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\begin{center}
\vskip 1cm{\LARGE\bf A ({\it p,q})-Analogue of the {\it r}-Whitney-Lah Numbers} \vskip 1cm
\large
Jos\'{e} L. Ram\'{\i}rez\\
Departamento de Matem\'{a}ticas\\
Universidad Sergio Arboleda\\
110221 Bogot\'{a}\\
Colombia\\
\href{mailto:josel.ramirez@ima.usergioarboleda.edu.co}{\tt josel.ramirez@ima.usergioarboleda.edu.co}\\
\ \\
Mark Shattuck\\
Department of Mathematics\\
University of Tennessee\\
Knoxville, TN 37996\\
USA\\
\href{mailto:shattuck@math.utk.edu}{\tt shattuck@math.utk.edu}\\
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\begin{abstract}
In this paper, we consider a $(p,q)$-generalization of the
$r$-Whitney-Lah numbers that reduces to these recently introduced
numbers when $p=q=1$. We develop a combinatorial interpretation for
our generalized numbers in terms of a pair of statistics on an
extension of the set of $r$-Lah distributions wherein certain elements
are assigned a color. We obtain generalizations of some earlier
results for the $r$-Whitney-Lah sequence, including explicit formulas
and various recurrences, as well as ascertain some new results for this
sequence. We provide combinatorial proofs of some additional formulas
in the case when $q=1$, among them one that generalizes an identity
expressing the $r$-Whitney-Lah numbers in terms of the $r$-Lah
numbers. Finally, we introduce the $(p,q)$-Whitney-Lah matrix and
study some of its properties.
\end{abstract}
%%-----------------------------------------------------------------
\section{Introduction}
Given variables $x$ and $m$ and a positive integer $k$, define the
generalized rising and falling factorials of order $k$ by
\begin{align*}
[x|m]_k &= x(x+m)\cdots(x+(k-1)m), \\
(x|m)_k & =x(x-m)\cdots(x-(k-1)m),
\end{align*}
with $[x|m]_0=(x|m)_0=1$. The \emph{$r$-Whitney-Lah numbers}, which we
denote by $L(n,k)=L(n,k;r,m)$, were introduced by Cheon and Jung
\cite{Ch} and are connection constants in the polynomial identities
\begin{equation}\label{Lintroe1}
[x+2r|m]_n=\sum_{k=0}^nL(n,k)(x|m)_k, \qquad n \geq 0.
\end{equation}
In the $m=1$ case, the numbers $L(n,k;r,1)$ coincide with the $r$-Lah numbers \cite{MiR,NR}, which are also known as \emph{unsigned $r$-restricted Lah numbers}; see sequence \seqnum{A143497} in OEIS \cite{Slo}. These numbers enumerate partitions of the set $[n+r]$ into $k+r$ ordered lists in which the members of $[r]$ belong to different lists. When $r=0$ and $m=1$, the $L(n,k;r,m)$ reduce to the classical Lah numbers \cite{Lah}, which occur as \seqnum{A008297} in OEIS \cite{Slo}.
The $r$-Whitney-Lah numbers may also be defined, equivalently, as
\begin{equation}\label{Lintroe2}
L(n,k)=\sum_{j=k}^nw(n,j)W(j,k), \qquad 0 \leq k \leq n,
\end{equation}
where $w(n,j)$ and $W(j,k)$ are the $r$-Whitney numbers of the first and second kind, respectively (see, e.g., \cite{Be,Ch,Mer,Mezo}). This generalizes a well-known formula for the classical Lah numbers in terms of the Stirling numbers of the first and second kind. See \cite[Theorem 3.11]{NR} for the comparable relation involving $r$-Lah numbers. Here, we consider a two-variable polynomial generalization $L_{p,q}(n,k)$ of the $r$-Whitney-Lah sequence which reduces to it when $p=q=1$. Starting with an extension of formula \eqref{Lintroe2}, we derive the recurrence for $L_{p,q}(n,k)$ and other properties that generalize results from \cite{Ch} for $L(n,k)$. Note that when $r=0$ and $m=p=1$, the $L_{p,q}(n,k)$ coincide with previously studied $q$-Lah numbers \cite{LMS}. See also \cite{CB,CM} for related generalized Stirling polynomials.
