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\begin{center}
\vskip 1cm{\LARGE\bf On $q$-Boson Operators and $q$-Analogues of the\\
\vskip .1in
$r$-Whitney and $r$-Dowling Numbers}
\vskip 1cm
\large    
Mahid M. Mangontarum\\
Department of Mathematics\\
Mindanao State University -- Main Campus\\
Marawi City 9700\\
Philippines \\
\href{mailto:mmangontarum@yahoo.com}{\tt mmangontarum@yahoo.com} \\
\href{mailto:mangontarum.mahid@msumain.edu.ph}{\tt mangontarum.mahid@msumain.edu.ph} \\
\ \\    
Jacob Katriel\\
Department of Chemistry\\
Technion -- Israel Institute of Technology\\
Haifa 32000\\
Israel\\
\href{mailto:jkatriel@technion.ac.il}{\tt jkatriel@technion.ac.il}\\
\end{center}

\vskip .2 in

\begin{abstract}
We define the $(q,r)$-Whitney numbers of the first and second kinds in terms of the $q$-Boson operators,
and obtain several fundamental properties such 
as recurrence formulas, orthogonality and inverse relations, and other interesting identities. 
As a special case, we obtain a $q$-analogue of the $r$-Stirling numbers of the first and second kinds. 
Finally, we define the $(q,r)$-Dowling polynomials in terms of sums of $(q,r)$-Whitney numbers of the second 
kind, and obtain some of their properties.
\end{abstract}

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\section{Introduction}
\label{sec:Introduction}

The investigation of $q$-analogues of combinatorial identities has proven to be a rich source of insight as well as of useful 
generalizations. Some examples of $q$-analogues are  the \textit{$q$-real number}, the \textit{$q$-factorial} and the \textit{$q$-falling factorial of order $r$}, respectively, given by 
\begin{equation*}
[x]_q=\frac{q^x-1}{q-1},\ [n]_q!=\prod_{i=1}^n[i]_q,\ [x]_{q,n}=\prod_{i=0}^{r-1}[x-i]_q,
\end{equation*}
for any real number $x$ and non-negative integers $n$ and $r$, and the \textit{$q$-binomial coefficients (also known as Gaussian polynomials)}
\begin{equation*}
\binom{n}{r}_q=\frac{[n]_q!}{[r]_q![n-r]_q!}=\frac{[n]_{q,r}}{[r]_q!}.
\end{equation*}
The formulation of $q$-analogues is not unique, but some choices appear to allow more productive generalizations than others.
In the present paper we apply the properties of the $q$-boson operators as a framework for the generation of $q$-deformations of 
a family of combinatorial identities involving the Whitney numbers.

A lattice $L$ in which every element is the join of elements $x$ and $y$ (in $L$) such that $x$ and $y$ cover the zero element $0$, and 
is semimodular, is called a \textit{geometric lattice}. Originally, if $L$ is a finite lattice of rank $n$, then the Whitney numbers 
$w(n,k)$ and $W(n,k)$ of the first and second kinds of $L$ are defined as the coefficients of the characteristic polynomial and as the number of elements of $L$ of corank $k$, respectively. Now, Dowling \cite{Dowl} defined a class of these geometric lattices, called \textit{Dowling lattice}, which is a generalization of the partition lattice. Let $Q_n(G)$ be the Dowling lattice of rank $n$ associated to a finite group $G$ of order $m&gt;0$. Benoumhani \cite{Ben1} defined the \textit{Whitney numbers of the first and second kind of $Q_n(G)$}, denoted by $w_m(n,k)$ and $W_m(n,k)$, respectively, in terms of the relations
\begin{equation}
\label{w01}
m^n(x)_n=\sum_{k=0}^{n}w_m(n,k)(mx+1)^k
\end{equation}
and
\begin{equation}
\label{W02}
(mx+1)^n=\sum_{k=0}^{n}m^kW_m(n,k)(x)_k,
\end{equation}
where $(x)_n=x(x-1)\cdots(x-n+1)$ is the falling factorial of $x$ of order $n$. Notice that if the group $G$ is the trivial group ($m=1$), 
multiplication of both equations \eqref{w01} and \eqref{W02} by $(x+1)$ yields the horizontal generating functions for the well-known Stirling numbers of the first and second kind \cite{Stir}, denoted by $\fstirling{n}{k}$ and $\sstirling{n}{k}$, respectively. Hence,
\begin{equation*}
w_1(n,k)=\fstirling{n+1}{k+1},\ W_1(n,k)=\sstirling{n+1}{k+1}.
\end{equation*}
We note that Benoumhani \cite{Ben1,Ben2} already established the fundamental properties of the numbers $w_m(n,k)$ and $W_m(n,k)$ while Dowling \cite{Dowl} gave a detailed discussion of geometric lattices. Other generalizations of the Stirling numbers $\fstirling{n}{k}$ and $\sstirling{n}{k}$ were already considered by several authors. For instance, Broder \cite{Bro} defined the $r$-Stirling numbers $\widehat{\fstirling{n+r}{k+r}}_r$ and 
$\widehat{\sstirling{n+r}{k+r}}_r$ of the first and second kind whose relation to the Whitney numbers is 
stated in equations \eqref{w7} and \eqref{w8} below. Belbachir and Bousbaa \cite{Bel} recently introduced the translated Whitney numbers $\widetilde{w}_{(\alpha)}(n,k)$ and $\widetilde{W}_{(\alpha)}(n,k)$ of the first and second kind, which are related to the Stirling numbers via 
\begin{equation*}
\widetilde{w}_{(\alpha)}(n,k)=\alpha^{n-k}\fstirling{n}{k},\ \widetilde{W}_{(\alpha)}(n,k)=\alpha^{n-k}\sstirling{n}{k}.
\end{equation*}
Furthermore, Mez\H{o} \cite{Mez1} defined the $r$-Whitney numbers 
$w_{m,r}(n,k)$ and $W_{m,r}(n,k)$ of the first and second kind as the coefficients in the expressions
\begin{equation}
m^n(x)_n=\sum_{k=0}^{n}w_{m,r}(n,k)(mx+r)^k\label{w1}
\end{equation}
and
\begin{equation}
(mx+r)^n=\sum_{k=0}^{n}m^kW_{m,r}(n,k)(x)_k.\label{w2}
\end{equation}
respectively. Further developement of the numbers $w_{m,r}(n,k)$ and $W_{m,r}(n,k)$ were due to Cheon and Jung \cite{Cheon}, Merca \cite{Merca}, Corcino et al.\ \cite{Tina2}, Corcino et al.\ \cite{CorMonSus}, C. B. Corcino and R. B. Corcino \cite{Tina1}, and R. B. Corcino and C. B Corcino \cite{CorTin1,CorTin2}. 

Corcino and Hererra \cite{CorHer} introduced the \textit{limit of the differences of the generalized factorial} 
\begin{equation}
F_{\alpha,\gamma}(n,k)=\lim_{\beta\rightarrow 0}\frac{\left[\Delta_t^k\left(\beta t+\gamma|\alpha\right)_n\right]_{t=0}}{k!\beta^k},\label{ldgf}
\end{equation}
where
\begin{equation}
\left(\beta t+\gamma|\alpha\right)_n=\prod_{j=0}^{n-1}\left(\beta t+\gamma-j\alpha\right),\ \left(\beta t+\gamma|\alpha\right)_0=1,
\end{equation}
which is a generalization of the Stirling numbers of the first kind. The numbers $F_{\alpha,-\gamma}(n,k)$ are actually the $r$-Whitney numbers of the first kind in \eqref{w1}. That is,
\begin{equation*}
F_{\alpha,-\gamma}(n,k)=w_{\alpha,\gamma}(n,k).
\end{equation*}
Similarly, Corcino \cite{CorRB} defined the \textit{$(r,\beta)$-Stirling numbers} $\rbstirling{n}{k}_{r,\beta}$ as coefficients in
\begin{equation}
t^n=\sum_{k=0}^n\binom{\frac{t-r}{\beta}}{k}\beta^{k}k!\rbstirling{n}{k}_{r,\beta}.\label{rbeta}
\end{equation}
The numbers $\rbstirling{n}{k}_{r,\beta}$ are found to be equivalent to the $r$-Whitney numbers of the second kind in \eqref{w2}. To be precise,
\begin{equation*}
\rbstirling{n}{k}_{r,\beta}=W_{\beta,r}(n,k).
\end{equation*}
Corcino et al.\ \cite{CorTinAl}, and Corcino and Aldema \cite{CorAl} further studied the numbers $\rbstirling{n}{k}_{r,\beta}$.

