# Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials

## Abstract

In this paper, we consider the Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. We present several explicit formulas and recurrence relations for these polynomials. Also, we establish a connection between our polynomials and several known families of polynomials.

## 1 Introduction

In this paper, we use umbral calculus techniques (see [1, 2]) to obtain several new and interesting identities of Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. To define the umbral calculus, let Π be the algebra of polynomials in a single variable x over $$\mathbb{C}$$ and $$\Pi^{*}$$ be the vector space of all linear functionals on Π. The action of a linear functional $$L\in\Pi^{*}$$ on a polynomial $$p(x)$$ is denoted by $$\langle L|p(x)\rangle$$, and linearly extended as $$\langle cL+dL'|p(x)\rangle=c\langle L|p(x)\rangle+d\langle L'|p(x)\rangle$$, where $$c,d\in\mathbb{C}$$. Define $$\mathcal{H}=\{f(t)=\sum_{k\geq0} a_{k}\frac{t^{k}}{k!}\mid a_{k}\in\mathbb{C}\}$$ to be the algebra of formal power series in a single variable t. The formal power series $$f(t)\in\mathcal{H}$$ defines a linear functional on Π by setting $$\langle f(t)|x^{n}\rangle=a_{n}$$ for all $$n\geq0$$. Thus, we have (see [1, 2])

$$\bigl\langle t^{k}|x^{n}\bigr\rangle =n!\delta_{n,k} \quad \mbox{for all }n,k\geq 0,$$
(1.1)

where $$\delta_{n,k}$$ is the Kronecker symbol. Let $$f_{L}(t)=\sum_{n\geq0}\langle L|x^{n}\rangle\frac{t^{n}}{n!}$$. By (1.1), we get that $$\langle f_{L}(t)|x^{n}\rangle=\langle L|x^{n}\rangle$$. Thus, the map $$L\mapsto f_{L}(t)$$ gives a vector space isomorphism from $$\Pi^{*}$$ onto $$\mathcal{H}$$. Therefore, $$\mathcal{H}$$ is thought of as a set of both formal power series and linear functionals, which is called the umbral algebra. The umbral calculus is the study of umbral algebra.

The order $$O(f(t))$$ of the non-zero power series $$f(t)$$ is defined to be k when $$f(t)=\sum_{n\geq k}a_{n}t^{n}$$ and $$a_{k}\neq0$$. Suppose that $$O(f(t))=1$$ and $$O(g(t))=0$$. Then there exists a unique sequence $$s_{n}(x)$$ of polynomials such that $$\langle g(t)f(t)^{k}|s_{n}(x)\rangle=n!\delta_{n,k}$$, where $$n,k\geq0$$. The sequence $$s_{n}(x)$$ is called the Sheffer sequence for $$(g(t),f(t))$$, and we write $$s_{n}(x)\sim(g(t),f(t))$$ (see [1, 2]). For $$f(t)\in\mathcal{H}$$ and $$p(x)\in\Pi$$, we have that $$\langle e^{yt}|p(x)\rangle=p(y)$$, $$\langle f(t)g(t)|p(x)\rangle=\langle g(t)|f(t)p(x)\rangle$$, $$f(t)=\sum_{n\geq0}\langle f(t)|x^{n}\rangle \frac {t^{n}}{n!}$$ and $$p(x)=\sum_{n\geq0}\langle t^{n}|p(x)\rangle\frac {x^{n}}{n!}$$. Therefore, $$\langle t^{k}|p(x)\rangle=p^{(k)}(0)$$, $$\langle1|p^{(k)}(x)\rangle=p^{(k)}(0)$$, where $$p^{(k)}(0)$$ denotes the kth derivative of $$p(x)$$ with respect to x at $$x=0$$. So, $$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}}{dx^{k}}p(x)$$ for all $$k\geq0$$ (see [1, 2]).

Let $$s_{n}(x)\sim(g(t),f(t))$$. Then we have

$$\frac{1}{g(\bar{f}(t))}e^{y\bar{f}(t)}=\sum_{n\geq0}s_{n}(y) \frac {t^{n}}{n!}$$
(1.2)

for all $$y\in\mathbb{C}$$, where $$\bar{f}(t)$$ is the compositional inverse of $$f(t)$$ (see [1, 2]). For $$s_{n}(x)\sim(g(t),f(t))$$ and $$r_{n}(x)\sim(h(t),\ell(t))$$, let $$s_{n}(x)=\sum_{k=0}^{n} c_{n,k}r_{k}(x)$$. Then we have

$$c_{n,k}=\frac{1}{k!} \biggl\langle \frac{h(\bar{f}(t))}{g(\bar {f}(t))}\bigl(\ell \bigl(\bar{f}(t)\bigr)\bigr)^{k}\Big|x^{n} \biggr\rangle$$
(1.3)

(see [1, 2]).

Throughout the paper, let $$r,s\in\mathbb{Z}_{>0}$$, and let $$\mathbf{a}=(a_{1},a_{2},\ldots,a_{r})$$, $$\mathbf{b}=(b_{1},b_{2},\ldots ,b_{s})$$ with $$a_{j},b_{i}\neq0$$ for all i, j. We define the Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})$$ (for other Barnes-types, see [35]) as

$$P_{r,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}D\mathcal{E}_{n}(\lambda ,x|\mathbf{a}; \mathbf{b})\frac{t^{n}}{n!},$$
(1.4)

where we define

$$P_{r,s}(t)=\prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda ((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} \biggr).$$

For $$x=0$$, $$D\mathcal{E}_{n}(\lambda|\mathbf{a};\mathbf{b})=D\mathcal {E}_{n}(\lambda,0|\mathbf{a};\mathbf{b})$$ are called the Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type numbers.