We develop a combinatorial interpretation of our $L_{p,q}(n,k)$ in terms of two statistics defined on extended $r$-Lah distributions in which certain elements are assigned a color. This allows one to ascertain further identities satisfied by $L_{p,q}(n,k)$ and to supply them with combinatorial explanations. When $p=q=1$, our proofs specialize to provide counting arguments of various identities involving $L(n,k)$ which were shown \cite{Ch} by algebraic methods using the Riordan group. In particular, we find a combinatorial proof of a generalization of the identity
\begin{equation}\label{Lintroe3}
L(n,k;r,m)=\sum_{i=k}^n\binom{n}{i}[2r(1-m)|m]_{n-i}m^{i-k}L(i,k;r,1), \qquad n,k \geq0,
\end{equation}
see \cite[Theorem 4.2]{Ch}. Note that formula \eqref{Lintroe3} expresses the $r$-Whitney-Lah numbers in terms of the $r$-Lah numbers. Due to the negative terms occurring in the sum (once its summands are expanded), to explain \eqref{Lintroe3}, we make use of a sign-changing involution defined on a certain extension of the set of $r$-Lah distributions whose set of survivors has cardinality given by $L(n,k;r,m)$.
The organization of this paper is as follows. In the next section, we define $L_{p,q}(n,k)$ as a convolution of generalized $r$-Whitney numbers and determine its defining recurrence. In the third section, we develop a combinatorial interpretation for $L_{p,q}(n,k)$ and prove additional properties of these polynomials. Further attention is paid to the special cases $q=0$ and $q=-1$. In the fourth section, additional identities are proven for $L_{p,q}(n,k)$ by combinatorial arguments in the case when $q=1$, including a generalization of \eqref{Lintroe3}. In the final section, we introduce the $(p,q)$-Whitney-Lah matrix and study some of its properties.
We make use of the following terminology and notation. A block within a partition of a finite set in which the elements may be written in any order is described as \emph{contents-ordered}. An element that is the smallest within a content-ordered block is said to be \emph{minimal}; all other elements are \emph{non-minimal}. If $m$ and $n$ are positive integers, then let $[m,n]=\{m,m+1,\ldots,n\}$ if $m \leq n$, with $[m,n]=\varnothing$ if $m>n$. The special case $[1,n]$
is denoted by $[n]$, with $[0]=\varnothing$. Throughout, we let $I$ stand for the set $[r+1,r+n]$. Given a variable $q$, let $[i]_q=1+q+\cdots+q^{i-1}$ for a positive integer $i$ and $[0]_q=0$. Finally, the $q$-binomial coefficient $\binom{n}{k}_q$ is defined as $\binom{n}{k}_q=\frac{[n]_q!}{[k]_q![n-k]_q!}$ for $0 \leq k \leq n$ and is zero otherwise, where $[k]_q!=\prod_{i=1}^k[i]_q$ if $k \geq 1$ and $[0]_q!=1$.
\section{Definition and recurrence}
The $(p,q)$-Whitney numbers of the first and second kind (see \cite{MRS,RS}, where they were introduced) are connection constants in the identities
\begin{equation}\label{Lintroe4}
m^n\prod_{i=0}^{n-1}(x+[i]_q)=\sum_{k=0}^nw_{p,q}(n,k)(mx-[r]_p)^k, \qquad n \geq 0,
\end{equation}
and
\begin{equation}\label{Lintroe5}
(mx+[r]_p)^n=\sum_{k=0}^nW_{p,q}(n,k)m^k\prod_{i=0}^{k-1}(x-[i]_q), \qquad n \geq 0.
\end{equation}
Note that when $p=q=1$, the $w_{p,q}(n,k)$ and $W_{p,q}(n,k)$ reduce, respectively, to the $r$-Whitney numbers of the first and second kind. See \cite{MK} for a related $q$-analogue studied from an algebraic standpoint.
In analogy to \eqref{Lintroe2}, define $L_{p,q}(n,k)=L_{p,q}(n,k;r,m)$ by
\begin{equation}\label{Leq1}
L_{p,q}(n,k):=\sum_{j=k}^n w_{p,q}(n,j)W_{p,q}(j,k), \qquad 0 \leq k \leq n.
\end{equation}
Note that the $L_{p,q}(n,k)$ reduce to $L(n,k)$ when $p=q=1$. Thus, one may refer to these numbers as \emph{generalized $r$-Whitney-Lah numbers}. The $L_{p,q}(n,k)$ satisfy the following two-term recurrence.
\begin{theorem}\label{Lth1}
If $n,k \geq 1$, then
\begin{equation}\label{Lth1e1}
L_{p,q}(n,k)=L_{p,q}(n-1,k-1)+(2[r]_p+m([n-1]_q+[k]_q))L_{p,q}(n-1,k),
\end{equation}
with $L_{p,q}(n,0)=\prod_{i=0}^{n-1}(2[r]_p+m[i]_q)$ and $L_{p,q}(0,k)=\delta_{k,0}$ for all $n,k \geq 0$.
