A Note on Type-Two Degenerate Poly-Changhee Polynomials of the Second Kind

Abstract

In this paper, we first define type-two degenerate poly-Changhee polynomials of the second kind by using modified degenerate polyexponential functions. We derive new identities and relations between type-two degenerate poly-Changhee polynomials of the second kind. Finally, we derive type-two degenerate unipoly-Changhee polynomials of the second kind and discuss some of their identities.

1. Introduction

As is well known, Changhee polynomials $C{h}_n\left(x\right)$ are defined by means of the following generating function

$${\int }_{{\mathbb{Z}}_p}{(1+t)}^{x+y}{d}_{{\mu }_{-1}}y=\frac{2}{2+t}{(1+t)}^x=\sum _{n=0}^{\infty }C{h}_n\left(x\right)\frac{t^n}{n!}$$

(1)

(see [1,2]).

In the case when $x=0$, $C{h}_n\left(0\right)=C{h}_n$ are called Changhee numbers.

The Euler polynomials are defined by the following generating function:

$${\int }_{{\mathbb{Z}}_p}{e}^{(x+y)t}{d}_{{\mu }_{-1}}y=\frac{2}{e^t+1}{e}^{xt}=\sum _{n=0}^{\infty }{E}_n\left(x\right)\frac{t^n}{n!}$$

(2)

(see [3]).

When $x=0$, $E_n\left(0\right)=E_n$ are called the Euler numbers.

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

$$C{h}_n\left(x\right)=\sum _{l=0}^n{E}_l\left(x\right)S_1(n,l),$$

and

$$E_n\left(x\right)=\sum _{l=0}^{n}C{h}_n\left(x\right)S_2(n,l),(n\ge 0)$$

(3)

(see [1]).

For any non-zero $\lambda \in \mathbb{R}$, the degenerate exponential functions are defined by (see [4,5,6]):

$$e_{\lambda }^x\left(t\right)={(1+\lambda t)}^{\frac{x}{\lambda }},e_{\lambda }\left(t\right)=e_{\lambda }^1\left(t\right)={(1+\lambda t)}^{\frac{1}{\lambda }}.$$

(4)

Here, we note that

$$e_{\lambda }^x\left(t\right)=\sum _{n=0}^{\infty }{\left(x\right)}_{n,\lambda }\frac{t^n}{n!},$$

(5)

where ${\left(x\right)}_{0,\lambda }=1,{\left(x\right)}_{n,\lambda }=x(x-\lambda )(x-2\lambda )\cdots (x-(n-1)\lambda ),(n\ge 1).$

In [7,8], Carlitz considered degenerate Bernoulli polynomials, which are given by

$$\frac{t}{e_{\lambda }\left(t\right)-1}{e}_{\lambda }^x\left(t\right)=\frac{t}{{(1+\lambda t)}^{\frac{1}{\lambda }}-1}{(1+\lambda t)}^{\frac{x}{\lambda }}=\sum _{n=0}^{\infty }{\beta }_{n,\lambda }\left(x\right)\frac{t^n}{n!}.$$

(6)

On setting $x=0$, ${\beta }_{n,\lambda }\left(0\right)={\beta }_{n,\lambda }$ are called degenerate Bernoulli numbers.

For $k\in \mathbb{Z}$, the modified degenerate polyexponential function [9] was defined by Kim and Kim to be

$${\mathrm{Ei}}_{k,\lambda }\left(x\right)=\sum _{n=1}^{\infty }\frac{{\left(1\right)}_{n,\lambda }{x}^n}{(n-1)!n^k},(\mid x\mid <1).$$

(7)

Note that

$${\mathrm{Ei}}_{1,\lambda }\left(x\right)=\sum _{n=1}^{\infty }\frac{{\left(1\right)}_{n,\lambda }{x}^n}{n!}=e_{\lambda }\left(x\right)-1.$$

(8)

In [9], Kim et al. introduced degenerate poly-Genocchi polynomials, which are given by

$$\frac{{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\infty }{G}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!},(k\in \mathbb{Z}).$$

(9)

In the case when $x=0$, $G_{n,\lambda }^{\left(k\right)}\left(0\right)=G_{n,\lambda }^{\left(k\right)}$ are called degenerate poly-Genocchi numbers.

