Theory and Modern Applications

# A note on $$(h,q)$$-Boole polynomials

## Abstract

Kim et al. (Appl. Math. Inf. Sci. 9(6):1-6, 2015) consider the q-extensions of Boole polynomials. In this paper, we consider Witt-type formula for the q-Boole polynomials with weights and derive some new interesting identities and properties of those polynomials and numbers from the Witt-type formula which are related to special polynomials and numbers.

## 1 Introduction

Let p be a fixed odd prime number. Throughout this paper, $$\mathbb {Z}_{p}$$, $$\mathbb{Q}_{p}$$, and $$\mathbb{C}_{p}$$ will, respectively, denote the ring of p-adic rational integers, the field of p-adic rational numbers, and the completions of algebraic closure of $$\mathbb{Q}_{p}$$. The p-adic norm is defined by $$|p|_{p} =\frac{1}{p}$$.

When one talks of q-extension, q is variously considered as an indeterminate, a complex $$q \in\mathbb{C}$$, or p-adic number $$q \in\mathbb{C}_{p}$$. If $$q \in\mathbb{C}$$, one normally assumes that $$|q|<1$$. If $$q \in\mathbb{C}_{p}$$, then we assume that $$|q-1|_{p} < p^{- \frac{1}{p-1}}$$ so that $$q^{x} = \exp(x\log q)$$ for each $$x \in\mathbb{Z}_{p}$$. Throughout this paper, we use the notation

$$[x]_{-q} =\frac{1- (-q)^{x}}{1-(-q)}.$$

Note that $$\lim_{q \rightarrow-1} [x]_{-q} = x$$ for each $$x \in \mathbb{Z}_{p}$$.

Let $$UD(\mathbb{Z}_{p})$$ be the space of uniformly differentiable functions on $$\mathbb{Z}_{p}$$. For $$f\in UD({\mathbb{Z}_{p}})$$, the p-adic invariant integral on $${\mathbb{Z}_{p}}$$ is defined by Kim as follows:

$$I_{-q}(f)= \int_{\mathbb{Z}_{p}}f(x)\,d\mu_{-q}(x) =\lim_{N \rightarrow\infty} \frac{1}{[p^{N}]_{-q}}\sum _{x=0}^{p^{N}-1} f(x) (-q)^{x}\quad\mbox{(see [1--5])}.$$
(1.1)

Let $$f_{1}$$ be the translation of f with $$f_{1} (x )=f (x+1 )$$. Then, by (1.1), we get

$$I_{-q}(f_{1})+I_{-q} (f)=[2]_{q} f(0).$$
(1.2)

As is well known, the Stirling number of the first kind is defined by

$$(x )_{n}=x (x-1 )\cdots (x-n+1 )=\sum _{l=0}^{n}S_{1} (n,l )x^{l},$$
(1.3)

and the Stirling number of the second kind is given by the generating function:

$$\bigl(e^{t}-1 \bigr)^{m}=m!\sum _{l=m}^{\infty}S_{2} (l,m )\frac {t^{l}}{l!}\quad \mbox{(see [6, 7])}.$$
(1.4)

It is well known that the $$(h,q)$$ -Euler polynomials are defined by the generating function:

$$\biggl(\frac{q+1}{q^{h}e^{t}+1} \biggr)e^{xt}=\sum _{n=0}^{\infty}E_{n,q} (x|h)\frac{t^{n}}{n!}\quad\mbox{(see [8])},$$
(1.5)

where h is an integer. When $$x=0$$ and $$h=0$$, $$E_{n,q} (0|h) =E_{n,q}(h)$$ are called the ordinary q-Euler numbers.

