfill \end{array}$$ 2.51

where $n\ge 1$.

We note that

$$\begin{array}{cc}& 〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\left({\mathit{\partial }}_t{\left(1+t\right)}^y\right)\right|x^{n-1}〉\hfill \\ \hfill & =y〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{y-1}\right|x^{n-1}〉\hfill \\ \hfill & =y{K}_{n-1,4}\left(y-1\mid \mathit{\lambda }\right).\hfill \end{array}$$ 2.52

Now, we observe that

$$\begin{array}{cc}& {\mathit{\partial }}_t\left(\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right)\hfill \\ \hfill & =\frac{\frac{\mathit{\lambda }}{1+\mathit{\lambda }t}\left({\left(1+t\right)}^{\mathit{\lambda }}-1\right)-log\left(1+\mathit{\lambda }t\right)\mathit{\lambda }{\left(1+t\right)}^{\mathit{\lambda }-1}}{{\left({\left(1+t\right)}^{\mathit{\lambda }}-1\right)}^2}\hfill \\ \hfill & =\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\frac{1}{t}\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{\mathit{\lambda }-1}\right\}.\hfill \end{array}$$ 2.53

Thus, by (2.53), we get

$$\begin{array}{cc}& 〈\left.\left({\mathit{\partial }}_t\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right){\left(1+t\right)}^y\right|x^{n-1}〉\hfill \\ & =〈\left.\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\frac{1}{t}\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{\mathit{\lambda }-1}\right\}{\left(1+t\right)}^y\right|x^{n-1}〉\hfill \\ & =\frac{1}{n}〈\left.\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{\mathit{\lambda }-1}\right\}{\left(1+t\right)}^y\right|x^n〉\hfill \\ & =\frac{1}{n}〈\left.\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{\mathit{\lambda }-1}\right\}{\left(1+t\right)}^y\right|\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{x}^n〉\hfill \\ & =\frac{1}{n}〈\left.\left\{\frac{1}{1+\mathit{\lambda }t}-\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{\mathit{\lambda }-1}\right\}{\left(1+t\right)}^y\right|\sum _{l=0}^{\infty }{K}_l\left(\mathit{\lambda }\right)\frac{t^l}{l!}{x}^n〉\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)〈\left.\frac{1}{1+\mathit{\lambda }t}{\left(1+t\right)}^y\right|x^{n-l}〉\hfill \\ & -〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1+t\right)}^{y+\mathit{\lambda }-1}\right|x^{n-l}〉\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{〈\left.{\left(1+t\right)}^y\right|\sum _{m=0}^{\infty }{\left(-\mathit{\lambda }t\right)}^m{x}^{n-l}〉-K_{n-1,4}\left(y+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m〈\left.{\left(1+t\right)}^y\right|x^{n-l-m}〉\right.\hfill \\ & \left.-K_{n-1,4}\left(y+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(y\right)}_{n-l-m}-K_{n-l,4}\left(y+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(y\right)}_{n-l-m}-K_{n-l,4}\left(y+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$ 2.54

From (2.51), (2.52) and (2.54), we have

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)-x{K}_{n-1,4}\left(x-1\mid \mathit{\lambda }\right)\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(x\right)}_{n-l-m}-K_{n-l,4}\left(x+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$ 2.55

Therefore, by (2.55), we obtain the following theorem.

Theorem 10

For$n\ge 1$, we have

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)-x{K}_{n-1,4}\left(x-1\mid \mathit{\lambda }\right)\hfill \\ & =\frac{1}{n}\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)K_l\left(\mathit{\lambda }\right)\left\{\sum _{m=0}^{n-l}{\left(-\mathit{\lambda }\right)}^m{\left(n-l\right)}_m{\left(x\right)}_{n-l-m}-K_{n-l,4}\left(x+\mathit{\lambda }-1\mid \mathit{\lambda }\right)\right\}.\hfill \end{array}$$

Let us consider the following two Sheffer sequences:

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right),{\left(x\right)}^{\left(n\right)}=x\left(x+1\right)\cdots \left(x+\left(n-1\right)\right)\sim \left(1,1-e^{-t}\right).\hfill \end{array}$$ 2.56