Let $\lambda \in {\mathbb{C}}_p$ with $\mid \lambda \mid \le 1$. The degenerate Changhee polynomials of the second kind $C{h}_{n,\lambda }\left(x\right)$ are defined by

$$\begin{array}{c}{\int }_{{\mathbb{Z}}_p}{(1+\lambda log(1+\lambda t))}^{{}^{\frac{x+y}{\lambda }}}{d}_{{\mu }_{-1}}y=\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}{(1+\lambda log(1+\lambda t))}^{\frac{x}{\lambda }}\\ =\sum _{n=0}^{\infty }C{h}_{n,\lambda }\left(x\right)\frac{t^n}{n!}\end{array}$$

(10)

(see [10]).

When $x=0$, $C{h}_{n,\lambda }\left(0\right)=C{h}_{n,\lambda }$ are called the degenerate Changhee numbers of the second kind.

In [11], the degenerate Daehee polynomials $D_{n,\lambda }\left(x\right)$ are defined by

$$\frac{{log}_{\lambda }(1+t)}{t}{(1+t)}^x=\sum _{n=0}^{\infty }{D}_{n,\lambda }\left(x\right)\frac{t^n}{n!},(\lambda \in \mathbb{R}).$$

(11)

For $x=0$, $D_{n,\lambda }\left(0\right)=D_{n,\lambda }$ are called degenerate Daehee numbers.

Note that ${lim}_{\lambda \to 0}{D}_{n,\lambda }\left(x\right)=D_n\left(x\right),(n\ge 0)$ (see [12]).

The degenerate Stirling numbers of the first kind are defined by

$$\frac{1}{k!}{\left({log}_{\lambda }(1+t)\right)}^k=\sum _{n=k}^{\infty }{S}_{1,\lambda }(n,k)\frac{t^n}{n!},(k\ge 0)$$

(12)

(see [6,13,14,15,16,17,18,19]).

Note here that ${lim}_{\lambda \to 0}{S}_{1,\lambda }(n,k)=S_1(n,k)$, where $S_1(n,k)$ are the Stirling numbers of the first kind given by

$$\frac{1}{k!}{(log(1+t))}^k=\sum _{n=k}^{\infty }{S}_1(n,k)\frac{t^n}{n!},(k\ge 0)$$

(13)

(see [5,20]).

The degenerate Stirling numbers of the second kind are given by

$$\frac{1}{k!}{\left(e_{\lambda }\left(t\right)-1\right)}^k=\sum _{n=l}^{\infty }{S}_{2,\lambda }(n,l)\frac{t^n}{n!}$$

(14)

(see [21]).

We note here that ${lim}_{\lambda \to 0}{S}_{2,\lambda }(n,k)=S_2(n,k)$, where $S_2(n,k)$ are the Stirling numbers of the second kind given by

$$\frac{1}{k!}{(e^t-1)}^k=\sum _{n=l}^{\infty }{S}_2(n,l)\frac{t^n}{n!}$$

(15)

(see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]).

In this article, we introduce type-two degenerate poly-Changhee polynomials of the second kind and derive explicit expressions and some identities of those polynomials. In addition, we introduce type-two degenerate unipoly-Changhee polynomials of the second kind and derive explicit multifarious properties.

2. Type-Two Degenerate Poly-Changhee Polynomials of the Second Kind

In this section, we define degenerate Changhee polynomials of the second kind by using the modified degenerate polyexponential function; these are called type-two degenerate poly-Changhee numbers and polynomials of the second kind in the following.

Let $\lambda \in \mathbb{C}$ and $k\in \mathbb{Z}$; we consider that the type-two degenerate poly-Changhee polynomials of the second kind are defined by

$$\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}=\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}.$$

(16)

In the special case, when $x=0$, $C{h}_{n,\lambda }^{\left(k\right)}\left(0\right)=C{h}_{n,\lambda }^{\left(k\right)}$ are called type-two degenerate poly-Changhee numbers of the second kind, where ${log}_{\lambda }\left(t\right)=\frac{1}{\lambda }(t^{\lambda }-1)$ is the compositional inverse of $e_{\lambda }\left(t\right)$ that satisfies

$${log}_{\lambda }\left(e_{\lambda }\left(t\right)\right)=e_{\lambda }\left({log}_{\lambda }\left(t\right)\right)=t.$$

For $k=1$ in (16), we get

$$\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}=\sum _{n=0}^{\infty }C{h}_{n,\lambda }\left(x\right)\frac{t^n}{n!},$$

(17)

where $C{h}_{n,\lambda }\left(x\right)$ are called degenerate Changhee polynomials of the second kind (see Equation (10)).

