Some Symmetric Identities for Degenerate Carlitz-type (p, q)-Euler Numbers and Polynomials

Abstract

In this paper we define the degenerate Carlitz-type $(p,q)$-Euler polynomials by generalizing the degenerate Euler numbers and polynomials, degenerate Carlitz-type q-Euler numbers and polynomials. We also give some theorems and exact formulas, which have a connection to degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials.

1. Introduction

Many researchers have studied about the degenerate Bernoulli numbers and polynomials, degenerate Euler numbers and polynomials, degenerate Genocchi numbers and polynomials, degenerate tangent numbers and polynomials (see [1,2,3,4,5,6,7]). Recently, some generalizations of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, tangent numbers and polynomials are provided (see [6,8,9,10,11,12,13]). In this paper we define the degenerate Carlitz-type $(p,q)$-Euler polynomials and numbers and study some theories of the degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials.

Throughout this paper, we use the notations below: $\mathbb{N}$ denotes the set of natural numbers, $\mathbb{Z}$ denotes the set of integers, ${\mathbb{Z}}_{+}=\mathbb{N}\cup \left\{0\right\}$ denotes the set of nonnegative integers. We remind that the classical degenerate Euler numbers ${\mathcal{E}}_n\left(\lambda \right)$ and Euler polynomials ${\mathcal{E}}_n(x,\lambda )$, which are defined by generating functions like $\left(1\right)$, and $\left(2\right)$ (see [1,2])

$$\frac{2}{{(1+\lambda t)}^{\frac{1}{\lambda }}+1}={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_n\left(\lambda \right)\frac{t^n}{n!},$$

(1)

and

$$\frac{2}{{(1+\lambda t)}^{\frac{1}{\lambda }}+1}{(1+\lambda t)}^{\frac{x}{\lambda }}={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_n(x,\lambda )\frac{t^n}{n!},$$

(2)

respectively.

Carlitz [1] introduced some theories of the degenerate Euler numbers and polynomials. We recall that well-known Stirling numbers of the first kind $S_1(n,k)$ and the second kind $S_2(n,k)$ are defined by this (see [2,7,14])

$${\left(x\right)}_n={\displaystyle \sum _{k=0}^n}{S}_1(n,k)x^k\mathrm{and}{x}^n={\displaystyle \sum _{k=0}^n}{S}_2(n,k){\left(x\right)}_k,$$

respectively. Here ${\left(x\right)}_n=x(x-1)\cdots (x-n+1)$. The numbers $S_2(n,m)$ is like this

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

We also have

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

The generalized falling factorial ${\left(x\right|\lambda )}_n$ with increment $\lambda$ is defined by

$${\left(x\right|\lambda )}_n={\displaystyle \prod _{k=0}^{n-1}}(x-\lambda k)$$

for positive integer n, with ${\left(x\right|\lambda )}_0=1$; as we know,

$${\left(x\right|\lambda )}_n={\displaystyle \sum _{k=0}^n}{S}_1(n,k){\lambda }^{n-k}{x}^k.$$

${\left(x\right|\lambda )}_n={\lambda }^n{\left({\lambda }^{-1}x\right|1)}_n$ for $\lambda \ne 0$. Clearly ${\left(x\right|0)}_n=x^n$. The binomial theorem for a variable x is

$${(1+\lambda t)}^{x/\lambda }={\displaystyle \sum _{n=0}^{\infty }}{\left(x\right|\lambda )}_n{\displaystyle \frac{t^n}{n!}}.$$

The $(p,q)$-number is defined as

$${\left[n\right]}_{p,q}=\frac{p^n-q^n}{p-q}=p^{n-1}+p^{n-2}q+p^{n-3}{q}^2+\cdots +p^2{q}^{n-3}+p{q}^{n-2}+q^{n-1}.$$

We begin by reminding the Carlitz-type $(p,q)$-Euler numbers and polynomials (see [9,10,11]).

Definition 1.

For $0<q<p\le 1$ and $h\in \mathbb{Z}$, the Carlitz-type $(p,q)$-Euler polynomials $E_{n,p,q}\left(x\right)$ and $(h,p,q)$-Euler polynomials $E_{n,p,q}^{\left(h\right)}\left(x\right)$ are defined like this

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{E}_{n,p,q}\left(x\right)\frac{t^n}{n!}={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{e}^{{[m+x]}_{p,q}t},\hfill \\ {\displaystyle \sum _{n=0}^{\infty }}{E}_{n,p,q}^{\left(h\right)}\left(x\right)\frac{t^n}{n!}={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{p}^{hm}{e}^{{[m+x]}_{p,q}t},\hfill \end{array}$$

(3)

respectively (see [9,10,11]).