We recall here that the polynomials $$D_{n,\lambda}(x|\mathbf{a})$$ given by

$$P_{r,0}(t) (1+\lambda t)^{\frac{x}{\lambda}} =\sum _{n\geq0}D_{n,\lambda}(x|\mathbf{a})\frac{t^{n}}{n!}$$

are called the Barnes-type Daehee polynomials with λ-parameter (see [6, 7]). Also, the polynomials $$\mathcal{E}_{n}(\lambda ,x|\mathbf{b})$$ given by

$$P_{0,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\mathcal{E}_{n}(\lambda ,x|\mathbf{b}) \frac{t^{n}}{n!}$$
(1.5)

are called the Barnes-type degenerate Euler polynomials which are studied in [811]. In the case $$x=0$$, we write $$\mathcal{E}_{n}(\lambda|\mathbf{b})=\mathcal{E}_{n}(\lambda,0|\mathbf {b})$$, which are called the Barnes-type degenerate Euler numbers. Note that $$\lim_{\lambda\rightarrow0}\mathcal{E}_{n}(\lambda,x|\mathbf {b})=E_{n}(x|\mathbf{b})$$ and $$\lim_{\lambda\rightarrow\infty}\lambda^{-n}\mathcal{E}_{n}(\lambda ,\lambda x|\mathbf{b})=(x)_{n}$$, where $$(x)_{n}=\prod_{i=0}^{n-1}(x-i)$$ with $$(x)_{0}=1$$ and $$E_{n}(x|\mathbf{b})$$ are the Barnes-type degenerate Euler polynomials given by

$$\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)e^{xt}=\sum_{n\geq 0}E_{n}(x| \mathbf{b})\frac{t^{n}}{n!}.$$

It is immediate from (1.2) and (1.4) to see that $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})$$ is the Sheffer sequence for the pair $$g(t)=\prod_{i=1}^{r} (\frac {e^{a_{i}t}-1}{t} )\prod_{i=1}^{s} (\frac{e^{b_{i}t}+1}{2} )$$ and $$f(t)=\frac{e^{\lambda t}-1}{\lambda}$$. Thus,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})\sim \Biggl(\prod_{i=1}^{r} \biggl( \frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl(\frac {e^{b_{i}t}+1}{2} \biggr),\frac{e^{\lambda t}-1}{\lambda} \Biggr).$$
(1.6)

The aim of the present paper is to present several new identities for Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials by the use of umbral calculus. For some of the related works, one is referred to the papers [1220].

## 2 Explicit formulas

In this section we suggest several explicit formulas for the Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. To do that, we recall that the Stirling numbers $$S_{1}(n,m)$$ of the first kind are defined as $$(x)_{n}=\sum_{m=0}^{n}S_{1}(n,m)x^{m}\sim(1,e^{t}-1)$$ or $$\frac{1}{j!}(\log(1+t))^{j}=\sum_{\ell\geq j}S_{1}(\ell,j)\frac {t^{\ell}}{\ell!}$$. Let $$(x|\lambda)_{n}$$ be the generalized falling factorials defined by $$(x|\lambda)_{n}=\prod_{i=0}^{n-1}(x-i\lambda)$$ with $$(x|\lambda)_{0}=1$$, namely $$(x|\lambda)_{n}=\lambda^{n}(x/\lambda)_{n}$$.

Let $$\mathit{BE}_{n}(x|\mathbf{a};\mathbf{b})$$ be the Barnes-type Bernoulli and Euler mixed-type polynomials given by

$$\prod_{i=1}^{r} \biggl( \frac{t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)e^{xt}=\sum _{n\geq0}\mathit{BE}_{n}(x|\mathbf {a};\mathbf{b}) \frac{t^{n}}{n!}.$$
(2.1)

Note that $$\mathit{BE}_{n}^{r,s}(x)$$ denotes the special case $$\mathit{BE}_{n}(x|\underbrace{1,1,\ldots,1}_{r};\underbrace{1,1,\ldots,1}_{s})$$ and was treated in [21, 22] by using p-adic integrals on $$\mathbb{Z}_{p}$$.

### Theorem 2.1

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}).$$

### Proof

By (1.6), we have that

$$\prod_{i=1}^{r} \biggl( \frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl(\frac{e^{b_{i}t}+1}{2} \biggr)D\mathcal{E}_{n}(\lambda,x|\mathbf {a}; \mathbf{b})\sim \biggl(1,\frac{e^{\lambda t}-1}{\lambda} \biggr).$$
(2.2)

Thus,

\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \prod_{i=1}^{r} \biggl(\frac {t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl( \frac{2}{e^{b_{i}t}+1} \biggr)x^{m} \\ &=\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}\mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}), \end{aligned}

as claimed. □

### Theorem 2.2

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{j=0}^{n} \Biggl(\sum_{\ell=j}^{n} \binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell-j}D \mathcal{E}_{n-\ell}(\lambda|\mathbf{a};\mathbf {b}) \Biggr) x^{j}.$$

### Proof

We proceed the proof by applying the conjugate representation: for $$s_{n}(x)\sim(g(t),f(t))$$, we have $$S_{n}(x)=\sum_{j=0}^{n}\frac{1}{j!}\langle g(\bar{f}(t))^{-1}\bar{f}(t)^{j}|x^{n}\rangle x^{j}$$. By (1.6), we obtain

\begin{aligned}& \bigl\langle g\bigl(\bar{f}(t)\bigr)^{-1}\bar{f}(t)^{j}|x^{n} \bigr\rangle \\& \quad = \biggl\langle P_{r,s}(t)\frac{\log^{j}(1+\lambda t)}{\lambda ^{j}}\Big|x^{n} \biggr\rangle =\lambda^{-j} \biggl\langle P_{r,s}(t)\Big|j!\sum _{\ell\geq j}S_{1}(\ell ,j)\frac{\lambda^{\ell}t^{\ell}}{\ell!}x^{n} \biggr\rangle \\& \quad =\lambda^{-j}j!\sum_{\ell=j}^{n} \binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell} \bigl\langle P_{r,s}(t)|x^{n-\ell} \bigr\rangle =\lambda^{-j}j!\sum _{\ell =j}^{n}\binom{n}{\ell}S_{1}( \ell,j)\lambda^{\ell}D\mathcal{E}_{n-\ell }(\lambda|\mathbf{a}; \mathbf{b}). \end{aligned}