Obviously,

$$\begin{array}{c}\underset{\lambda ⟶0}{lim}\left(\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\right)=\sum _{n=0}^{\infty }\underset{\lambda ⟶0}{lim}C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}\\ =\frac{2{\mathrm{Ei}}_k(log(1+t))}{t(2+t)}{(1+t)}^x=\sum _{n=0}^{\infty }C{h}_n^{\left(k\right)}\left(x\right)\frac{t^n}{n!},\end{array}$$

(18)

where $C{h}_n^{\left(k\right)}\left(x\right)$ are called type-two poly-Changhee polynomials.

By using Equations (7), (10), and (16), we observe that

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}=\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\\ =\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}\frac{1}{t}\sum _{m=0}^{\infty }\frac{{\left({log}_{\lambda }(1+t)\right)}^{m+1}}{(m+1)!{(m+1)}^{k-1}}\\ =\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}\frac{1}{t}\sum _{m=0}^{\infty }\frac{1}{{(m+1)}^{k-1}}\sum _{l=m+1}^{\infty }{S}_{1,\lambda }(l,m+1)\frac{t^l}{l!}\\ =\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}\frac{1}{t}\sum _{m=0}^{\infty }\frac{1}{{(m+1)}^{k-1}}\sum _{l=m}^{\infty }{S}_{1,\lambda }(l+1,m+1)\frac{t^l}{(l+1)!}\\ =\left(\sum _{s=0}^{\infty }C{h}_{n,\lambda }\frac{t^n}{n!}\right)\left(\sum _{l=0}^{\infty }\sum _{m=0}^l\frac{1}{{(m+1)}^{k-1}}\frac{S_{1,\lambda }(l+1,m+1)}{l+1}\frac{t^l}{l!}\right)\\ L.H.S=\sum _{n=0}^{\infty }\left(\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda }\frac{S_{1,\lambda }(l+1,m+1)}{l+1{(m+1)}^{k-1}}\right)\frac{t^n}{n!}.\end{array}$$

(19)

Therefore, using (19), we obtain the following theorem.

Theorem 1.

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

$$C{h}_{n,\lambda }^{\left(k\right)}=\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda }\frac{S_{1,\lambda }(l+1,m+1)}{l+1{(m+1)}^{k-1}}.$$

(20)

Corollary 1.

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

$$C{h}_{n,\lambda }^{\left(1\right)}=\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda }\frac{S_{1,\lambda }(l+1,m+1)}{l+1}.$$

(21)

From (16), we observe that

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}\\ =\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(\sum _{m=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\frac{x}{\lambda }}{m}\right){\lambda }^m{\left(log(1+t)\right)}^m\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(\sum _{m=0}^{\infty }{\left(x\right)}_{m,\lambda }\sum _{l=m}^{\infty }{S}_1(l,m)\frac{t^s}{s!}\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(\sum _{l=0}^{\infty }\sum _{m=0}^l{\left(x\right)}_{m,\lambda }{S}_1(l,m)\frac{t^l}{l!}\right)\\ \sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}=\sum _{n=0}^{\infty }\left(\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda }^{\left(k\right)}{\left(x\right)}_{m,\lambda }{S}_1(l,m)\right)\frac{t^n}{n!}.\end{array}$$

(22)

By comparing the coefficients on both sides of (22), we obtain the following theorem.

Theorem 2.

Let $n\ge 0$ and $k\in \mathbb{Z}$. Then, we have

$$C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)=\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda }^{\left(k\right)}{\left(x\right)}_{m,\lambda }{S}_1(l,m).$$

(23)

In [4], the degenerate Bernoulli polynomials of the second kind are defined by

$$\frac{t}{{log}_{\lambda }(1+t)}{(1+t)}^x=\sum _{n=0}^{\infty }{b}_{n,\lambda }\left(x\right)\frac{t^n}{n!}$$

(24)

(see [30]).

For $x=0$, $b_{n,\lambda }\left(0\right)=b_{n,\lambda }$ are called degenerate Bernoulli numbers of the second kind.