Recall that the classical \textit{Boson operators} $a$ and $a^\dagger$ satisfy the commutation relation
\begin{equation}
[a,a^\dagger]\equiv aa^\dagger-a^\dagger a=1.\label{bos}
\end{equation}
If we define the Fock space by the basis $\{\ket{s} \; ; \; s=0,1,2,\ldots\}$, to be referred to as Fock states, the relations 
$a\ket{s}=\sqrt{s}\ket{s-1}$ and $a^{\dagger}\ket{s}=\sqrt{s+1}\ket{s+1}$ form a representation that satisfies the commutation relation \eqref{bos}. The operator $\hat{n}\equiv a^\dagger a$, when acting on $\ket{s}$, yields
\begin{equation*}
a^\dagger a\ket{s}=s\ket{s},
\end{equation*}
and the operator $(a^\dagger)^k a^k$, when acting on the same state, yields
\begin{equation*}
(a^\dagger)^k a^k\ket{s}=(s)_k\ket{s}.
\end{equation*}
Let $\{\bra{s}\equiv (\ket{s})^{\dagger} \; ; \; s=0,1,2,\ldots\}$ denote the Fock basis of the dual space.
Requiring the normalization of the scalar product $\braket{0}{0}=1$ we note that 
\begin{equation*}
\braket{s+1}{s+1}=\frac{1}{s+1}\bra{s}aa^{\dagger}\ket{s}=\frac{1}{s+1}\Big(\bra{s}a^{\dagger}a\ket{s}+\braket{s}{s}\Big)=\braket{s}{s}.
\end{equation*}
Hence, from the normalization of $|0&gt;$ it follows that all the Fock states are normalized. Moreover, since
$\bra{s+1}a^{\dagger}\ket{s}=\sqrt{s+1}\braket{s+1}{s+1}$ and $(a\ket{s+1})^{\dagger}\ket{s}=\sqrt{s+1}\braket{s}{s}$, it follows that $a^{\dagger}$ is the 
Hermitian conjugate of $a$. That is, $a^{\dagger}a$ is Hermitian.
Orthogonality follows from the fact that the Fock states are eigenstates of $a^{\dagger}a$ with distinct eigenvalues.

Hence, the \textit{horizontal generating functions} of the Stirling numbers $\fstirling{n}{k}$ and $\sstirling{n}{k}$,
\begin{equation}
\label{standard}
(x)_n=\sum_{k=0}^n (-1)^{n-k}\fstirling{n}{k}x^k
\end{equation}
and
\begin{equation*}
x^n=\sum_{k=0}^n\sstirling{n}{k}(x)_k,
\end{equation*}
can be expressed as
\begin{equation*}
(a^\dagger)^n a^n=\sum_{k=0}^n (-1)^{n-k}\fstirling{n}{k}(a^\dagger a)^k
\end{equation*}
and
\begin{equation*}
(a^\dagger a)^n=\sum_{k=0}^n\sstirling{n}{k}(a^\dagger)^k a^k,
\end{equation*}
respectively \cite{Kat1}.

Now, the defining relations for the $r$-Whitney numbers, \eqref{w1} and \eqref{w2}, can be expressed as
\begin{equation}
m^n(a^\dagger)^n a^n=\sum_{k=0}^{n}w_{m,r}(n,k)(ma^\dagger a+r)^k\label{w3}
\end{equation}
and
\begin{equation}
(ma^\dagger a+r)^n=\sum_{k=0}^{n}m^kW_{m,r}(n,k)(a^\dagger)^k a^k.\label{w4}
\end{equation}

Making use of the $q$-Boson operators \cite{Arik} that satisfy
\begin{equation}
[a,a^\dagger]_q\equiv aa^\dagger-qa^\dagger a=1,\label{qbos}
\end{equation}
we have
\begin{equation*}
a\ket{s}=\sqrt{[s]_q}\ket{s-1},  \  a^{\dagger}\ket{s}=\sqrt{[s+1]_q}\ket{s+1},
\end{equation*}
hence,
\begin{equation*}
a^\dagger a\ket{s}=[s]_q\ket{s},
\end{equation*}
and 
\begin{equation*}
(a^\dagger)^k a^k\ket{s}=[s]_{q,k}\ket{s}.
\end{equation*}
\begin{remark}
Although we use the same notation for the boson and for the $q$-boson operators, no confusion should arise because
the meaning of these symbols should be clear from the context.
\end{remark}
In line with this, the defining relations for Carlitz's \cite{Car3} $q$-Stirling numbers of the first and second kind, $\fstirling{n}{k}_q$ and $\sstirling{n}{k}_q$,
can be written in the form \cite{Kat1}
\begin{equation}
(a^\dagger)^n a^n=\sum_{k=1}^n(-1)^{n-k}\fstirling{n}{k}_q(a^\dagger a)^k\label{c1}
\end{equation}
and
\begin{equation}
(a^\dagger a)^n=\sum_{k=1}^n\sstirling{n}{k}_q(a^\dagger)^k a^k,\label{c2}
\end{equation}
respectively. 

We define $q$-analogues for the Whitney numbers $w_{m,r}(n,k)$ and $W_{m,r}(n,k)$ via 
the same pattern as in \eqref{c1} and \eqref{c2}.

\section{$(q,r)$-Whitney numbers}
\label{sec:Whitney Numbers}

\begin{definition}
For non-negative integers $n$ and $k$ and complex numbers $r$ and $m$, the $(q,r)$-Whitney numbers of the 
first and second kind, denoted by $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$, respectively, are defined by
\begin{equation}
m^n(a^\dagger)^n a^n=\sum_{k=0}^{n}w_{m,r,q}(n,k)(ma^\dagger a+r)^k\label{qw1}
\end{equation}
and
\begin{equation}
(ma^\dagger a+r)^n=\sum_{k=0}^{n}m^kW_{m,r,q}(n,k)(a^\dagger)^k a^k\label{qw2}
\end{equation}
with initial conditions $w_{m,r,q}(0,0)=W_{m,r,q}(0,0)=1$ and $w_{m,r,q}(n,k)=W_{m,r,q}(n,k)=0$ for $k&gt;n$ and for $k&lt;0$, where 
the operators $a^\dagger$ and $a$ satisfy the relation in \eqref{qbos}.
\end{definition}