Recently, DS Kim and T Kim introduced the Changhee polynomials of the first kind are defined by the generating function:

$$\frac{2}{2+t}(1+t)^{x}=\sum _{n=0}^{\infty}Ch_{n}(x)\frac{t^{n}}{n!}\quad\mbox{(see [1, 9--11])},$$
(1.6)

and T Kim et al. defined the q-Changhee polynomials as follows:

$$\frac{[2]_{q}}{q(1+t)+1}(1+t)^{x}=\sum _{n=0} ^{\infty} Ch_{n,q}(x)\frac {t^{n}}{n!}\quad\mbox{(see [9, 11, 12])}.$$
(1.7)

As is well known, the Boole polynomials are defined by the generating function:

$$\sum_{n=0} ^{\infty}Bl_{n}(x|\lambda) \frac{t^{n}}{n!}=\frac {1}{1+(1+t)^{\lambda}}\quad\mbox{(see [7, 13])}.$$

When $$\lambda=1$$, $$2Bl_{n}(x|1)=Ch_{n}(x)$$ are Changhee polynomials. In [11], Kim et al. consider the q-analog of Boole polynomials, and found some new and interesting identities related to special polynomials, and Y Do and D Lim investigated the properties of $$(h,q)$$-Daehee numbers and polynomials, which are defined by

$$\int _{\mathbb {Z}_{p}}q^{-hy}(x+y)_{n}\,d\mu_{q}(y)\quad\mbox{(see [14])}.$$

In this paper, we consider Witt-type formula for the q-Boole polynomials with weights and derive some new interesting identities and properties of those polynomials and numbers from the Witt-type formula which are related to special polynomials and numbers.

## 2 q-Analog of Boole polynomials with weight

In this section, we assume that $$t\in{\mathbb{C}}_{p}$$ with $$|t|_{p}< p^{-\frac{1}{p-1}}$$, $$\lambda\in{\mathbb{Z}}_{p}$$ with $$\lambda \neq0$$ and $$h\in{\mathbb{Z}}$$. From (1.2), we have

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}(1+t)^{x+\lambda y}\,d\mu_{-q}(y)=\frac {1+q}{q^{h}(1+t)^{\lambda}+1}(1+t)^{x} =\sum_{n=0} ^{\infty}[2]_{q}Bl_{n,q}(x|h, \lambda)\frac{t^{n}}{n!},$$
(2.1)

where $$Bl_{n,q} (x|h,\lambda)$$ are the $$(h,q)$$-Boole polynomials which are defined by

$$\frac{1}{q^{h}(1+t)^{\lambda}+1}(1+t)^{x}=\sum _{n=0} ^{\infty }Bl_{n,q}(x|h,\lambda) \frac{t^{n}}{n!}.$$
(2.2)

By (2.1), we can derive the following equation:

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom{x+\lambda y}{n}\,d\mu_{-q}= \frac{1+q}{n!}Bl_{n,q} (x|h,\lambda).$$
(2.3)

In the special case $$x=0$$, $$Bl_{n,q} (0|h,\lambda)=Bl_{n,q}(h,\lambda)$$ are called the $$(h,q)$$ -Boole numbers.

Note that

\begin{aligned}[b] (1+t)^{x+\lambda y}&=e^{(x+\lambda y)\log(1+t)}\\ &=\sum_{n=0} ^{\infty}\frac{(x+\lambda y)^{n}}{n!}\bigl( \log(1+t)\bigr)^{n}\\ &=\sum_{n=0} ^{\infty}\frac{(x+\lambda y)^{n}}{n!}m!\sum _{m=n} ^{\infty }S_{1}(m,n) \frac{t^{m}}{m!}\\ &=\sum_{n=0} ^{\infty} \Biggl\{ \sum _{m=0} ^{n} (x+\lambda y)^{m}S_{1}(n,m) \Biggr\} \frac{t^{n}}{n!}. \end{aligned}
(2.4)

The $$(h,q)$$ -Euler polynomials are defined by the generating function:

$$\frac{1+q}{q^{h}e^{t}+1}e^{xt}=\sum _{n=0} ^{\infty}E_{n,q}(x|h)\frac{t^{n}}{n!}.$$
(2.5)

Note that $$\lim_{q\rightarrow1}E_{n,q}(x|1)=E_{n}(x)$$. When $$x=0$$, $$E_{n}(0|h)=E_{n,q}(h)$$ are called the $$(h,q)$$ -Euler numbers.

