**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 [1–7]). Recently, some generalizations of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, tangent numbers and polynomials are provided (see [6,8–13]). 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: N denotes the set of natural numbers, Z denotes the set of integers, Z+ = N ∪ {0} denotes the set of nonnegative integers. We remind that the classical degenerate Euler numbers E*n*(*λ*) and Euler polynomials E*n*(*<sup>x</sup>*, *<sup>λ</sup>*), which are defined by generating functions like (1), and (2) (see [1,2])

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

and

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

respectively.

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

$$(\mathbf{x})\_n = \sum\_{k=0}^n S\_1(n,k)\mathbf{x}^k \text{ and } \mathbf{x}^n = \sum\_{k=0}^n S\_2(n,k)(\mathbf{x})\_{k\prime}$$

respectively. Here (*x*)*n* = *x*(*x* − <sup>1</sup>)···(*<sup>x</sup>* − *n* + <sup>1</sup>). The numbers *<sup>S</sup>*2(*<sup>n</sup>*, *m*) is like this

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

.

.

We also have

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

The generalized falling factorial (*x*|*λ*)*n* with increment *λ* is defined by

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

for positive integer *n*, with (*x*|*λ*)0 = 1; as we know,

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

(*x*|*λ*)*n* = *<sup>λ</sup><sup>n</sup>*(*<sup>λ</sup>*−1*x*|1)*n* for *λ* = 0. Clearly (*x*|0)*n* = *<sup>x</sup>n*. The binomial theorem for a variable *x* is

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

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

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

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

**Definition 1.** *For* 0 < *q* < *p* ≤ 1 *and h* ∈ Z*, the Carlitz-type* (*p*, *q*)*-Euler polynomials En*,*p*,*<sup>q</sup>*(*x*) *and* (*h*, *p*, *q*)*-Euler polynomials E*(*h*) *<sup>n</sup>*,*p*,*<sup>q</sup>*(*x*) *are defined like this*

$$\begin{aligned} \sum\_{n=0}^{\infty} E\_{n,p,q}(\mathbf{x}) \frac{t^n}{n!} &= [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m e^{[m+x]\_{p,q}t}, \\ \sum\_{n=0}^{\infty} E\_{n,p,q}^{(h)}(\mathbf{x}) \frac{t^n}{n!} &= [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m p^{lm} e^{[m+x]\_{p,q}t}, \end{aligned} \tag{3}$$

*respectively (see [9–11]).*

Now we make the degenerate Carlitz-type (*p*, *q*)-Euler number <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*λ*) and (*p*, *q*)-Euler polynomials <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*<sup>x</sup>*, *<sup>λ</sup>*). 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* ≤ 1*, the degenerate Carlitz-type* (*p*, *q*)*-Euler numbers* <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*λ*) *and polynomials* <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*<sup>x</sup>*, *λ*) *are related to the generating functions*

$$F\_{p,q}(t,\lambda) = \sum\_{n=0}^{\infty} \mathcal{E}\_{n,p,q}(\lambda) \frac{t^n}{n!} = [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m (1+\lambda t) \stackrel{[m]\_{p,q}}{\lambda} \tag{4}$$

*and*

$$F\_{\mathbb{P}, \mathbb{q}}(t, \mathbf{x}, \boldsymbol{\lambda}) = \sum\_{n=0}^{\infty} \mathcal{E}\_{n, p, \mathbb{q}}(\mathbf{x}, \boldsymbol{\lambda}) \frac{t^n}{n!} = [2]\_{\mathbb{q}} \sum\_{m=0}^{\infty} (-1)^m q^m (1 + \boldsymbol{\lambda}t) \frac{[m + \mathbf{x}]\_{p, q}}{\boldsymbol{\lambda}} \tag{5}$$

*respectively.*

Let *p* = 1 in (4) and (5), we can ge<sup>t</sup> the degenerate Carlitz-type *q*-Euler number <sup>E</sup>*<sup>n</sup>*,*<sup>q</sup>*(*<sup>x</sup>*, *λ*) and *q*-Euler polynomials <sup>E</sup>*<sup>n</sup>*,*<sup>q</sup>*(*<sup>x</sup>*, *λ*) respectively. Obviously, if *p* = 1, then we have

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

When *p* = 1, we have

$$\lim\_{q \to 1} \mathcal{E}\_{\boldsymbol{n}, p, \boldsymbol{q}}(\boldsymbol{x}, \boldsymbol{\lambda}) = \mathcal{E}\_{\boldsymbol{\n}}(\boldsymbol{x}, \boldsymbol{\lambda}), \quad \lim\_{q \to 1} \mathcal{E}\_{\boldsymbol{n}, p, \boldsymbol{q}}(\boldsymbol{\lambda}) = \mathcal{E}\_{\boldsymbol{\n}}(\boldsymbol{\lambda}).$$