Therefore, $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{j=0}^{n} (\sum_{\ell=j}^{n}\binom{n}{\ell}S_{1}(\ell,j)\lambda ^{\ell-j}D\mathcal{E}_{n-\ell}(\lambda|\mathbf{a};\mathbf {b}) ) x^{j}$$, as claimed. □

### Theorem 2.3

For all $$n\geq1$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) =\sum _{\ell =0}^{n-1}\binom{n-1}{\ell}\lambda^{\ell}B_{\ell}^{(n)}\mathit{BE}_{n-\ell}(x|\mathbf{a}; \mathbf{b}),$$

where $$B_{\ell}^{(n)}$$ is the ℓth Bernoulli number of order n (see [23]).

### Proof

We proceed the proof by using the following transfer formula: for $$p_{n}(x)\sim(1,f(t))$$ and $$q_{n}(x)\sim(1,g(t))$$, we have that $$q_{n}(x)=x (\frac{f(t)}{g(t)} )^{n}x^{-1}p_{n}(x)$$ for all $$n\geq1$$. So, by the fact that $$x^{n}\sim(1,t)$$ and (2.2), we obtain

\begin{aligned}& \prod_{i=1}^{r} \biggl(\frac{e^{a_{i}t}-1}{t} \biggr)\prod_{i=1}^{s} \biggl( \frac{e^{b_{i}t}+1}{2} \biggr)D\mathcal{E}_{n}(\lambda,x|\mathbf {a}; \mathbf{b}) \\& \quad =x \biggl(\frac{\lambda t}{e^{\lambda t}-1} \biggr)^{n}x^{n-1}=x\sum _{\ell\geq0}B_{\ell}^{(n)} \frac{\lambda ^{\ell}t^{\ell}}{\ell!}x^{n-1}=\sum_{\ell=0}^{n-1} \binom{n-1}{\ell}\lambda ^{\ell}B_{\ell}^{(n)}x^{n-\ell}, \end{aligned}

which, by (2.1), implies

\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n-1}\binom{n-1}{\ell}\lambda^{\ell}B_{\ell}^{(n)}\prod_{i=1}^{r} \biggl(\frac{t}{e^{a_{i}t}-1} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{e^{b_{i}t}+1} \biggr)x^{n-\ell} \\ &=\sum_{\ell=0}^{n-1}\binom{n-1}{\ell} \lambda^{\ell}B_{\ell}^{(n)}\mathit{BE}_{n-\ell}(x| \mathbf{a};\mathbf{b}), \end{aligned}

as required. □

In order to state our next theorem, we recall the polynomials $$\beta\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})$$, which are called the Barnes-type degenerate Bernoulli and Euler mixed-type polynomials. They are defined as

$$Q_{r,s}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\beta\mathcal{E}_{n}(\lambda ,x|\mathbf{a} ; \mathbf{b})\frac{t^{n}}{n!},$$
(2.3)

where $$Q_{r,s}(t)=\prod_{i=1}^{r} (\frac{t}{(1+\lambda t)^{\frac{a_{i}}{\lambda}}-1} )\prod_{i=1}^{s} (\frac {2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} )$$, for example, see [3].

### Theorem 2.4

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) =\sum _{\ell =0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell+r}{r}}\lambda ^{\ell}S_{1}(\ell+r,r)\beta\mathcal{E}_{n-\ell}(\lambda,x|\mathbf {a}; \mathbf{b}).$$

### Proof

By (1.4), we have

\begin{aligned} D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b})&= \biggl\langle \sum _{\ell\geq0}D\mathcal {E}_{\ell}(\lambda,y| \mathbf{a};\mathbf{b})\frac{t^{\ell}}{\ell !}\Big|x^{n} \biggr\rangle = \bigl\langle P_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}|x^{n} \bigr\rangle \\ &= \biggl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}\Big| \frac{\log^{r}(1+\lambda t)}{\lambda^{r} t^{r}}x^{n} \biggr\rangle \\ &= \biggl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}\Big| r!\sum _{\ell\geq0}\frac{S_{1}(\ell+r,r)\lambda^{\ell}t^{\ell}}{(\ell +r)!}x^{n} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \biggl\langle \sum _{m\geq0}\beta\mathcal{E}_{m}(\lambda ,y|\mathbf{a} ; \mathbf{b})\frac{t^{m}}{m!}\Big|x^{n-\ell} \biggr\rangle , \end{aligned}

which, by (2.3), implies $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{\ell =0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell+r}{r}}\lambda ^{\ell}S_{1}(\ell+r,r)\beta\mathcal{E}_{n-\ell}(\lambda,x|\mathbf {a};\mathbf{b})$$, as required. □

In order to present our next theorem, we recall the polynomials $$\beta_{n}(\lambda,x|\mathbf{a})$$, which are called the Barnes-type degenerate Bernoulli polynomials. They are given by

$$Q_{r,0}(t) (1+\lambda t)^{\frac{x}{\lambda}}=\sum _{n\geq0}\beta_{n}(\lambda,x|\mathbf {a}) \frac{t^{n}}{n!},$$
(2.4)

for example, see [8, 9, 23].

