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\title{\bf Signed Excedance Enumeration in the Hyperoctahedral group}
\author{
Sivaramakrishnan Sivasubramanian\\
\small Department of Mathematics\\[-0.8ex]
\small Indian Institute of Technology, Bombay, India\\
\small \tt krishnan@math.iitb.ac.in 
}



\date{\dateline{Jul 6, 2013}{Mar 30, 2014}{Apr 16, 2014}\\
\small Mathematics Subject Classifications: 05A05, 05A30, 05A19
}

\begin{document}

\maketitle

\begin{abstract}
Several signed excedance-like statistics have nice formulae or 
generating functions
when summed over the symmetric group and over its subset of derangements.  
We give counterparts of some of these results when we sum over the 
hyperoctahedral group and its subset of derangements.  Our results 
motivate us to define and derive attractive bivariate formulae which 
generalise some of these results for the symmetric group.

\bigskip\noindent \textbf{Keywords:} Signed excedance;
  permutation group; hyperoctahedral group 
\end{abstract}

\section{Introduction}
For a positive integer $n$, let $[n] = \{1,2,\ldots,n \}$ and let
$\SSS_n$ be the set of permutations on $[n]$.  
Let $\BB_n$ be the set of permutations 
$\sigma$ of $\{-n, -(n-1), \ldots, -1, 1, 2, \ldots n\}$ satisfying
$\sigma(-i) = -\sigma(i)$.  Clearly any such $\sigma$ is well 
defined when given $\sigma(i)$ for $i \in [n]$.
$\BB_n$ is referred to as the hyperoctahedral group or the group
of signed permutations on $[n]$, though we do not need its group
structure in this work.  
Clearly, $|\SSS_n| = n!$ and $|\BB_n| = 2^n n!$.  
For $\sigma \in \BB_n$, from now on we denote $\sigma(i)$ instead 
as $\sigma_i$.  For $1 \leq k 
\leq n$, we also denote $-k$ alternatively as $\ol{k}$.

Let $\pi = (\pi_1,\pi_2,\ldots,\pi_n) \in \SSS_n$.  Define its excedance
set, $\ExcS(\pi)$ as $\{ i \in [n]: \pi_i > i\}$ and its number of 
excedances as $\exc(\pi) = |\ExcS(\pi)|$.  For $\pi \in \SSS_n$, define
its number of inversions as $\inva(\pi) = 
| \{ 1 \leq i < j \leq n : \pi_i > \pi_j \} |$
and its $n$-th position index $\np(\pi)$ as the index $i \in [n]$ 
such that $\pi_i = n$.  

For a positive integer $n \geq 1$, define the signed excedance enumerator as 
$\SE_n(q) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} q^{\exc(\pi)}$.
If $\DD_n$ is the
set of derangements on $[n]$, and if the number of signed derangements 
is defined as $\SDer_n = \sum_{\pi \in \DD_n} (-1)^{\inva(\pi)}$, 
then it is known that $\SDer_n = (-1)^{n-1}(n-1)$
(see Sivasubramanian \cite[Remark 5]{siva-exc-det}).
We recall that for a non-negative integer, $i$, its $q$-analogue is 
defined as $[i]_q = 1 + q + q^2 + \cdots + q^{i-1}$, where
$q$ is an indeterminate and $[0]_q = 0$.  For $n \geq 1$, define 
$\DSE_n(q) = \sum_{\pi \in \DD_n} (-1)^{\inva(\pi)} q^{\exc(\pi)}$
as the signed excedance enumerator over derangements.

In this work, we enumerate signed excedances in $\BB_n$.  Our motivation
comes from the following two beautiful results in $\SSS_n$.  


\begin{theorem}[Mantaci~\cite{mantaci-jcta-93}]
  \label{thm:mantaci-exc}
  Let $n \geq 1$ be a positive integer.  Then \[ \SE_n(q) = (1-q)^{n-1}.\]
\end{theorem}

\begin{theorem}[Mantaci and Rakotondrajao~\cite{mantaci-rak_exc_der}]
   \label{thm:der-sign-exc}
Let $n \geq 1$ be a positive integer.  Then
$$\DSE_n(q) = (-1)^{n-1}q \cdot[n-1]_q.$$
\end{theorem}

The original proofs of both results used sign reversing involutions
and both papers prove more detailed results as well.
An alternate proof of both these results was given by Sivasubramanian
in \cite{siva-exc-det}, where both $\SE_n(q)$ and $\DSE_n(q)$ were found 
out to be determinants of suitably defined  $n \times n$ matrices.  The 
connection between determinants and signed excedance enumeration
is a simple consequence of Leibniz's formula for the determinant.
Further, in 
\cite{siva-exc-det}, several other known and new excedance type 
statistics in $\SSS_n$ were also enumerated with signs, by evaluating determinants 
of similar $n \times n$ matrices.  In this work, we give counterparts of 
some of the results of \cite{siva-exc-det} in the case when enumeration is 
done over $\BB_n$, the set of signed permutations.
Since we do not have a determinant-type expansion involving
a signed sum over elements of the hyperoctahedral group, our 
proofs use sign-reversing involutions on $\BB_n$.   

It is easy to see that $\BB_n$  can be considered as a subgroup of $\SSS_{2n}$.  
Thus, it 
would be very interesting if the results in this paper on $\BB_n$ can be 
obtained by evaluating the determinant of a $2n \times 2n$ matrix, or 
even the determinant of a matrix with order polynomial in $n$,
as done in \cite{siva-exc-det}.

Our proof of these results %give us bivariate signed excedance enumeration results and this 
motivate us to define bivariate signed excedance enumerators in 
$\BB_n$ and hence in $\SSS_n$ as well.  Over $\SSS_n$, these bivariate versions 
give a sharpening of several known results including Theorem
\ref{thm:mantaci-exc} and Theorem \ref{thm:der-sign-exc} (see Theorems
\ref{thm:perm_sgn_exc_biv} and \ref{thm:der_sgn_exc_biv} respectively).



Several definitions of excedance exist in $\BB_n$ and we follow Brenti's
definition from \cite{brenti-q-eulerian-94}, which we recall.  
For $\sigma = (\sigma_1, \sigma_2, \ldots, \sigma_n) \in \BB_n$, define
$\ExcS(\sigma) = \{ i \in [n]: \sigma_{|\sigma_i |} > \sigma_i \} \cup
\{i \in [n]: \sigma_i = - i \}$ and let
$\exc(\sigma) = |\ExcS(\sigma)|$.
%|\{i \in [n]: \sigma_{|\sigma_i|} > \sigma_i \}| 
%+ |\{i \in [n]: \sigma_i = -i \}|$.  
We next give the definition of
inversions in $\BB_n$.  For $\sigma \in \BB_n$, let 
$\Negs(\sigma) = \{i \in [n]: \sigma_i < 0 \}$ be the set of indices where
$\sigma$ takes negative values and let 
$\nsum(\sigma) = - ( \sum_{i \in \Negs(\sigma)} \sigma_i)$ be the absolute 
value of the sum of the negative components of $\sigma$.  Define the 
number of type-A inversions of $\sigma$, as before
as $\inva(\sigma) = |\{ 1 \leq i < j \leq n : \sigma_i > \sigma_j \} |$.
Here, comparison is done with respect to the standard order on $\ZZ$.
Define the number of inversions of $\sigma \in \BB_n$ as 
$\invb(\sigma) = \nsum(\sigma) + \inva(\sigma)$.  This combinatorial 
definition of inversions in $\BB_n$ is also due to Brenti (see 
\cite [Proposition 3.1]{brenti-q-eulerian-94}).  If $\sigma \in \BB_n$, 
define $\np(\sigma)$ as the index $i \in [n]$ such that $| \sigma_i| = n$ and
$\onep(\sigma)$ as the index $i \in [n]$ such that $|\sigma_i| = 1$.
%\blue{
In some of our proofs, we need the cycles of $\sigma \in \BB_n$.
This is an alternate view of $\sigma$ and one can go from the one-line
notation of $\sigma$ to its cycle notation and vice versa.
We define the cycle notation and give an example of it below.  
Given $\sigma \in \BB_n$, we form a disjoint set of cycles ${\cal C_{\sigma}} = \{C_k\}_{k \in P_{\sigma}}$ 
where $P_{\sigma}$ is some index set such 
that each $\sigma_i$ for $i \in [n]$ is contained in exactly one cycle $C \in {\cal C_{\sigma}}$.  
Each $C$ is formed from the one-line notation of $\sigma$ as follows.
%such that if $C$ is a cycle with elements 
We write the elements of $C$ sequentially and if $C$ has elements 
$(x_1, x_2, \ldots, x_r)$, then in the one-line notation, we must have had 
$\sigma_{|x_i|} = x_{i+1}$ where $x_{r+1} = x_1$.
For example, given the one-line notation of $\sigma = (4,\ol{1},3,6,2,\ol{5}) \in \BB_6$, its
cycle notation is 
${\cal C_{\sigma}} = \{ (\ol{1},4,6,\ol{5},2); (3) \}$.   To get the one-line notation 
of $\sigma$ given ${\cal C_{\sigma}}$, we need to get $\sigma_i$ for each $i \in [n]$.
For each $i \in [n]$, either $i$ or $\ol{i}$ will be present in exactly one of the 
cycles $C_r \in {\cal C_{\sigma}}$.  Let $i$ occur in $C_r$ with sign $\epsilon$ where
$\epsilon = \pm 1$.  $\sigma_i$ is the element succeeding $\epsilon i$ in $C_r$ with the 
sign of the succeeding element taken into account.  Though we write the elements of $C_r$ in a 
sequence, as $C_r$ is a cycle 
the first element of $C_r$ is the succeeding element of the last element of $C_r$.
%}
%\blue{\bf give an example of cycle}.

Let $\sigma \in \BB_n$ and $\pi \in \SSS_n$.  We carefully distinguish between 
$\invb(\sigma)$ and $\inva(\pi)$, but make no such distinction between $\exc(\sigma)$ 
and $\exc(\pi)$, or between $\np(\sigma)$ and $\np(\pi)$, though the definitions are 
different.  This will not cause any problem.

