Open Access

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

  • Dmitry V Dolgy1,
  • Dae San Kim2Email author,
  • Taekyun Kim3 and
  • Toufik Mansour4
Journal of Inequalities and Applications20152015:154

https://doi.org/10.1186/s13660-015-0676-6

Received: 20 February 2015

Accepted: 24 April 2015

Published: 8 May 2015

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.

Keywords

Barnes-type Daehee with λ-parameter and degenerate Euler mixed-type polynomial umbral calculus

MSC

05A15 05A40 11B83

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

Declarations

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Institute of Natural Sciences, Far Eastern Federal University
(2)
Department of Mathematics, Sogang University
(3)
Department of Mathematics, Kwangwoon University
(4)
Department of Mathematics, University of Haifa

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© Dolgy et al.; licensee Springer. 2015