### Theorem 2.5

For all $$n\geq0$$,

\begin{aligned} D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n}\sum_{m=0}^{n-\ell } \frac{\binom{n}{\ell}\binom{n-\ell}{m}}{\binom{\ell +r}{r}}\lambda^{\ell}S_{1}(\ell+r,r) \mathcal{E}_{n-\ell-m}(\lambda|\mathbf{b})\beta _{m}(\lambda,x| \mathbf{a}) \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}\frac{\binom{n}{\ell}\binom {n-\ell }{m}}{\binom{\ell+r}{r}}\lambda^{\ell}S_{1}(\ell+r,r)\beta_{n-\ell-m}(\lambda|\mathbf{a})\mathcal {E}_{m}(\lambda,x|\mathbf{b}). \end{aligned}

### Proof

By the proof of Theorem 2.4, we have

\begin{aligned} D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b})&=\sum _{\ell =0}^{n}\frac{\binom{n}{\ell }}{\binom{\ell+r}{r}}\lambda^{\ell}S_{1}(\ell+r,r) \bigl\langle Q_{r,s}(t) (1+\lambda t)^{\frac{y}{\lambda}}|x^{n-\ell} \bigr\rangle \\ &=\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \bigl\langle Q_{0,s}(t)|Q_{r,0}(t) (1+\lambda t)^{\frac{y}{\lambda}} x^{n-\ell} \bigr\rangle . \end{aligned}

Thus, by (1.5) and (2.4), we obtain

\begin{aligned}& D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b}) \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r) \Biggl\langle Q_{0,s}(t)\bigg|\sum_{m=0}^{n-\ell} \binom {n-\ell }{m}\beta_{m}(\lambda,y|\mathbf{a})x^{n-\ell-m} \Biggr\rangle \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r)\sum _{m=0}^{n-\ell}\binom{n-\ell}{m}\beta_{m}( \lambda ,y|\mathbf{a} ) \bigl\langle Q_{0,s}(t)|x^{n-\ell-m} \bigr\rangle \\& \quad =\sum_{\ell=0}^{n}\frac{\binom{n}{\ell}}{\binom{\ell +r}{r}} \lambda^{\ell}S_{1}(\ell+r,r)\sum _{m=0}^{n-\ell}\binom{n-\ell}{m}\beta_{m}( \lambda ,y|\mathbf{a} )\mathcal{E}_{n-\ell-m}(\lambda|\mathbf{b}), \end{aligned}

which completes the proof of the first formula.

The second formula can be obtained by using very similar techniques. □

## 3 Recurrence relations

In this section, we present several recurrence relations for Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. Our first recurrence is based on the polynomials $$(x|\lambda)_{n}$$.

### Theorem 3.1

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x+y|\mathbf{a};\mathbf{b})=\sum _{j=0}^{n}\binom{n}{j}D\mathcal {E}_{j}(\lambda,x|\mathbf{a};\mathbf{b}) (y|\lambda)_{n-j}.$$

### Proof

Let $$p_{n}(x)=\prod_{i=1}^{r} (\frac{e^{a_{i}t}-1}{t} )\prod_{i=1}^{s} (\frac{e^{b_{i}t}+1}{2} )D\mathcal{E}_{n}(\lambda ,x|\mathbf{a};\mathbf{b})$$. By (2.2) we have that $$p_{n}(x)=(x|\lambda)_{n}\sim (1,\frac{e^{\lambda t}-1}{\lambda } )$$, which leads to the required recurrence. □

The second recurrence is obtained from the fact that $$f(t)s_{n}(x)=ns_{n-1}(x)$$ for all $$s_{n}(x)\sim(g(t),f(t))$$ (see [1, 2]).

### Theorem 3.2

For all $$n\geq1$$,

$$D\mathcal{E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})-D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b})=n\lambda D \mathcal{E}_{n-1}(\lambda,x|\mathbf {a};\mathbf{b}).$$

### Proof

By (1.6) and $$f(t)s_{n}(x)=ns_{n-1}(x)$$ whenever $$s_{n}(x)\sim (g(t),f(t))$$, we have

$$\frac{e^{\lambda t}-1}{\lambda}D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b} )=nD\mathcal{E}_{n-1}(\lambda,x|\mathbf{a};\mathbf{b}),$$

which implies $$D\mathcal{E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})-D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b})=n\lambda D\mathcal{E}_{n-1}(\lambda,x|\mathbf {a};\mathbf{b})$$, as required. □

The next result gives an explicit formula for $$\frac{d}{dx}D\mathcal {E}_{n}(\lambda,x+\lambda|\mathbf{a};\mathbf{b})$$.

### Theorem 3.3

For all $$n\geq1$$,

$$\frac{d}{dx}D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=n! \sum_{\ell=0}^{n-1}\frac {(-\lambda)^{n-\ell-1}}{\ell!(n-\ell)}D \mathcal{E}_{\ell}(\lambda ,x|\mathbf{a} ;\mathbf{b}).$$

### Proof

It is well known that for $$s_{n}(x)\sim(g(t),f(t))$$, $$\frac {d}{dx}s_{n}(x)=\sum_{\ell=0}^{n-1}\binom{n}{\ell}\langle\bar {f}(t)|x^{n-\ell}\rangle s_{\ell}(x)$$ (see [1, 2]). In our case, by (1.6), we have

\begin{aligned} \bigl\langle \bar{f}(t)|x^{n-\ell}\bigr\rangle &= \biggl\langle \frac {1}{\lambda}\log (1+\lambda t)\Big|x^{n-\ell} \biggr\rangle \\ &= \lambda^{-1} \biggl\langle \sum_{m\geq1} \frac {(-1)^{m-1}(m-1)!\lambda ^{m}t^{m}}{m!}\Big|x^{n-\ell} \biggr\rangle \\ &=\lambda^{-1}(-1)^{n-\ell-1}\lambda^{n-\ell}(n-\ell-1)! \\ &=(- \lambda )^{n-\ell-1}(n-\ell-1)!. \end{aligned}

Thus $$\frac{d}{dx}D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf {b})=n!\sum_{\ell =0}^{n-1}\frac{(-\lambda)^{n-\ell-1}}{\ell!(n-\ell)}D\mathcal {E}_{\ell}(\lambda,x|\mathbf{a};\mathbf{b})$$, as required. □

Another recurrence relation can be stated as follows.