From (7), we note that

$$\begin{array}{c}\frac{d}{dx}{\mathrm{Ei}}_{\mathrm{k},˘}\left({log}_{˘}(1+\mathrm{x})\right)=\frac{\mathrm{d}}{\mathrm{dx}}\sum _{\mathrm{n}=1}^{\infty }\frac{{\left(1\right)}_{\mathrm{n},˘}{\left({log}_{˘}(1+\mathrm{x})\right)}^{\mathrm{n}}}{(\mathrm{n}-1)!{\mathrm{n}}^{\mathrm{k}}}\\ =\frac{{(1+x)}^{\lambda -1}}{{log}_{\lambda }(1+x)}\sum _{n=1}^{\infty }\frac{{\left(1\right)}_{n,\lambda }{\left({log}_{\lambda }(1+x)\right)}^n}{(n-1)!n^{k-1}}=\frac{{(1+x)}^{\lambda -1}}{{log}_{\lambda }(1+x)}{\mathrm{Ei}}_{\mathrm{k}-1,˘}\left({log}_{\lambda }(1+x)\right).\end{array}$$

(25)

Thus, from (16) and (25), we have

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{x^n}{n!}=\frac{2}{x(1+{(1+\lambda log(1+x))}^{\frac{1}{\lambda }})}{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+x)\right)\\ =\frac{2}{x(1+{(1+\lambda log(1+x))}^{\frac{1}{\lambda }})}{\int }_0^x\underset{(k-2)-\mathrm{times}}{\underbrace{\frac{{(1+t)}^{\lambda -1}}{{log}_{\lambda }(1+t)}{\int }_0^t\cdots \frac{{(1+t)}^{\lambda -1}}{{log}_{\lambda }(1+t)}{\int }_0^t}}\frac{{(1+t)}^{\lambda -1}}{{log}_{\lambda }(1+t)}tdt\cdots dt\\ =\frac{2}{1+{(1+\lambda log(1+x))}^{\frac{1}{\lambda }}}\sum _{m=0}^{\infty }\sum _{m_1+\cdots +m_{k-1}=m}\left(\genfrac{}{}{0pt}{}{m}{m_1+\cdots +m_{k-1}}\right)\\ \times \frac{b_{m_1,\lambda }(\lambda -1)}{m_1+1}\frac{b_{m_2,\lambda }(\lambda -1)}{m_1+m_2+1}\cdots \frac{b_{m_{k-1},\lambda }(\lambda -1)}{m_1+\cdots +m_{k-1}+1}\frac{x^m}{m!}\\ \sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{x^n}{n!}=\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^n\left(\genfrac{}{}{0pt}{}{n}{m}\right)\sum _{m_1+\cdots +m_{k-1}=m}\left(\genfrac{}{}{0pt}{}{m}{m_1+\cdots +m_{k-1}}\right)\\ \times \frac{b_{m_1,\lambda }(\lambda -1)}{m_1+1}\frac{b_{m_2,\lambda }(\lambda -1)}{m_1+m_2+1}\cdots \frac{b_{m_{k-1},\lambda }(\lambda -1)}{m_1+\cdots +m_{k-1}+1}C{h}_{n-m,\lambda }\frac{x^n}{n!}.\end{array}$$

(26)

Therefore, using (26), we obtain the following theorem.

Theorem 3.

For $n\ge 0$, we have

$$\begin{array}{c}C{h}_{n,\lambda }^{\left(k\right)}=\frac{1}{2}\sum _{m=0}^n\left(\genfrac{}{}{0pt}{}{n}{m}\right)\sum _{m_1+\cdots +m_{k-1}=m}\left(\genfrac{}{}{0pt}{}{m}{m_1+\cdots +m_{k-1}}\right)\\ \times \frac{b_{m_1,\lambda }(\lambda -1)}{m_1+1}\frac{b_{m_2,\lambda }(\lambda -1)}{m_1+m_2+1}\cdots \frac{b_{m_{k-1},\lambda }(\lambda -1)}{m_1+\cdots +m_{k-1}+1}C{h}_{n-m,\lambda }.\end{array}$$

(27)

Corollary 2.

For $n\ge 0$, we have

$$C{h}_{n,\lambda }^{\left(2\right)}=\frac{1}{2}\sum _{m=0}^n\left(\genfrac{}{}{0pt}{}{n}{m}\right)\frac{b_{m,\lambda }(\lambda -1)}{m+1}C{h}_{n-m,\lambda }.$$

Let $k\ge 1$ be an integer. For $s\in \mathbb{C}$, we define the function ${\eta }_{k,\lambda }\left(s\right)$ as

$$\begin{array}{c}{\eta }_{k,\lambda }\left(s\right)=\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_0^{\infty }\frac{t^{s-1}}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)dt\\ =\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_0^1\frac{t^{s-1}}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)dt\\ +\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_1^{\infty }\frac{t^{s-1}}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)dt.\end{array}$$

