text:
\section{Introduction}
This article concerns a family of multivariate polynomials associated with ascents and descents on labeled rooted plane $k$-ary trees that was introduced by Gessel \cite{Gessel-unpublished} and studied by \cite{Drake-thesis,Kalikow}. Gessel further observed that various evaluations of these polynomials counted the number of regions in well-known deformations of Coxeter hyperplane arrangements, and asked for a bijective/combinatorial explanation \cite{Gessel-Oberwolfach}. In what follows, we describe progress in this direction.

We study this family of polynomials from a rook-theoretic perspective, drawing inspiration from recent work of Lewis-Morales \cite{Lewis-Morales} studying diagrams of permutations as well as the work of Leven-Rhoades-Wilson \cite{Leven-Rhoades-Wilson} relating rook placements and Shi-Ish arrangements.

To give a flavor of our main results, we establish some notation. Let $n\geq 1$. A local binary search tree on $n$ nodes is defined to be a labeled rooted plane binary tree on $n$ nodes with labels drawn from $[n]=\{1,\ldots, n\}$ such that the label of the left child (right child) is always smaller (respectively greater) than the parent. The Linial arrangement $\hyplin_{n-1}$ is the real hyperplane arrangement comprising the hyperplanes $x_i-x_j=1$ for $1\leq i<j\leq n$ considered in the ambient vector space $V_{n-1}$ consisting of points $(x_1,\ldots,x_n)\in \mathbb{R}^n$ such that $x_1+\cdots +x_n=0$.  A special case of one of our main theorems is the following result.
\begin{theorem}\label{thm: intro 1}
The number of bounded regions in the Linial arrangement $\hyplin_{n-1}$ equals the following quantities.
\begin{enumerate}
\item The number of local binary search trees on $n$ nodes such that the nodes labeled $1$ and $n$ are both leaves. Here the root node is not considered to be a leaf node.
\item The number of nonattacking rook placements of $n-1$ rooks on the skew Ferrers board of shape given by  $(2n-3,\ldots,n-1)/(n-1,\ldots,1)$.
\item The number of sequences $(x_1,\ldots, x_{n-1})$ of positive integers satisfying $1\leq x_i\leq n-2$ for $1\leq i\leq n-1$, and $x_i+i\neq x_j+j$ for $1\leq i\neq j\leq n-1$.
\end{enumerate}
\end{theorem}
A few remarks about the above theorem are in order. The question of finding a combinatorial meaning to the number of bounded regions of the Linial arrangement in the setting of labeled trees has been raised in \cite{Athanasiadis-Advances} (cf. discussion after Theorem 4.2 therein). To the best of the author's knowledge, there have not been any results in this direction.
Secondly, that local binary search trees have to do with regions in the Linial arrangement is no surprise. Indeed, Postnikov \cite{Postnikov-JCTA} proved that the number of regions in the Linial arrangement equals the number of local binary search trees (in fact, Postnikov worked with intransitive trees or alternating trees, but establishes a bijection between them and local binary search trees \cite[Section 4.1]{Postnikov-JCTA}). As the theorem above shows, the number of bounded regions in the Linial arrangement is counted by a fairly nicely described subset of the set of local binary search trees. Starting from $n=1$, the first few terms of this sequence (not present in the OEIS) are: $0$, $0$, $1$, $4$, $26$, $212$, $2108$, $24720$.