By (1.2), we can derive easily the following equation:

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}e^{(x+y)t}\,d\mu_{-q}(y)=\frac{1+q}{q^{h}e^{t}+1}e^{xt} =\sum_{n=0} ^{\infty}E_{n,q} (x|h) \frac{t^{n}}{n!}.$$
(2.6)

Since

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}e^{(x+y)t}\,d\mu_{-q}(y)=\sum _{n=0} ^{\infty} \int _{\mathbb {Z}_{p}}q^{(h-1)y}(x+y)^{n} \,d \mu_{-q}(y)\frac{t^{n}}{n!},$$

by (2.5), we have

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}(x+y)^{n}\,d\mu_{-q}(y)=E_{n,q} (x|h) \quad(n\geq0).$$
(2.7)

From (2.4) and (2.7), we get

\begin{aligned}[b] &\int _{\mathbb {Z}_{p}}q^{(h-1)y}(1+t)^{x+\lambda y}\,d\mu_{-q}(y)\\ &\quad=\sum_{n=0} ^{\infty} \Biggl\{ \sum _{m=0} ^{n} \int _{\mathbb {Z}_{p}}q^{(h-1)y}(x+\lambda y)^{m} \,d\mu_{-q}(y) S_{1}(n,m) \Biggr\} \frac{t^{n}}{n!}\\ &\quad=\sum_{n=0} ^{\infty} \Biggl\{ \sum _{m=0} ^{n}\lambda^{m} E_{m,q} \biggl(\frac{x}{\lambda} \vert h \biggr)S_{1}(n,m) \Biggr\} \frac{t^{n}}{n!}. \end{aligned}
(2.8)

Thus, by (2.2), (2.3), and (2.8), we obtain the following theorem.

### Theorem 2.1

For $$n \geq0$$, we have

$$Bl_{n,q}(x|h,\lambda)=\frac{1}{[2]_{q}}\sum _{m=0} ^{n} \lambda^{m}E_{m,q} \biggl(\frac{x}{\lambda}\Big| h \biggr)S_{1}(n,m)$$

and

$$\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom{x+\lambda y}{n}\,d\mu_{-q}=\frac{[2]_{q}}{n!}Bl_{n,q} (x|h,\lambda).$$

By TheoremÂ 2.1, we note that

$$Bl_{n,q}(x|h,\lambda)=\frac{1}{[2]_{q}}\int _{\mathbb {Z}_{p}}q^{(h-1)y}(x+\lambda y)_{n}\,d\mu_{-q}(y),$$

where $$(x)_{n}=x(x-1)\cdots(x-n+1)$$. When $$\lambda=1$$ and $$h=0$$, we have

$$Bl_{n,q}(x|0,1)=\frac{1}{[2]_{q}}\int _{\mathbb {Z}_{p}}q^{-1}(x+y)^{n}\,d\mu_{-q}(y).$$
(2.9)

In [13], Arici et al. defined the q-analog of Changhee polynomials by the generating function:

$$\sum_{n=0} ^{\infty}Ch_{n}(x|q) \frac{t^{n}}{n!}=\frac{[2]_{q}}{[2]_{t}+1}(1+t)^{x}.$$
(2.10)

By (2.10), we have

$$\int _{\mathbb {Z}_{p}}q^{-y}(1+t)^{x+y}\,d\mu_{-q}(y)= \frac{[2]_{q}}{[2]_{t}+1}(1+t)^{x}=\sum_{n=0} ^{\infty}Ch_{n}(x|q)\frac{t^{n}}{n!}.$$
(2.11)

By (1.6) and (2.10), we note that

$$\frac{[2]_{q}}{2}Ch_{n}(x)=Ch_{n}(x|q).$$
(2.12)

From (2.11), we get

$$\int _{\mathbb {Z}_{p}}q^{-1}(x+y)_{n}\,d\mu_{-q}(y)=Ch_{n}(x|q).$$
(2.13)

By (2.9), (2.12), and (2.13), we have

$$Bl_{n,q} (x|0,1)=\frac{1}{[2]_{q}}Ch_{n}(x|q)= \frac{1}{2}Ch_{n}(x).$$

By replacing t as $$e^{t}-1$$ in (2.1), we derive the following equations:

\begin{aligned}[b] \frac{1+q}{q^{h}e^{\lambda t}+1}e^{xt}&=\sum _{n=0} ^{\infty }[2]_{q}Bl_{n,q}(x|h, \lambda)\frac{1}{n!}\bigl(e^{t}-1\bigr)^{n}\\ &=\sum_{n=0} ^{\infty}[2]_{q}Bl_{n,q}(x|h, \lambda)\frac{1}{n!}n!\sum_{m=n} ^{\infty}S_{2}(m,n)\frac{t^{m}}{m!}\\ &=\sum_{n=0} ^{\infty}\sum _{m=0} ^{n} [2]_{q}Bl_{m,q}(x|h, \lambda )S_{2}(n,m)\frac{t^{n}}{n!} \end{aligned}
(2.14)

and

$$\frac{1+q}{q^{h}e^{\lambda t}+1}e^{xt}= \frac{1+q}{q^{h}e^{\lambda t}+1}e^{ (\frac{x}{\lambda} )\lambda t} =\sum_{n=0} ^{\infty}E_{n,q} \biggl( \frac{x}{\lambda}\Big| h \biggr)\lambda^{m}\frac{t^{m}}{m!}.$$
(2.15)

Hence, by (2.14) and (2.15), we obtain the following theorem.

### Theorem 2.2

For $$n \geq0$$, we have

$$\sum_{m=0} ^{n}Bl_{m,q}(x|h, \lambda)S_{2}(n,m)=\frac{\lambda ^{m}}{q+1}E_{n,q} \biggl( \frac{x}{\lambda}\Big| h \biggr).$$

From now on, we define the $$(h_{1},\ldots,h_{r},q)$$ -Boole numbers of the first kind as follows:

\begin{aligned}[b] &[2]_{q} ^{r} Bl_{n,q} ^{(h_{1},\ldots,h_{r})} (\lambda)\\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\bigl(\lambda(x_{1}+ \cdots+x_{r})\bigr)_{n}\,d\mu _{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \quad(n \geq0). \end{aligned}
(2.16)

By (2.16), we have

\begin{aligned} &[2]_{q} ^{r} \sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})} (\lambda) \frac{t^{n}}{n!} \\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}\sum_{n=0} ^{\infty} q^{h_{1}+\cdots+h_{r}-r}\binom{\lambda (x_{1}+\cdots+x_{r})}{n}t^{n}\,d\mu_{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\sum_{n=0} ^{\infty} \int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}(1+t)^{\lambda (x_{1}+\cdots+x_{k})}\,d\mu_{-q}(x_{1})\cdots \,d \mu_{-q}(x_{r}) \\ &\quad=\prod_{i=1} ^{r} \biggl( \frac{1+q}{q^{h_{i}}(1+t)^{\lambda}+1} \biggr) \\ &\quad=(1+q)^{r}\sum_{n=0} ^{\infty} \biggl(\sum_{l_{1}+\cdots+l_{r}=n}\binom {n}{l_{1},\ldots,l_{r}}B_{i_{1},q} (h, \lambda)\cdots B_{i_{r},q} (h,\lambda ) \biggr)\frac{t^{n}}{n!}. \end{aligned}
(2.17)

Thus, by (2.17), we obtain the following corollary.

### Corollary 2.3

For $$n \geq0$$, we have

$$Bl_{n,q} ^{(h_{1},\ldots,h_{r})} (\lambda)=\sum_{l_{1}+\cdots+l_{r}=n} \binom {n}{l_{1},\ldots,l_{r}}B_{i_{1},q} (h,\lambda)\cdots B_{i_{r},q} (h,\lambda).$$

The $$(h_{1},\ldots,h_{r},q)$$ -Euler polynomials are defined by the generating function to be

\begin{aligned} &\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}e^{(x_{1}+\cdots+x_{r}+x)t}\,d\mu _{-q}(x_{1})\cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\prod_{i=1} ^{r} \biggl( \frac{1+q}{q^{h_{i}}e^{t}+1} \biggr)e^{xt} \\ &\quad=\sum_{n=0} ^{\infty}E_{n,q}(x|h_{1},\ldots,h_{r}) \frac{t^{n}}{n!}. \end{aligned}
(2.18)

By (2.18), we have

$$\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}(x_{1}+\cdots+x_{r}+x)^{n}\,d \mu _{-q}(x_{1})\cdots \,d\mu_{-q}(x_{r})=E_{n,q} (x|h_{1},\ldots,h_{r}).$$

In the special case $$x=0$$, $$E_{n,q} (0|h_{1},\ldots,h_{r})=E_{n,q} (h_{1},\ldots,h_{r})$$ are called the $$(h_{1},\ldots,h_{r},q)$$ -Euler numbers.