(28)

The second integral converges absolutely for any $s\in \mathbb{C}$, and hence, the second term on the right-hand side vanishes at non-positive integers. That is,

$$\underset{s\to -m}{lim}\left|\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_1^{\infty }\frac{t^{s-1}}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)dt\right|\le \frac{1}{\mathsf{\Gamma }(-m)}M=0.$$

(29)

On the other hand, for $\Re \left(s\right)>0$, the first integral in (29) can be written as

$$\frac{1}{\mathsf{\Gamma }\left(s\right)}\sum _{l=0}^{\infty }\frac{C{h}_{l,\lambda }^{\left(k\right)}}{l!}\frac{1}{s+l},$$

which defines an entire function of s. Thus, we may conclude that ${\eta }_{k,\lambda }\left(s\right)$ can be continued to an entire function of s.

Further, from (28) and (29), we obtain

$$\begin{array}{c}{\eta }_{k,\lambda }(-m)=\underset{s\to -m}{lim}\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_0^1\frac{t^{s-1}}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)dt\\ =\underset{s\to -m}{lim}\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_0^1{t}^{s-1}\sum _{l=0}^{\infty }\frac{C{h}_{l,\lambda }^{\left(k\right)}{t}^l}{l!}dt=\underset{s\to -m}{lim}\frac{1}{\mathsf{\Gamma }\left(s\right)}\sum _{l=0}^{\infty }\frac{C{h}_{l,\lambda }^{\left(k\right)}}{s+l}\frac{1}{l!}\\ =\cdots +0+\cdots +0+\underset{s\to -m}{lim}\frac{1}{\mathsf{\Gamma }\left(s\right)}\frac{1}{s+m}\frac{C{h}_{m,\lambda }^{\left(k\right)}}{m!}+0+0+\cdots \\ =\underset{s\to -m}{lim}\frac{\left(\frac{\mathsf{\Gamma }(1-s)sin\pi s}{\pi }\right)}{s+m}\frac{C{h}_{m,\lambda }^{\left(k\right)}}{m!}=\mathsf{\Gamma }(1+m)cos\left(\pi m\right)\frac{C{h}_{m,\lambda }^{\left(k\right)}}{m!}\\ ={(-1)}^{m}C{h}_{m,\lambda }^{\left(k\right)}.\end{array}$$

(30)

Therefore, using (30), we obtain the following theorem.

Theorem 4.

Let $k\ge 1$ and $m\in \mathbb{N}\bigcup \left\{0\right\}$, $s\in \mathbb{C}$; we have

$${\eta }_{k,\lambda }(-m)={(-1)}^{m}C{h}_{m,\lambda }^{\left(k\right)}.$$

From (16), we note that

$$\begin{array}{c}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)=t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\\ =t\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(1+\sum _{m=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\frac{1}{\lambda }}{m}\right){\lambda }^m{(log(1+t))}^m\right)\\ =t\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(1+\sum _{m=0}^{\infty }{\left(1\right)}_{m,\lambda }\sum _{l=m}^{\infty }{S}_1(l,m)\frac{t^l}{l!}\right)\\ =t\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\frac{t^n}{n!}\right)\left(1+\sum _{l=0}^{\infty }\left(\sum _{m=0}^l{\left(1\right)}_{m,\lambda }{S}_1(l,m)\right)\frac{t^l}{l!}\right)\\ =t\sum _{n=0}^{\infty }\left(C{h}_{n,\lambda }^{\left(k\right)}+\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-l,\lambda }^{\left(k\right)}\right)\frac{t^n}{n!}\\ =\sum _{n=1}^{\infty }n\left(C{h}_{n-1,\lambda }^{\left(k\right)}+\sum _{l=0}^{n-1}\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-1-l,\lambda }^{\left(k\right)}\right)\frac{t^n}{n!}.\end{array}$$

(31)

On the other hand,

$$\begin{array}{c}2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)=2\sum _{m=1}^{\infty }\frac{{\left(1\right)}_{m,\lambda }{\left({log}_{\lambda }(1+t)\right)}^m}{(m-1)!m^k}\\ =2\sum _{m=1}^{\infty }\frac{{\left(1\right)}_{m,\lambda }{\left({log}_{\lambda }(1+t)\right)}^m}{(m-1)!m^k}\frac{m!}{m!}\\ =2\sum _{m=1}^{\infty }\frac{{\left(1\right)}_{m,\lambda }}{m^{k-1}}\sum _{n=m}^{\infty }{S}_{1,\lambda }(n,m)\frac{t^n}{n!}\\ L.H.S=2\sum _{n=1}^{\infty }\left(\sum _{m=1}^n\frac{{\left(1\right)}_{m,\lambda }{S}_{1,\lambda }(n,m)}{m^{k-1}}\right)\frac{t^n}{n!}.\end{array}$$

(32)

Therefore, using (31) and (32), we obtain the following theorem.