We see that

$$\begin{split} (1+\lambda t) \frac{[\mathbf{x}+\mathbf{y}]\_{p,q}}{\lambda} &= e \frac{[\mathbf{x}+\mathbf{y}]\_{p,q}}{\lambda} \log(1+\lambda t) \\ &= \sum\_{n=0}^{\infty} \left( \frac{[\mathbf{x}+\mathbf{y}]\_{p,q}}{\lambda} \right)^{n} \frac{(\log(1+\lambda t))^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{n} S\_{1}(n,m) \lambda^{n-m} [\mathbf{x}+\mathbf{y}]\_{p,q}^{m} \right) \frac{t^{n}}{n!} .\end{split} \tag{6}$$

By (5), it follows that

$$\begin{split} &\sum\_{n=0}^{\infty} \mathcal{E}\_{n,p,q}(x,\lambda) \frac{t^n}{n!} \\ &= [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m (1+\lambda t) \frac{[m+x]\_{p,q}}{\lambda} \\ &= [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m \\ &\quad \times \sum\_{n=0}^{\infty} \sum\_{l=0}^n S\_1(n,l) \lambda^{n-l} \frac{\sum\_{j=0}^l (\frac{l}{j})(-1)^j p^{(x+m)(l-j)} q^{(x+m)j}}{(p-q)^l} \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left( [2]\_q \sum\_{l=0}^n \sum\_{j=0}^l \frac{S\_1(n,l) \lambda^{n-l} \binom{l}{j} (-1)^j q^{rj} p^{x(l-j)}}{(p-q)^l} \frac{1}{1+q^{l+1}} \right) \frac{t^n}{n!} .\end{split} \tag{7}$$

By comparing the coefficients *tnn*!in the above equation, we have the following theorem.

**Theorem 1.** *For* 0 < *q* < *p* ≤ 1 *and n* ∈ Z+*, we have*

$$\begin{split} \mathcal{E}\_{n,p,q}(x,\lambda) &= [2]\_q \sum\_{l=0}^n \sum\_{j=0}^l \frac{S\_1(n,l)\lambda^{n-l} \binom{l}{j} (-1)^j q^{rj} p^{x(l-j)}}{(p-q)^l} \frac{1}{1+q^{j+1} p^{l-j}} \\ &= [2]\_q \sum\_{m=0}^\infty \sum\_{l=0}^n S\_1(n,l) \lambda^{n-l} (-1)^m q^m [x+m]\_{p,q}^l \\ \mathcal{E}\_{n,p,q}(\lambda) &= [2]\_q \sum\_{l=0}^n \sum\_{j=0}^l \frac{S\_1(n,l)\lambda^{n-l} \binom{l}{j} (-1)^j}{(p-q)^l} \frac{1}{1+q^{j+1} p^{l-j}} \\ &= [2]\_q \sum\_{m=0}^\infty \sum\_{l=0}^n S\_1(n,l) \lambda^{n-l} (-1)^m q^m [m]\_{p,q}^l. \end{split}$$

We make the degenerate Carlitz-type (*p*, *q*)-Euler number <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*λ*). Some cases are

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

We use *t* instead of *eλ<sup>t</sup>* − 1 *λ* in (5), we have

$$\begin{split} \sum\_{m=0}^{\infty} E\_{m,p,q}(\mathbf{x}) \frac{t^m}{m!} &= \sum\_{n=0}^{\infty} \mathcal{E}\_{n,p,q}(\mathbf{x},\boldsymbol{\lambda}) \left( \frac{e^{\lambda t} - 1}{\lambda} \right)^n \frac{1}{n!} \\ &= \sum\_{n=0}^{\infty} \mathcal{E}\_{n,p,q}(\mathbf{x},\boldsymbol{\lambda}) \boldsymbol{\lambda}^{-n} \sum\_{m=n}^{\infty} S\_2(m,n) \boldsymbol{\lambda}^m \frac{t^m}{m!} \\ &= \sum\_{m=0}^{\infty} \left( \sum\_{n=0}^m \mathcal{E}\_{n,p,q}(\mathbf{x},\boldsymbol{\lambda}) \boldsymbol{\lambda}^{m-n} S\_2(m,n) \right) \frac{t^m}{m!} . \end{split} \tag{8}$$

Thus we have the following theorem.