Before proceeding we note that from \eqref{qw1} and \eqref{qw2},
\begin{eqnarray}
w_{m,0,q}(n,k) &amp;=&amp; (-m)^{n-k}\fstirling{n}{k}_q,\label{w5} \\
W_{m,0,q}(n,k) &amp;=&amp; m^{n-k}\sstirling{n}{k}_q.\label{w6}
\end{eqnarray}
Similarly, the $r$-Stirling numbers $\widehat{\fstirling{n+r}{k+r}}_r$ and $\widehat{\sstirling{n+r}{k+r}}_r$ are specified by the horizontal generating functions
\begin{equation*}
(x-r)_n=\sum_{k=0}^n(-1)^{n-k}\widehat{\fstirling{n+r}{k+r}}_rx^k,
\end{equation*}
or, equivalently,
\begin{equation*}
(x)_n=\sum_{k=0}^n(-1)^{n-k}\widehat{\fstirling{n+r}{k+r}}_r (x+r)^k,
\end{equation*}
and
\begin{equation*}
(x+r)^n=\sum_{k=0}^n\widehat{\sstirling{n+r}{k+r}}_r(x)_r.
\end{equation*}
Hence, $\widehat{\fstirling{n+r}{k+r}}_{q,r}$ and $\widehat{\sstirling{n+r}{k+r}}_{q,r}$, the $q$-analogues of 
$\widehat{\fstirling{n+r}{k+r}}_r$ and $\widehat{\sstirling{n+r}{k+r}}_r$, respectively, are specified by the horizontal generating functions
\begin{eqnarray}
(a^{\dagger})^na^n &amp;=&amp; \sum_{k=0}^n(-1)^{n-k}\widehat{\fstirling{n+r}{k+r}}_{q,r}(a^{\dagger}a+r)^k,\label{qr1}\\
(a^{\dagger}a+r)^n &amp;=&amp; \sum_{k=0}^n\widehat{\sstirling{n+r}{k+r}}_{q,r}(a^{\dagger})^ka^k.\label{qr2}
\end{eqnarray}
It follows that
\begin{eqnarray}
w_{1,r,q}(n,k) &amp;=&amp; (-1)^{n-k}\widehat{\fstirling{n+r}{k+r}}_{q,r},\label{w7} \\
W_{1,r,q}(n,k) &amp;=&amp; \widehat{\sstirling{n+r}{k+r}}_{q,r}.\label{w8}
\end{eqnarray}
We will refer to the $q$-analogues in \eqref{qr1} and \eqref{qr2} as the $(q,r)$-Stirling numbers of the first and second kind, respectively.

\begin{theorem}
The $(q,r)$-Whitney numbers $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$ satisfy the following identities:
\begin{equation}
w_{m,r,q}(n,k)=(-1)^{n-k}\sum_{i=k}^n \binom{i}{k}r^{i-k}m^{n-i}\fstirling{n}{i}_q, \label{identity1}
\end{equation}
\begin{equation}
W_{m,r,q}(n,k)=\sum_{i=k}^n\binom{n}{i}r^{n-i}m^{i-k}\sstirling{i}{k}_q.\label{identity2}
\end{equation}
\end{theorem}

\begin{proof}
From Eq.~\eqref{c1}, we get
\begin{eqnarray*}
m^n(a^\dagger)^n a^n&amp;=&amp;m^n\sum_{i=0}^n(-1)^{n-i}\fstirling{n}{i}_q(a^\dagger a)^i\\
&amp;=&amp;m^n\sum_{i=0}^n(-1)^{n-i}\fstirling{n}{i}_q\left(\frac{\hat{z}-r}{m }\right)^i\\
&amp;=&amp;m^n\sum_{i=0}^n(-1)^{n-i}\fstirling{n}{i}_q\frac{1}{m^i}\sum_{k=0}^i\binom{i}{k}\hat{z}^k(-r)^{i-k}\\
&amp;=&amp;\sum_{k=0}^n (-1)^{n-k}\left\{\sum_{i=k}^n m^{n-i}\fstirling{n}{i}_q\binom{i}{k}r^{i-k}\right\}\hat{z}^k,
\end{eqnarray*}
where $\hat{z}=ma^\dagger a+r$. Furthermore, comparing the coefficient of $\hat{z}^k$ with that in equation \eqref{qw1} yields equation \eqref{identity1}.

To prove equation \eqref{identity2}, we write
\begin{eqnarray*}
(ma^\dagger a+r)^n&amp;=&amp;\sum_{i=0}^n\binom{n}{i}r^{n-i}m^i(a^\dagger a)^i\\
&amp;=&amp;\sum_{i=0}^n\binom{n}{i}r^{n-i}m^i\sum_{k=0}^i\sstirling{i}{k}_q(a^\dagger)^k a^k\\
&amp;=&amp;\sum_{k=0}^n\left\{\sum_{i=k}^nr^{n-i}m^i\sstirling{i}{k}_q\binom{n}{i}\right\}(a^\dagger)^k a^k.
\end{eqnarray*}
Comparing the coefficient of $(a^\dagger)^k a^k$ with that in equation \eqref{qw2} gives us \eqref{identity2}.
\end{proof}

\begin{remark}

\noindent
(a) As $q\rightarrow 1$, we have
\begin{equation*}
w_{m,r}(n,k)=\sum_{i=k}^n(-1)^{n-k}\binom{i}{k}r^{i-k}m^{n-i}\fstirling{n}{i};
\end{equation*}
\begin{equation*}
W_{m,r}(n,k)=\sum_{i=k}^n\binom{n}{i}r^{n-i}m^{i-k}\sstirling{i}{k}.
\end{equation*}

\noindent
(b) Note that of all the factors in equations \eqref{identity1} and \eqref{identity2} only the Stirling numbers are $q$-deformed.
\end{remark}
The following corollary is a direct consequence of the previous theorem.
\begin{corollary}The $(q,r)$-Stirling numbers are given by
\begin{equation}
\widehat{\fstirling{n+r}{k+r}}_{q,r}=\sum_{i=k}^n\binom{i}{k}r^{i-k}\fstirling{n}{i}_q;
\end{equation}
\begin{equation}
\widehat{\sstirling{n+r}{k+r}}_{q,r}=\sum_{i=k}^n\binom{n}{i}r^{n-i}\sstirling{i}{k}_q.
\end{equation}
\end{corollary}


\section{Some recurrence relations}
\label{sec:Recurrence}

In this section, we present some recurrence relations involving the $(q,r)$-Whitney numbers.  

We recall the $q$-boson identities
\begin{equation*}
[a,(a^{\dagger})^n]_{q^n}=[n]_q(a^{\dagger})^{n-1}
\end{equation*}
and
\begin{equation*}
[a^n,a^{\dagger}]_{q^n}=[n]_qa^{n-1},
\end{equation*}
that can be easily established by induction.
The latter can also be written in the form
\begin{equation*}
a^{\dagger}a^n=q^{-n}(a^na^{\dagger}-[n]_qa^{n-1}).
\end{equation*}

\begin{theorem}
The $(q,r)$-Whitney numbers $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$ satisfy the following triangular recurrence relations:
\begin{equation}
w_{m,r,q}(n+1,k)=q^{-n}\Big(w_{m,r,q}(n,k-1)-(m[n]_q+r)w_{m,r,q}(n,k)\Big), \label{identity4}
\end{equation}
\begin{equation}
W_{m,r,q}(n+1,k)=q^{k-1}W_{m,r,q}(n,k-1)+(m[k]_q+r)W_{m,r,q}(n,k).\label{identity5}
\end{equation}
\end{theorem}
\begin{proof}
From equation \eqref{qw1},
$\displaystyle\sum_{k=0}^{n+1}w_{m,r,q}(n+1,k)(ma^{\dagger}a+r)^k=m^{n+1}(a^{\dagger})^n(a^{\dagger}a^n)a$
\begin{eqnarray*}
&amp;=&amp;m^{n+1}(a^{\dagger})^nq^{-n}(a^na^{\dagger}-[n]_qa^{n-1})a\\
&amp;=&amp;m^{n+1}q^{-n}\left((a^{\dagger})^na^n\right)(a^{\dagger}a)-m^{n+1}q^{-n}[n]_q(a^{\dagger})^na^n\\
&amp;=&amp;q^{-n}\sum_{k=0}^nw_{m,r,q}(n,k)(ma^{\dagger}a+r)^k(ma^{\dagger}a+r-r)-mq^{-n}[n]_q\sum_{k=0}^nw_{m,r,q}(n,k)(ma^{\dagger}a+r)^k\\
&amp;=&amp;q^{-n}\sum_{k=1}^{n+1}w_{m,r,q}(n,k-1)(ma^{\dagger}a+r)^k-q^{-n}(m[n]_q+r)\sum_{k=0}^nw_{m,r,q}(n,k)(ma^{\dagger}a+r)^k\\
&amp;=&amp;q^{-n}\sum_{k=0}^{n+1} \left\{w_{m,r,q}(n,k-1)-(m[n]_q+r)w_{m,r,q}(n,k)\right\}(ma^{\dagger}a+r)^k.
\end{eqnarray*}
Equating coefficients of $(ma^{\dagger}a+r)^k$ gives us \eqref{identity4} and a similar derivation yields equation \eqref{identity5}.
\end{proof}