From (1.14) and (1.15), we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n{C}_{n,m}{\left(x\right)}^{\left(m\right)},\end{array}$$ 2.57

where

$$\begin{array}{cc}& C_{n,m}\hfill \\ \hfill & =\frac{1}{m!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(1-\frac{1}{1+t}\right)}^m\right|x^n〉\hfill \\ \hfill & =\frac{1}{m!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\sum _{l=0}^{\infty }{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{t^{m+l}}{l!}{x}^n〉\hfill \\ \hfill & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\frac{t}{log\left(1+t\right)}{x}^{n-m-l}〉\hfill \\ \hfill & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\sum _{k=0}^{\infty }{b}_k\frac{t^k}{k!}{x}^{n-m-l}〉\hfill \\ \hfill & =\sum _{l=0}^{n-m}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\left(\begin{array}{c}n\\ m+l\end{array}\right)\left(\begin{array}{c}m+l\\ m\end{array}\right)\hfill \\ \hfill & \times \sum _{k=0}^{n-m-l}\left(\begin{array}{c}n-m-l\\ k\end{array}\right)b_k〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-m-l-k}〉\hfill \\ \hfill & =\sum _{l=0}^{n-m}\sum _{k=0}^{n-m-l}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\left(\begin{array}{c}n\\ m+l\end{array}\right)\left(\begin{array}{c}m+l\\ m\end{array}\right)\hfill \\ \hfill & \times \left(\begin{array}{c}n-m-l\\ k\end{array}\right)b_k{D}_{n-m-l-k}{\mathit{\lambda }}^{n-m-l-k}\hfill \\ \hfill & =\sum _{l=0}^{n-m}\sum _{k=0}^{n-m-l}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\left(\begin{array}{c}n\\ m+l\end{array}\right)\left(\begin{array}{c}m+l\\ m\end{array}\right)\hfill \\ \hfill & \times \left(\begin{array}{c}n-m-l\\ k\end{array}\right)b_{n-m-l-k}{D}_k{\mathit{\lambda }}^k\hfill \end{array}$$ 2.58

Therefore, by (2.57) and (2.58), we obtain the following theorem.

Theorem 11

For $n\ge 0$, we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{m=0}^n\left\{\sum _{l=0}^{n-m}\sum _{k=0}^{n-m-l}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\left(\begin{array}{c}n\\ m+l\end{array}\right)\right.\hfill \\ & \times \left.\left(\begin{array}{c}m+l\\ m\end{array}\right)\left(\begin{array}{c}n-m-l\\ k\end{array}\right)b_{n-m-l-k}{D}_k{\mathit{\lambda }}^k\right\}{\left(x\right)}^{\left(m\right)}.\hfill \end{array}$$

For $K_{n,4}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{e^{\mathit{\lambda }t}-1}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)$, ${\left(x\right)}^{\left(n\right)}\sim \left(1,1-e^{-t}\right),$ we have

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n{C}_{n,m}{\left(x\right)}^{\left(m\right)},\end{array}$$ 2.59

where

$$\begin{array}{cc}& C_{n,m}\hfill \\ & =\frac{1}{m!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{\left(1-\frac{1}{1+t}\right)}^m\right|x^n〉\hfill \\ & =\frac{1}{m!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{{\left(1+t\right)}^{\mathit{\lambda }}-1}\right|\sum _{l=0}^{\infty }{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{t^{m+l}}{l!}{x}^n〉\hfill \\ & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\frac{\mathit{\lambda }t}{{\left(1+t\right)}^{\mathit{\lambda }}-1}{x}^{n-m-l}〉\hfill \\ & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|\sum _{k=0}^{\infty }{K}_k\left(\mathit{\lambda }\right)\frac{t^k}{k!}{x}^{n-m-l}〉\hfill \\ & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}\sum _{k=0}^{n-m-l}\left(\begin{array}{c}n-m-l\\ k\end{array}\right)K_k\left(\mathit{\lambda }\right)\hfill \\ & \times 〈\left.\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }t}\right|x^{n-m-l-k}〉\hfill \\ & =\frac{1}{m!}\sum _{l=0}^{n-m}{\left(-1\right)}^l{\left(m+l-1\right)}_l\frac{{\left(n\right)}_{m+l}}{l!}\hfill \\ & \times \sum _{k=0}^{n-m-l}\left(\begin{array}{c}n-m-l\\ k\end{array}\right)K_k\left(\mathit{\lambda }\right)D_{n-m-l-k}{\mathit{\lambda }}^{n-m-l-k}\hfill \\ & =\sum _{l=0}^{n-m}\sum _{k=0}^{n-m-l}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\left(\begin{array}{c}n\\ m+l\end{array}\right)\left(\begin{array}{c}m+l\\ m\end{array}\right)\hfill \\ & \times \left(\begin{array}{c}n-m-l\\ k\end{array}\right)K_{n-m-l-k}\left(\mathit{\lambda }\right)D_k{\mathit{\lambda }}^k.\hfill \end{array}$$ 2.60

Therefore, by (2.59) and (2.60), we obtain the following theorem.