Using Brenti's definition of excedance, Chen, Tang and Zhao 
\cite{chen-tang-zhao-typeB-der} show a binomial type equation that the 
type-B Eulerian polynomial and the type-B derangement 
polynomial enumerated by excedance satisfy, in the univariate unsigned 
case. %, satisfy a binomial type equation. 
In Subsection \ref{subsec:der_exc_enum}, we show that a very similar
binomial type equation is satisfied by our signed bivariate analogue 
when the exponent of $t$ in the term corresponding to $\sigma$ is 
either $\np(\sigma)$ or $\onep(\sigma)$,
see Corollaries \ref{cor:binom} and \ref{cor:binom_one}.


\section{The involution and a few lemmas}
\label{sec:two_lems}
We begin by proving a few lemmas about a sign-reversing involution which we 
use frequently.  We first define the involution.
Let $\sigma = (\sigma_1, \sigma_2, \ldots, \sigma_n) \in \BB_n$ and let $r$ 
be an index where $1 \leq r \leq n$.  We define an involution 
$\tau_r: \BB_n \ra \BB_n$.  For ease of notation, we write 
$\tau_r(\sigma) = \psi$ and $\psi = (\psi_1, \psi_2, \ldots, \psi_n)$.
The involution is as follows: if $\sigma_r = k > 0$, define $\psi_r = \ol{k}$ and if 
$\sigma_r = \ol{k} < 0$, let $\psi_r = k$.  For indices
$i \not= r$, define $\psi_i = \sigma_i$.   That is, $\tau_r$ negates the content at 
index $r$ and leaves the content at other indices unchanged.  It is evident
that $\tau_r$ is an involution for all $1 \leq r \leq n$.  
We prove the following property of $\tau_r$.

\begin{lemma}
  \label{lem:inv_rev}
  For all $1 \leq r \leq n$ and $\sigma \in \BB_n$, 
$\invb(\sigma) \not \equiv \invb(\tau_r(\sigma))$ (mod 2).
\end{lemma}
\begin{proof}
Fix $r$ and denote $\tau_r(\sigma)$ as $\psi = (\psi_1, \psi_2, \ldots, \psi_n)$.  
We assume $\sigma_r = k > 0$ without loss of generality.  
Consider the set of indices $S = \{i: |\sigma_i| < k \}$.  Clearly, 
$|S| = k-1$.
Denote the indices with elements of $S$ occuring before position 
$r$ as $S_1$.  i.e.  $S_1 = \{ 1 \leq i < r: -k < \sigma_i < k \}$.  
Similarly define $S_2 = \{ r \leq i \leq n: -k < \sigma_i < k \}$.  
Thus, $|S_1| + |S_2| = k-1$.  Recall $\invb(\sigma) = \nsum(\sigma) 
+ \inva(\sigma)$.  To show that $\invb(\sigma)$ and $\invb(\psi)$
have opposite parity, we count their differences.  As
$\sigma_r > 0$, there is no contribution to $\nsum(\sigma)$ and there
are $|S_2|$ type-A inversions due to $\sigma_r (= k)$.  As 
$\psi_r = \ol{k} < 0$, we have $k$ added to $\nsum(\psi)$ and there
exist $|S_1|$ type-A inversions in $\psi$ that do not occur in $\sigma$.  
As these are the only differences, 
$\invb(\sigma) - \invb(\psi) = |S_2| - (k + |S_1|) = -1-2|S_1|$ which is odd,
completing the proof.
\end{proof}

Just as $\tau_r$ negates index $r$, we need involutions $\tau_S$ for
$S \subseteq [n]$, which simultaneously negate all the indices 
in $S$ and leave the contents indexed by $[n] -S$ unchanged.  
By iterating the above lemma, we get the following simple corollary whose 
proof we omit.

\begin{corollary}
  \label{cor:odd_even}
  Let $\sigma \in \BB_n$ and let $S \subseteq [n]$. 
  Then, $\invb(\tau_S(\sigma)) \equiv (-1)^{|S|}  \cdot \invb(\sigma)$ (mod 2).
\end{corollary}


We move to our next lemma about excedances.  We need the following definitions
before we state the lemma.
For $\sigma \in \BB_n$, define $\ExcSa(\sigma) =  \{i \in [n]: 
\sigma_{|\sigma(i)|} > \sigma_i \}$, $\NFPS(\sigma) = \{i \in [n]: 
\sigma_i = -i \}$ and $\FPS(\sigma) = \{i \in [n]: \sigma_i = i\}$.  
Let $\fp(\sigma) = |\FPS(\sigma)|$.  
Clearly, $ \ExcS(\sigma) = \ExcSa(\sigma) \cup \NFPS(\sigma)$ is 
the set of indices where excedances occur in $\sigma$ and 
$\exc(\sigma) = |\ExcS(\sigma)|$.  For $\sigma \in \BB_n$, 
define $\WkNExcS(\sigma) = [n] - \ExcS(\sigma)$, 
$\wknexc(\sigma) = |\WkNExcS(\sigma) |$.  Define 
$\NExcS(\sigma) = \WkNExcS(\sigma) - \FPS(\sigma)$ 
and $\nexc(\sigma) = |\NExcS(\sigma) |$. 

\begin{lemma}
  \label{lem:cycle_min}
  Let $\sigma \in \BB_n$ and let $C$ be a cycle of $\sigma$ with at least 
two elements in it.  Let 
$m \in C$ be the minimum element of $C$ in absolute value (i.e. among
the elements $x \in C$, $m$ has least $|x|$ value).  Let $|\sigma_r| = |m|$ and
let $\psi = \tau_r(\sigma)$ (i.e. $\psi$ is the same as $\sigma$ except for
the sign of $m$).  Then, $\ExcS(\psi) = \ExcS(\sigma)$ and thus
$\exc(\psi) = \exc(\sigma)$.  Further, $\wknexc(\sigma) = \wknexc(\psi)$ 
and $\nexc(\sigma) = \nexc(\psi)$.
\end{lemma}
\begin{proof}  Assume that $m$ occurs in $\sigma$ with positive sign (the other
  case is proved identically).
Since $m$ is not in a cycle of length one, it has both a successor 
$\scc(m)$ defined as $\scc(m) = \sigma_m$ and a 
predecessor $\pred(m)$ defined as $\pred(m) = \sigma^{-1}(m)$ and for
brevity denote $k = \pred(m)$.  Define $\psi = \tau_k(\sigma)$ (i.e. $\psi$
is identical to $\sigma$ with the sole difference being that it has 
opposite sign for $m$).  
Clearly in both $\sigma$ and $\psi$ one of the possibilities
for a type-B excedance, namely an index $i$ such that $\sigma_i = -i$ will 
never occur at $r$ (as $\sigma_r = m$ and $\psi_r = \ol{m}$).  Since $m$ 
is the absolute value-wise smallest element of $C$ in $\sigma$, if there is an 
excedance 
due to $\pred(m)$ and $m$ in $\sigma$, then there will be an excedance
between $\pred(\ol{m})$ and $\ol{m}$ in $\psi$ and vice versa.  Likewise,
there will be an excedance in $\sigma$ between $m$ and $\scc(m)$ iff
there is an excedance between $\ol{m}$ and $\scc(\ol{m})$.

The argument above shows that $\ExcSa(\sigma) = \ExcSa(\psi)$.  Further,
since the sign of $m$ is the only difference between $\sigma$ and $\psi$,
$\NFPS(\sigma) = \NFPS(\psi)$.
Thus we have $\ExcS(\sigma) = \ExcS(\psi)$ and so  $\exc(\sigma) = \exc(\psi)$.
This implies that  $\WkNExcS(\sigma) = \WkNExcS(\psi)$.
and thus $\wknexc(\sigma) = \wknexc(\psi)$.  It is also clear that 
$\FPS(\sigma) = \FPS(\psi)$ and hence $\nexc(\sigma) = \nexc(\psi)$,
completing the proof.
\end{proof}

\vspace{1 mm}



Let $\sigma \in \BB_n$ for $n \geq 3$ have $r$ as a fixed point 
(i.e. $\sigma_r = r$) for some $1 \leq r \leq n$.  
Define $\rho \in \BB_{n-1}$ as follows: delete the occurence of 
$r$ in $\sigma$ and then replace $i$ by $i-1$ for 
all $r+1 \leq i \leq n$ with signs preserved (i.e. $\ol{(k+1)}$ will 
get changed to $\ol{k}$ if $k \geq r$).  We will use this operation
later on and for brevity, say $\rho$ is obtained
from $\sigma$ by  {\it ``shrinking $\sigma$ on $[r,n]$''}.
It is clear that the cycles of $\rho$ are identical to 
the cycles of $\sigma$ with just the following changes: 
the fixed point $r$ of $\sigma$ is removed and for $i \geq r$, all 
elements $i+1$ get changed to $i$ at their respective places in the 
cycles of $\sigma$ with the $i$'s in $\rho$ having the sign of $(i+1)$'s 
in $\sigma$.

\begin{lemma}
  \label{lem:same_inv_induct}
  Let $\sigma \in \BB_n$ have $r$ as a fixed point and let $\rho \in \BB_{n-1}$ 
be obtained by deleting $\sigma_r$ and then shrinking $\sigma$ on $[r,n]$.
Then, $\invb(\sigma) \equiv \invb(\rho)$ (mod 2),
$\exc(\sigma) = \exc(\rho)$,  $\wknexc(\sigma) \not \equiv \wknexc(\rho)$ (mod 2)
and $\nexc(\sigma) = \nexc(\rho)$.
\end{lemma}
\begin{proof}
  We first prove that $\invb(\sigma) \equiv \invb(\rho)$ (mod 2).  All elements to the left
  of $r$ (i.e. all $\sigma_k$ for $k < r$)
  that are larger than $r$ will contribute to $\inva(\sigma)$ but not to $\inva(\rho)$.
  Similarly, all elements smaller than $r$ to its right (i.e. all $\sigma_k$ for 
  $k > r$) will contribute to $\inva(\sigma)$ but not to $\inva(\rho)$.  Likewise, if 
  there are $p$ negative elements $\sigma_{i_1}, \sigma_{i_2}, \ldots, \sigma_{i_p}$ 
  with absolute value larger 
  than $r$, they will contribute $s=\sum_{j=1}^p |\sigma_{i_j}|$ to $\nsum(\sigma)$.  
  In $\rho$, they will contribute $s - p$ to $\nsum(\rho)$.