\end{theorem}
\begin{proof}
The initial condition $L_{p,q}(0,k)=\delta_{k,0}$ is clear. Replacing $x$ with $\frac{x+2[r]_p}{m}$ in \eqref{Lintroe4}, and using \eqref{Lintroe5}, gives
\begin{align}
\prod_{i=0}^{n-1}(x+2[r]_p+m[i]_q)&=\sum_{j=0}^nw_{p,q}(n,j)(x+[r]_p)^j\notag\\
&=\sum_{j=0}^nw_{p,q}(n,j)\sum_{k=0}^jW_{p,q}(j,k)\prod_{i=0}^{k-1}(x-m[i]_q)\notag\\
&=\sum_{k=0}^n\prod_{i=0}^{k-1}(x-m[i]_q)\sum_{j=k}^nw_{p,q}(n,j)W_{p,q}(j,k)\notag\\
&=\sum_{k=0}^nL_{p,q}(n,k)\prod_{i=0}^{k-1}(x-m[i]_q).\label{Lth1e2} \end{align}
For $n \geq 1$, we then have
\begin{align*}
&\sum_{k=0}^nL_{p,q}(n,k)\prod_{i=0}^{k-1}(x-m[i]_q)=\prod_{i=0}^{n-1}(x+2[r]_p+m[i]_q)\\
&\quad=(x+2[r]_p+m[n-1]_q)\prod_{i=0}^{n-2}(x+2[r]_p+m[i]_q)\\
&\quad=(x+2[r]_p+m[n-1]_q)\sum_{k=0}^{n-1}L_{p,q}(n-1,k)\prod_{i=0}^{k-1}(x-m[i]_q)\\
&\quad=\sum_{k=0}^{n-1}L_{p,q}(n-1,k)\prod_{i=0}^{k}(x-m[i]_q)\\
&\quad\quad+\sum_{k=0}^{n-1}(2[r]_p+m([n-1]_q+[k]_q)L_{p,q}(n-1,k)\prod_{i=0}^{k-1}(x-m[i]_q)\\
&\quad=\sum_{k=0}^{n}(L_{p,q}(n-1,k-1)+(2[r]_p+m([n-1]_q+[k]_q))L_{p,q}(n-1,k))\prod_{i=0}^{k-1}(x-m[i]_q),
\end{align*}
upon replacing $k$ by $k-1$ in the first sum. Equating coefficients of $\prod_{i=0}^{k-1}(x-m[i]_q)$ gives \eqref{Lth1e1}.
Taking $x=0$ in \eqref{Lth1e2} gives the formula for $L_{p,q}(n,0)$.
\end{proof}
Let $[x|m,q]_k=x(x+m[1]_q)\cdots(x+m[k-1]_q)$ and $(x|m,q)_k=x(x-m[1]_q)\cdots(x-m[k-1]_q)$ denote the generalized $q$-rising and $q$-falling factorials of order $k$. Then the proof of the prior theorem shows the following result, which generalizes \eqref{Lintroe1}.
\begin{corollary}\label{Lth1c1}
The $L_{p,q}(n,k)$ are connection constants in the polynomial identities
\begin{equation}\label{Lth1c1e1}
[x+2[r]_p|m,q]_n=\sum_{k=0}^nL_{p,q}(n,k)(x|m,q)_k, \qquad n \geq 0.
\end{equation}
\end{corollary}
\section{Combinatorial interpretation and properties}
We now provide a combinatorial interpretation for the array $L_{p,q}(n,k)$. To do so, we first need to define a property possessed by certain elements within a contents-ordered block.
\begin{definition}\label{Ldef1}
Suppose that a contents-ordered block contains a single element $s \in [r]$ along with some members of $I$. Then $x \in I$ is said to satisfy the nearness property if there exists no $y \in I$ with $yk \geq 0$, then
\begin{equation}\label{Lth3e1}
L_{p,q}(n,k)=\sum_{j=k-1}^{n-1}\prod_{i=j+1}^{n-1}(2[r]_p+m([k]_q+[i]_q))L_{p,q}(j,k-1)
\end{equation}
and
\begin{equation}\label{Lth3e2}
L_{p,q}(n,k)=\sum_{j=0}^k(2[r]_p+m([j]_q+[n-k+j-1]_q))L_{p,q}(n-k+j-1,j).