Equations \eqref{identity4} and \eqref{identity5} are useful in computing the first few values of $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$,
using the initial values specified above. 

\begin{remark}

\noindent
(a) From \eqref{identity4} we obtain the explicit expression
$$w_{m,r,q}(n,0)=(-1)^n q^{-\frac{n(n-1)}{2}} \prod_{i=0}^{n-1}(m[i]_q+r).$$
On the other hand, the relation \eqref{identity1} yields
$$w_{m,r,q}(n,0)=(-1)^n\sum_{i=0}^n r^im^{n-i}\fstirling{n}{i}_q.$$
Equating these expressions and substituting $x=\frac{r}{m}$ we obtain
$$\sum_{i=0}^n\fstirling{n}{i}_q x^i = q^{-\frac{n(n-1)}{2}}\prod_{i=0}^{n-1}([i]_q+x).$$
This is a horizontal generating function for the $q$-Stirling numbers of the first kind in terms of a $q$-analogue
of the rising factorial. Indeed, replacing $x$ by $-[s]_q$, and noting that 
$$[s]_q-[i]_q=q^{-i}[s-i]_q$$
and 
$$\prod_{i=0}^{n-1}q^i = q^{\binom{n}{2}},$$
we obtain
$$\sum_{i=0}^n \fstirling{n}{i}_q(-1)^i[s]_q^i =(-1)^n\prod_{i=0}^{n-1}[s-i]_q.$$

\noindent
(b) From \eqref{identity5} $W_{m,r,q}(n+1,0)=rW_{m,r,q}(n,0)$, hence $W_{m,r,q}(n,0)=r^n$.
The same result is obtained from \eqref{identity2}. That is, 
$$W_{m,r,q}(n,0)=\sum_{i=0}^n\binom{n}{i}r^{n-i}m^i\delta_{i,0}=r^n.$$

\noindent
(c) As $q\rightarrow 1$, we have
\begin{equation*}
w_{m,r}(n+1,k)=w_{m,r}(n,k-1)-(mn+r)w_{m,r}(n,k);
\end{equation*}
\begin{equation*}
W_{m,r}(n+1,k)=W_{m,r}(n,k-1)+(mk+r)W_{m,r}(n,k).
\end{equation*}
This confirms that $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$ are proper $q$-analogues of $w_{m,r}(n,k)$ and $W_{m,r}(n,k)$, respectively.
\end{remark}

As a consequence of the previous theorem, when $m=1$ we have
\begin{corollary}
The $(q,r)$-Stirling numbers satisfy the following triangular recurrence relations:
\begin{eqnarray*}
\widehat{\fstirling{n+1+r}{k+r}}_{q,r}&amp;=&amp;q^{-n}\widehat{\fstirling{n+r}{k-1+r}}_{q,r}+([n]_q+r)q^{-n}\widehat{\fstirling{n+r}{k+r}}_{q,r},\\
\widehat{\sstirling{n+1+r}{k+r}}_{q,r}&amp;=&amp;q^{k-1}\widehat{\sstirling{n+r}{k-1+r}}_{q,r}+([k]_q+r)\widehat{\sstirling{n+r}{k+r}}_{q,r}.
\end{eqnarray*}
\end{corollary}
We can use these recurrence relations to compute the first few values of the $(q,r)$-Stirling numbers of the first and second kind, respectively.

\begin{theorem}
The $(q,r)$-Whitney numbers satisfy the following recurrence relations
\begin{equation}
\label{T1}
w_{m,r+1,q}(n,\ell) = \sum_{k=\ell}^n\binom{k}{\ell}(-1)^{k-\ell}w_{m,r,q}(n,k),
\end{equation}
\begin{equation}
\label{T2}
W_{m,r+1,q}(n,k) = \sum_{\ell=k}^n\binom{n}{\ell}W_{m,r,q}(\ell,k).
\end{equation}
\end{theorem}
\begin{proof}
From equation \eqref{qw1}, we have
\begin{eqnarray*}
m^n(a^{\dagger})^na^n &amp;=&amp; \sum_{k=0}^n w_{m,r,q}(n,k)(ma^{\dagger}a+r)^k \\
                     &amp;=&amp; \sum_{k=0}^n w_{m,r,q}(n,k)\Big( (ma^{\dagger}a+r+1)-1\Big)^k  \\
                     &amp;=&amp; \sum_{k=0}^n w_{m,r,q}(n,k)\sum_{\ell=0}^k\binom{k}{\ell}(-1)^{k-\ell}(ma^{\dagger}a+r+1)^{\ell} \\
                     &amp;=&amp; \sum_{\ell=0}^n(ma^{\dagger}a+r+1)^{\ell}\sum_{k=\ell}^n\binom{k}{\ell}(-1)^{k-\ell}w_{m,r,q}(n,k). 
\end{eqnarray*}
On the other hand,
$$m^n(a^{\dagger})^na^n = \sum_{\ell=0}^n w_{m,r+1,q}(n,\ell)(ma^{\dagger}a+r+1)^{\ell}.$$
Hence, by comparing the coefficients of $(ma^{\dagger}a+r+1)^{\ell}$ we obtain equation \eqref{T1}.

Similarly, from equation \eqref{qw2} 
\begin{equation*}
 (ma^{\dagger}a+r+1)^n=\sum_{k=0}^n m^k W_{m,r+1,q}(n,k)(a^{\dagger})^ka^k,
\end{equation*}
and since
\begin{eqnarray*}
(ma^{\dagger}a+r+1)^n &amp;=&amp; \sum_{\ell=0}^n\binom{n}{\ell} (ma^{\dagger}a+r)^{\ell}\\
	&amp;=&amp; \sum_{\ell=0}^n\binom{n}{\ell}\sum_{k=0}^{\ell}m^kW_{m,r,q}(\ell,k)(a^{\dagger})^ka^k \\
        &amp;=&amp; \sum_{k=0}^n m^k(a^{\dagger})^ka^k \sum_{\ell=k}^n\binom{n}{\ell}W_{m,r,q}(\ell,k),
\end{eqnarray*}
we obtain equation \eqref{T2}.
\end{proof}

When $m=1$, the theorem reduces to the recursion formulas for $(q,r)$-Stirling numbers. That is, 
\begin{corollary}
\begin{eqnarray*}
\widehat{\fstirling{n+r+1}{l+r+1}}_{q,r+1}&amp;=&amp;\sum_{k=l}^n(-1)^{l-k}\binom{k}{l}\widehat{\fstirling{n+r}{k+r}}_{q,r},\\
\widehat{\sstirling{n+r+1}{k+r+1}}_{q,r+1}&amp;=&amp;\sum_{l=k}^n\binom{n}{l}\widehat{\sstirling{l+r}{k+r}}_{q,r}.
\end{eqnarray*}
\end{corollary}

\section{Orthogonality and inverse relations}

\begin{theorem}
The $(q,r)$-Whitney numbers $w_{m,r,q}(n,k)$ and $W_{m,r,q}(k,j)$ satisfy the following orthogonality relations:
\begin{equation}
\label{O1}
\sum_{k=j}^{n}W_{m,r,q}(n,k)w_{m,r,q}(k,j)=\delta_{jn},
\end{equation}
and
\begin{equation}
\label{O2}
\sum_{k=j}^{n}w_{m,r,q}(n,k)W_{m,r,q}(k,j)=\delta_{jn},
\end{equation}
where $\delta_{jn}$ is the Kronecker delta.
\end{theorem}
\begin{proof}
Using equation \eqref{qw1} we substitute $m^k(a^\dagger)^k a^k$ in \eqref{qw2}, obtaining 
\begin{eqnarray*}
(ma^\dagger a+r)^n&amp;=&amp;\sum_{k=0}^{n}W_{m,r,q}(n,k)\sum_{j=0}^{k}w_{m,r,q}(k,j)(ma^\dagger a+r)^j\\
&amp;=&amp;\sum_{j=0}^{n}\left\{\sum_{k=j}^{n}W_{m,r,q}(n,k)w_{m,r,q}(k,j)\right\}(ma^\dagger a+r)^j.
\end{eqnarray*}
Comparing the coefficients of $(ma^\dagger a+r)^j$ yields equation \eqref{O1}.
Equation \eqref{O2} is obtained similarly.
\end{proof}