### Theorem 3.4

For all $$n\geq1$$,

\begin{aligned}& D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b}) \\& \quad = \Biggl(x-\sum_{i=1}^{r}a_{i}- \sum_{j=1}^{s}b_{j} \Biggr)D \mathcal {E}_{n-1}(\lambda,x-\lambda|\mathbf{a};\mathbf{b}) + \frac{r}{n}\sum_{\ell=0}^{n} \binom{n}{\ell}\lambda^{\ell}\mathfrak {b}_{\ell}D \mathcal{E}_{n-\ell}(\lambda,x-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \sum_{\ell=0}^{n}\binom{n}{\ell}\lambda ^{\ell}\mathfrak{b}_{\ell}D\mathcal{E}_{n-\ell}( \lambda,x-\lambda |a_{i},a_{1},\ldots ,a_{r}; \mathbf{b}) \\& \qquad {} +\frac{1}{2}\sum_{j=1}^{s} b_{j}D\mathcal{E}_{n-1}(\lambda,x-\lambda |\mathbf{a} ;b_{j},b_{1},\ldots,b_{s}), \end{aligned}

where $$\mathfrak{b}_{n}$$ is the nth Bernoulli number of the second kind, which is defined by $$\frac{t}{\log(1+t)}=\sum_{n\geq 0}\mathfrak {b}_{n}\frac{t^{n}}{n!}$$.

### Proof

Let $$n\geq1$$. Then

\begin{aligned}& D\mathcal{E}_{n}(\lambda,y|\mathbf{a};\mathbf{b}) \\& \quad = \biggl\langle \sum _{\ell\geq0}D\mathcal {E}_{\ell}(\lambda,y| \mathbf{a};\mathbf{b})\frac{t^{\ell}}{\ell !}\Big|x^{n} \biggr\rangle \\& \quad = \bigl\langle P_{r,s}(t) (1+\lambda t)^{y/\lambda}|x^{n} \bigr\rangle = \biggl\langle \frac{d}{dt} \bigl(P_{r,s}(t) (1+ \lambda t)^{y/\lambda } \bigr)\Big|x^{n-1} \biggr\rangle \\& \quad = \Biggl\langle \frac{d}{dt}\prod_{i=1}^{r} \biggl(\frac{\log (1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda }}+1} \biggr) (1+\lambda t)^{y/\lambda}\bigg|x^{n-1} \Biggr\rangle \end{aligned}
(3.1)
\begin{aligned}& \qquad {}+ \Biggl\langle \prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda ((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr)\frac{d}{dt}\prod_{i=1}^{s} \biggl(\frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda}}+1} \biggr) (1+\lambda t)^{y/\lambda}\bigg|x^{n-1} \Biggr\rangle \end{aligned}
(3.2)
\begin{aligned}& \qquad {}+ \biggl\langle P_{r,s}(t)\frac{d}{dt}(1+\lambda t)^{y/\lambda }\Big|x^{n-1} \biggr\rangle . \end{aligned}
(3.3)

By (1.6), the term in (3.3) equals

$$y \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda }|x^{n-1} \bigr\rangle =yD\mathcal{E}_{n-1}(\lambda,y-\lambda |\mathbf{a}; \mathbf{b} ).$$
(3.4)

For the term in (3.2), we observe that

$$\frac{d}{dt}\prod_{i=1}^{s} \biggl( \frac{2}{(1+\lambda t)^{\frac {b_{i}}{\lambda}}+1} \biggr) =\prod_{i=1}^{s} \biggl( \frac{2}{(1+\lambda t)^{\frac{b_{i}}{\lambda }}+1} \biggr)\sum_{i=1}^{s} \biggl(\frac{-b_{i}}{1+\lambda t}+\frac {b_{i}}{2(1+\lambda t)}\frac{2}{(1+\lambda t)^{b_{i}/\lambda}+1} \biggr).$$

So the term in (3.2) is

\begin{aligned}& -\sum_{j=1}^{s}b_{j} \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda}|x^{n-1} \bigr\rangle +\frac{1}{2}\sum_{j=1}^{s}b_{j} \biggl\langle P_{r,s}(t)\frac{2(1+\lambda t)^{(y-\lambda)/\lambda}}{(1+\lambda t)^{b_{j}/\lambda}+1}\Big|x^{n-1} \biggr\rangle \\& \quad =-\sum_{j=1}^{s}b_{j}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) + \frac{1}{2}\sum_{j=1}^{s}b_{j}D \mathcal{E}_{n-1}(\lambda,y-\lambda |\mathbf{a} ;b_{j},b_{1}, \ldots,b_{s}). \end{aligned}
(3.5)

For the term in (3.1), we note that

\begin{aligned}& (1+\lambda t)\frac{d}{dt}\prod_{i=1}^{r} \biggl(\frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr) \\& \quad =\prod_{i=1}^{r} \biggl( \frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{\frac{a_{i}}{\lambda}}-1)} \biggr) \Biggl(-\sum_{i=1}^{r}a_{i} +\frac {1}{t}\sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda }-1} \biggr) \Biggr), \end{aligned}

where $$\frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1}$$ has order at least 1. Thus, the term in (3.1) equals

\begin{aligned}& -\sum_{i=1}^{r}a_{i} \bigl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda}|x^{n-1} \bigr\rangle \\& \qquad {} + \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\bigg| \frac {1}{t}\sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1} \biggr)x^{n-1} \Biggr\rangle \\& \quad =-\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {} +\frac{1}{n} \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda )/\lambda }\bigg| \sum_{i=1}^{r} \biggl( \frac{\lambda t}{\log(1+\lambda t)}-\frac{a_{i} t}{(1+\lambda t)^{a_{i}/\lambda}-1} \biggr)x^{n} \Biggr\rangle \\& \quad =-\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b}) \\& \qquad {}+ \frac {r}{n} \Biggl\langle P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\bigg| \sum_{\ell\geq0}^{r}\mathfrak{b}_{\ell}\frac{\lambda^{\ell}t^{\ell}}{\ell !}x^{n} \Biggr\rangle \\& \qquad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \Biggl\langle \frac{\log(1+\lambda t)}{\lambda((1+\lambda t)^{a_{i}/\lambda}-1)}P_{r,s}(t) (1+\lambda t)^{(y-\lambda)/\lambda}\Bigg| \sum_{\ell\geq0}^{r}\mathfrak{b}_{\ell}\frac {\lambda^{\ell}t^{\ell}}{\ell!}x^{n} \Biggr\rangle , \end{aligned}