From (1.5) and (2.16), we note that

\begin{aligned} &(1+q)^{r}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda) \\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\bigl(\lambda(x_{1}+ \cdots+x_{r})\bigr)_{n}\,d\mu _{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\sum_{l=0} ^{n} S_{1}(n,l)\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\lambda ^{l}(x_{1}+ \cdots+x_{r})^{l}\,d\mu_{-q}(x_{1})\cdots \,d \mu_{-q}(x_{r}) \\ &\quad=\sum_{l=0} ^{n} S_{1}(n,l) \lambda^{l}E_{l,q}(h_{1},\ldots,h_{r}). \end{aligned}
(2.19)

Therefore, by (2.19), we obtain the following theorem.

### Theorem 2.4

For $$n \geq0$$, we get

$$Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda)=\frac{1}{(1+q)^{r}}\sum _{l=0} ^{n} S_{1}(n,l)\lambda^{l}E_{l,q}(h_{1}, \ldots,h_{r}).$$

By replacing t by $$e^{t}-1$$ in (2.17), we have

\begin{aligned}{} [2]_{q} ^{r}\sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda)\frac {(e^{t}-1)^{n}}{n!}&=\prod_{i=1} ^{r} \biggl(\frac{1+q}{q^{h_{i}}e^{\lambda t}+1} \biggr) \\ &=\sum_{n=0} ^{\infty}E_{n,q} (h_{1},\ldots,h_{r})\lambda^{n}\frac{t^{n}}{n!} \end{aligned}
(2.20)

and

\begin{aligned}{} [2]_{q} ^{r}\sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda)\frac {1}{n!}\bigl(e^{t}-1 \bigr)^{n} &=[2]_{q} ^{r}\sum_{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda)\sum _{m=n} ^{\infty}S_{2}(m,n)\frac{t^{m}}{m!} \\ &=[2]_{q} ^{r}\sum_{n=0} ^{\infty} \Biggl\{ \sum_{m=0} ^{n} Bl_{m,q} ^{(h_{1},\ldots,h_{r})}(\lambda)S_{2}(n,m) \Biggr\} \frac{t^{n}}{n!}. \end{aligned}
(2.21)

Hence, by (2.20) and (2.21), we obtain the following theorem.

### Theorem 2.5

For $$n \geq0$$, we have

$$\frac{\lambda^{n}}{[2]_{q} ^{r}}E_{n,q} (h_{1},\ldots,h_{r})= \sum_{m=0} ^{n} Bl_{m,q} ^{(h_{1},\ldots,h_{r})}(\lambda)S_{2}(n,m).$$

Let us define the $$(h_{1},\ldots,h_{r},q)$$ -Boole polynomials of the first kind as follows:

\begin{aligned} &[2]_{q} ^{r}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda) \\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\bigl(\lambda(x_{1}+ \cdots+x_{r})+x\bigr)_{n}\,d\mu _{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}), \end{aligned}
(2.22)

where $$n \geq0$$ and $$r\in{\mathbb{N}}$$. By (2.22), we can derive the generating function of the $$(h_{1},\ldots ,h_{r},q)$$-Boole polynomials of the first kind as follows:

\begin{aligned} &[2]_{q} ^{r}\sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda )\frac{t^{n}}{n!} \\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}(1+t)^{\lambda(x_{1}+\cdots+x_{r})+x}\,d\mu _{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\prod_{i=1} ^{r} \biggl( \frac{1+q}{q^{h_{i}}(1+t)^{\lambda}+1} \biggr) (1+t)^{x}. \end{aligned}
(2.23)