\end{equation}
\end{theorem}
\begin{proof}
To show \eqref{Lth3e1}, consider the size $r+j+1$ of the smallest element of the last non-special block within a member of $\mathcal{L}(n,k)$, where $k-1\leq j \leq n-1$. Then there are $L_{p,q}(j,k-1)$ possibilities concerning the placement of the elements of $[r+j]$, with $r+j+1$ starting a new block. For each $i \in [j+2,n]$, there are $2[r]_p+m([k]_q+[i-1]_q)$ possibilities concerning placement of the element $r+i$, whence there are $\prod_{i=j+1}^{n-1}(2[r]_p+m([k]_q+[i]_q))$ possibilities for all such elements. Summing over $j$ gives \eqref{Lth3e1}.
To show \eqref{Lth3e2}, consider the size, $r+n-k+j$, of the largest element of $I$ not occupying its own block. Note that we must have $0\leq j \leq k$. Then each element of $[r+n-k+j+1,r+n]$ within a member of $\mathcal{L}(n,k)$ must belong to its own block, leaving $k-(k-j)=j$ non-special blocks. Thus, there are $L_{p,q}(n-k+j-1,j)$ possibilities concerning arrangement of the elements of $[r+n-k+j-1]$. At the time that the element $r+n-k+j$ is placed, there are $j$ non-special blocks currently occupied. Since this element is not to be the smallest within its block, there are $2[r]_p+m([j]_q+[n-k+j-1]_q)$ possibilities concerning its placement. Summing over all $j$ gives \eqref{Lth3e2}.
\end{proof}
Note that the $p=q=m=1$ case of \eqref{Lth3e1} occurs as \cite[Theorem 3.3]{NR}. Let $(a_i)_{i\geq0}$ and $(b_i)_{i \geq0}$ be sequences of complex numbers (or indeterminates), with the $b_i$ distinct. Let the array $\{u(n,k)\}_{n,k\geq0}$ be defined by the recurrence
$$u(n,k)=u(n-1,k-1)+(a_{n-1}+b_k)u(n-1,k), \qquad n,k \geq 1,$$
with boundary conditions $u(n,0)=\prod_{i=0}^{n-1}(a_i+b_0)$ and $u(0,k)=\delta_{k,0}$ for all $n,k \geq0$. By \cite[Theorem 1.1]{MMS}, we have the formula
\begin{equation}\label{Lprelim}
u(n,k)=\sum_{j=0}^k \left(\frac{\prod_{i=0}^{n-1}(a_i+b_j)}{\prod_{\substack{i=0\\ i\neq j}}^k(b_j-b_i)}\right), \qquad n,k \geq0.
\end{equation}
From this, one can obtain an explicit formula for $L_{p,q}(n,k)$.
\begin{theorem}\label{Lth4}
If $n,k \geq 0$, then
\begin{equation}\label{Lth4e1}
L_{p,q}(n,k)=\frac{1}{m^kq^{\binom{k}{2}}[k]_q!}\sum_{j=0}^k(-1)^{k-j}q^{\binom{k-j}{2}}\binom{k}{j}_q\prod_{i=0}^{n-1}(m[i]_q+m[j]_q+2[r]_p).
\end{equation}
\end{theorem}
\begin{proof}
Taking $a_i=m[i]_q+2[r]_p$ and $b_i=m[i]_q$ in \eqref{Lprelim} implies
\begin{align*}
L_{p,q}(n,k)&=\sum_{j=0}^k\frac{\prod_{i=0}^{n-1}(m[i]_q+m[j]_q+2[r]_p)}{\prod_{i=0}^{j-1}m([j]_q-[i]_q)\cdot\prod_{i=j+1}^km([j]_q-[i]_q)}\\
&=\sum_{j=0}^j\frac{\prod_{i=0}^{n-1}(m[i]_q+m[j]_q+2[r]_p)}{m^jq^{\binom{j}{2}}\prod_{i=0}^{j-1}[j-i]_q\cdot(-m)^{k-j}q^{j(k-j)}\prod_{i=j+1}^k[i-j]_q}\\
&=\frac{1}{m^k[k]_q!}\sum_{j=0}^k\frac{(-1)^{k-j}}{q^{\binom{j}{2}+j(k-j)}}\binom{k}{j}_q\prod_{i=0}^{n-1}(m[i]_q+m[j]_q+2[r]_p),
\end{align*}
which gives \eqref{Lth4e1}, by the fact $\binom{k}{2}=\binom{j}{2}+\binom{k-j}{2}+j(k-j)$.
\end{proof}
Formula \eqref{Lth4e1} may be simplified further when $q=1$.
\begin{corollary}\label{Lth4c1}
If $n,k \geq 0$, then
\begin{equation}\label{Lth4c1e1}
L_{p,1}(n,k)=\frac{m^{n-k}n!}{k!}\binom{n+\frac{2[r]_p}{m}-1}{n-k}.