The classical \textit{binomial inversion formula} given by
\begin{equation}
f_k=\sum_{j=0}^k\binom{k}{j}g_j\Leftrightarrow g_k=\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}f_j\label{binominv}
\end{equation}
can be a useful tool in deriving the explicit formula of the classical Stirling numbers of the second kind. The $q$-analogue of \eqref{binominv} is given by \cite{Comt}
\begin{equation}
f_k=\sum_{j=0}^k\binom{k}{j}_qg_j\Leftrightarrow g_k=\sum_{j=0}^k(-1)^{k-j}q^{\binom{k-j}{2}}\binom{k}{j}_qf_j,\label{qbinominv}
\end{equation}
The next theorem presents an inverse relation for the $(q,r)$-Whitney numbers $w_{m,r,q}(n,k)$ and $W_{m,r,q}(k,j)$.
\begin{theorem}
The $(q,r)$-Whitney numbers $w_{m,r,q}(n,\ell)$ and $W_{m,r,q}(n,\ell)$ satisfy the following inverse relation:
\begin{equation}
f_n=\sum_{\ell=0}^nw_{m,r,q}(n,\ell)g_{\ell}\Leftrightarrow g_n=\sum_{\ell=0}^nW_{m,r,q}(n,\ell)f_{\ell}.
\end{equation}
\end{theorem}
\begin{proof}
By the hypothesis,
\begin{eqnarray*}
\sum_{\ell=0}^n W_{m,r,q}(n,\ell)f_{\ell}&amp;=&amp;\sum_{\ell=0}^n W_{m,r,q}(n,\ell)\sum_{k=0}^{\ell} w_{m,r,q}(\ell,k)g_k\\
&amp;=&amp;\sum_{k=0}^n\left\{\sum_{\ell=k}^n W_{m,r,q}(n,\ell)w_{m,r,q}(\ell,k)\right\}g_k\\
&amp;=&amp;\sum_{k=0}^n\left\{\delta_{kn}\right\}g_k\\
&amp;=&amp;g_n.
\end{eqnarray*}
The converse can be shown similarly.
\end{proof}

The next theorem can be deduced in a similar way, from the orthogonality relations
\begin{theorem}
The $(q,r)$-Whitney numbers $w_{m,r,q}(n,\ell)$ and $W_{m,r,q}(n,\ell)$ satisfy the following inverse relation:
\begin{equation}
f_{\ell}=\sum_{n=\ell}^{\infty}w_{m,r,q}(n,\ell)g_n\Leftrightarrow g_{\ell}=\sum_{n=\ell}^{\infty}W_{m,r,q}(n,\ell)f_n.
\end{equation}
\end{theorem}

\section{$(q,r)$-Dowling polynomials and numbers}

Cheon and Jung \cite{Cheon} defined the \textit{$r$-Dowling polynomials}, denoted by $D_{m,r}(n,x)$, 
in terms of sums of $r$-Whitney numbers of the second kind. That is,  
\begin{equation}
D_{m,r}(n,x)=\sum_{k=0}^nW_{m,r}(n,k)x^k.\label{rdowling}
\end{equation}
When $x=1$, we obtain the \textit{$r$-Dowling numbers} $D_{m,r}(n)\equiv D_{m,r}(n,1)$. The polynomials \eqref{rdowling} are actually equivalent to the $(r,\beta)$-Bell polynomials $G_{n,\beta,r}(x)$ of R. B. Corcino and C. B. Corcino \cite{Cor}. That is,
\begin{equation*}
D_{\beta,r}(n,x)=G_{n,\beta,r}(x).
\end{equation*}
Moreover, 
\begin{itemize}
\item when $m=1$ and $r=1$, we recover the classical \textit{Dowling polynomials} $D(n,x)\equiv D_{1,1}(n,x)$;
\item when $m=1$ and $r=0$, we recover the classical \textit{Bell polynomials} $B_n(x)\equiv D_{1,0}(n,x)$;
\item when $m=1$, we recover Mez\H{o}'s \cite{Mez} \textit{$r$-Bell polynomials} $B_{n,r}(x)$. That is, $D_{1,r}(n,x)=B_{n,r}(x)$; and
\item when $m=\alpha$ and $r=0$, we recover the \textit{translated Dowling polynomials} $\widetilde{D}_{(\alpha)}(n;x)$ by Mangontarum et al.\ \cite{Mah1}. That is, $D_{\alpha,0}(n,x)=\widetilde{D}_{(\alpha)}(n;x)$. 
\end{itemize}
\smallskip
Taking these into consideration, the next definition seems to be natural.
\begin{definition}
For non-negative integers $n$ and $k$, and complex numbers $m$ and $r$, the $(q,r)$-Dowling polynomials, 
denoted by $D_{m,r,q}(n,x)$, are defined by
\begin{equation}
D_{m,r,q}(n,x)=\sum_{k=0}^{n}W_{m,r,q}(n,k)x^k\label{qDP}
\end{equation}
and the $(q,r)$-Dowling numbers, denoted by $D_{m,r,q}(n)$, are defined by
\begin{equation}
D_{m,r,q}(n)=D_{m,r,q}(n,1).\label{qDN}
\end{equation}
\end{definition}
\smallskip

The \textit{coherent states}
\begin{equation}
\ket{\gamma}=\hbox{exp}\left(-\frac{|\gamma|^2}{2}\right)\sum_{k\geq 0}\frac{\gamma^k}{\sqrt{k!}}\ket{k}, \label{coherent}
\end{equation}
where $\gamma$ is an arbitrary (complex-valued) constant, satisfy $a\ket{\gamma}=\gamma\ket{\gamma}$ and 
$\braket{\gamma}{\gamma}=1$. Katriel \cite{Kat2} gave an illustration on how \eqref{coherent} can be a very useful tool in the derivation of certain Dobinski-type formulas. The \textit{q-coherent states} corresponding to the $q$-Boson operators 
were defined as 
\begin{equation}
\ket{\gamma}_q=\left(\widehat{e}_q(-|\gamma|^2)\right)^{\frac{1}{2}}\sum_{k\geq 0}\frac{\gamma^k}{\sqrt{[k]_q!}}\ket{k}\label{qCS}
\end{equation}
which satisfy $a\ket{\gamma}=\gamma\ket{\gamma}$. Here, $\widehat{e}_q(x)$ is the type 2 $q$-exponential function given by
\begin{equation}
\widehat{e}_q(x)=\prod_{i=1}^{\infty}(1+(1-q)q^{i-1}x)=\sum_{i\geq 0}q^{\binom{i}{2}}\frac{x^i}{[i]_q!},\label{type2qexp}
\end{equation}
which is the inverse of the type 1 $q$-exponential function
\begin{equation}
e_q(x)=\prod_{i=1}^{\infty}(1-(1-q)q^{i-1}x)^{-1}=\sum_{i\geq 0} \frac{x^i}{[i]_q!}.\label{type1qexp}
\end{equation}
That is, $e_q(x)\widehat{e}_q(-x)=1$.