Theorem 5.

Let $k\ge 1$ and $m\in \mathbb{N}\bigcup \left\{0\right\}$, $s\in \mathbb{C}$; we have

$$\sum _{m=1}^n\frac{{\left(1\right)}_{m,\lambda }{S}_{1,\lambda }(n,m)}{m^{k-1}}$$

$$=\frac{n}{2}\left(C{h}_{n-1,\lambda }^{\left(k\right)}+\sum _{l=0}^{n-1}\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-1-l,\lambda }^{\left(k\right)}\right).$$

For $k=1$ in Theorem 5, we get the following corollary.

Corollary 3.

For $m\in \mathbb{N}\bigcup \left\{0\right\}$, $s\in \mathbb{C}$, we have

$$\sum _{m=1}^n{\left(1\right)}_{m,\lambda }{S}_{1,\lambda }(n,m)$$

$$=\frac{n}{2}\left(C{h}_{n-1,\lambda }+\sum _{l=0}^{n-1}\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-1-l,\lambda }\right).$$

From (16), we note that

$$\begin{array}{c}\sum _{n=0}^{\infty }\left(C{h}_{n,\lambda }^{\left(k\right)}(x+1)+C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\right)\frac{t^n}{n!}\\ =\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\left({(1+\lambda log(1+t))}^{\frac{1}{\lambda }}\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}\right)\left(\sum _{m=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\frac{1}{\lambda }}{m}\right){\lambda }^m{(log(1+t))}^m\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}\right)\left(\sum _{m=0}^{\infty }{\left(1\right)}_{m,\lambda }\sum _{l=m}^{\infty }{S}_1(l,m)\frac{t^l}{l!}\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}\right)\left(\sum _{l=0}^{\infty }\left(\sum _{m=0}^l{\left(1\right)}_{m,\lambda }{S}_1(l,m)\right)\frac{t^l}{l!}\right)\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-l,\lambda }^{\left(k\right)}\left(x\right)\right)\frac{t^n}{n!}.\end{array}$$

(33)

By comparing the coefficients on both sides of (33), we get the following theorem.

Theorem 6.

Let $k\in \mathbb{Z}$ and $n\ge 0$; we have

$$C{h}_{n,\lambda }^{\left(k\right)}(x+1)+C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)=\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right){\left(1\right)}_{m,\lambda }{S}_1(l,m)C{h}_{n-l,\lambda }^{\left(k\right)}\left(x\right).$$

3. Type-Two Degenerate Unipoly-Changhee Polynomials of the Second Kind

The unipoly function $u_k\left(x\right|p)$ is defined by Kim and Kim to be (see [20]):

$$u_k\left(x\right|p)=\sum _{n=1}^{\infty }\frac{p\left(n\right)}{n^k}{x}^n,(k\in \mathbb{Z}),$$

(34)

where p is any arithmetic function that is a real or complex valued function defined on the set of positive integers $\mathbb{N}$.

Moreover,

$$u_k\left(x\right|1)=\sum _{n=1}^{\infty }\frac{x^n}{n^k}={\mathrm{Li}}_k\left(x\right)$$

(35)

(see [22,23,28]) is the ordinary polylogarithm function.

In this paper, we consider the degenerate unipoly function attached to polynomials $p\left(x\right)$ as follows:

$$u_{k,\lambda }\left(x\right|p)=\sum _{i=1}^{\infty }p\left(i\right)\frac{{\left(1\right)}_{i,\lambda }{x}^i}{i^k}.$$

(36)

It is worth noting that

$$u_{k,\lambda }\left(x|\frac{1}{\mathsf{\Gamma }}\right)={\mathrm{Ei}}_{k,\lambda }\left(x\right)$$

(37)

is the modified degenerate polyexponential function.