Now we make the degenerate Carlitz-type $(p,q)$-Euler number ${\mathcal{E}}_{n,p,q}\left(\lambda \right)$ and $(p,q)$-Euler polynomials ${\mathcal{E}}_{n,p,q}(x,\lambda )$. In the next section, we introduce the degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials. We will study some their properties after introduction.

2. Degenerate Carlitz-Type $(p,q)$-Euler Polynomials

In this section, we define the degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials and make some of their properties.

Definition 2.

For $0<q<p\le 1$, the degenerate Carlitz-type $(p,q)$-Euler numbers ${\mathcal{E}}_{n,p,q}\left(\lambda \right)$ and polynomials ${\mathcal{E}}_{n,p,q}(x,\lambda )$ are related to the generating functions

$$F_{p,q}(t,\lambda )={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q}\left(\lambda \right)\frac{t^n}{n!}={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{(1+\lambda t)}^{{\displaystyle \frac{{\left[m\right]}_{p,q}}{\lambda }}},$$

(4)

and

$$F_{p,q}(t,x,\lambda )={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q}(x,\lambda )\frac{t^n}{n!}={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{(1+\lambda t)}^{{\displaystyle \frac{{[m+x]}_{p,q}}{\lambda }}},$$

(5)

respectively.

Let $p=1$ in (4) and (5), we can get the degenerate Carlitz-type q-Euler number ${\mathcal{E}}_{n,q}(x,\lambda )$ and q-Euler polynomials ${\mathcal{E}}_{n,q}(x,\lambda )$ respectively. Obviously, if $p=1$, then we have

$${\mathcal{E}}_{n,p,q}(x,\lambda )={\mathcal{E}}_{n,q}(x,\lambda ),{\mathcal{E}}_{n,p,q}\left(\lambda \right)={\mathcal{E}}_{n,q}\left(\lambda \right).$$

When $p=1$, we have

$$\underset{q\to 1}{lim}{\mathcal{E}}_{n,p,q}(x,\lambda )={\mathcal{E}}_n(x,\lambda ),\underset{q\to 1}{lim}{\mathcal{E}}_{n,p,q}\left(\lambda \right)={\mathcal{E}}_n\left(\lambda \right).$$

We see that

$$\begin{array}{cc}\hfill {(1+\lambda t)}^{{\displaystyle \frac{{[x+y]}_{p,q}}{\lambda }}}& =e^{{\displaystyle \frac{{[x+y]}_{p,q}}{\lambda }}log(1+\lambda t)}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}{\left({\displaystyle \frac{{[x+y]}_{p,q}}{\lambda }}\right)}^n{\displaystyle \frac{{(log(1+\lambda t))}^n}{n!}}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \sum _{m=0}^n}{S}_1(n,m){\lambda }^{n-m}{[x+y]}_{p,q}^m\right){\displaystyle \frac{t^n}{n!}}.\hfill \end{array}$$

(6)

By (5), it follows that

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q}(x,\lambda )\frac{t^n}{n!}\hfill \\ ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{(1+\lambda t)}^{{\displaystyle \frac{{[m+x]}_{p,q}}{\lambda }}}\hfill \\ ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m\hfill \\ \times {\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{\displaystyle \frac{{\sum }_{j=0}^l\left(\genfrac{}{}{0pt}{}{l}{j}\right){(-1)}^j{p}^{(x+m)(l-j)}{q}^{(x+m)j}}{{(p-q)}^l}}\frac{t^n}{n!}\hfill \\ ={\displaystyle \sum _{n=0}^{\infty }}\left({\left[2\right]}_q{\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{j=0}^l}{\displaystyle \frac{S_1(n,l){\lambda }^{n-l}\left(\genfrac{}{}{0pt}{}{l}{j}\right){(-1)}^j{q}^{xj}{p}^{x(l-j)}}{{(p-q)}^l}}{\displaystyle \frac{1}{1+q^{j+1}{p}^{l-j}}}\right)\frac{t^n}{n!}\hfill \end{array}.$$

(7)

By comparing the coefficients $\frac{t^n}{n!}$ in the above equation, we have the following theorem.

Theorem 1.

For $0<q<p\le 1$ and $n\in Z_{+}$, we have

$$\begin{array}{cc}\hfill {\mathcal{E}}_{n,p,q}(x,\lambda )& ={\left[2\right]}_q{\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{j=0}^l}{\displaystyle \frac{S_1(n,l){\lambda }^{n-l}\left(\genfrac{}{}{0pt}{}{l}{j}\right){(-1)}^j{q}^{xj}{p}^{x(l-j)}}{{(p-q)}^l}}{\displaystyle \frac{1}{1+q^{j+1}{p}^{l-j}}}\hfill \\ & ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{(-1)}^m{q}^m{[x+m]}_{p,q}^l,\hfill \\ \hfill {\mathcal{E}}_{n,p,q}\left(\lambda \right)& ={\left[2\right]}_q{\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{j=0}^l}{\displaystyle \frac{S_1(n,l){\lambda }^{n-l}\left(\genfrac{}{}{0pt}{}{l}{j}\right){(-1)}^j}{{(p-q)}^l}}{\displaystyle \frac{1}{1+q^{j+1}{p}^{l-j}}}\hfill \\ & ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{(-1)}^m{q}^m{\left[m\right]}_{p,q}^l.\hfill \end{array}$$