  Let $X = \{ j : j < r \mbox{ and } \sigma_j > r (= \sigma_r) \}$ be the set of 
  indices smaller than $r$ whose image under $\sigma$ is larger than $r$ and let $\alpha = |X|$.
  Similarly, let $Y = \{ j: j > r \mbox{ and } \sigma_j < r\}$ be the set of %elements with 
  indices larger than $r$ whose image under $\sigma$ is smaller than $r$ and let $\beta = |Y|$.  Let 
  $Z = \{j: 1\leq j \leq n: \sigma_j < 0 \mbox{ and } |\sigma_j| > r\}$ and let 
  $\gamma = |Z|$.  

  From the argument in the first paragraph, the difference between $\invb(\sigma)$ 
  and $\invb(\rho)$ is $\alpha + \beta + \gamma$.  We show that $\alpha + \beta + 
  \gamma \equiv 0$ (mod 2). To see this, let $d$ be the number of negative elements 
  in $\sigma$.  We first consider the case
  when $d = 0$.  Recall $\sigma_r = r$. As there are $\alpha$ elements to the left 
  of index $r$ that are larger, there are $(r-1-\alpha)$ elements to $r$'s left that are smaller
  than $r$.  Thus there are $\alpha$ elements to the right of $r$ that are smaller than $r$.
  Hence, $\beta = \alpha$.  Since there are no negative elements, $\gamma = 0$.  Thus
  $\alpha + \beta + \gamma =  2 \alpha \equiv 0$ (mod 2).   We record this as follows:
  \begin{equation}
		\label{eqn:mod2_sign}
		\mbox{when $\Negs(\sigma) = \Negs(\rho)= \emptyset$, } \invb(\sigma) \equiv \invb(\rho) \mbox{ (mod 2)  }
  \end{equation}

  Hence, when $\sigma$ has no negative elements and if $\sigma_r = r$, the $\rho$ 
  obtained by shrinking $\sigma$ on $[r,n]$
  %which have no negative elements 
  has the same sign as $\sigma$.
 Moving onto the general case, if $\sigma$ has $d$ elements that are negative,
  then let $|\sigma| \in \BB_n$  be the signed permutation obtained by replacing 
  $\sigma_i$ by $|\sigma_i|$ for all $i$, 
  and likewise obtain $|\rho|$ from $\rho$.  By Corollary 
  \ref{cor:odd_even}, $\invb(\sigma) \equiv (-1)^d \invb(|\sigma|)$ (mod 2) and 
  $\invb(\rho) \equiv (-1)^d \invb(|\rho|)$ (mod 2).  Since both $|\sigma|$ and $|\rho|$ have
  no zero elements, by (\ref{eqn:mod2_sign}), we get $\invb(|\sigma|) \equiv \invb(|\rho|)$ (mod 2).
  Thus, $\invb(\sigma) \equiv \invb(\rho)$ (mod 2).

  
  From the cycles of $\sigma$ and $\rho$, it is easy to see that $\exc(\sigma) = \exc(\rho)$ and 
  since $\wknexc(\sigma) = n- \exc(\sigma)$ and
  $\wknexc(\rho) = n-1 - \exc(\rho)$, it is clear that they have opposite parities.  Since $\fp(\sigma) =
  \fp(\rho) + 1$, we get $\nexc(\sigma) = 
  n - \exc(\sigma) - \fp(\sigma) -1 =
  \nexc(\rho)$, completing the proof.
\end{proof}

\begin{lemma}
  \label{lem:minus_same_inv_induct}
  Let $\sigma \in \BB_n$ have $\sigma_r = \ol{r}$ and let $\rho \in \BB_{n-1}$ be obtained by deleting
$\ol{r}$ and then shrinking $\sigma$ on $[r,n]$.  Then, $\invb(\sigma) \not \equiv \invb(\rho)$ (mod 2).  Further,
$\exc(\sigma) = \exc(\rho) + 1$, $\wknexc(\sigma) = \wknexc(\rho)$ and 
$\nexc(\sigma) \equiv \nexc(\rho)$ (mod 2).
\end{lemma}
\begin{proof}
  Obtain $\psi$ from $\sigma$ by changing the sign of $r$ (thus $\psi_r = r$).  By Lemma \ref{lem:inv_rev},
  $(-1)^{\invb(\psi)} \not\equiv (-1)^{\invb(\sigma)}$ (mod 2) and by Lemma \ref{lem:same_inv_induct}
  we get $(-1)^{\invb(\psi)} \equiv (-1)^{\invb(\rho)}$ (mod 2).  
  It is easy to see that $\exc(\sigma) = \exc(\rho) + 1$ and $\wknexc(\sigma) = \wknexc(\rho)$.  Moreover, 
  as $\fp(\sigma) = \fp(\rho)$, we get $\nexc(\sigma) \equiv \nexc(\rho)$ (mod 2), completing
  the proof. 
\end{proof}

\section{Signed excedance enumeration in $\BB_n$}
\label{sec:sgnexcBn}

For $n \geq 1$ and $1 \leq i \leq n$, let 
$\BSE_n^i(q,t) = \sum_{\sigma \in \BB_n}(-1)^{\invb(\sigma)}q^{\exc(\sigma)}t^{\ip(\sigma)}$
be a bivariate signed enumerator counting excedance and the index where $i$ up to sign occurs.
Define $\BSE_0^0(q,t) = 1$.  
We begin with the following bivariate analogue of Theorem \ref{thm:mantaci-exc} to the 
hyperoctahedral group.

\begin{theorem}
  \label{thm:sgn_exc_hyp_i}
For $n \geq 1$ and $1 \leq i \leq n$, $\BSE_n^i(q,t) = t^i(1-q)^n$.
\end{theorem}
\begin{proof}
  The result can be readily checked when $n=1,2$ and so we assume $n \geq 3$.
  Let $X_i = \{ \sigma \in \BB_n: \ip(\sigma) = i \}$.  Consider the map $\tau_k : (\BB_n - X_i) \mapsto 
  (\BB_n - X_i)$ defined as follows: let $\sigma \in \BB_n -X_i$.  Since $i$ is not in a cycle by itself,
  it is in a cycle $C$ of  length at least two.  Let $m$ be the minimum element of $C$ in absolute value
  and let $|\sigma(k)| = |m|$.  Define $\psi = \tau_k(\sigma)$.  Thus in $\psi$, $m$ occurs with opposite
  sign as compared to $\sigma$.  By Lemma \ref{lem:inv_rev}, $\invb(\psi) \not \equiv  \invb(\sigma)$ 
  (mod 2) and by Lemma \ref{lem:cycle_min}, $\exc(\psi) = \exc(\sigma)$.  Further, 
  $\ip(\psi)= \ip(\sigma)$ and thus 
  $\sum_{\sigma \in \BB_n - X_i} (-1)^{\invb(\sigma)} q^{\exc(\sigma)} t^{\ip(\sigma)} = 0$.  
  
  Thus, 
  $ \BSE_n^i(q,t)  = \sum_{\sigma \in X_i} 
  (-1)^{\invb(\sigma)} q^{\exc(\sigma)} t^{\ip(\sigma)}$.  We can thus pull out $t^i$ from each term in
  the summation. 
  We will show that $\sum_{\sigma \in X_i} (-1)^{\invb(\sigma)} q^{\exc(\sigma)} = (1-q)^n$. 
  Let $X_i^+ = \{\sigma \in X_i: \sigma_i = +i \}$ and $X_i^- = X_i - X_i^+$.   For each $\sigma \in X_i^+$, 
  deleting $i$ and then shrinking $\sigma$ on $[i,n]$ to get $\rho$ will clearly give us all $\rho 
  \in \BB_{n-1}$.   An identical statement is again true when we get $\rho \in 
  \BB_{n-1}$ from 
  $\sigma \in X_i^-$ by deleting $\sigma_i$ and shrinking $\sigma$ from $[i,n]$.  
  
  Let $a_n^+ = \sum_{\sigma \in X_i^+} (-1)^{\invb(\sigma)} q^{\exc(\sigma)}$ and 
  $a_n^- = \sum_{\sigma \in X_i^-}  (-1)^{\invb(\sigma)} q^{\exc(\sigma)}$.   We will show that $a_n^+ + a_n- =
  (1-q)^n$ by induction on $n$.     The result is clear when $n = 1,2$ and so let $n > 2$.  
  Recall that $\rho$ is obtained by shrinking $\sigma$ on $[i,n]$.
  By Lemma \ref{lem:same_inv_induct}, $\invb(\sigma) = \invb(\rho)$ and $\exc(\sigma) = \exc(\rho)$.
  Thus by induction, $a_n^+ = (1-q)^{n-1}$.  To evaluate $a_n^-$, we get by Lemma
  \ref{lem:minus_same_inv_induct} that $\invb(\sigma) \not \equiv \invb(\rho)$ (mod 2) and
  $\exc(\sigma) = \exc(\rho) + 1$.  Thus by induction, $a_n^- = -q(1-q)^{n-1}$.  Adding the
  two completes the proof.
\end{proof}