\end{equation}
\end{corollary}
\begin{proof}
Letting $q=1$ in \eqref{Lth4e1} gives
\begin{align*}
L_{p,1}(n,k)&=\frac{1}{m^kk!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\prod_{i=0}^{n-1}(mi+mj+2[r]_p)\\
&=\frac{m^{n-k}}{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}\prod_{i=0}^{n-1}\left(i+j+\frac{2[r]_p}{m}\right)\\
&=\frac{m^{n-k}n!}{k!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\binom{j+n+\frac{2[r]_p}{m}-1}{n}=\frac{m^{n-k}n!}{k!}\binom{n+\frac{2[r]_p}{m}-1}{n-k},
\end{align*}
where we have made use of the binomial identity \cite[Identity 5.24]{GKP}.
\end{proof}
\begin{remark} If $m,n \geq 1$ and $p,r\geq 0$, one can show using \eqref{Lth4c1e1} that the sequence $L_{p,1}(n,k)$ for $0 \leq k \leq n$ is strictly log-concave (and therefore unimodal). Taking $p=m=1$ in \eqref{Lth4c1e1} yields \cite[Theorem 3.7]{NR}, which was given a bijective proof.
\end{remark}
Using \eqref{Lth4c1e1}, one can obtain the following exponential generating function formula.
\begin{corollary}\label{Lth4c2}
If $k\geq 0$, then
\begin{equation}\label{Lth4c2e1}
\sum_{n\geq k}L_{p,1}(n,k)\frac{x^n}{n!}=\frac{1}{k!}\left(\frac{x}{1-mx}\right)^k\left(\frac{1}{1-mx}\right)^{\frac{2[r]_p}{m}}.
\end{equation}
\end{corollary}
We have the following explicit expression in the case when $q=0$.
\begin{proposition}\label{Lprq0}
If $n \geq k \geq 1$, then
\begin{equation}\label{Lprq0e1}
L_{p,0}(n,k)=\binom{n-1}{k-1}(2[r]_p+2m)^{n-k}+2[r]_p\sum_{i=k}^{n-1}\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}(2[r]_p+m)^{n-i-1}.
\end{equation}
\end{proposition}
\begin{proof}
Taking $q=0$ in \eqref{Lth1e1}, and noting $[i]_q\mid_{q=0}=[i>0]$, gives the recurrence
\begin{equation}\label{Lprq0e2}
L_{p,0}(n,k)=L_{p,0}(n-1,k-1)+2([r]_p+m)L_{p,0}(n-1,k), \qquad n>k \geq 1,
\end{equation}
with $L_{p,0}(n,0)=2[r]_p(2[r]_p+m)^{n-1}$ for $n>0$ and $L_{p,0}(n,n)=1$ for $n \geq 0$. If $k \geq 0$, then let $f_k(x)=\sum_{n \geq k}L_{p,0}(n,k)x^n$. Note that $f_0(x)=\frac{1-mx}{1-(2[r]_p+m)x}$, by the initial values. Multiplying both sides of \eqref{Lprq0e2} by $x^n$, and summing over $n>k$, yields
$$f_k(x)-x^k=x(f_{k-1}(x)-x^{k-1})+2([r]_p+m)xf_k(x),$$
whence
$$f_k(x)=\frac{x}{1-2([r]_p+m)x}f_{k-1}(x), \qquad k \geq 1.$$
This implies
\begin{align*}
f_k(x)&=\frac{1-mx}{1-(2[r]_p+m)x}\cdot\frac{x^k}{(1-2([r]_p+m)x)^{k}}\\
&=\frac{1-mx}{1-(2[r]_p+m)x}\sum_{i\geq k}\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}x^i\\
&=(1-mx)\sum_{n \geq k}x^n\sum_{i=k}^n\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}(2[r]_p+m)^{n-i}.
\end{align*}
Extracting the coefficient of $x^n$ gives
\begin{align*}
&L_{p,0}(n,k)=[x^n]f_k(x)\\
&\quad=\sum_{i=k}^{n}\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}(2[r]_p+m)^{n-i}\\
&\quad\quad-m\sum_{i=k}^{n-1}\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}(2[r]_p+m)^{n-i-1}\\
&\quad=\binom{n-1}{k-1}(2[r]_p+2m)^{n-k}+2[r]_p\sum_{i=k}^{n-1}\binom{i-1}{k-1}(2[r]_p+2m)^{i-k}(2[r]_p+m)^{n-i-1},
\end{align*}
as desired.