**Theorem 2.** *For m* ∈ Z+*, we have*

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

Use *t* instead of log(1 + *λt*)1/*<sup>λ</sup>* in (3), we have

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

$$\begin{aligned} &= [2]\_q \sum\_{m=0}^{\infty} (-1)^m q^m (1+\lambda t) \frac{[m+x]\_{p,q}}{\lambda} \\ &= \sum\_{m=0}^{\infty} \mathcal{E}\_{m,p,q}(x\_\prime \lambda) \frac{t^m}{m!} \end{aligned} \tag{9}$$

and

$$\begin{split} &\sum\_{n=0}^{\infty} E\_{n,p,q}(\mathbf{x}) \left( \log(1+\lambda t)^{1/\lambda} \right)^{n} \frac{1}{n!} \\ &= \sum\_{m=0}^{\infty} \left( \sum\_{n=0}^{m} E\_{n,p,q}(\mathbf{x}) \lambda^{m-n} S\_1(m,n) \right) \frac{t^{m}}{m!} . \end{split} \tag{10} $$

Thus we have the below theorem from (9) and (10).

**Theorem 3.** *For m* ∈ Z+*, we have*

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

We have the degenerate Carlitz-type (*p*, *q*)-Euler polynomials <sup>E</sup>*<sup>n</sup>*,*p*,*<sup>q</sup>*(*<sup>x</sup>*, *<sup>λ</sup>*). some cases are

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

We introduce a (*p*, *q*)-analogue of the generalized falling factorial (*x*|*λ*)*n* with increment *λ*. The generalized (*p*, *q*)-falling factorial ([*x*]*<sup>p</sup>*,*<sup>q</sup>*|*λ*)*<sup>n</sup>* with increment *λ* is defined by

$$([\mathfrak{x}]\_{\mathcal{P}\mathcal{A}}|\lambda)\_{\mathfrak{n}} = \prod\_{k=0}^{n-1} ([\mathfrak{x}]\_{\mathcal{P}\mathcal{A}} - \lambda k)^{\mathfrak{n}}$$

for positive integer *n*, where ([*x*]*<sup>p</sup>*,*<sup>q</sup>*|*λ*)<sup>0</sup> = 1. By (4) and (5), we ge<sup>t</sup>

$$\begin{aligned} & [2]\_{\mathfrak{q}}(-1)^{n}q^{n}\sum\_{l=0}^{\infty}(-1)^{l}q^{l}(1+\lambda t)\frac{[l+n]\_{p,\mathfrak{q}}}{\lambda} \\ & + [2]\_{\mathfrak{q}}\sum\_{l=0}^{\infty}(-1)^{l}q^{l}(1+\lambda t)\frac{[l+n]\_{p,\mathfrak{q}}}{\lambda} \\ & = [2]\_{\mathfrak{q}}\sum\_{l=0}^{n-1}(-1)^{l}q^{l}(1+\lambda t)\frac{[l]\_{p,\mathfrak{q}}}{\lambda} .\end{aligned}$$

Hence we have

$$\begin{split} &((-1)^{n+1}q^n \sum\_{m=0}^{\infty} \mathcal{E}\_{m,p,q}(u,\lambda) \frac{t^m}{m!} + \sum\_{m=0}^{\infty} \mathcal{E}\_{m,p,q}(\lambda) \frac{t^m}{m!} \\ &= \sum\_{m=0}^{\infty} \left( [2]\_q \sum\_{l=0}^{n-1} (-1)^l q^l ([l]\_{p,q}|\lambda)\_m \right) \frac{t^m}{m!} . \end{split} \tag{11}$$

By comparing the coefficients of *tmm*!on both sides of (11), we have the following theorem.