\vspace{1 mm}

We state two corollaries of Theorem \ref{thm:sgn_exc_hyp_i} when in the cases when 
$t=1$ and $t=n$ as we only
state counterparts of these two when enumeration is done in the symmetric group.
For $n \geq 1$, let 
$\BSE_n^1(q,t) = \sum_{\sigma \in \BB_n}(-1)^{\invb(\sigma)}q^{\exc(\sigma)}t^{\onep(\sigma)}$
be a bivariate signed enumerator where the exponent of $t$ in the term corresponding to $\sigma$ 
is $\onep(\sigma)$ 
%excedance and the index where $1$ up to sign occurs 
and let 
$\BSE_n^n(q,t) = \sum_{\sigma \in \BB_n}(-1)^{\invb(\sigma)}q^{\exc(\sigma)}t^{\np(\sigma)}$
be a bivariate signed enumerator where the exponent of $t$ in the term corresponding to 
$\sigma$ is 
%counting excedance and the index where $n$ up to sign occurs.
$\np(\sigma)$.


\begin{corollary}
  \label{cor:sgn_exc_hyp}
For $n \geq 1$, $\BSE_n^1(q,t) = t(1-q)^n$ and $\BSE_n^n(q,t) = t^n(1-q)^n$.
\end{corollary}

Bivariate enumerators as in Corollary \ref{cor:sgn_exc_hyp} are not known
in $\SSS_n$.   
This motivates us to define a bivariate signed excedance enumerator 
for the permutation group $\SSS_n$.  Towards finding such bivariate analogues of 
Theorem \ref{thm:mantaci-exc}, 
we consider the following $n \times n$ matrices
 $$
 M_n = \left( 
  \begin{array}{ccccc}
	1 & q & q & \cdots & qt \\
	1 & 1 & q & \cdots & qt^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	1 & 1 & 1 & \cdots & t^n \\
  \end{array}
  \right)  
  \mbox{ and } 
 M_n^1 = \left( 
  \begin{array}{ccccc}
	t & q & q & \cdots & q \\
	t^2 & 1 & q & \cdots & q \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	t^n & 1 & 1 & \cdots & 1 \\
  \end{array}
  \right)$$


  i.e. if $M_n = (m_{i,j})_{1 \leq i,j \leq n}$, then $m_{i,j}$ is 
obtained by the following procedure: set $m_{i,j} = q$ if 
$i < j$, and $m_{i,j} = 1$ otherwise.  After this, if $j=n$, set 
$m_{i,j} = m_{i,j} \cdot t^i$.  Likewise, if $M_n^1 = (m_{i,j}')$, then,
set $m_{i,j}' =q$ if $i < j$ and $m_{i,j}' = 1$ otherwise.  After 
this, if $j=1$, set $m_{i,j}' = m_{i,j}' \cdot  t^i$.


For $n \geq 1$, define $\SE_n^n(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} q^{\exc(\pi)} 
t^{\np(\pi)}$.  Thus, $\SE_1^1(q,t) = t$. We get the following bivariate generalization of Theorem 
\ref{thm:mantaci-exc}.

\begin{theorem}
  \label{thm:perm_sgn_exc_biv}
  For $n \geq 2$, $\SE_n^n(q,t) = t^{n-1}(1-q)^{n-2}(t-q)$.
\end{theorem}
\begin{proof}
Our proof is similar to the proof of Theorem 1 of \cite{siva-exc-det}.  Consider the
matrix $M_n$.  Clearly, $\det(M_n) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} \prod_{i=1}^n m_{i, \pi_i}$.
For a $\pi \in \SSS_n$, let $T_{\pi} = \prod_{i=1}^n m_{i,\pi_i}$ be up to sign, the term
corresponding to $\pi$ occuring in the determinant expansion.  Since $m_{i,j} = q$ if $i < j$ and 
$m_{i,j} = 1$ otherwise, and since $m_{i,n}$ also contributes a factor of $t^i$, we get 
$T_{\pi} = q^{\exc(\pi)} t^{\np(\pi)}$. 
Thus, $\SE_n(q,t) = \det(M_n)$.  Throughout this work, columns of $n\times n$ matrices are 
numbered from $1 \ldots n$ and $\Col_i$ refers to the $i$-th column.  We induct on $n$ to show
the result.  The case when $n=2$ is easy to check and thus
assume $n \geq 3$.  By performing the elementary column operation $\Col_1 := \Col_1 - \Col_2$
and then evaluating $\det(M_n)$, it is easy to see that 
$\det(M_n) = t^{n-1} (1-q)^{n-2} (t-q)$, completing the proof.
\end{proof}


For $n \geq 1$, define $\SE_n^1(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} q^{\exc(\pi)} 
t^{\onep(\pi)}$.  We get the following different generalization of Theorem 
\ref{thm:mantaci-exc}. 


\begin{theorem}
  \label{thm:perm_sgn_exc_biv_prime}
  For $n \geq 2$, $\SE_n^1(q,t) = t (1-q)^{n-2}(1-qt^{n-1})$.
\end{theorem}
\begin{proof}
  Arguing as in the proof Theorem \ref{thm:perm_sgn_exc_biv},  we get $\SE_n^1(q,t) = \det(M_n^1)$.
  Performing the elementary row operation $\Row_1 := \Row_1 - \Row_2$, and
  then evaluating the determinant makes it easy to see that 
  $\det(M_n^1) = t (1-q)^{n-2}(1-qt^{n-1})$, 
  completing the proof.
\end{proof}

\subsection{Adding quantities to the sign}

In this subsection, we enumerate similar signed excedance statistics.  A difference here is
that the exponent of -1 is a sum of several statistics of $\sigma$.   
For $n \geq 1$ and $1\leq i \leq n$, define 
$\BSWkSkE_n^i(q,t) = \sum_{\sigma \in \BB_n}(-1)^{\invb(\sigma) + \wknexc(\sigma)}
q^{\exc(\sigma)}t^{\ip(\sigma)}$ 
as the bivariate weak skew signed excedance enumerator with the exponent of $t$ being
the index in $\sigma$ where $i$ up to sign occurs.


\begin{theorem}
  \label{thm:sgnwkskexc_hyp_i}
For $n \geq 1$ and $1 \leq i \leq n$, $\BSWkSkE_n^i(q,t) = t^i (-1)^n(1+q)^n$.
\end{theorem}
\begin{proof}
  This proof is very similar to the proof of Theorem \ref{thm:sgn_exc_hyp_i}.
  Let $\sigma = (\sigma_1,\sigma_2,\ldots,\sigma_n) \in \BB_n$ and as before, 
  %Let $\sigma \in \BB_n$ and as before, 
let $X_i = \{\sigma \in \BB_n: \ip(\sigma) = i \}$.  Consider the map $\tau_k:
(\BB_n - X_i) \mapsto (\BB_n - X_i)$ defined as in the proof of Theorem
\ref{thm:sgn_exc_hyp_i} (i.e. change the sign of the smallest element in absolute
value in the cycle containing $i$).
Let $\psi = \tau_k(\sigma)$.  Clearly, $\ip(\sigma) =  \ip(\psi)$ and further, by 
Lemma \ref{lem:cycle_min}, $\wknexc(\sigma) = \wknexc(\psi)$.  Thus, as before, 
we get 
$$\sum_{\sigma \in \BB_n - X_i}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\exc(\sigma)}t^{\ip(\sigma)} = 0.$$

Hence,
\[ \BSSkE_n(q,t) = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\exc(\sigma)}
t^{\ip(\sigma)}\]
and thus, a factor of $t^i$ can be removed from all terms.
We will show that if $a_n = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \wknexc(\sigma)}
q^{\exc(\sigma)}$, then $a_n = (-1)^n(1+q)^n$.  Again, we induct on $n$ with
the base case when $n=1,2$ being easy. Thus, assume $n > 2$ and define 
$X_i^+ = \{ \sigma \in X_i: \sigma_i = +i \}$ and $X_i^- = X_i - X_i^+$.

Let $a_n^+ = \sum_{\sigma \in X_i^+}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\exc(\sigma)}$ 
and 
$a_n^- = \sum_{\sigma \in X_i^-}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\exc(\sigma)}$.
We compute $a_n^+$ first.  Let $\sigma \in X_i^+$ and let $\rho \in \BB_{n-1}$ be obtained by
deleting $i$ and shrinking $\sigma$ on $[i,n]$.  By Lemma \ref{lem:same_inv_induct}, 
$\invb(\sigma) + \wknexc(\sigma) \not \equiv \invb(\rho) + \wknexc(\rho)$ (mod 2) while
$\exc(\sigma) = \exc(\rho)$.  Thus, by induction, $a_n^+ = - (1+q)^{n-1}$.

We next compute $a_n^-$.  Let $\sigma \in X_i^-$ and $\rho$ be obtained by deleting
$\sigma_i$ and shrinking $\sigma$ on $[i,n]$.  By Lemma \ref{lem:minus_same_inv_induct},
$\invb(\sigma) + \wknexc(\sigma) \not \equiv \invb(\rho) + \wknexc(\rho)$ (mod 2) while
$\exc(\sigma) = \exc(\rho) + 1$.  By induction, $a_n^- = -q \cdot(1+q)^{n-1}$.
Adding $a_n^+$ and $a_n^-$ completes the proof.
\end{proof}

We define similar bivariate generating functions for the permutation group $\SSS_n$.  
For 
$n \geq 1$, define $\SWkSkE_n^n(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi) + \wknexc(\pi)} 
q^{\exc(\pi)} t^{\np(\pi)}$.  Likewise, for 
$n \geq 1$, define $\SWkSkE_n^1(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi) + \wknexc(\pi)} 
q^{\exc(\pi)} t^{\onep(\pi)}$.  Consider the following $n \times n$ matrices.