Theorem 12

For$n\ge 0$, we have

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{m=0}^n\left\{\sum _{l=0}^{n-m}\sum _{k=0}^{n-m-l}{\left(-1\right)}^{l}l!\left(\begin{array}{c}m+l-1\\ l\end{array}\right)\right.\hfill \\ & \times \left.\left(\begin{array}{c}n\\ m+l\end{array}\right)\left(\begin{array}{c}m+l\\ m\end{array}\right)\left(\begin{array}{c}n-m-l\\ k\end{array}\right)K_{n-m-l-k}\left(\mathit{\lambda }\right)D_k{\mathit{\lambda }}^k\right\}{\left(x\right)}^{\left(m\right)}.\hfill \end{array}$$

Let us consider the following two Sheffer sequences:

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right),B_n^{\left(s\right)}\left(x\right)\sim \left({\left(\frac{e^t-1}{t}\right)}^s,t\right).\end{array}$$

Note that

$$\begin{array}{c}\hfill \sum _{n=0}^{\infty }{B}_n^{\left(s\right)}\left(x\right)={\left(\frac{t}{e^t-1}\right)}^s{e}^{xt},\left(\text{see Kim et al. 2015; Sen et al. 2013; Ustinov et al. 2002}\right),\end{array}$$

where $B_n^{\left(s\right)}\left(x\right)$ are called the higher-order Bernoulli polynomials.

From (1.14) and (1.15), we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n{C}_{n,m}{B}_m^{\left(s\right)}\left(x\right),\end{array}$$ 2.61

where

$$\begin{array}{cc}& C_{n,m}\hfill \\ & =\frac{1}{m!}〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(log\left(1+t\right)\right)}^m\right|x^n〉\hfill \\ & =〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\frac{1}{m!}{\left(log\left(1+t\right)\right)}^m{x}^n〉\hfill \\ & =〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\sum _{l=m}^{\infty }{S}_1\left(l,m\right)\frac{t^l}{l!}{x}^n〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\right|\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{x}^{n-l}〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\right|\sum _{k=0}^{\infty }{K}_{k,3}\left(\mathit{\lambda }\right)\frac{t^k}{k!}{x}^{n-l}〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{k,3}\left(\mathit{\lambda }\right)〈\left.{\left(\frac{t}{log\left(1+t\right)}\right)}^s\right|x^{n-l-k}〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{k,3}\left(\mathit{\lambda }\right)b_{n-l-k}^{\left(s\right)}\hfill \\ & =\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)S_1\left(l,m\right)K_{k,3}\left(\mathit{\lambda }\right)b_{n-l-k}^{\left(s\right)}.\hfill \end{array}$$ 2.62

Here, the Bernoulli numbers of the second kind of order s are defined by the generating function

$$\begin{array}{c}\hfill {\left(\frac{t}{log\left(1+t\right)}\right)}^s=\sum _{j=0}^{\infty }{b}_j^{\left(s\right)}\frac{t^j}{j!},\left(\text{see Kim et al. 2015; Roman 1984}\right).\end{array}$$

Therefore, by (2.61) and (2.62), we obtain the following theorem.

Theorem 13

For$n\ge 0$, we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n\left(\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)S_1\left(l,m\right)K_{k,3}\left(\mathit{\lambda }\right)b_{n-l-k}^{\left(s\right)}\right)B_m^{\left(s\right)}\left(x\right).\end{array}$$

Remark

In a similar manner, one shows that, for $n\ge 0$,

$$\begin{array}{c}\hfill K_{n,4}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n\left(\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)S_1\left(l,m\right)K_{k,4}\left(\mathit{\lambda }\right)b_{n-l-k}^{\left(s\right)}\right)B_m^{\left(s\right)}\left(x\right).\end{array}$$

For $\mathit{\mu }\in \mathbb{C}$ with $\mathit{\mu }\ne 1$, $s\in \mathbb{N}$, the Frobenius-Euler polynomials of order s are defined by the generating function

$$\begin{array}{c}\hfill {\left(\frac{1-\mathit{\mu }}{e^t-\mathit{\mu }}\right)}^s{e}^{xt}=\sum _{n=0}^{\infty }{H}_n^{\left(s\right)}\left(x\mid \mathit{\mu }\right)\frac{t^n}{n!},\left(\text{see [1,16]}\right).\end{array}$$