**Theorem 4.** *For n* ∈ Z+*, we have*

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

We ge<sup>t</sup> that

$$\begin{split} & \quad (1+\lambda t)\_{p,q} \\ &= (1+\lambda t)\_{\overline{\lambda}} \frac{p^{y}[\mathbf{x}]\_{p,q}}{\lambda} (1+\lambda t)\_{\overline{\lambda}} \frac{q^{x}[y]\_{p,q}}{\lambda} \\ &= \sum\_{m=0}^{\infty} (p^{y}[\mathbf{x}]\_{p,q}|\lambda)\_{m} \frac{t^{m}}{m!} e^{\mathbf{q}\cdot[\mathbf{1}+\lambda t]} \frac{q^{x}[\mathbf{y}]\_{p,q}}{\lambda} \\ &= \sum\_{m=0}^{\infty} (p^{y}[\mathbf{x}]\_{p,q}|\lambda)\_{m} \frac{t^{m}}{m!} \sum\_{l=0}^{m} \left(\frac{q^{x}[\mathbf{y}]\_{p,q}}{\lambda}\right)^{l} \frac{\log(1+\lambda t)^{l}}{l!} \\ &= \sum\_{m=0}^{\infty} (p^{y}[\mathbf{x}]\_{p,q}|\lambda)\_{m} \frac{t^{m}}{m!} \sum\_{l=0}^{m} \left(\frac{q^{x}[\mathbf{y}]\_{p,q}}{\lambda}\right)^{l} \sum\_{k=l}^{\infty} S\_{1}(k,l) \lambda^{k} \frac{t^{k}}{k!} \\ &= \sum\_{n=0}^{\infty} \left(\sum\_{k=0}^{n} \sum\_{l=0}^{k} \binom{n}{k} (p^{y}[\mathbf{x}]\_{p,q}|\lambda)\_{n-k} \lambda^{k-l} q^{x}[y]\_{p,q}^{l} S\_{1}(k,l)\right) \frac{t^{n}}{n!} .\end{split} \tag{12}$$

By (5) and (12), we ge<sup>t</sup>

$$\begin{split} &\sum\_{n=0}^{\infty} \mathcal{E}\_{n,p,q} (\mathbf{x}, \boldsymbol{\lambda}) \frac{\mathbf{f}^{n}}{n!} \\ &= [2]\_{q} \sum\_{m=0}^{\infty} (-1)^{m} q^{m} (1+\lambda t) \frac{[m+\mathbf{x}]\_{p,q}}{\lambda} \\ &= [2]\_{q} \sum\_{m=0}^{\infty} (-1)^{m} q^{m} \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{n} \sum\_{l=0}^{k} \binom{n}{k} (p^{m} [\mathbf{x}]\_{p,q} | \lambda)\_{n-k} \lambda^{k-l} q^{xl} [m]\_{p,q}^{l} \mathbf{S}\_{1} (k,l) \right) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \left( [2]\_{q} \sum\_{m=0}^{\infty} \sum\_{k=0}^{n} \sum\_{l=0}^{k} \binom{n}{k} (-1)^{m} q^{m} (p^{m} [\mathbf{x}]\_{p,q} | \lambda)\_{n-k} \lambda^{k-l} q^{xl} \mathbf{S}\_{1} (k,l) \right) \frac{t^{n}}{n!} .\end{split}$$

By comparing the coefficients of *tnn*!in the above equation, we have the theorem below.

**Theorem 5.** *For* 0 < *q* < *p* ≤ 1 *and n* ∈ Z+*, we have*

$$\mathcal{E}\_{\boldsymbol{n},\boldsymbol{p},\boldsymbol{q}}(\boldsymbol{x},\boldsymbol{\lambda}) = [2]\_{\boldsymbol{q}} \sum\_{m=0}^{\infty} \sum\_{k=0}^{n} \sum\_{l=0}^{k} \binom{n}{k} (-1)^{m} q^{m} (p^{m}[\boldsymbol{x}]\_{p,\boldsymbol{q}}|\boldsymbol{\lambda})\_{n-k} \boldsymbol{\lambda}^{k-l} q^{xl} \mathcal{S}\_{1}(k,l).$$

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

In this section, we are going to ge<sup>t</sup> 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 [*xy*]*<sup>p</sup>*,*<sup>q</sup>* = [*x*]*py*,*q<sup>y</sup>* [*y*]*<sup>p</sup>*,*<sup>q</sup>* for any *x*, *y* ∈ C.

By using *w*1*x* + *<sup>w</sup>*1*i w*2 instead of *x* in Definition 2, use *p* by *p<sup>w</sup>*<sup>2</sup> , use *q* by *q<sup>w</sup>*<sup>2</sup> and use *λ* by *λ* [*<sup>w</sup>*2]*<sup>p</sup>*,*<sup>q</sup>* , respectively, we can ge<sup>t</sup>