  $$
  T_n = \left(
  \begin{array}{ccccc}
	-1 & q & q & \cdots & qt \\
	-1 & -1 & q & \cdots & qt^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	-1 & -1 & -1 & \cdots & -t^n \\
  \end{array}
  \right)
  \mbox{ and } 
  T_n^1 = \left(
  \begin{array}{ccccc}
	-t & q & q & \cdots & q \\
	-t^2 & -1 & q & \cdots & q \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	-t^n & -1 & -1 & \cdots & -1 \\
  \end{array}
  \right)
  $$  
If $T_n = (t_{i,j})_{1 \leq i,j \leq n}$, then $t_{i,j}$ is 
obtained as follows: set $t_{i,j} = q$ if $i < j$, and $t_{i,j} = -1$
otherwise.  Then, if $j = n$, set $t_{i,j} = t_{i,j} \cdot t^i$.  
Similarly, if $T_n^1 = (t'_{i,j})_{1 \leq i,j \leq n}$, then $t'_{i,j}$ 
obtained as follows: set $t'_{i,j} = q$ if $i < j$, and $t'_{i,j} = -1$
otherwise.  Then, if $j = 1$, set $t_{i,j} = t_{i,j} \cdot t^i$.  
We get the following generalizations of Theorem 2 of \cite{siva-exc-det}.
  
\begin{theorem}
  \label{thm:perm_sgn_wkskexc_biv}
  For $n \geq 2$, $\SWkSkE_n^n(q,t) = (-1)^n t^{n-1}(1+q)^{n-2}(t+q)$.
  Similarly, for $n \geq 2$, $\SWkSkE_n^1(q,t) = (-1)^n t(1+q)^{n-2}(1+qt^{n-1})$.
\end{theorem}
\begin{proof}
Arguing as in the proof of Theorem  \ref{thm:perm_sgn_exc_biv}, we get 
%\blue{$\SSkE_n^n(q,t) = \det(T_n)$. - is this a wrong macro used} 
$\SWkSkE_n^n(q,t) = \det(T_n)$.
Performing the elementary 
column operation $\Col_1 := \Col_1 - \Col_2$ and then evaluating $\det(T_n)$, it can
be checked that $\det(T_n) = (-1)^n t^{n-1} (1+q)^{n-2} (t+q)$.
Similarly, it is easy to see that $\SWkSkE_n^1(q,t) = \det(T_n^1)$.  
Evaluating the determinant after performing the column operation $\Col_n := \Col_n - \Col_{n-1}$ 
shows that $\det(T_n^1) = (-1)^n t (1+q)^{n-2} (1+qt^{n-1})$, completing the proof.
\end{proof}

\vspace{2 mm}

\noindent
We next move on to type-B analogues of Theorem 5 of \cite{siva-exc-det}.
For $n \geq 1$ and for $1 \leq i \leq n$, let 
$\BSSkE_n^i(q,t) = \sum_{\sigma \in \BB_n}(-1)^{\invb(\sigma) + \nexc(\pi)}
q^{\exc(\sigma)}t^{\ip(\sigma)}$
be a bivariate skew signed enumerator counting excedance and the index where 
$i$ up to sign occurs.  

\begin{theorem}
  \label{thm:sgnskexc}
For $n \geq 1$ and $1 \leq i \leq n$, $\BSSkE_n^i(q,t) = t^i(1-q)^n$.
\end{theorem}
\begin{proof}
  This proof is very similar to the proof of Theorem 
  \ref{thm:sgnwkskexc_hyp_i}, and hence we just mention the 
differences.  
As before, define
$X_i = \{ \sigma \in \BB_n: \ip(\sigma) = i \}$.  It is easy to show that 
$\sum_{\sigma \in \BB_n - X_i}
(-1)^{\invb(\sigma) + \nexc(\sigma)}q^{\exc(\sigma)}t^{\ip(\sigma)} = 0$.

Hence,
$\BSSkE_n^i(q,t) = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \nexc(\sigma)}q^{\exc(\sigma)}
t^{\ip(\sigma)}$
and thus a factor of $t^i$ can be removed from all terms.  
We will show that if $a_n = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \nexc(\sigma)}
q^{\exc(\sigma)}$, then $a_n = (1-q)^n$.  Again, define 
$X_i^+ = \{ \sigma \in X_i: \sigma_i = +i \}$ and $X_i^- = X_i - X_i^+$.  Let
$a_n^+ = \sum_{\sigma \in X_i^+}(-1)^{\invb(\sigma) + \nexc(\sigma)}q^{\exc(\sigma)}$ 
and let $a_n^- = \sum_{\sigma \in X_i^-}(-1)^{\invb(\sigma) + \nexc(\sigma)}q^{\exc(\sigma)}$.

Let $\sigma \in \BB_n$ and $\rho$ be obtained by deleting $i$ and by shrinking $\sigma$ on $[i,n]$.
By induction and Lemma \ref{lem:same_inv_induct}, as $\invb(\sigma) + \nexc(\sigma) =
\invb(\rho) + \nexc(\rho)$ and as $\exc(\sigma) = \exc(\rho)$, we get 
$a_n^+ = a_{n-1}$.  Similarly, using Lemma \ref{lem:minus_same_inv_induct}, 
we get $a_n^- = -q a_{n-1}$.  Adding the two completes the 
proof.
\end{proof}


%\begin{conjecture}
%  \label{conj:sgnskexc_hyp_i}
%For $n \geq 1$, $\BSSkE_n^i(q,t) = t^i(1-q)^n$.
%\end{conjecture}

%% Check this.  This is quite interesting


\noindent  
We consider bivariate analogues of Theorem \ref{thm:sgnskexc} when enumeration is 
done in $\SSS_n$.  For $n \geq 1$, let $\SSkE_n(q,t) = \sum_{\pi \in \SSS_n} 
(-1)^{ \inva(\pi)  + \nexc(\pi) } q^{\exc(\pi)} t^{\np(\pi)}$ be its bivariate signed 
skew excedance enumerator and let $\SSkE_0(q,t) = 1$.
Consider the following $n \times n$ matrices

%\blue{\bf Write this properly as a Theorem.  Also state that the recurrence is an
%analogue of the earlier recurrence. }

 $$U_n = \left( 
  \begin{array}{ccccc}
	1 & q & q & \cdots & qt \\
	-1 & 1 & q & \cdots & qt^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	-1 & -1 & -1 & \cdots & t^n \\
  \end{array}
  \right) % \mbox{ and } 
  %U_n^1 = \left(
  %\begin{array}{ccccc}
	%t & q & q & \cdots & q \\
	%-t^2 & 1 & q & \cdots & q \\
	%\vdots & \vdots & \vdots & \ddots & \vdots \\
	%-t^n & -1 & -1 & \cdots & 1 \\
  %\end{array}
  %\right) 
  $$  
  

  i.e. if $U_n = (u_{i,j})_{1 \leq i,j \leq n}$, then $u_{i,j}$ is 
obtained as follows: set $u_{i,j} = q$ if $i < j$, $u_{i,j} = 1$ 
if $i=j$ and $u_{i,j} = -1$ otherwise.  After this, if $j=n$, set 
$u_{i,j} = u_{i,j} \cdot t^i$.   
%Similarly, if $U_n^1 = (u'_{i,j})_{1 \leq i,j \leq n}$, then $u'_{i,j}$ is 
%obtained as follows: set $u'_{i,j} = q$ if $i < j$, $u'_{i,j} = 1$ 
%if $i=j$ and $u'_{i,j} = -1$ otherwise.  After this, if $j=1$, set 
%$u'_{i,j} = u'_{i,j} \cdot t^i$.

\vspace{1 mm}

We give the following bivariate analogue of Theorem 5 of \cite{siva-exc-det}.
First consider the sequence of bivariate polynomials defined by the the following
recurrence:

\begin{equation}
  \label{eqn:qt_u_n}
  d_n(q,t) = (3-q)t \cdot d_{n-1}(q,t) + 2t^2(q-1) \cdot d_{n-2}(q,t) + 2^{n-3}qt(1-t).
\end{equation}

\noindent
with $d_0(q,t) = 1$ and $d_1(q,t) = t$.  A few initial $d_n(q,t)$'s are 
given below.

\begin{eqnarray*}
d_0(q,t) & = & 1 \\
d_1(q,t) & = & t \\
d_2(q,t) & = & t^2 + tq \\
d_3(q,t) & = & t^3 + q(t^3+t^2+2t) - q^2t^2 \\
d_4(q,t) & = & t^4 + q(4t^4+t^3+2t^2+4t) - q^2(t^4+2t^3+2t^2 )
+ q^3t^3 \\
d_5(q,t) & = & t^5 + q(11t^5 + t^4 + 2t^3 + 4t^2 + 8t)
- q^2(5t^5 - 3t^4 - 4t^3 - 4t^2) \\
 & & + q^3(t^5 + 3t^4 + 2t^3 ) - q^4t^4 
\end{eqnarray*}

\begin{theorem}
  \label{thm:biv_sk_exc}
For $n \geq 0$,  $\SSkE_n(q,t) = d_n(q,t)$.
\end{theorem}



\begin{proof}
Let $U_n$ be the matrix defined above.  By using arguments as done before, we
see that $\det(U_n) = \SSkE_n(q,t)$.  We claim that $\det( U_n) = d_n(q,t)$. 
By performing the elementary column operation $\Col_1 := \Col_1 - \Col_2$ 
and then evaluating $\det U_n$, we get the following recurrence.  We note
that on setting $t=1$, we recover the recurrence given in the proof of 
Theorem 5 of \cite{siva-exc-det}.
%\textcolor{red}{\bf Should we derive the recurrence?  }

\begin{equation}
  \label{eqn:det_qt_u_n}
  \det U_n  = (3-q)t \cdot \det U_{n-1}  + 2t^2(q-1) \cdot \det U_{n-2} + 2^{n-3}qt(1-t).
\end{equation}