For $K_{n,3}\left(x\mid \mathit{\lambda }\right)\sim \left(\frac{\mathit{\lambda }t}{log\left(1+\mathit{\lambda }\left(e^t-1\right)\right)},e^t-1\right)$, $H_n^{\left(s\right)}\left(x\mid \mathit{\mu }\right)\sim \left({\left(\frac{e^t-\mathit{\mu }}{1-\mathit{\mu }}\right)}^s,t\right)$, we have

$$\begin{array}{c}\hfill K_{n,3}\left(x\mid \mathit{\lambda }\right)=\sum _{m=0}^n{C}_{n,m}{H}_m^{\left(s\right)}\left(x\mid \mathit{\mu }\right),\end{array}$$ 2.63

where

$$\begin{array}{cc}\hfill C_{n,m}& =\frac{1}{m!}〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{\left(log\left(1+t\right)\right)}^m\right|x^n〉\hfill \\ & =〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\frac{1}{m!}{\left(log\left(1+t\right)\right)}^m{x}^n〉\hfill \\ & =〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}\right|\sum _{l=m}^{\infty }{S}_1\left(l,m\right)\frac{t^l}{l!}{x}^n〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\right|\frac{log\left(1+\mathit{\lambda }t\right)}{\mathit{\lambda }log\left(1+t\right)}{x}^{n-l}〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\right|\sum _{k=0}^{\infty }{K}_{k,3}\left(\mathit{\lambda }\right)\frac{t^k}{k!}{x}^{n-l}〉\hfill \\ & =\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{k,3}\left(\mathit{\lambda }\right)〈\left.{\left(\frac{1-\mathit{\mu }+t}{1-\mathit{\mu }}\right)}^s\right|x^{n-l-k}〉\hfill \\ & =\frac{1}{{\left(1-\mathit{\mu }\right)}^s}\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{k,3}\left(\mathit{\lambda }\right)\hfill \\ & \times \sum _{j=0}^s\left(\begin{array}{c}s\\ j\end{array}\right){\left(1-\mathit{\mu }\right)}^{s-j}〈\left.t^j\right|x^{n-l-k}〉\hfill \\ & =\frac{1}{{\left(1-\mathit{\mu }\right)}^s}\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{n-l-k,3}\left(\mathit{\lambda }\right)\hfill \\ & \times \sum _{j=0}^s\left(\begin{array}{c}s\\ j\end{array}\right){\left(1-\mathit{\mu }\right)}^{s-j}〈\left.t^j\right|x^k〉\hfill \\ & =\frac{1}{{\left(1-\mathit{\mu }\right)}^s}\sum _{l=m}^n\left(\begin{array}{c}n\\ l\end{array}\right)S_1\left(l,m\right)\sum _{k=0}^{n-l}\left(\begin{array}{c}n-l\\ k\end{array}\right)K_{n-l-k,3}\left(\mathit{\lambda }\right)k!\left(\begin{array}{c}s\\ k\end{array}\right){\left(1-\mathit{\mu }\right)}^{s-k}\hfill \\ & =\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)\left(\begin{array}{c}s\\ k\end{array}\right)\frac{k!}{{\left(1-\mathit{\mu }\right)}^k}{S}_1\left(l,m\right)K_{n-l-k,3}\left(\mathit{\lambda }\right).\hfill \end{array}$$ 2.64

Therefore, by (2.63) and (2.64), we obtain the following theorem.

Theorem 14

For$n\ge 0$, we have

$$\begin{array}{cc}& K_{n,3}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{m=0}^n\left(\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)\left(\begin{array}{c}s\\ k\end{array}\right)\frac{k!}{{\left(1-\mathit{\mu }\right)}^k}{S}_1\left(l,m\right)K_{n-l-k,3}\left(\mathit{\lambda }\right)\right)H_m^{\left(s\right)}\left(x\mid \mathit{\mu }\right).\hfill \end{array}$$

Remark

Proceeding similarly to the above, one can show that, for $n\ge 0$,

$$\begin{array}{cc}& K_{n,4}\left(x\mid \mathit{\lambda }\right)\hfill \\ & =\sum _{m=0}^n\left(\sum _{l=m}^n\sum _{k=0}^{n-l}\left(\begin{array}{c}n\\ l\end{array}\right)\left(\begin{array}{c}n-l\\ k\end{array}\right)\left(\begin{array}{c}s\\ k\end{array}\right)\frac{k!}{{\left(1-\mathit{\mu }\right)}^k}{S}_1\left(l,m\right)K_{n-l-k,4}\left(\mathit{\lambda }\right)\right)H_m^{\left(s\right)}\left(x\mid \mathit{\mu }\right).\hfill \end{array}$$

Conclusion

The first degenerate version of the Bernoulli polynomials of the second kind appeared in the paper by Korobov (1996, 2001). Here, we study two degenerate versions of the Bernoulli polynomials of the second kind which will be called Korobov polynomials of third kind and of the fourth kind. Some properties, identities, recurrence relations and connections with other polynomials are investigated by using umbral calculus.