$$\begin{split} &\sum\_{n=0}^{\infty} \left( [2]\_{q^{w\_1}} [w\_2]\_{p,q}^n \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \mathcal{E}\_{n,p^{w\_2},q^{w\_2}} \left( w\_1 x + \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \right) \frac{t^n}{n!} \\ &= [2]\_{q^{w\_1}} \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \sum\_{n=0}^{\infty} \mathcal{E}\_{n,p^{w\_2},q^{w\_2}} \left( w\_1 x + \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \frac{([w\_2]\_{p,q} t)^n}{n!} \\ &= [2]\_{q^{w\_1}} \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} [2]\_{q^{w\_2}} \sum\_{n=0}^{\infty} (-1)^n q^{w\_2 n} \\ & \qquad \times \left( 1 + \frac{\lambda}{[w\_2]\_{p,q}} [w\_2]\_{p,q} t \right) \frac{[w\_1 x + \frac{w\_1 i}{w\_2} + n]\_{p^{w\_2}, q^{w\_2}}}{\frac{\lambda}{[w\_2]\_{p,q}}} \\ &= [2]\_{q^{w\_1}} \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} [2]\_{q^{w\_2}} \sum\_{n=0}^{\infty} (-1)^n q^{w\_2 n} \\ & \qquad \times (1 + \lambda t) \frac{[w\_1 w\_2 x + w\_1 i + n w\_2]\_{p,q}}{\lambda} . \end{split}$$

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 ≤ *j* ≤ *w*1 − 1. So this can be written as

[2]*q<sup>w</sup>*<sup>1</sup> [2]*q<sup>w</sup>*<sup>2</sup> *<sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*<sup>i</sup>* ∞ ∑ *<sup>n</sup>*=0 (−<sup>1</sup>)*nqw*2*<sup>n</sup>* × (1 + *λt*) [*<sup>w</sup>*1*w*2*<sup>x</sup>* + *<sup>w</sup>*1*i* + *nw*2]*<sup>p</sup>*,*<sup>q</sup> λ* . = [2]*q<sup>w</sup>*<sup>1</sup> [2]*q<sup>w</sup>*<sup>2</sup> *<sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*<sup>i</sup>* ∞ ∑ *<sup>w</sup>*1*r*+*j*=0 0≤*j*≤*<sup>w</sup>*1−1 (−<sup>1</sup>)*<sup>w</sup>*1*r*<sup>+</sup>*jqw*2(*<sup>w</sup>*1*r*+*j*) × (1 + *λt*) [*<sup>w</sup>*1*w*2*<sup>x</sup>* + *<sup>w</sup>*1*i* + (*<sup>w</sup>*1*<sup>r</sup>* + *j*)*<sup>w</sup>*2]*<sup>p</sup>*,*<sup>q</sup> λ* . = [2]*q<sup>w</sup>*<sup>1</sup> [2]*q<sup>w</sup>*<sup>2</sup> *<sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*<sup>i</sup> <sup>w</sup>*1−1 ∑ *j*=0 ∞ ∑ *<sup>r</sup>*=0 (−<sup>1</sup>)*<sup>w</sup>*1*<sup>r</sup>*(−<sup>1</sup>)*jqw*2*w*1*rqw*<sup>2</sup> *j* × (1 + *λt*) [*<sup>w</sup>*1*w*2*<sup>x</sup>* + *<sup>w</sup>*1*i* + *w*1*w*2*r* + *w*2 *j*]*<sup>p</sup>*,*<sup>q</sup> λ* = [2]*q<sup>w</sup>*<sup>1</sup> [2]*q<sup>w</sup>*<sup>2</sup> *<sup>w</sup>*2−1 ∑ *i*=0 *<sup>w</sup>*1−1 ∑ *j*=0 ∞ ∑ *<sup>r</sup>*=0 (−<sup>1</sup>)*<sup>i</sup>*(−<sup>1</sup>)*r*(−<sup>1</sup>)*jqw*1*iqw*2*w*1*rqw*<sup>2</sup> *j* × (1 + *λt*) [*<sup>w</sup>*1*w*2*<sup>x</sup>* + *<sup>w</sup>*1*i* + *w*1*w*2*r* + *w*2 *j*]*<sup>p</sup>*,*<sup>q</sup> λ* .

We have the below formula using the above formula

$$\begin{split} \sum\_{n=0}^{\infty} \left( [2]\_{q^{w\_2}} [w\_2]\_{p,q}^n \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \mathcal{E}\_{w, p^{w\_2}, q^{w\_2}} \left( w\_1 \mathbf{x} + \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \right) \frac{t^n}{n!} \\ = [2]\_{q^{w\_1}} [2]\_{q^{w\_2}} \sum\_{i=0}^{w\_2-1} \sum\_{j=0}^{w\_1-1} \sum\_{r=0}^{\infty} (-1)^i (-1)^r (-1)^j q^{w\_1 i} q^{w\_2 w\_1 r} q^{w\_2 j} \\ \times \frac{[w\_1 w\_2 \mathbf{x} + w\_1 i + w\_1 w\_2 r + w\_2 j]\_{p,q}}{\lambda} \end{split} \tag{13}$$