We make the degenerate Carlitz-type $(p,q)$-Euler number ${\mathcal{E}}_{n,p,q}\left(\lambda \right)$. Some cases are

$$\begin{array}{cc}\hfill {\mathcal{E}}_{0,p,q}\left(\lambda \right)=& 1,\hfill \\ \hfill {\mathcal{E}}_{1,p,q}\left(\lambda \right)=& {\displaystyle \frac{{\left[2\right]}_q}{(p-q)(1+pq)}}-{\displaystyle \frac{{\left[2\right]}_q}{(p-q)(1+q^2)}},\hfill \\ \hfill {\mathcal{E}}_{2,p,q}\left(\lambda \right)=& -{\displaystyle \frac{{\left[2\right]}_q\lambda }{(p-q)(1+pq)}}+{\displaystyle \frac{{\left[2\right]}_q}{{(p-q)}^2(1+p^{2}q)}}+{\displaystyle \frac{{\left[2\right]}_q\lambda }{(p-q)(1+q^2)}}\hfill \\ & -{\displaystyle \frac{2{\left[2\right]}_q}{{(p-q)}^2(1+p{q}^2)}}+{\displaystyle \frac{{\left[2\right]}_q}{{(p-q)}^2(1+q^3)}},\hfill \\ \hfill {\mathcal{E}}_{3,p,q}\left(\lambda \right)=& {\displaystyle \frac{2{\left[2\right]}_q{\lambda }^2}{(p-q)(1+pq)}}-{\displaystyle \frac{3{\left[2\right]}_q\lambda }{{(p-q)}^2(1+p^{2}q)}}+{\displaystyle \frac{{\left[2\right]}_q}{{(p-q)}^3(1+p^{3}q)}}\hfill \\ & -{\displaystyle \frac{2{\left[2\right]}_q{\lambda }^2}{(p-q)(1+q^2)}}+{\displaystyle \frac{6{\left[2\right]}_q\lambda }{{(p-q)}^2(1+p{q}^2)}}-{\displaystyle \frac{3{\left[2\right]}_q}{{(p-q)}^3(1+p^2{q}^2)}}\hfill \\ & -{\displaystyle \frac{3{\left[2\right]}_q\lambda }{{(p-q)}^2(1+q^3)}}+{\displaystyle \frac{3{\left[2\right]}_q}{{(p-q)}^3(1+p{q}^3)}}-{\displaystyle \frac{{\left[2\right]}_q}{{(p-q)}^3(1+q^4)}}.\hfill \end{array}$$

We use t instead of ${\displaystyle \frac{e^{\lambda t}-1}{\lambda }}$ in (5), we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{m=0}^{\infty }}{E}_{m,p,q}\left(x\right){\displaystyle \frac{t^m}{m!}}& ={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q}(x,\lambda ){\left({\displaystyle \frac{e^{\lambda t}-1}{\lambda }}\right)}^n{\displaystyle \frac{1}{n!}}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q}(x,\lambda ){\lambda }^{-n}{\displaystyle \sum _{m=n}^{\infty }}{S}_2(m,n){\lambda }^m{\displaystyle \frac{t^m}{m!}}\hfill \\ & ={\displaystyle \sum _{m=0}^{\infty }}\left({\displaystyle \sum _{n=0}^m}{\mathcal{E}}_{n,p,q}(x,\lambda ){\lambda }^{m-n}{S}_2(m,n)\right){\displaystyle \frac{t^m}{m!}}.\hfill \end{array}$$

(8)

Thus we have the following theorem.

Theorem 2.

For $m\in Z_{+}$, we have

$$E_{m,p,q}\left(x\right)={\displaystyle \sum _{n=0}^m}{\mathcal{E}}_{n,p,q}(x,\lambda ){\lambda }^{m-n}{S}_2(m,n).$$

Use t instead of $log{(1+\lambda t)}^{1/\lambda }$ in (3), we have

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{E}_{n,p,q}\left(x\right){\left(log{(1+\lambda t)}^{1/\lambda }\right)}^n{\displaystyle \frac{1}{n!}}\hfill \\ ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{(1+\lambda t)}^{{\displaystyle \frac{{[m+x]}_{p,q}}{\lambda }}}\hfill \\ ={\displaystyle \sum _{m=0}^{\infty }}{\mathcal{E}}_{m,p,q}(x,\lambda ){\displaystyle \frac{t^m}{m!}},\hfill \end{array}$$

(9)

and

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{E}_{n,p,q}\left(x\right){\left(log{(1+\lambda t)}^{1/\lambda }\right)}^n{\displaystyle \frac{1}{n!}}\hfill \\ ={\displaystyle \sum _{m=0}^{\infty }}\left({\displaystyle \sum _{n=0}^m}{E}_{n,p,q}\left(x\right){\lambda }^{m-n}{S}_1(m,n)\right){\displaystyle \frac{t^m}{m!}}.\hfill \end{array}$$

(10)

Thus we have the below theorem from $\left(9\right)$ and $\left(10\right)$.