\end{proof}
It is possible to give a bijective proof of the prior result using our combinatorial interpretation for $L_{p,q}(n,k)$.
\bigskip
\noindent \textbf{Combinatorial proof of Proposition \ref{Lprq0}.}
\medskip
We argue directly that the right-hand side of \eqref{Lprq0e1} gives the weight of all members of $\mathcal{L}(n,k)$ having $\beta$ statistic value zero. In order for a member of $\mathcal{L}(n,k)$ to have zero $\beta$ value, both its $\beta_1$ and $\beta_2$ values must be zero. For that to occur, each element of $I$ either (i) is minimal, (ii) satisfies the nearness property, (iii) has $r+1$ as its predecessor, or (iv) is non-minimal, but is a left-to-right minimum in the leftmost non-special block.
Note that exactly $k$ elements of $I$ must satisfy condition (i) within any member of $\mathcal{L}(n,k)$. First suppose that the element $r+1$ is among them. In that case, there are $\binom{n-1}{k-1}$ ways in which to choose the other elements that satisfy (i). For each of the remaining $n-k$ elements $x$ of $I$, there are $2[r]_p$ possibilities if (ii) is to be satisfied, since in this case $x$ comes either directly after or before some member of $[r]$ at the time it is placed (we assume elements are arranged one-by-one in ascending order). If (iii) is to be satisfied, the element $x$ at the time it is placed must directly follow $r+1$ within its block. In order for (iv) to hold, the element $x$ must be inserted at the very beginning of the leftmost non-special block (which in this case contains $r+1$). Note that the last two options entail that $x$ be assigned one of $m$ possible colors. Thus, there are $2[r]_p+2m$ ways to place each $x$ and hence $(2[r]_p+2m)^{n-k}$ ways in all to position the remaining $n-k$ unchosen members of $I$.
Now assume that $r+1$ is not minimal and suppose that all elements of $J=[r+1,r+n-i]$ for some $i$ belong to special blocks, with $r+n-i+1$ being the smallest minimal element of $I$. Note that $k \leq i \leq n-1$ since $r+1$ is not minimal. There are $2[r]_p$ choices concerning the placement of $r+1$ since it may directly follow or precede any member of $[r]$ and $2[r]_p+m$ choices for each of the other members of $J$ since they may also directly follow $r+1$, whence there are $2[r]_p(2[r]_p+m)^{n-i-1}$ possibilities in all for the members of $J$. Concerning the other $k-1$ minimal elements, they can be picked arbitrarily from the set $[r+n-i+2,r+n]$, which can be done in $\binom{i-1}{k-1}$ ways. For each of the remaining $i-k$ elements belonging to this set, there are $2[r]_p+2m$ possibilities since each one may satisfy conditions (ii), (iii) or (iv), whence there are $(2[r]_p+2m)^{i-k}$ possibilities concerning their arrangement. Summing over $i$ gives the weight of all members of $\mathcal{L}(n,k)$ having $\beta$ value zero in which the element $r+1$ is not minimal. This yields the second term on the right-hand side of \eqref{Lprq0e1} and completes the proof. \hfill \qed
\medskip
Note that \eqref{Lprq0e1} does not follow from \eqref{Lprelim} since the $b_i=[i]_q$ terms are not distinct when $q=0$. Formula \eqref{Lprelim} also does not apply in the case when $q=-1$ for the same reason, which we now consider. Note that the sign-balance of the $\beta$ statistic on $\mathcal{L}_{n,k}$ can be obtained by evaluating $L_{p,q}(n,k)$ at $q=-1$. Let $H(x,y)=\sum_{n,k\geq0}L_{p,-1}(n,k)x^ny^k$. There is the following explicit formula for $H(x,y)$.
\begin{proposition}\label{minus1}
We have
\begin{equation}\label{minus1e1}
H(x,y)=\frac{1+x(y+c-m)-x^2((y-c)^2+cm)-x^3(y-c-m)(y^2-c^2+cm)}{1-2x^2(y^2+c^2)+x^4(y^2-c^2-cm)(y^2-c^2+cm)},
\end{equation}
where $c=2[r]_p+m$.