From a similar approach, we can have that

$$\sum\_{n=0}^{\infty} \left( [2]\_{q^{w\_2}} [w\_1]\_{p,q}^n \sum\_{i=0}^{w\_1 - 1} (-1)^i q^{w\_2 i} \mathcal{E}\_{n, p^{w\_1} q^{w\_1}} \left( w\_2 \mathbf{x} + \frac{w\_2 i}{w\_1}, \frac{\lambda}{[w\_1]\_{p,q}} \right) \right) \frac{t^n}{n!} $$

$$= [2]\_{q^{w\_1}} [2]\_{q^{w\_2}} \sum\_{i=0}^{w\_1 - 1} \sum\_{j=0}^{w\_2 - 1} \sum\_{r=0}^{\infty} (-1)^i (-1)^r (-1)^j q^{w\_2 i} q^{w\_1 w\_1 r} q^{w\_1 j} \tag{14}$$

$$\times \left( 1 + \lambda t \right) \frac{[w\_1 w\_2 \mathbf{x} + w\_2 i + w\_1 w\_2 r + w\_1 j]\_{p,q}}{\lambda}$$

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*

$$\begin{aligned} & [2]\_{q^{w\_1}} [w\_2]\_{p,q}^{\
u} \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \mathcal{E}\_{n,p^{w\_2},q^{w\_2}} \left( w\_1 \mathbf{x} + \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \\ &= [2]\_{q^{w\_2}} [w\_1]\_{p,q}^{\
u} \sum\_{j=0}^{w\_1-1} (-1)^j q^{w\_2 j} \mathcal{E}\_{n,p^{w\_1},q^{w\_1}} \left( w\_2 \mathbf{x} + \frac{w\_2 j}{w\_1}, \frac{\lambda}{[w\_1]\_{p,q}} \right) . \end{aligned}$$

Letting *λ* → 0 in Theorem 6, we can immediately obtain the symmetric identities for Carlitz-type (*p*, *q*)-Euler polynomials (see [10])

$$\begin{aligned} & [2]\_{q^{w\_1}} [w\_2]\_{p,q}^n \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} E\_{n,p^{w\_2},q^{w\_2}} \left( w\_1 \mathbf{x} + \frac{w\_1 i}{w\_2} \right) \\ &= [2]\_{q^{w\_2}} [w\_1]\_{p,q}^n \sum\_{j=0}^{w\_1-1} (-1)^j q^{w\_2 j} E\_{n,p^{w\_1},q^{w\_1}} \left( w\_2 \mathbf{x} + \frac{w\_2 j}{w\_1} \right) . \end{aligned}$$

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*

$$\mathcal{E}\_{n,p,q}(\mathbf{x},\lambda) = \frac{[2]\_q [w\_1]\_{p,q}^n}{[2]\_{q^{w\_1}}} \sum\_{j=0}^{w\_1 - 1} (-1)^j q^j \mathcal{E}\_{n,p^{w\_1}, q^{w\_1}} \left( \frac{\mathbf{x} + j}{w\_1}, \frac{\lambda}{[w\_1]\_{p,q}} \right) . \tag{15}$$

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

$$\mathcal{E}\_{n,q}(\mathbf{x},\lambda) = \frac{[2]\_q [w\_1]\_q^n}{[2]\_{q^{w\_1}}} \sum\_{j=0}^{w\_1 - 1} (-1)^j q^j \mathcal{E}\_{n,q}{}^{w\_1} \left( \begin{array}{c} \mathbf{x} + j \\ \overline{w}\_1 \end{array}; \frac{\lambda}{[\overline{w}\_1]\_q} \right) . \tag{16}$$

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

$$\mathcal{E}\_n(\mathbf{x}, \boldsymbol{\lambda}) = w\_1^{\boldsymbol{w}} \sum\_{j=0}^{\boldsymbol{w}\_1 - 1} (-1)^j \mathcal{E}\_n \left( \frac{\mathbf{x} + i}{\boldsymbol{w}\_1}, \frac{\boldsymbol{\lambda}}{\boldsymbol{w}\_1} \right). \tag{17}$$

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

$$E\_n(\mathbf{x}) = w\_1^n \sum\_{j=0}^{w\_1 - 1} (-1)^j E\_n \left( \frac{\mathbf{x} + i}{w\_1} \right).$$