Theorem 3.

For $m\in Z_{+}$, we have

$${\mathcal{E}}_{m,p,q}(x,\lambda )={\displaystyle \sum _{n=0}^m}{E}_{n,p,q}\left(x\right){\lambda }^{m-n}{S}_1(m,n).$$

We have the degenerate Carlitz-type $(p,q)$-Euler polynomials ${\mathcal{E}}_{n,p,q}(x,\lambda )$. some cases are

$$\begin{array}{cc}\hfill {\mathcal{E}}_{0,p,q}(x,\lambda )& =1,\hfill \\ \hfill {\mathcal{E}}_{1,p,q}(x,\lambda )& ={\displaystyle \frac{{\left[2\right]}_q{p}^x}{(p-q)(1+pq)}}-{\displaystyle \frac{{\left[2\right]}_q{q}^x}{(p-q)(1+q^2)}},\hfill \\ \hfill {\mathcal{E}}_{2,p,q}(x,\lambda )& =-{\displaystyle \frac{{\left[2\right]}_q\lambda p^x}{(p-q)(1+pq)}}+{\displaystyle \frac{{\left[2\right]}_q{p}^{2x}}{{(p-q)}^2(1+p^{2}q)}}+{\displaystyle \frac{{\left[2\right]}_q\lambda q^x}{(p-q)(1+q^2)}}\hfill \\ & -{\displaystyle \frac{2{\left[2\right]}_q{p}^x{q}^x}{{(p-q)}^2(1+p{q}^2)}}+{\displaystyle \frac{{\left[2\right]}_q{q}^{2x}}{{(p-q)}^2(1+q^3)}},\hfill \\ \hfill {\mathcal{E}}_{3,p,q}(x,\lambda )& ={\displaystyle \frac{2{\left[2\right]}_q{\lambda }^2{p}^x}{(p-q)(1+pq)}}-{\displaystyle \frac{3{\left[2\right]}_q\lambda p^{2x}}{{(p-q)}^2(1+p^{2}q)}}+{\displaystyle \frac{{\left[2\right]}_q{p}^{3x}}{{(p-q)}^3(1+p^{3}q)}}\hfill \\ & -{\displaystyle \frac{2{\left[2\right]}_q{\lambda }^2{q}^x}{(p-q)(1+q^2)}}+{\displaystyle \frac{6{\left[2\right]}_q\lambda p^x{q}^x}{{(p-q)}^2(1+p{q}^2)}}-{\displaystyle \frac{3{\left[2\right]}_q{p}^{2x}{q}^x}{{(p-q)}^3(1+p^2{q}^2)}}\hfill \\ & -{\displaystyle \frac{3{\left[2\right]}_q\lambda q^{2x}}{{(p-q)}^2(1+q^3)}}+{\displaystyle \frac{3{\left[2\right]}_q{p}^x{q}^{2x}}{{(p-q)}^3(1+p{q}^3)}}-{\displaystyle \frac{{\left[2\right]}_q{q}^{3x}}{{(p-q)}^3(1+q^4)}}.\hfill \end{array}$$

We introduce a $(p,q)$-analogue of the generalized falling factorial ${\left(x\right|\lambda )}_n$ with increment $\lambda$. The generalized $(p,q)$-falling factorial ${\left({\left[x\right]}_{p,q}\right|\lambda )}_n$ with increment $\lambda$ is defined by

$${\left({\left[x\right]}_{p,q}\right|\lambda )}_n={\displaystyle \prod _{k=0}^{n-1}}({\left[x\right]}_{p,q}-\lambda k)$$

for positive integer n, where ${\left({\left[x\right]}_{p,q}\right|\lambda )}_0=1$.