By using (36), we define type-two degenerate unipoly-Changhee polynomials of the second kind by

$$\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}=\sum _{n=0}^{\infty }C{h}_{n,\lambda ,p}^{\left(k\right)}\left(x\right)\frac{t^n}{n!}.$$

(38)

In the case when $x=0,C{h}_{n,\lambda ,p}^{\left(k\right)}\left(0\right)=C{h}_{n,\lambda ,p}^{\left(k\right)}$ are called type-two degenerate unipoly-Changhee numbers of the second kind. Let us take $p\left(n\right)=\frac{1}{\mathsf{\Gamma }\lambda }$. Then, we have

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda ,\frac{1}{\mathsf{\Gamma }}}^{\left(k\right)}\left(x\right)\frac{t^n}{n!}=\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|\frac{1}{\mathsf{\Gamma }}p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\\ =\frac{2}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=1}^{\infty }\frac{{\left({log}_{\lambda }(1+t)\right)}^m}{m^k(m+1)!}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\\ =\frac{2{\mathrm{Ei}}_{k,\lambda }\left({log}_{\lambda }(1+t)\right)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\\ =\sum _{n=0}^{\infty }C{h}_{n,\lambda }^{\left(k\right)}\left(x\right)\frac{t^n}{n!}.\end{array}$$

(39)

Thus, using (39), we have the following theorem.

Theorem 7.

Let $n\ge 0$ and $k\in \mathbb{Z}$, and let $\mathsf{\Gamma }n$ be a Gamma function. Then, we have

$$C{h}_{n,\lambda ,\frac{1}{\mathsf{\Gamma }}}^{\left(k\right)}\left(x\right)=C{h}_{n,\lambda }^{\left(k\right)}\left(x\right).$$

(40)

From (38), we get

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda ,p}^{\left(k\right)}\frac{t^n}{n!}=\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\\ =\frac{2}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=1}^{\infty }\frac{p\left(m\right){\left(1\right)}_{m,\lambda }}{m^k}{\left({log}_{\lambda }(1+t)\right)}^m\\ =\frac{2}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=0}^{\infty }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }(m+1)!}{{(m+1)}^k}\sum _{l=m+1}^{\infty }{S}_{1,\lambda }(m+1,l)\frac{t^l}{l!}\\ =\left(\sum _{j=0}^{\infty }C{h}_{j,\lambda }\frac{t^j}{j!}\right)\left(\sum _{m=0}^{\infty }\sum _{l=0}^m\frac{p(m+1){\left(1\right)}_{m+1,\lambda }(m+1)!}{{(m+1)}^k}\frac{S_{1,\lambda }(m+1,l+1)}{l+1}\frac{t^l}{l!}\right)\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^{\infty }\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)\frac{p(m+1){\left(1\right)}_{m+1,\lambda }(m+1)!S_{1,\lambda }(m+1,l+1)C{h}_{n-l,\lambda }}{{(m+1)}^k(l+1)}\right)\frac{t^n}{n!}.\end{array}$$

(41)

Therefore, by comparing the coefficients on both sides of (41), we obtain the following theorem.

Theorem 8.

Let $n\in \mathbb{N}$ and $k\in \mathbb{Z}$. Then, we have

$$C{h}_{n,\lambda ,p}^{\left(k\right)}=\sum _{l=0}^{\infty }\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)\frac{p(m+1){\left(1\right)}_{m+1,\lambda }(m+1)!S_{1,\lambda }(m+1,l+1)C{h}_{n-l,\lambda }}{{(m+1)}^k(l+1)}.$$

(42)

In particular,

$$C{h}_{n,\lambda ,\frac{1}{\mathsf{\Gamma }}}^{\left(k\right)}=C{h}_{n,\lambda }^{\left(k\right)}=\sum _{l=0}^{\infty }\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)\frac{S_{1,\lambda }(m+1,l+1)C{h}_{n-l,\lambda }}{{(m+1)}^{k-1}(l+1)}.$$

(43)