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*

$$\begin{aligned} & [2]\_{q^{w\_1}} [w\_2]\_{p,q}^n \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \mathcal{E}\_{n,p^{w\_2},q^{w\_2}} \left( \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \\ & = [2]\_{q^{w\_2}} [w\_1]\_{p,q}^n \sum\_{j=0}^{w\_1-1} (-1)^j q^{w\_2 j} \mathcal{E}\_{n,p^{w\_1},q^{w\_1}} \left( \frac{w\_2 j}{w\_1}, \frac{\lambda}{[w\_1]\_{p,q}} \right) . \end{aligned}$$

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

**Theorem 7.** *Let w*1 *and w*2 *be odd positive integers. Then*

$$\begin{split} &\sum\_{l=0}^{n} S\_{1}(n,l) \lambda^{w-l} [w\_{2}]\_{p,q}^{l} [2]\_{q^{w\_{1}}} \sum\_{i=0}^{w\_{2}-1} (-1)^{i} q^{w\_{1}i} E\_{l,p^{w\_{2}},q^{w\_{2}}} \left( \frac{w\_{1}}{w\_{2}} i \right) \\ &= \sum\_{l=0}^{n} S\_{1}(n,l) \lambda^{n-l} [w\_{1}]\_{p,q}^{l} [2]\_{q^{w\_{2}}} \sum\_{j=0}^{w\_{1}-1} (-1)^{j} q^{w\_{2}j} E\_{l,p^{w\_{1}},q^{w\_{1}}} \left( \frac{w\_{2}}{w\_{1}} j \right) . \end{split}$$

We ge<sup>t</sup> another result by applying the addition theorem about the Carlitz-type (*p*, *q*)-Euler polynomials *En*,*p*,*<sup>q</sup>*(*x*).

**Theorem 8.** *Let w*1 *and w*2 *be odd positive integers. Then we have*

$$\begin{split} &\sum\_{l=0}^{n}\sum\_{k=0}^{l} \binom{l}{k} \mathcal{S}\_{1}(\boldsymbol{n},l)\lambda^{n-l}p^{w\_{1}w\_{2}\ge k}[2]\_{q^{w\_{1}}}[w\_{1}]\_{p,q}^{k}[w\_{2}]\_{q,q}^{l-k}E\_{l-k,p^{w\_{2}},q^{w\_{2}}}^{(k)}(w\_{1}\boldsymbol{x})\mathcal{S}\_{l,k,p^{w\_{1}},q^{w\_{1}}}(w\_{2})\\ &=\sum\_{l=0}^{n}\sum\_{k=0}^{l} \binom{l}{k} \mathcal{S}\_{1}(\boldsymbol{n},l)\lambda^{n-l}p^{w\_{1}w\_{2}\ge k}[2]\_{q^{w\_{2}}}[w\_{2}]\_{p,q}^{k}[w\_{1}]\_{p,q}^{l-k}E\_{l-k,p^{w\_{1}},q^{w\_{1}}}^{(k)}(w\_{2}\boldsymbol{x})\mathcal{S}\_{l,k,p^{w\_{2}},q^{w\_{2}}}(w\_{1}),\end{split}$$

*where Sl*,*k*,*p*,*<sup>q</sup>*(*<sup>w</sup>*1) = ∑*<sup>w</sup>*1−<sup>1</sup> *i*=0 (−<sup>1</sup>)*iq*(*l*−*k*+<sup>1</sup>)*<sup>i</sup>*[*i*]*kp*,*<sup>q</sup> is called as the* (*p*, *q*)*-sums of powers.*