\end{proposition}
\begin{proof}
Note that $H(x,y)=\sum_{n\geq0}a(n;y)x^n$, where $a(n;y)=\sum_{k=0}^nL_{p,-1}(n,k)y^k$. We first write a recurrence for $a(n;y)$. Letting $q=-1$ in \eqref{Lth1e1}, and observing $[i]_q\mid_{q=-1}=[i \text{ is odd}]$, yields for $n\geq 1$ the formulas
$$L_{p,-1}(2n,k)=L_{p,-1}(2n-1,k-1)+(2[r]_p+m)L_{p,-1}(2n-1,k), \qquad k~ \text{ even},$$
and
$$L_{p,-1}(2n,k)=L_{p,-1}(2n-1,k-1)+(2[r]_p+2m)L_{p,-1}(2n-1,k), \qquad k~ \text{ odd}.$$
Multiplying the last two recurrences by $y^k$, summing over even and odd values of $k$, respectively, and adding the resulting equations implies
\begin{align}
a(2n;y)&=(y+2[r]_p+m)a(2n-1;y)+m\sum\limits_{\substack{k=0\\k \text{ odd}}}^nL_{p,-1}(2n-1,k)y^k\notag\\
&=(y+2[r]_p+m)a(2n-1;y)+\frac{m}{2}(a(2n-1;y)-a(2n-1;-y))\notag\\
&=\left(y+2[r]_p+\frac{3m}{2}\right)a(2n-1;y)-\frac{m}{2}a(2n-1;-y), \qquad n \geq 1.\label{minus1e2}
\end{align}
By similar reasoning, we also have
\begin{equation}\label{minus1e3}
a(2n-1;y)=\left(y+2[r]_p+\frac{m}{2}\right)a(2n-2;y)-\frac{m}{2}a(2n-2;-y), \qquad n \geq 1.
\end{equation}
Multiplying \eqref{minus1e2} and \eqref{minus1e3} by $x^{2n}$ and $x^{2n-1}$, respectively, and summing over all $n \geq 1$, implies
\begin{align*}
H(x,y)-1&=\sum_{n\geq 1}a(n;y)x^n\\
&=x\left(y+2[r]_p+\frac{m}{2}\right)H(x,y)-\frac{mx}{2}H(x,-y)+mx\sum_{n\geq1}a(2n-1;y)x^{2n-1}\\
&=x\left(y+2[r]_p+\frac{m}{2}\right)H(x,y)-\frac{mx}{2}H(x,-y)+\frac{mx}{2}(H(x,y)-H(-x,y)),
\end{align*}
which gives the functional equation
\begin{equation}\label{minus1e4}
(1-xy-cx)H(x,y)+\frac{mx}{2}H(-x,y)+\frac{mx}{2}H(x,-y)=1.
\end{equation}
Replacing $(x,y)$ by $(-x,y)$, $(x,-y)$, and $(-x,-y)$ in \eqref{minus1e4}, and solving the resulting linear system in four unknowns using Cramer's rule, yields \eqref{minus1e1}.
\end{proof}
\begin{remark} Note that $H(x,1)$ is the generating function for $u_n:=\sum_{k=0}^nL_{p,-1}(n,k)$, which gives the sign balance of the $\beta$ statistic over all $(r,m)$-Lah distributions of size $n+r$. From \eqref{minus1e1} when $y=1$, this quantity is seen to satisfy the fourth-order recurrence
$$u_n=2(c^2+1)u_{n-2}-(c^2+cm-1)(c^2-cm-1)u_{n-4}, \qquad n \geq 4.$$
Similarly, using \eqref{minus1e1}, one may deduce a more complicated linear recurrence satisfied by the array $L_{p,-1}(n,k)$.
\end{remark}
\section{Further formulas in the case {\it q~}=~1}
We have the following further results for $L_{p,q}(n,k)$ when $q=1$. The first formula in the next theorem generalizes \cite[Corollary 4.3]{Ch} and reduces to this result when $p=1$.
\begin{theorem}\label{Lth1q1}
If $n,k \geq 0$ and $r \geq s \geq 0$, then
\begin{equation}\label{Lth1q1e1}
L_{p,1}(n,k;r,m)=\sum_{i=k}^n \binom{n}{i}[2p^s[r-s]_p|m]_{n-i}L_{p,1}(i,k;s,m)
\end{equation}
and
\begin{equation}\label{Lth1q1e2}
L_{p,1}(n,k;r,m)=\sum_{i=k}^n\sum_{j=k}^i\binom{n}{i}\binom{i}{j}m^{j-k}[2[r-s]_p|m]_{n-i}[2p^{r-s}[s]_p|m]_{i-j}L(j,k;0,1).