By (4) and (5), we get

$$\begin{array}{l}-{\left[2\right]}_q{(-1)}^n{q}^n{\displaystyle \sum _{l=0}^{\infty }}{(-1)}^l{q}^l{(1+\lambda t)}^{{\displaystyle \frac{{[l+n]}_{p,q}}{\lambda }}}\\ +{\left[2\right]}_q{\displaystyle \sum _{l=0}^{\infty }}{(-1)}^l{q}^l{(1+\lambda t)}^{{\displaystyle \frac{{[l+n]}_{p,q}}{\lambda }}}\\ ={\left[2\right]}_q{\displaystyle \sum _{l=0}^{n-1}}{(-1)}^l{q}^l{(1+\lambda t)}^{{\displaystyle \frac{{\left[l\right]}_{p,q}}{\lambda }}}.\end{array}$$

Hence we have

$$\begin{array}{l}{(-1)}^{n+1}{q}^n{\displaystyle \sum _{m=0}^{\infty }}{\mathcal{E}}_{m,p,q}(n,\lambda ){\displaystyle \frac{t^m}{m!}}+{\displaystyle \sum _{m=0}^{\infty }}{\mathcal{E}}_{m,p,q}\left(\lambda \right){\displaystyle \frac{t^m}{m!}}\\ ={\displaystyle \sum _{m=0}^{\infty }}\left({\left[2\right]}_q{\displaystyle \sum _{l=0}^{n-1}}{(-1)}^l{q}^l{\left({\left[l\right]}_{p,q}\right|\lambda )}_m\right){\displaystyle \frac{t^m}{m!}}.\end{array}$$

(11)

By comparing the coefficients of $\frac{t^m}{m!}$ on both sides of (11), we have the following theorem.

Theorem 4.

For $n\in Z_{+}$, we have

$${\displaystyle \sum _{l=0}^{n-1}}{(-1)}^l{q}^l{\left({\left[l\right]}_{p,q}\right|\lambda )}_m={\displaystyle \frac{{(-1)}^{n+1}{q}^n{\mathcal{E}}_{m,p,q}(n,\lambda )+{\mathcal{E}}_{m,p,q}\left(\lambda \right)}{{\left[2\right]}_q}}.$$

We get that

$$\begin{array}{l}{\left(1+\lambda t\right)}^{{\displaystyle \frac{{[x+y]}_{p,q}}{\lambda }}}\\ ={(1+\lambda t)}^{{\displaystyle \frac{p^y{\left[x\right]}_{p,q}}{\lambda }}}{(1+\lambda t)}^{{\displaystyle \frac{q^x{\left[y\right]}_{p,q}}{\lambda }}}\\ ={\displaystyle \sum _{m=0}^{\infty }}{\left(p^y{\left[x\right]}_{p,q}\right|\lambda )}_m{\displaystyle \frac{t^m}{m!}}{e}^{log{(1+\lambda t)}^{{\displaystyle \frac{q^x{\left[y\right]}_{p,q}}{\lambda }}}}\\ ={\displaystyle \sum _{m=0}^{\infty }}{\left(p^y{\left[x\right]}_{p,q}\right|\lambda )}_m{\displaystyle \frac{t^m}{m!}}{\displaystyle \sum _{l=0}^{\infty }}{\left({\displaystyle \frac{q^x{\left[y\right]}_{p,q}}{\lambda }}\right)}^l{\displaystyle \frac{log{(1+\lambda t)}^l}{l!}}\\ ={\displaystyle \sum _{m=0}^{\infty }}{\left(p^y{\left[x\right]}_{p,q}\right|\lambda )}_m{\displaystyle \frac{t^m}{m!}}{\displaystyle \sum _{l=0}^{\infty }}{\left({\displaystyle \frac{q^x{\left[y\right]}_{p,q}}{\lambda }}\right)}^l{\displaystyle \sum _{k=l}^{\infty }}{S}_1(k,l){\lambda }^k{\displaystyle \frac{t^k}{k!}}\\ ={\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \sum _{k=0}^n}{\displaystyle \sum _{l=0}^k}\left(\genfrac{}{}{0pt}{}{n}{k}\right){\left(p^y{\left[x\right]}_{p,q}\right|\lambda )}_{n-k}{\lambda }^{k-l}{q}^{xl}{\left[y\right]}_{p,q}^l{S}_1(k,l)\right){\displaystyle \frac{t^n}{n!}}.\end{array}$$

(12)

By (5) and (12), we get

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p,q\zeta }(x,\lambda )\frac{t^n}{n!}\\ ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{(1+\lambda t)}^{{\displaystyle \frac{{[m+x]}_{p,q}}{\lambda }}}\\ ={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{(-1)}^m{q}^m{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \sum _{k=0}^n}{\displaystyle \sum _{l=0}^k}\left(\genfrac{}{}{0pt}{}{n}{k}\right){\left(p^m{\left[x\right]}_{p,q}\right|\lambda )}_{n-k}{\lambda }^{k-l}{q}^{xl}{\left[m\right]}_{p,q}^l{S}_1(k,l)\right){\displaystyle \frac{t^n}{n!}}\\ ={\displaystyle \sum _{n=0}^{\infty }}\left({\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{k=0}^n}{\displaystyle \sum _{l=0}^k}\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^m{q}^m{\left(p^m{\left[x\right]}_{p,q}\right|\lambda )}_{n-k}{\lambda }^{k-l}{q}^{xl}{S}_1(k,l)\right){\displaystyle \frac{t^n}{n!}}.\end{array}$$

By comparing the coefficients of $\frac{t^n}{n!}$ in the above equation, we have the theorem below.