**Proof.** From (3), Theorems 3 and 6, we have

[2]*q<sup>w</sup>*<sup>1</sup> [*<sup>w</sup>*2]*np*,*<sup>q</sup> <sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*i*E*<sup>n</sup>*,*p<sup>w</sup>*<sup>2</sup> ,*q<sup>w</sup>*<sup>2</sup> *w*1*x* + *<sup>w</sup>*1*i w*2 , *λ* [*<sup>w</sup>*2]*<sup>p</sup>*,*<sup>q</sup>* = [2]*q<sup>w</sup>*<sup>1</sup> [*<sup>w</sup>*2]*np*,*<sup>q</sup> <sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*<sup>i</sup> n* ∑ *l*=0 *El*,*p<sup>w</sup>*<sup>2</sup> ,*q<sup>w</sup>*<sup>2</sup> *w*1*x* + *<sup>w</sup>*1*i w*2 *λ* [*<sup>w</sup>*2]*<sup>p</sup>*,*<sup>q</sup> <sup>n</sup>*−*l <sup>S</sup>*1(*<sup>n</sup>*, *l*) = [2]*q<sup>w</sup>*<sup>1</sup> *n* ∑ *l*=0 *<sup>S</sup>*1(*<sup>n</sup>*, *<sup>l</sup>*)*λ<sup>n</sup>*−*<sup>l</sup>*[*<sup>w</sup>*2]*lp*,*<sup>q</sup> <sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*<sup>i</sup> l* ∑ *k*=0 *q<sup>w</sup>*1(*l*−*k*)*<sup>i</sup> p<sup>w</sup>*1*w*2*xk* × *E*(*k*) *l*−*k*,*p<sup>w</sup>*<sup>2</sup> ,*q<sup>w</sup>*<sup>2</sup> (*<sup>w</sup>*1*<sup>x</sup>*) [*<sup>w</sup>*1]*<sup>p</sup>*,*<sup>q</sup>* [*<sup>w</sup>*2]*<sup>p</sup>*,*<sup>q</sup> k* [*i*]*kpw*1 ,*q<sup>w</sup>*<sup>1</sup> = [2]*q<sup>w</sup>*<sup>1</sup> *n* ∑ *l*=0 *<sup>S</sup>*1(*<sup>n</sup>*, *<sup>l</sup>*)*λ<sup>n</sup>*−*<sup>l</sup> l* ∑ *k*=0 *lk <sup>p</sup><sup>w</sup>*1*w*2*xk*[*<sup>w</sup>*1]*kp*,*<sup>q</sup>*[*<sup>w</sup>*2]*<sup>l</sup>*−*<sup>k</sup> p*,*q p<sup>w</sup>*1*w*2*xlE*(*k*) *l*−*k*,*p<sup>w</sup>*<sup>2</sup> ,*q<sup>w</sup>*<sup>2</sup> (*<sup>w</sup>*1*<sup>x</sup>*) × *<sup>w</sup>*2−1 ∑ *i*=0 (−<sup>1</sup>)*iqw*1*iq*(*l*−*k*)*<sup>w</sup>*1*<sup>i</sup>*[*i*]*kpw*1 ,*q<sup>w</sup>*<sup>1</sup> .

Therefore, we induce that

$$\begin{split} \left[ \mathbb{1} \right]\_{q^{w\_1}} [w\_2]\_{p,q}^{w\_2-1} \sum\_{i=0}^{w\_2-1} (-1)^i q^{w\_1 i} \mathcal{E}\_{n,p^{w\_2},q^{w\_2}} \left( w\_1 \mathbf{x} + \frac{w\_1 i}{w\_2}, \frac{\lambda}{[w\_2]\_{p,q}} \right) \\ = \sum\_{l=0}^n \sum\_{k=0}^l \binom{l}{k} \mathcal{S}\_1(n,l) \lambda^{n-l} p^{w\_1 w\_2 \ge k} [2]\_{q^{w\_1}} [w\_1]\_{p,q}^k [w\_2]\_{p,q}^{l-k} p^{w\_1 w\_2 \ge l} \\ \quad \times E\_{l-k,p^{w\_2},q^{w\_2}}^{(k)} (w\_1 \mathbf{x}) \mathcal{S}\_{l,k,p^{w\_1},q^{w\_1}} (w\_2), \end{split} \tag{18}$$

and

$$\begin{split} \left[ \mathbb{1} \right]\_{q^{w\_2}} [w\_1]\_{p,q}^w \sum\_{j=0}^{w\_1-1} (-1)^j q^{w\_2 j} \mathcal{E}\_{n, p^{w\_1}, q^{w\_1}} \left( w\_2 x + \frac{w\_2 j}{w\_1}, \frac{\lambda}{[w\_1]\_{p,q}} \right) \\ = \sum\_{l=0}^n \sum\_{k=0}^l \binom{l}{k} S\_1(n, l) \lambda^{n-l} p^{w\_1 w\_2 \ge k} [2]\_{q^{w\_2}} [w\_2]\_{p,q}^k [w\_1]\_{p,q}^{l-k} \\ \qquad \times E\_{l-k, p^{w\_1}, q^{w\_1}}^{(k)} (w\_2 \ge) S\_{l, k, p^{w\_2}, q^{w\_2}} (w\_1) . \end{split} \tag{19}$$

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

**Author Contributions:** All authors contributed equally in writing this article. All authors read and approved the final manuscript.

**Funding:** This work was supported by the Dong-A university research fund.

**Acknowledgments:** The authors would like to thank the referees for their valuable comments, which improved the original manuscript in its present form.

**Conflicts of Interest:** The authors declare no conflict of interest.