Taking the expectation value of both sides of \eqref{qw2} with respect to $\ket{\gamma}$ yields
\begin{equation}
\bra{\gamma}(ma^\dagger a+r)^n\ket{\gamma}=\sum_{k=0}^{n}m^kW_{m,r,q}(n,k)|\gamma|^{2k}.
\end{equation}
The left-hand-side can be evaluated using the $q$-coherent states in \eqref{qCS}, yielding
\begin{equation}
\bra{\gamma}(ma^\dagger a+r)^n\ket{\gamma}=\widehat{e}_q\left(-|\gamma|^2\right)\sum_{k\geq 0}\frac{|\gamma|^{2k}}{[k]_q!}(m[k]_q+r)^n.
\end{equation}
Defining $x=m|\gamma|^2$ we obtain
\begin{equation}
\sum_{k=0}^{n}W_{m,r,q}(n,k)x^k=\widehat{e}_q\left(-\frac{x}{m}\right)\sum_{k\geq 0}\left(\frac{x}{m}\right)^k\frac{(m[k]_q+r)^n}{[k]_q!}.
\end{equation}
Using \eqref{qDP}, the following theorem is easily observed.
\begin{theorem}
The $(q,r)$-Dowling polynomials $D_{m,r,q}(n,x)$ and the $(q,r)$-Dowling numbers $D_{m,r,q}(n)$ have the following explicit formulas:
\begin{equation}
D_{m,r,q}(n,x)=\widehat{e}_q\left(-\frac{x}{m}\right)\sum_{k\geq 0}\left(\frac{x}{m}\right)^k\frac{(m[k]_q+r)^n}{[k]_q!},\label{qDPe}
\end{equation}
and
\begin{equation}
D_{m,r,q}(n)=\widehat{e}_q\left(-m^{-1}\right)\sum_{k\geq 0}\frac{(m[k]_q+r)^n}{m^k[k]_q!}.\label{qDNe}
\end{equation}
\end{theorem}
\begin{proof}
\eqref{qDNe} can be obtained by letting $x=1$ in \eqref{qDPe}.
\end{proof}

Katriel \cite{Kat2} defined the $q$-Bell polynomial as
\begin{equation}
\sum_{\ell=0}^k\sstirling{k}{\ell}_qx^{\ell}=\widehat{e}_q(x)\sum_{m=1}^{\infty}x^m\frac{[m]_q^k}{[m]_q!}.
\end{equation}
Expanding the right-hand side using \eqref{type2qexp} yields
\begin{equation}
\sum_{\ell=0}^k\sstirling{k}{\ell}_qx^{\ell}
 =\sum_{\ell=0}^{\infty}\frac{x^{\ell}}{[\ell]_q!}\sum_{j=0}^{\ell}(-1)^{\ell-j}q^{\binom{\ell-j}{2}}\binom{\ell}{j}_q[j]_q^k.
\end{equation}
Equating coefficients of equal powers of $x$ gives us
\begin{equation}
\sstirling{k}{\ell}_q=\frac{1}{[\ell]_q!}\sum_{j=0}^{\ell}(-1)^{\ell-j}q^{\binom{\ell-j}{2}}\binom{\ell}{j}_q[j]^k_q.\label{c3}
\end{equation}
Notice that as $q\rightarrow 1$, \eqref{c3} reduces to the well-known explicit formula of $\sstirling{k}{j}$. That is
\begin{equation}
\lim_{q\rightarrow 1}\sstirling{k}{\ell}_q=\frac{1}{{\ell}!}\sum_{j=0}^{\ell}(-1)^{\ell-j}\binom{\ell}{j}j^k.
\end{equation}
In the following theorem, we will present an expression analogous to \eqref{c3} for the $q$-analogue $W_{m,r,q}(n,k)$.
\begin{theorem}
The $(q,r)$-Whitney numbers of the second kind, $W_{m,r,q}(n,k)$, have the following explicit formula:
\begin{equation}
W_{m,r,q}(n,\ell)=\frac{1}{m^{\ell}[\ell]_q!}\sum_{k=0}^{\ell}(-1)^{\ell-k}q^{\binom{\ell-k}{2}}\binom{\ell}{k}_q(m[k]_q+r)^n.\label{identity3}
\end{equation}
\end{theorem}
\begin{proof}
Substituting $y=\frac{x}{m}$ in \eqref{qDPe} gives us
\begin{eqnarray*}
\sum_{k=0}^nm^kW_{m,r,q}(n,k)y^k&amp;=&amp;\sum_{i\geq 0}q^{\binom{i}{2}}\frac{(-y)^i}{[i]_q!}\sum_{k\geq 0}y^k\frac{(m[k]_q+r)^n}{[k]_q!}\\
&amp;=&amp;\sum_{\ell\geq 0}\frac{y^{\ell}}{[\ell]_q!}\sum_{k=0}^{\ell}(-1)^{\ell-k}q^{\binom{\ell-k}{2}}\binom{\ell}{k}_q(m[k]_q+r)^n.
\end{eqnarray*}
Equating the coefficients of equal powers of $y$ on both sides of this equation we obtain equation \eqref{identity3}.
\end{proof}
Note that as $q\rightarrow 1$, we have
\begin{eqnarray*}
\lim_{q\rightarrow 1}W_{m,r,q}(n,\ell)&amp;=&amp;\frac{1}{m^{\ell}{\ell}!}\sum_{k=0}^{\ell}(-1)^{\ell-k}\binom{\ell}{k}(mk+r)^n\\
&amp;=&amp;W_{m,r}(n,\ell).
\end{eqnarray*}
Furthermore,
\begin{equation*}
\lim_{q\rightarrow 1}W_{m,1,q}(n,l)=W_m(n,l).
\end{equation*}

\begin{remark}
We can also prove \eqref{identity3} in the following manner: First, we write \eqref{qw2} as 
\begin{eqnarray*}
(m[\ell]_q+r)^n&amp;=&amp;\sum_{k=0}^{n}m^kW_{m,r,q}(n,k)[\ell]_{q,k}\\
&amp;=&amp;\sum_{k=0}^{\ell}\binom{\ell}{k}_q\left\{\frac{m^kW_{m,r,q}(n,k)[\ell]_{q,k}}{\binom{\ell}{k}_q}\right\}.
\end{eqnarray*}
Next, we apply the $q$-binomial inversion formula in \eqref{qbinominv} which gives us
\begin{equation*}
\frac{m^{\ell}W_{m,r,q}(n,\ell)[\ell]_{q,\ell}}{\binom{k}{k}_q}
         =\sum_{k=0}^{\ell}(-1)^{\ell-k}q^{\binom{\ell-k}{2}}\binom{l}{k}_q(m[k]_q+r)^n.
\end{equation*}
This is precisely the explicit formula obtained in the previous theorem.
\end{remark}

Now, using \eqref{identity3}, 
\begin{eqnarray*}
\sum_{n\geq 0}W_{m,r,q}(n,k)\frac{t^n}{[n]_q!}&amp;=&amp;\sum_{n\geq 0}\sum_{j=0}^k\frac{(-1)^{k-j}}{m^k[k]_q!}q^{\binom{k-j}{2}}\binom{k}{j}_q(m[j]_q+r)^n\frac{t^n}{[n]_q!}\\
&amp;=&amp;\frac{1}{m^k[k]_q!}\sum_{j=0}^k(-1)^{k-j}q^{\binom{k-j}{2}}\binom{k}{j}_qe_q\left[(m[j]_q+r)t\right],
\end{eqnarray*}
where $e_q(x)$ is the type 1 $q$-exponential function in \eqref{type1qexp}. Thus, we have the following theorem.
\begin{theorem}
The $(q,r)$-Whitney numbers of the second kind satisfy the following exponential generating function:
\begin{equation}
\sum_{n\geq 0}W_{m,r,q}(n,k)\frac{t^n}{[n]_q!}=\frac{1}{m^k[k]_q!}\sum_{j=0}^n(-1)^{k-j}q^{\binom{k-j}{2}}\binom{k}{j}_{\! q} e_q\left[(m[j]_q+r)t\right].\label{identity6}
\end{equation}
\end{theorem}
\begin{remark}
As $q\rightarrow 1$, we have
\begin{equation*}
\lim_{q\rightarrow 1}\sum_{n\geq 0}W_{m,r,q}(n,k)\frac{t^n}{[n]_q!}=\frac{e^{rt}}{k!}\left(\frac{e^{mt}-1}{m}\right)^k,
\end{equation*}
which is the exponential generating function of the $r$-Whitney numbers of the second kind.
\end{remark}

The $q$-difference operator \cite{Kim} can be written in the form
\begin{equation}
\Delta^k_{q}f(x)=\sum_{j=0}^k(-1)^{k-j}q^{\binom{k-j}{2}}\binom{k}{j}_{\! q}f(x+j).\label{qDiffOp}
\end{equation}
We are now ready to state the next theorem.