which is equal to

\begin{aligned}& -\sum_{i=1}^{r}a_{i}D \mathcal{E}_{n-1}(\lambda,y-\lambda|\mathbf {a};\mathbf{b})+ \frac {r}{n}\sum_{\ell=0}^{n} \binom{n}{\ell}\lambda^{\ell}\mathfrak {b}_{\ell}D \mathcal{E}_{n-\ell}(\lambda,y-\lambda|\mathbf{a};\mathbf {b}) \\& \quad {} -\frac{1}{n}\sum_{i=1}^{r}a_{i} \sum_{\ell=0}^{n}\binom{n}{\ell}\lambda ^{\ell}\mathfrak{b}_{\ell}D\mathcal{E}_{n-\ell}( \lambda,y-\lambda |a_{i},a_{1},\ldots ,a_{r}; \mathbf{b}). \end{aligned}
(3.6)

By using (3.4), (3.5) and (3.6) instead of (3.3), (3.2) and (3.1), respectively, we complete the proof. □

### Theorem 3.5

For all $$n\geq0$$,

\begin{aligned}& D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b}) \\& \quad =xD\mathcal{E}_{n}(\lambda,x-\lambda|\mathbf{a};\mathbf{b})-\sum _{i=1}^{r}a_{i}\sum _{m=0}^{n}S_{1}(n,m)\lambda^{n-m} \mathit{BE}_{m}(x-\lambda|\mathbf{a};\mathbf{b}) \\& \qquad {} -\sum_{m=0}^{n}\sum _{\ell=0}^{m}S_{1}(n,m)\lambda^{n-m} \binom {n}{m} \Biggl(\frac{B_{\ell+1}}{\ell+1}\sum_{i=1}^{r}a_{i}^{\ell+1}+ \frac{E_{\ell}(1)}{2}\sum_{j=1}^{s}b_{j}^{\ell+1} \Biggr) \\& \qquad {}\times\mathit{BE}_{m-\ell}(x-\lambda |\mathbf{a};\mathbf{b}), \end{aligned}

where $$B_{\ell}$$ is the ℓth Bernoulli number and $$E_{\ell}(1)$$ is the ℓth Euler polynomial evaluated at 1.

### Proof

It is well known that for $$s_{n}(x)\sim(g(t),f(t))$$, $$s_{n+1}(x)=(x-g'(t)/g(t))\frac{1}{f'(t)} s_{n}(x)$$ (see [1, 2]). In our case, by (1.6), we have

$$D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b})=xD\mathcal {E}_{n}(\lambda,x-\lambda |\mathbf{a};\mathbf{b})-e^{-\lambda t} \frac{g'(t)}{g(t)}D\mathcal {E}_{n}(\lambda,x|\mathbf{a} ;\mathbf{b}),$$

and by Theorem 2.1, we obtain

\begin{aligned} D\mathcal{E}_{n+1}(\lambda,x|\mathbf{a};\mathbf{b}) =&xD \mathcal {E}_{n}(\lambda,x-\lambda |\mathbf{a};\mathbf{b}) \\ &{}-\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}e^{-\lambda t}\frac {g'(t)}{g(t)}\mathit{BE}_{m}(x| \mathbf{a};\mathbf{b}). \end{aligned}
(3.7)

Note that

\begin{aligned} \frac{g'(t)}{g(t)} &=\bigl(\log g(t)\bigr)'=\sum _{i=1}^{r} \frac{a_{i}e^{a_{i}t}}{e^{a_{i}t}-1}-\frac {r}{t}+\sum _{j=1}^{s}\frac{b_{j}e^{b_{j}t}}{e^{b_{j}t}+1} \\ &=\sum_{i=1}^{r}a_{i}+ \frac{1}{t}\sum_{i=1}^{r} \biggl( \frac {a_{i}t}{e^{a_{i}t}-1}-1 \biggr)+\frac{1}{2}\sum_{j=1}^{s} \frac {2b_{j}e^{b_{j}t}}{e^{b_{j}t}+1} \\ &=\sum_{i=1}^{r}a_{i}+ \frac{1}{t}\sum_{i=1}^{r}\sum _{\ell\geq0}\beta _{\ell}a_{i}^{\ell}\frac{t^{\ell}}{\ell!}+\frac{1}{2}\sum_{j=1}^{s} \sum_{\ell\geq 0}E_{\ell}(1)b_{j}^{\ell+1} \frac{t^{\ell}}{\ell!} \\ &=\sum_{i=1}^{r}a_{i}+\sum _{\ell\geq0}\frac{\beta_{\ell+1}}{(\ell +1)!}\sum _{i=1}^{r}a_{i}^{\ell+1}t^{\ell}+\frac{1}{2}\sum_{\ell\geq0}\frac{E_{\ell}(1)}{\ell!}\sum _{j=1}^{s}b_{j}^{\ell +1}t^{\ell}. \end{aligned}

So

\begin{aligned} \frac{g'(t)}{g(t)}\mathit{BE}_{m}(x|\mathbf{a};\mathbf{b}) =&\sum _{i=1}^{r}a_{i}BE_{m}(x| \mathbf{a};\mathbf{b})+\sum_{\ell=0}^{m} \binom{m}{\ell}\frac{\beta_{\ell+1}}{\ell+1}\sum_{i=1}^{r}a_{i}^{\ell+1} \mathit{BE}_{m-\ell}(x|\mathbf{a};\mathbf{b}) \\ &{}+\frac{1}{2}\sum_{\ell=0}^{m} \binom{m}{\ell}E_{\ell}(1)\sum_{j=1}^{s}b_{j}^{\ell +1} \mathit{BE}_{m-\ell}(x|\mathbf{a};\mathbf{b}). \end{aligned}

Hence, by substituting into (3.7), we complete the proof. □

## 4 Relations with other families of polynomials

In this section, we establish a connection between Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials and several known families of polynomials.