Theorem 5.

For $0<q<p\le 1$ and $n\in Z_{+}$, we have

$${\mathcal{E}}_{n,p,q}(x,\lambda )={\left[2\right]}_q{\displaystyle \sum _{m=0}^{\infty }}{\displaystyle \sum _{k=0}^n}{\displaystyle \sum _{l=0}^k}\left(\genfrac{}{}{0pt}{}{n}{k}\right){(-1)}^m{q}^m{\left(p^m{\left[x\right]}_{p,q}\right|\lambda )}_{n-k}{\lambda }^{k-l}{q}^{xl}{S}_1(k,l).$$

3. Symmetric Properties about Degenerate Carlitz-Type $(p,q)$-Euler Numbers and Polynomials

In this section, we are going to get the main results of degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials. We also make some symmetric identities for degenerate Carlitz-type $(p,q)$-Euler numbers and polynomials. Let $w_1$ and $w_2$ be odd positive integers. Remind that ${\left[xy\right]}_{p,q}={\left[x\right]}_{p^y,q^y}{\left[y\right]}_{p,q}$ for any $x,y\in \mathbb{C}$.

By using $w_{1}x+\frac{w_{1}i}{w_2}$ instead of x in Definition 2, use p by $p^{w_2}$, use q by $q^{w_2}$ and use $\lambda$ by ${\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}},$ respectively, we can get

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\right){\displaystyle \frac{t^n}{n!}}\\ ={\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{n=0}^{\infty }}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right){\displaystyle \frac{{\left({\left[w_2\right]}_{p,q}t\right)}^n}{n!}}\\ ={\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{q}^{w_{2}n}\\ \times {\left(1+{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}{\left[w_2\right]}_{p,q}t\right)}^{{\displaystyle \frac{{[w_{1}x+\frac{w_{1}i}{w_2}+n]}_{p^{w_2},q^{w_2}}}{\frac{\lambda }{{\left[w_2\right]}_{p,q}}}}}\\ ={\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{q}^{w_{2}n}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+n{w}_2]}_{p,q}}{\lambda }}}.\end{array}$$

Since for any non-negative integer n and odd positive integer $w_1$, there is the unique non-negative integer r such that $n=w_{1}r+j$ with $0\le j\le w_1-1$. So this can be written as

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{n=0}^{\infty }}{(-1)}^n{q}^{w_{2}n}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+n{w}_2]}_{p,q}}{\lambda }}}.\\ ={\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{\begin{array}{c}{w}_{1}r+j=0\\ 0\le j\le w_1-1\end{array}}^{\infty }}{(-1)}^{w_{1}r+j}{q}^{w_2(w_{1}r+j)}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+(w_{1}r+j)w_2]}_{p,q}}{\lambda }}}.\\ ={\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{j=0}^{w_1-1}}{\displaystyle \sum _{r=0}^{\infty }}{(-1)}^{w_{1}r}{(-1)}^j{q}^{w_2{w}_{1}r}{q}^{w_{2}j}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+w_1{w}_{2}r+w_{2}j]}_{p,q}}{\lambda }}}\\ ={\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_2-1}}{\displaystyle \sum _{j=0}^{w_1-1}}{\displaystyle \sum _{r=0}^{\infty }}{(-1)}^i{(-1)}^r{(-1)}^j{q}^{w_{1}i}{q}^{w_2{w}_{1}r}{q}^{w_{2}j}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+w_1{w}_{2}r+w_{2}j]}_{p,q}}{\lambda }}}.\end{array}$$

We have the below formula using the above formula

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\left[2\right]}_{q^{w_2}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\right){\displaystyle \frac{t^n}{n!}}\\ ={\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_2-1}}{\displaystyle \sum _{j=0}^{w_1-1}}{\displaystyle \sum _{r=0}^{\infty }}{(-1)}^i{(-1)}^r{(-1)}^j{q}^{w_{1}i}{q}^{w_2{w}_{1}r}{q}^{w_{2}j}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{1}i+w_1{w}_{2}r+w_{2}j]}_{p,q}}{\lambda }}}.\end{array}$$

(13)

From a similar approach, we can have that

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\left[2\right]}_{q^{w_2}}{\left[w_1\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_1-1}}{(-1)}^i{q}^{w_{2}i}{\mathcal{E}}_{n,p^{w_1},q^{w_1}}\left(w_{2}x+\frac{w_{2}i}{w_1},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_{p,q}}}\right)\right){\displaystyle \frac{t^n}{n!}}\\ ={\left[2\right]}_{q^{w_1}}{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{i=0}^{w_1-1}}{\displaystyle \sum _{j=0}^{w_2-1}}{\displaystyle \sum _{r=0}^{\infty }}{(-1)}^i{(-1)}^r{(-1)}^j{q}^{w_{2}i}{q}^{w_1{w}_{1}r}{q}^{w_{1}j}\\ \times {\left(1+\lambda t\right)}^{{\displaystyle \frac{{[w_1{w}_{2}x+w_{2}i+w_1{w}_{2}r+w_{1}j]}_{p,q}}{\lambda }}}.\end{array}$$

(14)

Thus, we have the following theorem from (13) and (14).