\noindent
where $\det U_0 = 1, \det U_1 = t$ and $\det U_2 = t^2 + tq$ are easy to observe.  
It is clear that
recurrence (\ref{eqn:det_qt_u_n}) is identical to the recurrence (\ref{eqn:qt_u_n})
which completes this proof.
%Further, it is easy to
%see that the coefficient of $q$ in $d_n(q,t)$ for $n \geq 3$ is $A_{n,2}t^n + 
%\sum_{k=0}^{n-2}2^k t^{n-k}$, where $A_{n,2}$ is an Eulerian number. 
%\blue{\bf Check this.}
\end{proof}

\begin{remark}
The coefficient of $q$ in $d_n(q,1)$ is the Eulerian number 
$A_{n,2}$ (see the proof of Theorem 5 of \cite{siva-exc-det}).  
In the  bivariate analogue obtained, when $t$ is a variable, the 
coefficient of $q$, gives a polynomial analogue of $A_{n,2}$ in the 
variable $t$.   The Online Encyclopedia of Integer Sequences (see sequence id:A054123) 
mentions right Fibonacci row-sum arrays where this polynomial refinement of 
$A_{n,2}$ occurs in even rows.  This sequence thus has occurred while enumerating a
different problem and appears again here.  We do not know any connection between the two problems.
%A052509 on Knights-move
%Pascal triangle is also similar. 
\end{remark}


\subsection{Adding quantities to $q$'s exponent as well}

In this subsection, our generating function has a slightly modified exponent of $q$.
For $n \geq 1$,  and for $1 \leq i \leq n$, define
\[ \BSWkSkWE_n^i(q,t) = \sum_{\sigma \in \BB_n} (-1)^{\invb(\sigma) + \wknexc(\sigma)} 
q^{\wkexc(\sigma)} t^{\ip(\sigma)}.\]  Thus the fixed points of $\sigma$ contribute 
to the exponents of both -1 and $q$.

\begin{theorem}
  \label{thm:sgnwkskwkexc_i}
  For $n \geq 1$ and $1 \leq i \leq n$, $\BSWkSkWE_n^i(q,t) = t^i (-2q)^n$.
\end{theorem}
\begin{proof}
This proof is very similar to earlier proofs and thus, we just mention the 
differences.  As before, we get 
$\sum_{\sigma \in \BB_n - X_i}
(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\wkexc(\sigma)}t^{\ip(\sigma)} = 0$
where $X_i = \{ \sigma \in \BB_n: \ip(\sigma) = i \}$.

Hence $\BSWkSkWE_n(q,t) = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\wkexc(\sigma)}
t^{\ip(\sigma)}$ 
and thus a factor of $t^i$ can be removed from all terms.   
If $a_n = \sum_{\sigma \in X_i}(-1)^{\invb(\sigma) + \wknexc(\sigma)}
q^{\wkexc(\sigma)}$, then we need to show that $a_n = (-2q)^n$.  
We do this by induction on $n$.  The base case when $n=1,2$ are easy. Thus, assume
$n > 2$ and as in earlier proofs, define $X_n^+$ and $X_n^-$. 
Let $a_n^+ = \sum_{\sigma \in X_i^+}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\wkexc(\sigma)}$ 
and  
let $a_n^- = \sum_{\sigma \in X_i^-}(-1)^{\invb(\sigma) + \wknexc(\sigma)}q^{\wkexc(\sigma)}$.


For $\sigma \in X_i^+$, define $\rho \in \BB_{n-1}$ by deleting $i$ and shrinking
$\sigma$ on $[i,n]$.  By Lemma \ref{lem:same_inv_induct} as 
$\invb(\sigma) + \wknexc(\sigma) \not \equiv \invb(\rho) + \wknexc(\rho)$ and as
the fixed point $i$ contributes 1 to the exponent of $q$, we get 
$a_n^+ = -q(a_{n-1})$.  For $\sigma \in X_i^-$, define $\rho \in \BB_{n-1}$ by
deleting $\ol{i}$ and shrinking $\sigma$ on $[i,n]$.
By Lemma \ref{lem:minus_same_inv_induct} as $\invb(\sigma) + \wknexc(\sigma) \not \equiv
\invb(\rho) + \wknexc(\rho)$, we get a negative sign, and as $\sigma_i = \ol{i}$, we get 
$\wkexc(\sigma) = \wkexc(\rho) + 1$ and so we get $a_n^- = -q(a_{n-1})$.  Adding $a_n^+$
and $a_n^-$ completes the proof.
\end{proof}

We derive similar bivariate results when enumeration is done in the symmetric group
$\SSS_n$.  Define $\SWkSkWE_n(q,t) = \sum_{\pi \in \SSS_n} 
(-1)^{\inva(\pi) + \wknexc(\pi) } q^{\wkexc(\pi)} t^{\np(\pi)}$
as the bivariate signed weak-skew weak-excedance enumerator.   Let 
$\SWkSkWE_0(q,t) = 1$.  Consider the following $n \times n$ matrix.

$$ F_n = \left(
  \begin{array}{ccccc}
	-q & q & q & \cdots & qt \\
	-1 & -q & q & \cdots & qt^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	-1 & -1 & -1 & \cdots & -qt^n \\
  \end{array}
  \right)
  $$  
  
Likewise if $F_n = (f_{i,j})_{1 \leq i,j \leq n}$, then $f_{i,j}$ is 
obtained as follows: set $f_{i,j} = q$ if $i < j$, $f_{i,j} = -q$
if $i = j$ and $f_{i,j} = -1$
otherwise.  Then, if $j = n$, set $f_{i,j} = f_{i,j} \cdot t^i$.  
Consider the sequence of bivariate polynomials defined as follows:

\begin{equation}
  \label{eqn:qt_f_n}
	f_n = (1-3q)t \cdot f_{n-1} + 2qt^2(q-1) f_{n-2} + (-1)^n (q-1)^{n-2}(qt - qt^2)
\end{equation}

\noindent
with $f_0(q,t) = 1$ and $f_1(q,t) = -tq$.  We tabulate $f_n(q,t)$ for a few values
of $n$ below.

\begin{eqnarray*}
f_0(q,t) & = & 1 \\
f_1(q,t) & = & -tq \\
f_2(q,t) & = & tq + t^2q^2\\
f_3(q,t) & = & tq - (t^3 + 2t^2 + t)q^2 - t^3q^3 \\
f_4(q,t) & = & tq -(t^4+ 2t^2 + 2t )q^2 + (4t^4 + 4t^3 + 2t^2 + t)q^3 + t^4q^4 \\
f_5(q,t) & = & tq -(t^5+ 2t^2 + 3t )q^2 + (5t^5 + 4t^3 + 4t^2 + 3t)q^3 \\
 & & - (11t^4 + 8t^4 + 4t^3 + + 2t^2 + t)q^4 - t^5q^5
\end{eqnarray*}



\begin{theorem}
  \label{thm:biv_sgn_wk_sk_wk_exc}
For $n \geq 0$, $\BSWkSkWE_0(q,t) = f_n(t,q)$.
\end{theorem}
\begin{proof}
  Consider the matrix $F_n$ given above.  We get $\BSWkSkWE_n(q,t) = \det(F_n)$ 
by arguing as done earlier.  We claim that $\det( F_n) = f_n(q,t)$. 
Performing the column operation $\Col_1 := \Col_1 - \Col_2$ and then evaluating the determinant
gives us the following recurrence:

\begin{equation}
  \label{eqn:det_qt_f_n}
	\det F_n = (1-3q)t \cdot \det F_{n-1} + 2qt^2(q-1) \det F_{n-2} + (-1)^n (q-1)^{n-2}(qt - qt^2)
\end{equation}

We note that recurrence (\ref{eqn:det_qt_f_n}) is identical to recurrence (\ref{eqn:qt_f_n}).
Since the initial values of $f_n(q,t)$ and $\det F_n$ are identical and they satisfy the
same recurrence, they are identical for all $n$, completing the proof.
\end{proof}

\section{Enumerating excedance like statistics}


For a $\sigma \in \BB_n$, define its {\sl excedance-sum} as $\iexc(\sigma) = 
\sum_{i \in \ExcS(\sigma)}i$.
For the next result which is an analogue of Theorem 8 of \cite{siva-exc-det}, we recall
the following notation used in $q$-series theory.  Let $q$ be a variable and 
for a non negative integer $n$, define 

\begin{equation}
   \label{eqn:qn}
	(q;q)_n = \left\{ 
	\begin{array}{ll}
	    1 & \mbox{ if } n = 0 \\
		   \prod_{i=1}^n (1-q^i) & \mbox{ if } n > 0
		 \end{array}
		 \right.
\end{equation}

For a non-negative integer $i$, we also recall that 
$[i]_q = 1 + q + q^2 + \cdots + q^{i-1}$, and that $[0]_q = 0$.
For $n \geq 1$ and $1 \leq i \leq n$, define the bivariate signed 
excedance-sum enumerator to be 
$\BSI_n^i(q) = \sum_{\sigma \in \BB_n} (-1)^{\invb(\sigma)} 
q^{\iexc(\sigma)} t^{\ip(\sigma)}$.  


\begin{theorem}
  \label{thm:sign_excsum}
For $n \geq 1$ and $1 \leq i \leq n$, $\BSI_n^i(q) = t^i \cdot (q;q)_n$.
\end{theorem}
\begin{proof}
  Let $X_i = \{ \sigma \in \BB_n: \ip(\sigma) = i \}$.  
For $\sigma \in \BB_n - X_i$, the cycle $C$ containing $i$ has
length at least two. 
Let $\tau_k: ( \BB_n - X_i) \mapsto ( \BB_n - X_i)$ be 
the map which flips the sign of  the minimum element of $C$ in absolute
value.  Let $\psi = \tau_k(\sigma)$.
Hence, for all indices $i \not= k$, we have $\sigma_i = \psi_i$ while
we have $\sigma_k = i = - \psi_k$.
Lemma \ref{lem:cycle_min} shows that 
$\ExcS(\sigma) = \ExcS(\psi)$ which implies that 
$\iexc(\sigma) = \iexc(\psi)$.  Further, $\ip(\sigma) = \ip(\psi)$ and
$\invb(\sigma) \not \equiv \invb(\psi)$ (mod 2) by Lemma
\ref{lem:inv_rev}. Thus,
$\sum_{\sigma \in \BB_n - X_i} (-1)^{\invb(\sigma)} q^{\iexc(\sigma)}t^{\ip(\sigma)} = 0$.