### Theorem 4.1

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n}\binom{n}{m}D\mathcal {E}_{n-m}(\lambda|\mathbf{a};\mathbf{b}) (x|\lambda)_{m}.$$

### Proof

Note that $$(x|\lambda)_{n}\sim(1,\frac{e^{\lambda t}-1}{\lambda})$$. Let $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}(x|\lambda)_{m}$$. By (1.3) and (1.6), we have

\begin{aligned} c_{n,m} &=\frac{1}{m!} \bigl\langle P_{r,s}(t)|t^{m}x^{n} \bigr\rangle =\binom {n}{m} \bigl\langle P_{r,s}(t)|x^{n-m} \bigr\rangle \\ &=\binom {n}{m}D\mathcal {E}_{n-m}(\lambda|\mathbf{a}; \mathbf{b}), \end{aligned}

which completes the proof. □

For the following, we note that $$B_{n}^{(\alpha)}(x)\sim(\frac {(e^{t}-1)^{\alpha}}{t^{\alpha}},t)$$.

### Theorem 4.2

For all $$n\geq0$$, the polynomial $$D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})$$ is given by

$$\sum_{m=0}^{n} \Biggl(\sum _{\ell=m}^{n}\sum_{k=0}^{n-\ell} \sum_{q=0}^{n-\ell -k}\sum _{p=0}^{q} \frac{\binom{n}{\ell}\binom{n-\ell}{k}\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}a_{\ell,k,q,p}D \mathcal{E}_{n-\ell-k-q}(\lambda |\mathbf{a} ;\mathbf{b}) \Biggr)B^{(\alpha)}_{m}(x),$$

where $$a_{\ell,k,q,p}=S_{1}(\ell,m)S_{1}(q+\alpha,q-p+\alpha)S_{2}(q-p+\alpha ,\alpha )\lambda^{k+\ell+p-m}b_{\ell}^{(\alpha)}$$ and $$b_{\ell}^{(\alpha)}$$ is the ℓth Bernoulli number of the second kind of order α given by $$(\frac{t}{\log(1+t)})^{\alpha}=\sum_{\ell\geq0}b_{\ell }^{(\alpha)}\frac {t^{\ell}}{k!}$$.

### Proof

Let $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}B_{m}^{(\alpha )}(x)$$. By (1.3) and (1.6), we have

\begin{aligned} c_{n,m}&=\frac{1}{m!\lambda^{m}} \biggl\langle P_{r,s}(t) \biggl( \frac {(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}\biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{\alpha}\Big|\bigl(\log(1+\lambda t)\bigr)^{m}x^{n} \biggr\rangle \\ &=\frac{1}{\lambda^{m}}\sum_{\ell=m}^{n} \binom{n}{\ell}\lambda^{\ell }S_{1}(\ell,m) \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda }-1}{t} \biggr)^{\alpha}\bigg| \biggl( \frac{\lambda t}{\log(1+\lambda t)} \biggr)^{\alpha}x^{n-\ell} \biggr\rangle \\ &=\frac{1}{\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\binom {n}{\ell } \binom{n-\ell}{k}S_{1}(\ell,m)\lambda^{\ell+k}b_{k}^{(\alpha)} \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}\Big|x^{n-\ell-k} \biggr\rangle . \end{aligned}

One can show that

\begin{aligned} \biggl(\frac{(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{\alpha}&= \biggl(\frac {e^{\frac{1}{\lambda}\log(1+\lambda t)}-1}{t} \biggr)^{\alpha}\\ &=\sum_{q\geq0}\sum_{p=0}^{q} \binom{q+\alpha}{\alpha }^{-1}S_{1}(q+\alpha ,q-p+ \alpha)S_{2}(q-p+\alpha,\alpha)\lambda^{p} \frac{t^{q}}{q!}, \end{aligned}

where $$S_{2}(n,m)$$ is the Stirling number of the second kind. Thus,

\begin{aligned}& \biggl\langle P_{r,s}(t) \biggl(\frac{(1+\lambda t)^{1/\lambda }-1}{t} \biggr)^{\alpha}\Big|x^{n-\ell-k} \biggr\rangle \\& \quad =\sum_{q=0}^{n-\ell-k}\sum _{p=0}^{q}\frac{\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}S_{1}(q+ \alpha,q-p+\alpha)S_{2}(q-p+\alpha,\alpha )\lambda ^{p}\bigl\langle P_{r,s}(t)|x^{n-\ell-k-q}\bigr\rangle , \end{aligned}

where $$\langle P_{r,s}(t)|x^{n-\ell-k-q}\rangle=D\mathcal{E}_{n-\ell -k-q}(\lambda|\mathbf{a};\mathbf{b})$$. Hence,

$$c_{n,m} =\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{q=0}^{n-\ell -k} \sum_{p=0}^{q} \frac{\binom{n}{\ell}\binom{n-\ell}{k}\binom{n-\ell -k}{q}}{\binom {q+\alpha}{\alpha}}a_{\ell,k,q,p}D \mathcal{E}_{n-\ell-k-q}(\lambda |\mathbf{a} ;\mathbf{b}),$$

which completes the proof. □

By similar techniques as in the proof of the last theorem, we can express our polynomials $$D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})$$ in terms of the degenerate Bernoulli polynomials $$\beta_{n}^{(\alpha)}(\lambda,x)$$ of order α. These polynomials are the Sheffer sequence which is given by $$\beta_{n}^{(\alpha)}(\lambda,x)\sim((\frac{\lambda (e^{t}-1)}{e^{\lambda t}-1})^{\alpha},\frac{e^{\lambda t}-1}{\lambda})$$.