Theorem 6.

Let $w_1$ and $w_2$ be odd positive integers. Then one has

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\\ ={\left[2\right]}_{q^{w_2}}{\left[w_1\right]}_{p,q}^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^{w_{2}j}{\mathcal{E}}_{n,p^{w_1},q^{w_1}}\left(w_{2}x+\frac{w_{2}j}{w_1},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_{p,q}}}\right).\end{array}$$

Letting $\lambda \to 0$ in Theorem 6, we can immediately obtain the symmetric identities for Carlitz-type $(p,q)$-Euler polynomials (see [10])

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{E}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2}\right)\\ ={\left[2\right]}_{q^{w_2}}{\left[w_1\right]}_{p,q}^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^{w_{2}j}{E}_{n,p^{w_1},q^{w_1}}\left(w_{2}x+\frac{w_{2}j}{w_1}\right).\end{array}$$

It follows that we show some special cases of Theorem 6. Let $w_2=1$ in Theorem 6, we have the multiplication theorem for the degenerate Carlitz-type $(p,q)$-Euler polynomials.

Corollary 1.

Let $w_1$ be odd positive integer. Then

$${\mathcal{E}}_{n,p,q}(x,\lambda )={\displaystyle \frac{{\left[2\right]}_q{\left[w_1\right]}_{p,q}^n}{{\left[2\right]}_{q^{w_1}}}}{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^j{\mathcal{E}}_{n,p^{w_1},q^{w_1}}\left({\displaystyle \frac{x+j}{w_1}},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_{p,q}}}\right).$$

(15)

Let $p=1$ in (15). This leads to the multiplication theorem about the degenerate Carlitz-type q-Euler polynomials

$${\mathcal{E}}_{n,q}(x,\lambda )={\displaystyle \frac{{\left[2\right]}_q{\left[w_1\right]}_q^n}{{\left[2\right]}_{q^{w_1}}}}{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^j{\mathcal{E}}_{n,q^{w_1}}\left({\displaystyle \frac{x+j}{w_1}},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_q}}\right).$$

(16)

Giving $q\to 1$ in (16) induce to the multiplication theorem about the degenerate Euler polynomials

$${\mathcal{E}}_n(x,\lambda )=w_1^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{\mathcal{E}}_n\left({\displaystyle \frac{x+i}{w_1}},{\displaystyle \frac{\lambda }{w_1}}\right).$$

(17)

If $\lambda$ approaches to 0 in (17), this leads to the multiplication theorem about the Euler polynomials(see [15])

$$E_n\left(x\right)=w_1^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{E}_n\left({\displaystyle \frac{x+i}{w_1}}\right).$$

Let $x=0$ in Theorem 6, then we have the following corollary.

Corollary 2.

Let $w_1$ and $w_2$ be odd positive integers. Then it has

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\\ ={\left[2\right]}_{q^{w_2}}{\left[w_1\right]}_{p,q}^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^{w_{2}j}{\mathcal{E}}_{n,p^{w_1},q^{w_1}}\left(\frac{w_{2}j}{w_1},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_{p,q}}}\right).\end{array}$$

By Theorem 3 and Corollary 2, we have the below theorem.

Theorem 7.

Let $w_1$ and $w_2$ be odd positive integers. Then

$$\begin{array}{l}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{\left[w_2\right]}_{p,q}^l{\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{E}_{l,p^{w_2},q^{w_2}}\left(\frac{w_1}{w_2}i\right)\\ ={\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{\left[w_1\right]}_{p,q}^l{\left[2\right]}_{q^{w_2}}{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^{w_{2}j}{E}_{l,p^{w_1},q^{w_1}}\left(\frac{w_2}{w_1}j\right).\end{array}$$

We get another result by applying the addition theorem about the Carlitz-type $(p,q)$-Euler polynomials $E_{n,p,q}\left(x\right)$.

Theorem 8.