Hence, only elements of $X_i$ contribute to the sum concerned.  Since
each $\sigma \in X_i$ has $\ip(\sigma_i) = i$, we can pull out the term $t^i$ 
from the sum.  Actually, the above argument shows that whenever 
some $\sigma \in \BB_n$ has a 
cycle $C$ of length at least two, we can negate its minimum element in
absolute value to get $\psi$ 
and cancel out terms arising from $\sigma$ by assigning the $2^n$ signs 
to $\sigma$ (as $\sigma$ and $\psi$ will have opposite parity but have 
$\ExcS(\sigma) = \ExcS(\psi)$).

Thus, the only terms that survive the cancellations 
come from assigning signs to the identity permutation 
$e= 1,2,\ldots,n$.  
We are left with finding the signed excedance-sum over the $2^n$ 
signed permutations of $e$.  
Applying Corollary \ref{cor:odd_even} to the identity permutation, it is 
easy to see 
that $\invb(\sigma)$ has the same parity as the number of negative 
entries in $\sigma$ and $\iexc(\sigma)$ is the sum of the 
negative entries in $\sigma$.  Thus, we get 
$\BSI_n^i(q) = \sum_{S \subseteq [n]} (-1)^{|S|} 
q^{\sum_{x \in S} x} = \prod_{i=1}^n(1-q^i) = (q;q)_n$. Multiplication by $t^i$
completes the proof.  
\end{proof}

\vspace{2 mm}

We give a bivariate analogue of Theorem 8 of \cite{siva-exc-det} below.
Consider the following $n \times n$ matrices 

$P_n =  \left( \begin{array}{ccccc}
1 & q & q & \cdots & qt \\
1 & 1 & q^2 & \cdots & (qt)^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & (qt)^{n-1} \\
1 & 1 & 1 & \cdots & t^n
\end{array} \right)  \mbox{ and }
S_n =  \left( \begin{array}{ccccc}
1 & q & q^2 & \cdots & q^{n-1}t \\
1 & 1 & q & \cdots & q^{n-2}t^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & qt^{n-1} \\
1 & 1 & 1 & \cdots & t^n
\end{array} \right) 
$
																													  
\noindent
i.e. if $P_n = (p_{i,j})_{1 \leq i,j \leq n}$, then $p_{i,j}$ is 
obtained by the following procedure: set $p_{i,j} = q^i$ if
$i<j$ and $p_{i,j} = 1$ otherwise.  Then, set $p_{i,j} = p_{i,j} \cdot t^i$
if $j=n$.  Similarly, if  $S_n = (s_{i,j})_{1 \leq i,j \leq n}$, then $s_{i,j}$
is obtained as follows:  set $s_{i,j} = q^{j-i}$ if $i < j$
and $s_{i,j} = 1$ otherwise.  After this, set $s_{i,j} = s_{i,j} \cdot t^i$
if $j=n$.

For a $\pi \in \SSS_n$, define its excedance set as $\ExcS(\pi) = \{i \in [n]: 
\pi_i > i\}$.  Define its {\it excedance-sum} as 
$\iexc(\pi) = \sum_{i \in \ExcS(\pi)} i$ and its {\it excedance-length} as 
$\lexc(\pi) = \sum_{i \in \ExcS(\pi)} (\pi_i - i)$.  For $n \geq 1$, let
$\SI_n(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} q^{\iexc(\pi)}
t^{\np(\pi)}$ to be the bivariate signed excedance-sum enumerator.  
Similarly, define $\SL_n(q,t) = \sum_{\pi \in \SSS_n} (-1)^{\inva(\pi)} 
q^{\lexc(\pi)} t^{\np(\pi)}$ as the bivariate signed excedance-length enumerator.

\begin{theorem}
\label{thm:perm-sign-inexc}
For integers $n \geq 1$,
$\SI_n(q,t) = t^{n-1} (t-q^{n-1}) \prod_{i=1}^{n-2} (1-q^i) $.
\end{theorem}
\begin{proof}
  It is clear that $\SI_n(q,t) = \det(P_n)$.  Let $\Row_i$ denote the $i$-th row of 
  any $n\times n$ matrix for $1 \leq i \leq n$ (the matrix will be clear from 
  the context).  Perform the elementary row operation $\Row_n := \Row_n - \Row_{n-1}$ 
  and then evaluate the determinant to see that 
  $\det(P_n) = t^{n-1} (t-q^{n-1}) \prod_{i=1}^{n-2} (1-q^i)$,   completing the proof.
\end{proof}

\begin{theorem}
\label{thm:perm-sign-lenexc}
For integers $n \geq 1$, 
$\SL_n(q,t) = t^{n-1} (t-q) (1-q)^{n-2} $.
\end{theorem}
\begin{proof}
  It is clear that $\SL_n(q,t) = \det(S_n)$ and performing as before the operation
  $\Row_n := \Row_n - \Row_{n-1}$ and then evaluating the determinant gives us
$\det(S_n) = t^{n-1} (t-q) (1-q)^{n-2}$, completing the proof.
\end{proof}

\section{Signed excedance enumeration in $\BD_n$}
\label{sec:der_exc_enum}

Let $\BD_n$ be the set of derangements of $\BB_n$.  $\BD_n$ consists of 
the $\sigma \in \BB_n$ such that $\sigma_i \not= i$ for all $i \in [n]$.
%(see id:A000354 in OEIS). 
For $n \geq 1$ and $1 \leq i \leq n$, define \[ \BDSE_n^i(q,t) = 
\sum_{\sigma \in \BD_n} (-1)^{\invb(\sigma)} q^{\exc(\sigma)} t^{\ip(\sigma)}.\]
Define $\BDSE_0^0(q,t) = 1$ and $\BDSE_0^1(q,t) = t$.  
%\red{\bf Check these initial values.}



\begin{theorem}
\label{thm:bdsgnexc}
For $n \geq 1$ and $1 \leq i \leq n$, $\BDSE_n^i(q,t) = t^i(-q)^n$.
\end{theorem}
\begin{proof}
  This proof is very similar to that of Theorem \ref{thm:sgn_exc_hyp_i}
  and thus we only mention the differences.
Since $\sigma$ is a derangement, for
all $1\leq i \leq n$, $\sigma_i \not= i$.  As before, let $X_i = 
\{ \sigma \in \BB_n: \sigma_i = -i \}$.  Since for all 
$\sigma \in \BB_n - X_i$ the element $i$ is in a cycle $C$ of length
at least two, consider the involution on $\BB_n - X_i$ defined by
changing the sign of the minimum element in absolute value of $C$.
Let $\psi$ be the resulting signed permutation.  Clearly, $\psi \in \BD_n$.
By similar arguments, $\sum_{\sigma \in \BB_n - X_i} 
(-1)^{\invb(\sigma)} t^{\ip(\sigma)} q^{\exc(\sigma)} = 0$.  

Thus, only derangements with $\sigma_i = -i$ contribute to the sum.
The argument actually shows that if $\sigma \in \BB_n$ has a cycle 
of length at least two, it will get cancelled.  Thus we see that 
the unique derangement which contributes to the sum has $\sigma(i) = -i$ for
all $1 \leq i \leq n$.  Since $\ip(\sigma) = i$, $\exc(\sigma) = n$ and
$\invb(\sigma) = n^2 \equiv n$ (mod 2), we have 
$\BDSE_n^i(q,t) = t^i(-q)^n$, completing the proof.
\end{proof}

\vspace{2 mm}

We record two simple corollaries of Theorem \ref{thm:bdsgnexc} in the cases
when $i=1,n$.  We will use them in the next subsection.  
Define 
$\BDSE_n^n(q,t) = \sum_{\sigma \in \BD_n} 
(-1)^{\invb(\sigma)} q^{\exc(\sigma)} t^{\np(\sigma)}$ and
$\BDSE_n^1(q,t) = \sum_{\sigma \in \BD_n} 
(-1)^{\invb(\sigma)} q^{\exc(\sigma)} t^{\onep(\sigma)}$.



\begin{corollary}
  \label{cor:derb_sgn_exc_spl}
For $n \geq 1$, 
$\BDSE_n^n(q,t) = (-qt)^n$  and $\BDSE_n^1(q,t) = t(-q)^n$.
\end{corollary}

We give a similar bivariate generalization of this result to $\SSS_n$.  
Consider the following matrices.

\vspace{2 mm}

$
D_{M_n} = \left( 
  \begin{array}{ccccc}
	0 & q & q & \cdots & qt \\
	1 & 0 & q & \cdots & qt^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	1 & 1 & 1 & \cdots & 0 \\
  \end{array}
  \right) \hspace{1 cm}
D_{M_n}^1 = \left( 
  \begin{array}{ccccc}
	0 & q & q & \cdots & q \\
	 t^2& 0 & q & \cdots & q \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	t^n & 1 & 1 & \cdots & 0 \\
  \end{array}
\right)
$

\vspace{2 mm}

We obtain $D_{M_n} = (d_{i,j})_{1 \leq i,j \leq n}$ by the following procedure.
Set $D_{M_n} = M_n$, where recall that
%$d_{i,j}$ is identical to $M_n = (m_{i,j})$ (
the matrix $M_n$ was defined in Section \ref{sec:sgnexcBn}.  After this, set 
$d_{i,i} = 0$ for all $1 \leq i \leq n$.  Similarly, we get 
the entries of $D_{M_n}^1$ from $M_n^1$ by making a copy and then changing all diagonal 
elements to zero.