From (38), we observe that

$$\begin{array}{c}\sum _{n=0}^{\infty }{C}_{n,\lambda }^{(k,p)}\left(x\right)\frac{t^n}{n!}=\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}{(1+\lambda log(1+t))}^{\frac{x}{\lambda }}\\ =\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\frac{x}{\lambda }}{m}\right){\lambda }^m{\left(log(1+\lambda t)\right)}^m\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda ,p}^{\left(k\right)}\frac{t^n}{n!}\right)\left(\sum _{m=0}^{\infty }{\left(x\right)}_{m,\lambda }\sum _{l=m}^{\infty }{S}_1(l,m)\frac{t^l}{l!}\right)\\ =\left(\sum _{n=0}^{\infty }C{h}_{n,\lambda ,p}^{\left(k\right)}\frac{t^n}{n!}\right)\left(\sum _{l=0}^{\infty }\sum _{m=0}^l{\left(x\right)}_{m,\lambda }{S}_1(l,m)\frac{t^l}{l!}\right)\\ L.H.S=\sum _{n=0}^{\infty }\left(\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda ,p}^{\left(k\right)}{\left(x\right)}_{m,\lambda }{S}_1(l,m)\right)\frac{t^n}{n!}.\end{array}$$

(44)

From (44), we obtain the following theorem.

Theorem 9.

Let $n\ge 0$ and $k\in \mathbb{Z}$. Then, we have

$$C{h}_{n,\lambda ,p}^{\left(k\right)}\left(x\right)=\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)C{h}_{n-l,\lambda ,p}^{\left(k\right)}{\left(x\right)}_{m,\lambda }{S}_1(l,m).$$

(45)

From (38), we observe that

$$\begin{array}{c}\sum _{n=0}^{\infty }C{h}_{n,\lambda ,p}^{\left(k\right)}\frac{t^n}{n!}=\frac{2u_{k,\lambda }\left({log}_{\lambda }(1+t)\right|p)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\\ =\frac{2}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=0}^{\infty }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^{k}m!}{\left({log}_{\lambda }(1+t)\right)}^{m+1}\\ =\frac{2{log}_{\lambda }(1+t)}{t(1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }})}\sum _{m=0}^{\infty }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^{k}m!}{\left({log}_{\lambda }(1+t)\right)}^m\\ =\frac{{log}_{\lambda }(1+t)}{t}\frac{2}{1+{(1+\lambda log(1+t))}^{\frac{1}{\lambda }}}\sum _{m=0}^{\infty }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^k}\sum _{l=m}^{\infty }{S}_{1,\lambda }(l,m)\frac{t^l}{l!}\\ =\left(\sum _{s=0}^{\infty }{D}_{s,\lambda }\frac{t^s}{s!}\right)\left(\sum _{a=0}^{\infty }C{h}_{a,\lambda }\frac{t^a}{a!}\right)\left(\sum _{l=0}^{\infty }\sum _{m=0}^n\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^k}{S}_{1,\lambda }(l,m)\frac{t^l}{l!}\right)\\ =\left(\sum _{b=0}^{\infty }\sum _{a=0}^b\left(\genfrac{}{}{0pt}{}{b}{a}\right)D_{b-a,\lambda }C{h}_{a,\lambda }\frac{t^b}{b!}\right)\left(\sum _{l=0}^{\infty }\sum _{m=0}^n\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^k}{S}_{1,\lambda }(l,m)\frac{t^l}{l!}\right)\\ L.H.S=\sum _{n=0}^{\infty }\left(\sum _{l=0}^n\sum _{a=0}^{n-l}\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)D_{n-l-a,\lambda }{C}_{a,\lambda }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^k}{S}_{1,\lambda }(l,m)\right)\frac{t^n}{n!}.\end{array}$$

(46)

By comparing the coefficients on both sides of (46), we obtain the following theorem.

Theorem 10.

Let $n\ge 0$ and $k\in \mathbb{Z}$. Then, we have

$$C{h}_{n,\lambda ,p}^{\left(k\right)}=\sum _{l=0}^n\sum _{a=0}^{n-l}\sum _{m=0}^l\left(\genfrac{}{}{0pt}{}{n}{l}\right)D_{n-l-a,\lambda }C{h}_{a,\lambda }\frac{p(m+1){\left(1\right)}_{m+1,\lambda }m!}{{(m+1)}^k}{S}_{1,\lambda }(l,m).$$

(47)

4. Conclusions

In this article, we introduced type-two degenerate poly-Changhee polynomials of the second kind and derived some beautiful identities and relations between type-two degenerate poly-Changhee numbers of the second kind and Stirling numbers of first and second kind. In addition, we gave the relation between degenerate Bernoulli polynomials of the second kind and type-two degenerate poly-Changhee numbers of the second kind. Again, we defined type-two degenerate unipoly-Changhee polynomials of the the second kind and obtained some properties and relationships of degenerate unipoly-Changhee numbers of the second kind and the Daehee numbers.