### Theorem 4.3

For all $$n\geq0$$, the polynomial $$D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})$$ is given by

$$\sum_{m=0}^{n}\binom{n}{m}c_{n,m} \beta^{(\alpha)}_{m}(\lambda,x),$$

where $$c_{n,m}=\sum_{q=0}^{n-m}\sum_{p=0}^{q} \frac{\binom{n-m}{q}}{\binom{q+\alpha}{\alpha}}S_{1}(q+\alpha ,q-p+\alpha )S_{2}(q-p+\alpha,\alpha)\lambda^{p}D\mathcal{E}_{n-m-q}(\lambda |\mathbf{a};\mathbf{b})$$.

Now we are interested in expressing our polynomials in terms of $$H_{n}^{(\alpha)}(x|\mu)$$ which are called the Frobenius-Euler polynomials of order α. Note that $$H_{n}^{(\alpha)}(x|\mu)\sim ( (\frac{e^{t}-\mu}{1-\mu} )^{\alpha},t )$$ (see [10, 24]).

### Theorem 4.4

For all $$n\geq0$$,

$$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum _{m=0}^{n} \biggl(\frac {a_{n,m}}{(1-\mu)^{\alpha}\lambda^{m}} \biggr)H^{(\alpha)}_{m}(x|\mu),$$

where

$$a_{n,m} =\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{p=0}^{\alpha}\binom {n}{\ell }\binom{n-\ell}{k}\binom{\alpha}{p}S_{1}(\ell,m) \lambda^{\ell}(-\mu )^{\alpha-p}D\mathcal{E}_{k}(\lambda| \mathbf{a};\mathbf{b}) (p|\lambda)_{n-\ell-k}.$$

### Proof

Let $$D\mathcal{E}_{n}(\lambda,x|\mathbf{a};\mathbf{b})=\sum_{m=0}^{n}c_{n,m}H^{(\alpha )}_{m}(x|\mu)$$. By (1.3) and (1.6), we have

\begin{aligned} c_{n,m}&=\frac{1}{m!(1-\mu)^{\alpha}\lambda^{m}} \bigl\langle P_{r,s}(t) \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|\bigl(\log(1+\lambda t)\bigr)^{m}x^{n} \bigr\rangle \\ &=\frac{1}{m!(1-\mu)^{\alpha}\lambda^{m}} \biggl\langle P_{r,s}(t) \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}\Big|m!\sum_{\ell\geq m}S_{1}( \ell ,m)\frac{\lambda^{\ell}}{\ell!}t^{\ell}x^{n} \biggr\rangle \\ &=\frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n} \binom{n}{\ell }S_{1}(\ell,m)\lambda^{\ell}\bigl\langle \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|P_{r,s}(t)x^{n-\ell} \bigr\rangle \\ &=\frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell }\binom{n}{\ell} \binom{n-\ell}{k}S_{1}(\ell,m)\lambda^{\ell}D\mathcal {E}_{k}(\lambda|\mathbf{a};\mathbf{b})w_{n,\ell,k}, \end{aligned}

where

\begin{aligned} w_{n,\ell,k}&= \bigl\langle \bigl((1+\lambda t)^{1/\lambda}-\mu \bigr)^{\alpha}|x^{n-\ell-k} \bigr\rangle \\ &= \Biggl\langle \sum_{p=0}^{\alpha}\binom{\alpha}{p}(-\mu)^{\alpha -p}(1+\lambda t)^{p/\lambda}\bigg|x^{n-\ell-k} \Biggr\rangle \\ &=\sum_{p=0}^{\alpha}\binom{\alpha}{p}(- \mu)^{\alpha-p} \biggl\langle \sum_{q\geq0}(p| \lambda)_{q}\frac{t^{q}}{q!}\Big|x^{n-\ell-k} \biggr\rangle \\ &=\sum_{p=0}^{\alpha}\binom{\alpha}{p}(- \mu)^{\alpha-p} (p|\lambda )_{n-\ell-k}. \end{aligned}

Thus, the constants $$c_{n,m}$$ are given by

$$\frac{1}{(1-\mu)^{\alpha}\lambda^{m}}\sum_{\ell=m}^{n}\sum _{k=0}^{n-\ell}\sum_{p=0}^{\alpha}\binom{n}{\ell}\binom{n-\ell}{k}\binom{\alpha }{p}S_{1}(\ell ,m) \lambda^{\ell}(-\mu)^{\alpha-p}D\mathcal{E}_{k}(\lambda| \mathbf {a};\mathbf{b}) (p|\lambda)_{n-\ell-k},$$

which completes the proof. □

Now we are interested in expressing our polynomials in terms of $$\mathcal{E}_{n}^{(\alpha)}(\lambda,x)$$ which are called the degenerate Euler polynomials of order α. Note that

$$\mathcal{E}_{n}^{(\alpha)}(\lambda,x)\sim \biggl( \biggl( \frac {e^{t}+1}{2} \biggr)^{\alpha},\frac{e^{\lambda t}-1}{\lambda} \biggr)$$

(see [10]). Using similar techniques as in the proof of the above theorem, we obtain the following relation.

### Theorem 4.5

For all $$n\geq0$$, the polynomial $$D\mathcal{E}_{n}(\lambda,x|\mathbf {a};\mathbf{b})$$ is given by

$$\frac{1}{2^{\alpha}}\sum_{m=0}^{n} \binom{n}{m} \Biggl(\sum_{q=0}^{n-m} \sum_{p=0}^{\alpha}\binom{n-m}{q} \binom{\alpha}{p}(p|\lambda)_{q}D\mathcal {E}_{n-m-q}(\lambda| \mathbf{a};\mathbf{b}) \Biggr)\mathcal {E}^{(\alpha)}_{m}( \lambda,x).$$

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Correspondence to Dae San Kim.

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Dolgy, D.V., Kim, D.S., Kim, T. et al. Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomials. J Inequal Appl 2015, 154 (2015). https://doi.org/10.1186/s13660-015-0676-6