By (2.23), we can see easily

\begin{aligned} &\prod_{i=1} ^{r} \biggl(\frac{1+q}{q^{h_{i}}(1+t)^{\lambda}+1} \biggr) (1+t)^{x} \\ &\quad=[2]_{q} ^{r} \Biggl(\sum_{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(\lambda )\frac{t^{n}}{n!} \Biggr) \Biggl(\sum_{m=0} ^{\infty} \binom{x}{m}t^{m} \Biggr) \\ &\quad=[2]_{q} ^{r} \sum_{n=0} ^{\infty} \Biggl(\sum_{m=0} ^{n} m! \binom{x}{m}\frac {n!}{(n-m)!m!}Bl_{n-m,q} ^{(h_{1},\ldots,h_{r})}(\lambda) \Biggr)\frac {t^{n}}{n!} \\ &\quad=[2]_{q} ^{r} \sum_{n=0} ^{\infty} \Biggl(\sum_{m=0} ^{n} m! \binom{x}{m}\binom {n}{m}Bl_{n-m,q} ^{(h_{1},\ldots,h_{r})}(\lambda) \Biggr)\frac{t^{n}}{n!} \\ &\quad=[2]_{q} ^{r} \sum_{n=0} ^{\infty} \Biggl(\sum_{m=0} ^{n} \binom {n}{m}Bl_{n-m,q} ^{(h_{1},\ldots,h_{r})}(\lambda) (x)_{m} \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(2.24)

By (2.23) and (2.24), we obtain the following theorem.

### Theorem 2.6

For $$n \geq0$$, we have

$$Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda)=\sum_{m=0} ^{n} \binom {n}{m}Bl_{n-m,q} ^{(h_{1},\ldots,h_{r})}(\lambda) (x)_{m}.$$

Replacing t as $$e^{t}-1$$ in (2.23), we get

\begin{aligned}{} [2]_{q} ^{r}\sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda)\frac {1}{n!}\bigl(e^{t}-1 \bigr)^{n}&=\prod_{i=1} ^{n} \biggl(\frac{1+q}{q^{h_{i}}e^{\lambda t}+1} \biggr)e^{xt} \\ &=\sum_{n=0} ^{\infty}E_{n,q} ^{(h_{1},\ldots,h_{r})} \biggl(\frac{x}{\lambda } \biggr)\lambda^{n} \frac{t^{n}}{n!} \end{aligned}
(2.25)

and

\begin{aligned} &[2]_{q} ^{r}\sum _{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda )\frac{(e^{t}-1)^{n}}{n!} \\ &\quad=[2]_{q} ^{r}\sum_{n=0} ^{\infty} \Biggl(\sum_{m=0} ^{n} Bl_{m,q} ^{(h_{1},\ldots ,h_{r})}(x|\lambda)S_{2}(n,m) \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(2.26)

Hence, by (2.25) and (2.26), we obtain the following theorem.

### Theorem 2.7

For $$n \geq0$$, we have

$$\sum_{m=0} ^{n} Bl_{m,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda)S_{2}(n,m)=\frac {\lambda^{n}}{[2]_{q} ^{r}}E_{n,q} ^{(h_{1},\ldots,h_{r})} \biggl(\frac{x}{\lambda } \biggr).$$

From (2.23), we get

\begin{aligned} &[2]_{q} ^{r}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda) \\ &\quad=\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\bigl(\lambda(x_{1}+ \cdots+x_{r})+x\bigr)_{n}\,d\mu _{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\sum_{l=0} ^{n} S_{1}(n,l) \int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{h_{1}+\cdots+h_{r}-r}\bigl(\lambda(x_{1}+\cdots +x_{r})+x\bigr)^{l}\,d\mu_{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\sum_{l=0} ^{n} S_{1}(n,l) \lambda^{l}E_{n,q} ^{(h_{1},\ldots,h_{r})} \biggl(\frac {x}{\lambda} \biggr). \end{aligned}
(2.27)

Thus, by (2.27), we obtain the following theorem.

### Theorem 2.8

For $$n\geq0$$, we have

$$Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda)=\frac{1}{[2]_{q} ^{r}}\sum _{l=0} ^{n} S_{1}(n,l)\lambda^{l}E_{n,q} ^{(h_{1},\ldots,h_{r})} \biggl(\frac{x}{\lambda } \biggr).$$

Now, we define the $$(h,q)$$ -Boole polynomials of the second kind as follows:

$${\widehat{Bl}}_{n,q}(x|h,\lambda)=\frac{1}{[2]_{q}}\int _{\mathbb {Z}_{p}}q^{(h-1)y}(-\lambda y+x)_{n}\,d\mu_{-q}(y) \quad(n \geq0).$$
(2.28)