\begin{theorem}
The $(q,r)$-Whitney numbers of the second kind satisfy the following identity:
\begin{equation}
\sum_{n\geq 0}W_{m,r,q}(n,k)\frac{t^n}{[n]_q!}=\left\{\Delta^k_q\left(\frac{e_q[(m[x]_q+r)t]}{m^k[k]_q!}\right)\right\}_{x=0}.\label{identity7}
\end{equation}
\end{theorem}
\begin{proof}
\eqref{identity7} follows directly from \eqref{identity6} and \eqref{qDiffOp}.
\end{proof}
 The next corollary is easily verified.
\begin{corollary}
The $(q,r)$-Whitney numbers of the second kind can be expressed explicitly as
\begin{equation}
W_{m,r,q}(n,k)=\left\{\Delta^k_q\left(\frac{(m[x]_q+r)^n}{m^k[k]_q!}\right)\right\}_{x=0}.
\end{equation}
\end{corollary}

\section{Further identities for the $(q,r)$-Whitney numbers}

Graham et al.\ \cite{Graham} presented a useful set of Stirling number identities while Katriel \cite{Kat1} presented the $q$-analogues of all but two of them. Three of these identities are generalized to the $(q,r)$-Whitney numbers using appropriate modifications of the procedures presented by Katriel \cite{Kat1}. Their derivation requires the following.
\begin{lemma}
For $f(x)$ a polynomial, the operator identity 
\begin{equation}\label{L1}
a^{\dagger}f(1+qa^{\dagger}a)a=a^{\dagger}af(a^{\dagger}a), 
\end{equation}
holds.
\end{lemma}
\begin{proof}
We write the $q$-commutation relation, equation \eqref{qbos}, in the form $aa^{\dagger}=1+qa^{\dagger}a$.
It follows that
$$(a^{\dagger}a)(a^{\dagger}a)^k=a^{\dagger}(aa^{\dagger})^ka=a^{\dagger}(1+qa^{\dagger}a)^ka.$$
For $f(x)=\sum_{k}c_kx^k$ we obtain
\begin{eqnarray*}
a^{\dagger}af(a^{\dagger}a) &amp;=&amp; \sum_{k}c_k(a^{\dagger}a)(a^{\dagger}a)^k \\
                             &amp;=&amp; \sum_k c_ka^{\dagger}(1+qa^{\dagger}a)^ka = a^{\dagger}\left(\sum_k c_k(1+qa^{\dagger}a)^k\right)a \\
                             &amp;=&amp; a^{\dagger}f(1+qa^{\dagger}a)a.
\end{eqnarray*} 
\end{proof}
\begin{remark}
The lemma can also be written in the form 
\begin{equation}
\label{L2}
a^{\dagger}g(a^{\dagger}a)a=a^{\dagger}a g\left(\frac{1}{q}(a^{\dagger}a-1)\right),
\end{equation}
where $g(x)$ is a polynomial.
\end{remark}


\begin{theorem}[Identity 1]
The $(q,r)$-Whitney numbers of the second kind satisfy
\begin{equation*}
W_{m,r,q}(n+1,k)-rW_{m,r,q}(n,k)=\sum_{\ell=k-1}^n\binom{n}{\ell}q^{\ell}(m+r(1-q))^{n-\ell}W_{m,r,q}(\ell,k-1).
\end{equation*}
\end{theorem}
\begin{proof}
In terms of the identity \eqref{L1} and with the aid of \eqref{qw2}
\begin{eqnarray*}
a^{\dagger}\Big(m(1+qa^{\dagger}a)+r\Big)^na &amp;=&amp; a^{\dagger}a(ma^{\dagger}a+r)^n \\
       &amp;=&amp; \frac{1}{m}(ma^{\dagger}a+r-r)(ma^{\dagger}a+r)^n\\ 
       &amp;=&amp; \frac{1}{m}(ma^{\dagger}a+r)^{n+1}-\frac{r}{m}(ma^{\dagger}a+r)^n \\
       &amp;=&amp; \sum_{k=0}^{n+1}m^{k-1}(a^{\dagger})^ka^k\Big(W_{m,r,q}(n+1,k)-rW_{m,r,q}(n,k)\Big).
\end{eqnarray*}
On the other hand, defining $\alpha=m+r(1-q)$ (which will hold throught the present section),
\begin{eqnarray*}
a^{\dagger}\Big(m(1+qa^{\dagger}a)+r\Big)^na &amp;=&amp; a^{\dagger}\Big(q(ma^{\dagger}a+r)+\alpha\Big)^na \\
   &amp;=&amp; a^{\dagger}\left(\sum_{\ell=0}^n\binom{n}{\ell}q^{\ell}\alpha^{n-\ell}(ma^{\dagger}a+r)^{\ell}\right)a \\
   &amp;=&amp; \sum_{\ell=0}^n\binom{n}{\ell}q^{\ell}\alpha^{n-{\ell}}\sum_{k=0}^{\ell}m^kW_{m,r,q}(\ell,k)(a^{\dagger})^{k+1}a^{k+1} \\
   &amp;=&amp; \sum_{k=1}^{n+1}m^{k-1}(a^{\dagger})^ka^k\sum_{\ell=k-1}^n\binom{n}{\ell}q^{\ell}\alpha^{n-\ell}W_{m,r,q}(\ell,k-1).
\end{eqnarray*}
Equating coefficients of $m^{k-1}(a^{\dagger})^ka^k$ the theorem follows.
\end{proof}
For $r=0$ this identity reduces to the $q$-Stirling numbers identity \cite[identity\ 1]{Kat1}
\begin{equation*}
W_{m,0,q}(n+1,k)=\sum_{\ell=k-1}^n\binom{n}{\ell}q^{\ell}m^{n-\ell}W_{m,0,q}(\ell,k-1).
\end{equation*}
The following corollary is an immediate consequence of the previous theorem.
\begin{corollary}
As $q\rightarrow 1$,
\begin{equation*}
W_{m,r}(n+1,k)-rW_{m,r}(n,k)=\sum_{\ell=k-1}^n\binom{n}{\ell}m^{n-\ell}W_{m,r}(\ell,k-1).
\end{equation*}
\end{corollary}


\begin{theorem}[Identity 2]
The $(q,r)$-Whitney numbers of the first kind satisfy
\begin{eqnarray*}
w_{m,r,q}(n+1,\ell) &amp;=&amp;
 \sum_{k=\ell-1}^n\frac{1}{q^k}w_{m,r,q}(n,k) \Big(-(m+r(1-q)\Big)^{k-\ell}\\
         &amp; &amp;\ \cdot\left(\binom{k}{\ell-1}(-(m+r(1-q)))-r\binom{k}{\ell}\right).
\end{eqnarray*}
\end{theorem}
\begin{proof}
We note that from \eqref{qw1},
\begin{equation*}
m^{n+1}(a^{\dagger})^{n+1}a^{n+1} = \sum_{\ell=0}^{n+1}w_{m,r,q}(n+1,\ell)(ma^{\dagger}a+r)^{\ell}.
\end{equation*} 