\vspace{2 mm}

Motivated by the above result, define 
$\DSE_n^n(q,t) = \sum_{\pi \in \DD_n}  
(-1)^{\inva(\pi)} t^{\np(\pi)} q^{\exc(\pi)}$
as the bivariate signed excedance enumerator summed over derangements in $\SSS_n$.
Similarly, define $\DSE_n^1(q,t) = \sum_{\pi \in \DD_n}  
(-1)^{\inva(\pi)} t^{\onep(\pi)} q^{\exc(\pi)}$
as the bivariate signed excedance enumerator with respect to position of 1,
summed over derangements.  
We recall that for a positive integer $i$, $[i]_{qt} = 1 + qt + (qt)^2 + 
\cdots + (qt)^{i-1}$.  Define $[0]_{qt} = 0$.  We show the following bivariate 
generalizations of Theorem \ref{thm:der-sign-exc}.

\begin{theorem}
  \label{thm:der_sgn_exc_biv}
  For all positive integers $n \geq 1$, $\DSE_n^n(q,t) = (-1)^{n-1} qt \cdot [n-1]_{qt}$ and
similarly,  $\DSE_n^1(q,t) = (-1)^{n-1} qt^2 \cdot [n-1]_{qt}$.
\end{theorem}
\begin{proof}
  It is clear that $\DSE_n(q,t) = \det(D_{M_n})$.  Perform the row operation
  $\Row_n := \Row_n - \Row_{n-1}$ and then evaluate the determinant.
  It is easy by using induction on $n$ to see that 
  $\det(D_{M_n}) = (-1)^{n-1} qt \cdot [n-1]_{qt}$.  Similarly, it 
  is clear that 
  $\DSE_n^1(q,t) = \det( D_{M_n^1} )$.  
  Performing the column operation
  $\Col_n := \Col_n - \Col_{n-1}$ and then evaluating the determinant, it 
  is easy to see that 
  $\det( D_{M_n^1} ) = (-1)^{n-1} qt^2 [n-1]_{qt}$, completing the proof.
\end{proof}

\subsection{A binomial type equation}
\label{subsec:der_exc_enum}

Chen, Tang and Zhao in \cite{chen-tang-zhao-typeB-der} considered the 
following polynomials 
$\displaystyle B_n(q) = \sum_{\sigma \in \BB_n} q^{\exc(\sigma)}$ and  
$\displaystyle d_n^B(q) =  \sum_{\sigma \in \BD_n} q^{\exc(\sigma)} $,
where, they also use Brenti's definition for excedance in $\BB_n$.
They showed (see 
Equation (3.6) of \cite{chen-tang-zhao-typeB-der}) that 

\begin{equation}
  \label{eqn:binom_type}
  B_n(q) = \sum_{k=0}^n \binom{n}{k} d_k^B(q)
\end{equation}

Recall that $\BDSE_1^1(q,t) = -qt$ in contrast to the $\SSS_n$ case 
where there are no derangements when $n = 1$.  
The following 
corollary follows immediately from Corollaries \ref{cor:sgn_exc_hyp} 
and \ref{cor:derb_sgn_exc_spl}.  Note that when $t=1$, Corollary 
\ref{cor:binom} is identical to (\ref{eqn:binom_type}).

\begin{corollary}
  \label{cor:binom}
  $\displaystyle \BSE_n^n(q,t) = \sum_{k=0}^n \binom{n}{k} t^{n-k} \BDSE_k^k(q,t)$. %+ nt^{n-1}$.
\end{corollary}

Recall $\BSE_n^1(q,t)$, the bivariate signed excedance enumerator in $\BB_n$ defined in 
Section \ref{sec:sgnexcBn}.
%, where the exponent of $t$ in the term corresponding to $\sigma$ is $1\_pos(\sigma)$.  
Our next corollary is again immediate from 
Corollaries \ref{cor:sgn_exc_hyp} and \ref{cor:derb_sgn_exc_spl}.
Corollary \ref{cor:binom_one} is identical to (\ref{eqn:binom_type}). 

\begin{corollary}
  \label{cor:binom_one}
  $\displaystyle \BSE_n^1(q,t) = \sum_{k=0}^n \binom{n}{k} \BDSE_k^1(q,t)$. %+ nt^{n-1}$.
\end{corollary}


%\begin{proof}
%  The proof is immediate, except for the term $n t^{n-1}$ which is a minor adjustment for 
%  $\BDSE_1(q,t) = 0$.
%\end{proof}

%We do not know if a similar equation is true for either the signed univariate or the 
%signed bivariate excedance enumerator in $\SSS_n$.
%\blue{\bf Is this binomial type eqn true for the signed bivaritate excedance in $X_n$??
%Is this also true when $\onep$ is used?}



\subsection{Signed Excedance Sum}

In this last subsection, we give hyperoctahedral analogues of Theorem
19 of \cite{siva-exc-det}. For $n \geq 1$ and $1 \leq i \leq n$,
define $\BDSI_n^i(q,t) = \sum_{\sigma \in \BD_n} 
(-1)^{\invb(\sigma)} q^{\iexc(\sigma)} t^{\ip(\sigma)}$.

\begin{theorem}
  \label{thm:bdsgnexcsum}
  For $n \geq 1$ and $1 \leq i \leq n$, $\BDSI_n^i(q,t) = t^i (-1)^n q^{ \binom{n+1}{2} }$.
\end{theorem}
\begin{proof}
  This proof is very similar to the proof of Theorem \ref{thm:der_sgn_exc_biv} 
  and thus we only mention the differences.
  Let $X_i = \{ \sigma \in \BD_n: \sigma_i = -i \}$.  Consider the 
  involution $\tau_k$ on $\BD_n - X_i$ defined
  in the proof of Theorem \ref{thm:sign_excsum} (i.e. changing the sign of
  the minimum element in absolute value in the cycle containing $i$).   
	The rest of the argument is similar and we omit the 
  details.  By induction, the unique derangement that contributes has 
  $\sigma(i) = -i$ for
  all $1 \leq i \leq n$.  Since $\ip(\sigma) = i$, $\iexc(\sigma) = 
  \sum_{i=1}^n i = \binom{n+1}{2} $ 
  and  $\invb(\sigma) = n^2 \equiv n$ (mod 2), we have
  $\BDSI_n^i(q,t) = t^i(-1)^n q^{ \binom{n+1}{2} }$, completing the proof.
\end{proof}


We give a similar bivariate generalization of this result to $\SSS_n$.  
Consider the following matrices.

\vspace{2 mm}

$
D_{P_n} = \left( 
  \begin{array}{ccccc}
	0 & q & q & \cdots & qt \\
	1 & 0 & q^2 & \cdots & q^2t^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	1 & 1 & 1 & \cdots & 0 \\
  \end{array}
  \right) \hspace{1 cm}
D_{P_n}^1 = \left( 
  \begin{array}{ccccc}
	0 & q & q & \cdots & q \\
	 t^2& 0 & q^2 & \cdots & q^2 \\
	\vdots & \vdots & \vdots & \ddots & \vdots \\
	t^n & 1 & 1 & \cdots & 0 \\
  \end{array}
\right)
$

\vspace{2 mm}

i.e. if $D_{P_n} = (p_{i,j})_{1 \leq i,j \leq n}$, then $p_{i,j}$ is 
obtained by the following procedure: set $p_{i,j} = q^i$ if 
$i < j$, $p_{i,j} = 0$ if $i=j$ and $p_{i,j} = 1$ otherwise.  After this, if $j=n$, set 
$p_{i,j} = p_{i,j} \cdot t^i$.  Likewise, if $D_{P_n}^1 = (p_{i,j}')$, then,
set $p_{i,j}' =q^i$ if $i < j$, $p_{i,j}' = 0$ if $i=j$ and $p_{i,j}' = 1$ otherwise.  After 
this, if $j=1$, set $p_{i,j}' = p_{i,j}' \cdot  t^i$.


\vspace{2 mm}

Define
$\DSI_n^n(q,t) = \sum_{\pi \in \DD_n}  (-1)^{\inva(\pi)} q^{\iexc(\pi)} t^{\np(\pi)}$
as the bivariate signed excedance-sum enumerator over derangements in $\SSS_n$.
Likewise, define \[ \DSI_n^1(q,t) = \sum_{\pi \in \DD_n}  
(-1)^{\inva(\pi)} t^{\onep(\pi)} q^{\exc(\pi)}\]
as the bivariate signed excedance enumerator with respect to position of 1,
summed over derangements.  
We show the following bivariate generalizations 
of Theorem 19 of \cite{siva-exc-det}.

\begin{theorem}
  \label{thm:der_sgn_exc_biv1}
  Let $n \geq 2$.  Then, $\DSI_n^n(q,t) = (-1)^{n-1} ( \sum_{i=1}^{n-1} t^i q^{\binom{i+1}{2}} )$
  and similarly, $\DSI_n^1(q,t) = (-1)^{n-1} (\sum_{i=2}^n t^i q^{ \binom{i}{2}} )$.
\end{theorem}
\begin{proof}
  It is clear that 
  $\DSI_n^n(q,t) = \det(D_{P_n})$.  
  After performing the row operation
  $\Row_n := \Row_n - \Row_{n-1}$ and then evaluating the determinant, 
  it is easy to see by induction on $n$ that 
  $\det(D_{P_n}) = (-1)^{n-1} ( \sum_{i=1}^{n-1} t^i q^{ \binom{i+1}{2} } )$.  
  Similarly, it  is clear that $\DSI_n^1(q,t) = \det( D_{P_n^1} )$.  
  Performing the column operation
  $\Col_n := \Col_n - \Col_{n-1}$ and then evaluating the determinant, it 
  is easy to see that 
  \[ \det( D_{P_n^1} ) = (-1)^{n-1} \, \sum_{i=2}^n t^i q^{ \binom{i}{2} } ,\]
 completing the proof.
\end{proof}



\subsection*{Acknowledgement}
%Some Theorems in this work were in conjecture form, tested 
%using the computer package ``Sage''.  We thank the authors for generously 
%releasing Sage as an open-source package.  
The {\it Online Encyclopedia of Integer Sequences} was consulted for 
the results in Theorems~\ref{thm:biv_sk_exc} and~\ref{thm:biv_sgn_wk_sk_wk_exc}.
We are grateful to Professor N.J.A. Sloane for putting his searchable 
encyclopedia on the web.  
Support from project grant P07 IR052, given by IIT Bombay is acknowledged.  We 
thank the referee for a careful reading and for pointing out several places
for improvement in this work.

%\bibliographystyle{acm}
%\bibliography{main}

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\end{document}