By (2.28), we have

\begin{aligned} {\widehat{Bl}}_{n,q}(x|h, \lambda)&=\frac{1}{[2]_{q}}\sum_{l=0} ^{n}(- \lambda )^{l} S_{1}(n,l)\int _{\mathbb {Z}_{p}}\biggl(y-\frac{x}{\lambda} \biggr)^{l}\,d\mu_{-q}(y) \\ &=\frac{1}{[2]_{q}}\sum_{l=0} ^{n}(- \lambda)^{l} S_{1}(n,l)E_{l,q} \biggl(- \frac {x}{\lambda} \biggr). \end{aligned}
(2.29)

In the special case $$x=0$$, $${\widehat{Bl}}_{n,q} (0|h,\lambda)={\widehat {Bl}}_{n,q}(h,\lambda)$$ are called the $$(h,q)$$ -Boole numbers of the second kind. From (2.29), we can derive the generating function of $${\widehat{Bl}}_{n,q}(x|\lambda)$$ as follows:

\begin{aligned} \sum_{n=0} ^{\infty}{\widehat{Bl}}_{n,q}(x|h,\lambda)\frac {t^{n}}{n!}&= \frac{1}{[2]_{q}}\int _{\mathbb {Z}_{p}}q^{(h-1)y}(1+t)^{-\lambda y+x}\,d\mu _{-q}(y) \\ &=\frac{(1+t)^{\lambda}}{q^{h}+(1+t)^{\lambda}}(1+t)^{x}. \end{aligned}
(2.30)

By replacing t by $$e^{t}-1$$ in (2.30), we have

\begin{aligned} \sum_{n=0} ^{\infty}{\widehat{Bl}}_{n,q}(x|h,\lambda)\frac {(e^{t}-1)^{n}}{n!}&= \frac{e^{\lambda t}}{q^{h}+e^{\lambda t}}e^{xt} \\ &=\frac{1}{1+q}\sum_{n=0} ^{\infty}(- \lambda)^{n}E_{n,q} \biggl(-\frac {\lambda}{x}\Big| h \biggr)\frac{t^{n}}{n!} \end{aligned}
(2.31)

and

$$\sum_{n=0} ^{\infty}{ \widehat{Bl}}_{n,q}(x|h,\lambda)\frac{(e^{t}-1)^{n}}{n!} =\sum _{n=0} ^{\infty} \Biggl(\sum_{m=0} ^{n} {\widehat {Bl}}_{m,q}(x|h,\lambda)S_{2}(n,m) \Biggr)\frac{t^{n}}{n!}.$$
(2.32)

By (2.31) and (2.32), we obtain the following theorem.

### Theorem 2.9

For $$n \geq0$$, we have

$${\widehat{Bl}}_{n,q}(x|h,\lambda)=\frac{1}{[2]_{q}}\sum _{l=0} ^{n}(-\lambda )^{l} S_{1}(n,l)E_{l,q} \biggl(-\frac{x}{\lambda} \biggr)$$

and

$$\frac{1}{[2]_{q}}(-\lambda)^{n}E_{n,q} \biggl(- \frac{\lambda}{x}\Big| h \biggr)=\sum_{m=0} ^{n} {\widehat{Bl}}_{m,q}(x|h,\lambda)S_{2}(n,m).$$

For $$h_{1},\ldots,h_{r}\in{\mathbb{Z}}$$, we define the $$(h_{1},\ldots ,h_{r},q)$$ -Boole polynomials of the second kind as follows:

\begin{aligned} &{\widehat{Bl}}_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda) \\ &\quad=\frac{1}{q+1}\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}q^{(h_{1}+\cdots+h_{r}-r)y} \bigl(-\lambda (x_{1}+\cdots+x_{r})+x \bigr)_{n}\,d \mu_{-q}(x_{1})\cdots \,d\mu_{-q}(x_{r}). \end{aligned}
(2.33)