\end{equation}
\end{theorem}
\begin{proof}
We argue that the right-hand side of \eqref{Lth1q1e1} gives the sum of the $p$-weights of all members of $\mathcal{L}_{r,m}(n,k)$ according to the number, $n-i$, of elements of $I$ belonging to the \emph{last} $r-s$ special blocks. Once these elements have been chosen in $\binom{n}{i}$ ways, there are $2(p^s+\cdots+p^{r-1})=2p^s[r-s]_p$ possibilities concerning the choice of position for the smallest chosen element. After this smallest element is positioned, it is seen that there are $2p^s[r-s]_p+m$ possibilities for the position of the second smallest chosen element. Continuing in this manner, it follows that there are $[2p^s[r-s]_p|m]_{n-i}$ possibilities concerning the positions of the elements in the final $r-s$ special blocks. The remaining unselected $i$ members of $I$, together with the elements of $[s]$, then constitute a configuration in $\mathcal{L}_{s,m}(i,k)$, whence the factor of $L_{p,1}(i,k;s,m)$. Summing over all $i$ gives \eqref{Lth1q1e1}.
To show \eqref{Lth1q1e2}, we consider instead the number, $n-i$, of members of $I$ belonging to the \emph{first} $r-s$ special blocks. Note that there are $\binom{n}{i}[2[r-s]_p|m]_{n-i}$ ways in which to choose and arrange these elements. Concerning the remaining $i$ members of $I$, we must specify how many of them belong to the final $s$ special blocks, say $i-j$. There are $\binom{i}{j}[2p^{r-s}[s]_p|m]_{i-j}$ ways in which to choose and arrange these elements. The remaining $j$ members of $I$ then go in the non-special blocks and constitute a Lah distribution having $k$ blocks, wherein the $j-k$ non-minimal elements are each assigned one of $m$ possible colors. Summing over all possible $i$ and $j$ gives \eqref{Lth1q1e2}.
\end{proof}
The $p=1$ case of formula \eqref{Lth1q1e2} does not seem to have been previously noted. On the other hand, the $p=1$ case of the next result occurs as \cite[Theorem 4.2]{Ch} and was shown by algebraic methods using the Riordan group. Here, we provide a combinatorial proof using the interpretation developed above for $L_{p,q}(n,k)$.
\begin{theorem}\label{Lidt1}
If $n,k \geq 0$, then
\begin{equation}\label{Liden1}
L_{p,1}(n,k;r,m)=\sum_{i=k}^n\binom{n}{i}[2[r]_p(1-m)|m]_{n-i}m^{i-k}L_{p,1}(i,k;r,1).
\end{equation}
\end{theorem}
\begin{proof}
To give a bijective proof of this result, within each $(r,m)$-Lah distribution, we distinguish further certain elements, assigning some of these elements different weights. We then consider the sum of the weights of configurations containing three types of elements defined as follows. A \emph{critical element} (ce) is one that satisfies the nearness property and is either underlined (type 1) or overlined (type 2). A \emph{sub-critical element} (sce) is a member of $I$ having a predecessor that is either a ce or another sce.
Critical elements of type 1 are assigned no color, whereas ce's of type 2 and sce's are assigned one of $m$ colors. Finally, a \emph{non-critical element} (nce) is a member of $I$ that is neither critical nor sub-critical. By a \emph{distinguished non-critical element} (dcne), we mean an nce that satisfies the nearness property. Note that a dnce is to be neither underlined nor overlined. All nce's except those that are the smallest within their block are assigned one of $m$ colors (including dnce's, which did not receive a color in the definition of $\mathcal{L}(n,k)$ above). Let us refer to members of $\mathcal{L}_{r,m}(n,k)$ marked and colored as described above as \emph{extended $(r,m)$-Lah distributions}.
Given $i\geq0$, let $\mathcal{R}^{(i)}(n,k)=\mathcal{R}_{r,m}^{(i)}(n,k)$ denote the set of all extended $(r,m)$-Lah distributions containing exactly $i$ nce's (including dnce's). Note that $\mathcal{R}^{(i)}(n,k)$ is empty unless $k \leq i \leq n$ since ce's and sce's must belong to special blocks, by definition. Let $\mathcal{R}(n,k)=\cup_{i=k}^n\mathcal{R}^{(i)}(n,k)$. We define the $p$-weight of $\lambda \in \mathcal{R}(n,k)$ just as we did above for $\mathcal{L}(n,k)$ (that is, per Definitions \ref{Ldef3} and \ref{Ldef6}), with all elements satisfying the nearness property contributing (namely, the dnce's and ce's). Define the sign of $\lambda \in \mathcal{R}(n,k)$ to be $(-1)^{\nu(\lambda)}$, where $\nu(\lambda)$ denotes the number of ce's of type 2.
We now argue that the $i$-th term on the right-hand side of \eqref{Liden1} gives the (signed) weight of all members of $\mathcal{R}^{(i)}(n,k)$. Let $S=\{s_1