Let $w_1$ and $w_2$ be odd positive integers. Then we have

$$\begin{array}{l}{\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{k=0}^l}\left(\genfrac{}{}{0pt}{}{l}{k}\right)S_1(n,l){\lambda }^{n-l}{p}^{w_1{w}_{2}xk}{\left[2\right]}_{q^{w_1}}{\left[w_1\right]}_{p,q}^k{\left[w_2\right]}_{p,q}^{l-k}{E}_{l-k,p^{w_2},q^{w_2}}^{\left(k\right)}\left(w_{1}x\right)S_{l,k,p^{w_1},q^{w_1}}\left(w_2\right)\\ ={\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{k=0}^l}\left(\genfrac{}{}{0pt}{}{l}{k}\right)S_1(n,l){\lambda }^{n-l}{p}^{w_1{w}_{2}xk}{\left[2\right]}_{q^{w_2}}{\left[w_2\right]}_{p,q}^k{\left[w_1\right]}_{p,q}^{l-k}{E}_{l-k,p^{w_1},q^{w_1}}^{\left(k\right)}\left(w_{2}x\right)S_{l,k,p^{w_2},q^{w_2}}\left(w_1\right),\end{array}$$

where $S_{l,k,p,q}\left(w_1\right)={\sum }_{i=0}^{w_1-1}{(-1)}^i{q}^{(l-k+1)i}{\left[i\right]}_{p,q}^k$ is called as the $(p,q)$-sums of powers.

Proof. 

From (3), Theorems 3 and 6, we have

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\\ ={\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{l=0}^n}{E}_{l,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2}\right){\left({\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)}^{n-l}{S}_1(n,l)\\ ={\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{\left[w_2\right]}_{p,q}^l{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\displaystyle \sum _{k=0}^l}{q}^{w_1(l-k)i}{p}^{w_1{w}_{2}xk}\\ \times E_{l-k,p^{w_2},q^{w_2}}^{\left(k\right)}\left(w_{1}x\right){\left(\frac{{\left[w_1\right]}_{p,q}}{{\left[w_2\right]}_{p,q}}\right)}^k{\left[i\right]}_{p^{w_1},q^{w_1}}^k\\ ={\left[2\right]}_{q^{w_1}}{\displaystyle \sum _{l=0}^n}{S}_1(n,l){\lambda }^{n-l}{\displaystyle \sum _{k=0}^l}\left(\genfrac{}{}{0pt}{}{l}{k}\right)p^{w_1{w}_{2}xk}{\left[w_1\right]}_{p,q}^k{\left[w_2\right]}_{p,q}^{l-k}{p}^{w_1{w}_{2}xl}{E}_{l-k,p^{w_2},q^{w_2}}^{\left(k\right)}\left(w_{1}x\right)\\ \times {\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{q}^{(l-k)w_{1}i}{\left[i\right]}_{p^{w_1},q^{w_1}}^k.\end{array}$$

Therefore, we induce that

$$\begin{array}{l}{\left[2\right]}_{q^{w_1}}{\left[w_2\right]}_{p,q}^n{\displaystyle \sum _{i=0}^{w_2-1}}{(-1)}^i{q}^{w_{1}i}{\mathcal{E}}_{n,p^{w_2},q^{w_2}}\left(w_{1}x+\frac{w_{1}i}{w_2},{\displaystyle \frac{\lambda }{{\left[w_2\right]}_{p,q}}}\right)\\ ={\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{k=0}^l}\left(\genfrac{}{}{0pt}{}{l}{k}\right)S_1(n,l){\lambda }^{n-l}{p}^{w_1{w}_{2}xk}{\left[2\right]}_{q^{w_1}}{\left[w_1\right]}_{p,q}^k{\left[w_2\right]}_{p,q}^{l-k}{p}^{w_1{w}_{2}xl}\\ \times E_{l-k,p^{w_2},q^{w_2}}^{\left(k\right)}\left(w_{1}x\right)S_{l,k,p^{w_1},q^{w_1}}\left(w_2\right),\end{array}$$

(18)

and

$$\begin{array}{l}{\left[2\right]}_{q^{w_2}}{\left[w_1\right]}_{p,q}^n{\displaystyle \sum _{j=0}^{w_1-1}}{(-1)}^j{q}^{w_{2}j}{\mathcal{E}}_{n,p^{w_1},q^{w_1}}\left(w_{2}x+\frac{w_{2}j}{w_1},{\displaystyle \frac{\lambda }{{\left[w_1\right]}_{p,q}}}\right)\\ ={\displaystyle \sum _{l=0}^n}{\displaystyle \sum _{k=0}^l}\left(\genfrac{}{}{0pt}{}{l}{k}\right)S_1(n,l){\lambda }^{n-l}{p}^{w_1{w}_{2}xk}{\left[2\right]}_{q^{w_2}}{\left[w_2\right]}_{p,q}^k{\left[w_1\right]}_{p,q}^{l-k}\\ \times E_{l-k,p^{w_1},q^{w_1}}^{\left(k\right)}\left(w_{2}x\right)S_{l,k,p^{w_2},q^{w_2}}\left(w_1\right).\end{array}$$

(19)

By (18) and (19), we make the desired symmetric identity. □