By (2.33), we can derive the generating function of the $$(h_{1},\ldots ,h_{r},q)$$-Boole polynomials of the second kind as follows:

\begin{aligned} &\sum_{n=0} ^{\infty}{\widehat{Bl}}_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda ) \frac{t^{n}}{n!} \\ &\quad=\frac{1}{(1+q)^{r}}\int _{\mathbb {Z}_{p}}\cdots \int _{\mathbb {Z}_{p}}(1+t)^{-\lambda x_{1}-\cdots-\lambda x_{r}+x}\,d\mu_{-q}(x_{1}) \cdots \,d\mu_{-q}(x_{r}) \\ &\quad=\prod_{i=1} ^{r} \biggl( \frac{(1+t)^{\lambda}}{q^{h_{i}}+(1+t)^{\lambda }} \biggr) (1+t)^{x} \\ &\quad=\prod_{i=1} ^{r} \biggl( \frac{1}{q^{h_{i}}(1+t)^{-\lambda}+1} \biggr) (1+t)^{x} \\ &\quad=\sum_{n=0} ^{\infty}Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|-\lambda)\frac{t^{n}}{n!}. \end{aligned}
(2.34)

Hence, by (2.34), we obtain the following proposition.

### Proposition 2.10

For $$n \geq0$$, we have

$${\widehat{Bl}}_{n,q} ^{(h_{1},\ldots,h_{r})}(x|\lambda)=Bl_{n,q} ^{(h_{1},\ldots,h_{r})}(x|-\lambda).$$

Note that

\begin{aligned} \frac{(-1)^{n}[2]_{q}}{n!}Bl_{n,q} (x|h, \lambda)&=(-1)^{n}\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom {x+\lambda y}{n}\,d\mu_{-q}(y) \\ &=\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom{-x-\lambda y+n-1}{n}\,d\mu_{-q}(y) \\ &=\int _{\mathbb {Z}_{p}}q^{(h-1)y}\sum_{m=0} ^{n} \binom{-x-\lambda y}{m}\binom{n-1}{n-m}\,d\mu _{-q}(y) \\ &=\sum_{m=0} ^{n} \binom{n-1}{n-m}\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom{-x-\lambda y}{m}\,d\mu_{-q}(y) \\ &=[2]_{q}\sum_{m=0} ^{n} \binom{n-1}{n-m}\frac{{\widehat{Bl}}_{m,q} (-x|h,\lambda)}{m!}, \end{aligned}
(2.35)

and, by a similar method, we get

\begin{aligned} \frac{(-1)^{n}[2]_{q}}{n!}{\widehat{Bl}}_{n,q} (x|h,\lambda)&=(-1)^{n}\int _{\mathbb {Z}_{p}}q^{(h-1)y}\binom{x-\lambda y}{n}\,d\mu_{-q}(y) \\ &=[2]_{q}\sum_{m=0} ^{n} \binom{n-1}{n-m}\frac{Bl_{m,q} (-x|h,\lambda)}{m!}. \end{aligned}
(2.36)

By (2.35) and (2.36), we obtain the following theorem.

### Theorem 2.11

For $$n \geq0$$, we have

$$\frac{(-1)^{n}}{n!}Bl_{n,q} (x|h,\lambda)=\sum _{m=0} ^{n} \binom {n-1}{n-m}\frac{{\widehat{Bl}}_{m,q} (-x|h,\lambda)}{m!}$$

and

$$\frac{(-1)^{n}}{n!}{\widehat{Bl}}_{n,q} (x|h,\lambda)=\sum _{m=0} ^{n} \binom {n-1}{n-m}\frac{Bl_{m,q} (-x|h,\lambda)}{m!}.$$

By TheoremÂ 2.11, we obtain the following corollary.

### Corollary 2.12

For $$n \geq0$$, we have

$$Bl_{n,q} (x|h,\lambda)=\sum_{m=0} ^{n} \sum_{k=0} ^{m} (-1)^{n+m}\binom {n}{n-m,m-k,k}(n-1)_{l-1}Bl_{k,q}(x|h, \lambda)$$

where $$\binom{n}{p,q,r}=\frac{n!}{p!q!r!}$$, $$p+q+r=n$$.

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## Acknowledgements

The authors are grateful for the valuable comments and suggestions of the referees.

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Correspondence to Jin-Woo Park.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

The authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kwon, J., Park, JW. A note on $$(h,q)$$-Boole polynomials. Adv Differ Equ 2015, 198 (2015). https://doi.org/10.1186/s13662-015-0536-1