On the other hand, using \eqref{L2},
\begin{eqnarray*}
m^{n+1}(a^{\dagger})^{n+1}a^{n+1}
   &amp;=&amp; ma^{\dagger}\Big(m^n(a^{\dagger})^na^n\Big)a\\
   &amp;=&amp; ma^{\dagger}\Big(\sum_{k=0}^n w_{m,r,q}(n,k)(ma^{\dagger}a+r)^k\Big)a \\
   &amp;=&amp; ma^{\dagger}a\sum_{k=0}^n w_{m,r,q}(n,k)\Big(\frac{m}{q}(a^{\dagger}a-1)+r \Big)^k \\
   &amp;=&amp; ma^{\dagger}a\sum_{k=0}^n w_{m,r,q}(n,k)\frac{1}{q^k}\Big((ma^{\dagger}a+r)-\alpha \Big)^k\\
   &amp;=&amp; ((ma^{\dagger}a+r)-r)\sum_{k=0}^n w_{m,r,q}(n,k)\frac{1}{q^k}
       \sum_{\ell=0}^k\binom{k}{\ell}(ma^{\dagger}a+r)^{\ell}(-\alpha)^{k-\ell} \\
   &amp;=&amp; \sum_{k=0}^n w_{m,r,q}(n,k)\frac{1}{q^k}\sum_{\ell=0}^k\binom{k}{\ell}(ma^{\dagger}a+r)^{\ell+1}(-\alpha)^{k-\ell}\\
   &amp; &amp; -r\sum_{k=0}^n w_{m,r,q}(n,k)\frac{1}{q^k}\sum_{\ell=0}^n\binom{k}{\ell}(ma^{\dagger}a+r)^{\ell}(-\alpha)^{k-\ell}\\
   &amp;=&amp; \sum_{\ell=0}^{n+1}(ma^{\dagger}a+r)^{\ell}\sum_{k=\ell-1}^n\frac{1}{q^k}w_{m,r,q}(n,k)(-\alpha)^{k-\ell}\cdot \\
   &amp; &amp; \;\;\;\;\; \cdot \left(\binom{k}{\ell-1}(-\alpha)-r\binom{k}{\ell}\right).
\end{eqnarray*}
Equating the coefficients of equal powers of $ma^{\dagger}a+r$ we obtain the theorem.
\end{proof}
For $r=0$, we recover the $q$-Stirling numbers identity \cite[identity\ 2]{Kat1}
\begin{equation*}
w_{m,0,q}(n+1,\ell)=\sum_{k=\ell-1}^n\frac{1}{q^k}w_{m,0,q}(n,k)(-m)^{k-\ell+1}\binom{k}{\ell-1},
\end{equation*}
Moreover, we have the following corollary:
\begin{corollary}
As $q\rightarrow 1$,
\begin{equation*}
w_{m,r}(n+1,\ell)=-\sum_{k=\ell-1}^nw_{m,r}(n,k)(-m)^{k-\ell}\left(m\binom{k}{\ell-1}+r\binom{k}{\ell}\right).
\end{equation*}
\end{corollary}


\begin{theorem}[Identity 3]
The $(q,r)$-Whitney numbers of the second kind satisfy
\begin{equation*}
W_{m,r,q}(n,k-1) = \frac{1}{q^n}\sum_{\ell=k}^{n+1} (-m-r(1-q))^{n-\ell}\left(\binom{n}{\ell-1}(-m-r(1-q))-\binom{n}{\ell}r\right)W_{m,r,q}(\ell,k).
\end{equation*}
\end{theorem}
\begin{proof}
Note that
\begin{eqnarray*}
a^{\dagger}(ma^{\dagger}a+r)^na&amp;=&amp;\sum_{k=0}^n m^kW_{m,r,q}(n,k)(a^{\dagger})^{k+1}a^{k+1}\\
&amp;=&amp;\sum_{k=1}^{n+1} m^{k-1}W_{m,r,q}(n,k-1)(a^{\dagger})^ka^k,
\end{eqnarray*}
and on the other hand, using \eqref{L2},
\begin{eqnarray*}
 a^{\dagger}(ma^{\dagger}a+r)^na &amp;=&amp; a^{\dagger}a\Big(\frac{m}{q}(a^{\dagger}a-1)+r\Big)^n 
   =a^{\dagger}a\frac{1}{q^n}(ma^{(\dagger}a+r)-\alpha)^n \\
	 &amp;=&amp; \frac{1}{m} ((ma^{\dagger}a+r)-r)\frac{1}{q^n}\sum_{\ell=0}^n\binom{n}{\ell}(ma^{\dagger}a+r)^{\ell}(-\alpha))^{n-\ell} \\
  &amp;=&amp; \frac{1}{mq^n} \sum_{\ell=1}^{n+1} (ma^{\dagger}a+r)^{\ell}(-\alpha)^{n-\ell}\cdot\left( \binom{n}{\ell-1}(-\alpha) -r\binom{n}{\ell}\right) \\
  &amp;=&amp; \frac{1}{mq^n}\sum_{k=0}^{n+1} m^k(a^{\dagger})^ka^k\sum_{\ell=k}^{n+1} (-\alpha)^{n-\ell}\left(\binom{n}{\ell-1}(-\alpha)-\binom{n}{\ell}r\right)W_{m,r,q}(\ell,k).
\end{eqnarray*}
Equating the coefficients of $(a^{\dagger})^ka^k$ we obtain the theorem.
\end{proof}
For $r=0$ this theorem reduces to
\begin{equation*}
W_{m,0,q}(n,k-1)=\frac{1}{q^n}\sum_{\ell=k}^{n+1}(-m)^{n+1-\ell}\binom{n}{\ell-1}W_{m,0,q}(\ell,k).
\end{equation*}
Using equation \eqref{w6}, we can verify that this is just the $q$-Stirling numbers identity \cite[identity\ 3]{Kat1}. The next corollary is easily verified.
\begin{corollary}
As $q\rightarrow 1$,
\begin{equation*}
W_{m,r}(n,k-1)=\sum_{\ell=k}^{n+1}(-m)^{n-\ell}\left[\binom{n}{\ell-1}(-m)-\binom{n}{\ell}r\right]W_{m,r}(\ell,k).
\end{equation*}
\end{corollary}

Presently, much is yet to be learnt regarding the $(q,r)$-Whitney numbers. The classical $r$-Whitney and 
Stirling numbers are known to have various applications in different fields. It is tempting to explore 
applications for the $(q,r)$-Whitney numbers. 

To close this section, Corcino and Hererra \cite{CorHer} defined the $q$-analogue of the limit of the differences of the generalized factorial $F_{\alpha,\gamma}(n,k)$ 
in \eqref{ldgf}, denoted by $\phi_{\alpha,\gamma}[n,k]_q$. $\phi_{\alpha,\gamma}[n,k]_q$ can be defined in terms of the relation
\begin{equation}
\sum_{k=0}^n\phi_{\alpha,\gamma}[n,k]_qt^k=\left\langle t+[\gamma]_q|[\alpha]_q\right\rangle^q_n,
\end{equation}
where
\begin{equation}
\left\langle t+[\gamma]_q|[\alpha]_q\right\rangle^q_n=\prod_{j=0}^{n-1}\left(t+[\gamma]_q-[j\alpha]_q\right).
\end{equation}
The numbers $\phi_{\alpha,\gamma}[n,k]_q$ are actually $q$-analogues of the numbers $w_{m,r}(n,k)$. Similarly, Corcino and Montero \cite{CorMon} defined the $q$-analogue $\sigma[n,k]^{\beta,r}_q$ of the Rucinski-Voigt numbers in terms of the reccurence relation
\begin{equation}
\sigma[n,k]^{\beta,r}_q=\sigma[n-1,k-1]^{\beta,r}_q+\left([k\beta]_q+[r]_q\right)\sigma[n-1,k]^{\beta,r}_q.
\end{equation}
$\sigma[n,k]^{\beta,r}_q$ is also a $q$-analogue of the numbers $\rbstirling{n}{k}_{r,\beta}$ and $W_{m,r}(n,k)$. However, by comparing the defining relations for $\phi_{\alpha,\gamma}[n,k]_q$ and $\sigma[n,k]^{\beta,r}_q$ with those of the$(q,r)$-Whitney numbers $w_{m,r,q}(n,k)$ and $W_{m,r,q}(n,k)$, respectively, we note that they represent distinctly motivated $q$-analogues that cannot be simply related to one another.  

\section{Acknowledgments}
The authors would like to thank the editor-in-chief for his helpful comments and suggestions, and the referee 
for carefully reading the manuscript and giving invaluable recommendations which helped improve this paper.

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\noindent 2010 {\it Mathematics Subject Classification}: 
Primary 11B83; Secondary 11B73, 05A30.

\noindent \emph{Keywords:} Whitney number, Stirling number, Boson operator, $q$-analogue.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000110},
\seqnum{A003575},
\seqnum{A008275}, and
\seqnum{A008277}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 13 2015;
revised version received,  September 1 2015.
Published in {\it Journal of Integer Sequences}, September 7 2015.

\bigskip
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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