Our second main result relates the factorial polynomial of a specific skew Ferrers board to the characteristic polynomial of the Linial arrangement $\hyplin_{n-1}$.
\begin{theorem}\label{thm: intro 2}
Let $B$ denote the skew Ferrers board of shape $(2n-3,\ldots,n-1)/(n-1,\ldots,1)$.
Let $R(t,B)$ denote the factorial polynomial of $B$, and let $\chi_{\hyplin_{n-1}}(t)$ denote the characteristic polynomial of the arrangement $\hyplin_{n-1}$. Then we have the following equality.
\begin{align*}
R(t,B)=(-1)^{n-1}\chi_{\hyplin_{n-1}}(1-t)
\end{align*}
\end{theorem}
The outline of this article is as follows: We begin by defining notions related to hyperplane arrangements in Section \ref{sec: hyp arrangements}. We follow it up by defining rook-theoretic notions in Section \ref{sec:rook placements}. We also introduce certain special rectangular boards and define generalized excedance and subcedance statistics associated with maximal nonattacking rook placements on these boards. In Section \ref{sec:trees}, we introduce labeled rooted plane $k$-ary trees amongst other notions. We then define ascent-descent statistic on these trees and proceed to give our bijection $\Psi$ that takes maximal nonattacking rook placements on $(n-1)\times kn$ boards to labeled rooted plane $k$-ary trees on $[n]$. This bijection further has the property that it takes the $(\exc, \sub)$ associated with the rook placement to the $(\asc,\dsc)$ pair associated with the corresponding tree. This given, we define the Gessel polynomials. We use our bijection  to connect rook placements on certain skew Ferrers boards with the number of regions in extended Linial arrangements. In Section \ref{sec: chromatic poly} we identify graphs whose chromatic polynomials are related to characteristic polynomials of extended Linial arrangements. Finally, we conclude with a  brief discussion on future avenues in Section \ref{sec:final}.

\section{Hyperplane arrangements}\label{sec: hyp arrangements}
We begin by defining some basic notions concerning hyperplane arrangements. Our exposition will be brief; the reader is referred to \cite{Orlik-Terao,Stanley-notes} for more details. We will be faithful to the terminology used in \cite{Postnikov-Stanley} to a large extent.

A hyperplane arrangement is a finite collection of affine hyperplanes in a vector space. Given a finite-dimensional vector space $V$ over $\mathbb{R}$, let $\mathcal{A}$ be a hyperplane arrangement therein. We will denote the number of regions of $\mathcal{A}$, that is, the number of connected components of $V-\cup_{H\in \mathcal{A}}H$, by $r(\mathcal{A})$. Furthermore, the number of relatively bounded regions will be denoted by $b(\mathcal{A})$.

The poset that connects the study of hyperplane arrangements to combinatorics is the \emph{intersection poset } $L_{\mathcal{A}}$ associated with a hyperplane arrangement $\mathcal{A}$.
It consists of all nonempty intersections $H_{i_1}\cap \cdots \cap H_{i_k}$ of hyperplanes in $A$, ordered by reverse inclusion. Clearly, the unique  minimum element $\hat{0}$ in $L_{\mathcal{A}}$ is the ambient vector space $V$. Given $L_{\mathcal{A}}$, we can define the \emph{characteristic polynomial} of $\mathcal{A}$, denoted by $\chi_{\mathcal{A}}(q)$, as shown below.
\begin{align*}
\chi_{\mathcal{A}}(q)=\sum_{z\in L_{\mathcal{A}}}\mu(\hat{0},z)q^{\dim(z)}
\end{align*}
Here $\mu$ denotes the M\"obius function of the poset $L_\mathcal{A}$.

Another polynomial encoding topological information about the hyperplane arrangement $\mathcal{A}$ is the \emph{Poincar\'e polynomial} $\mathrm{Poin}_{A}(q)$ defined below.
\begin{align*}
\mathrm{Poin}_{\mathcal{A}}(q)=\sum_{m\geq 0}\dim H^{k}(C_{\mathcal{A}};\mathbb{C})q^k
\end{align*}
Here $C_{\mathcal{A}}$ denotes the complement of the complexified arrangement $\mathcal{A}_{\mathbb{C}}$ in the complexified vector space $V_{\mathbb{C}}$. Furthermore, $H^{k}(\text{ };\mathbb{C})$ denotes the singular cohomology with coefficients belonging to $\mathbb{C}$. The Poincar\'e polynomial $\mathrm{Poin}_{\mathcal{A}}(q)$ is a $q$-analogue of $r(\mathcal{A})$ as is made precise by the following theorem of Orlik-Solomon \cite{Orlik-Solomon-Inventiones}.
\begin{theorem}
We have that $r(\mathcal{A})=\mathrm{Poin}_{\mathcal{A}}(1)$ and $b(\mathcal{A})=\mathrm{Poin}_{\mathcal{A}}(-1)$.
\end{theorem}
Compare the theorem above with the following seminal result proved by Zaslavsky \cite{Zaslavsky}.
\begin{theorem}\label{thm: Zaslav}
Consider a hyperplane arrangement $\mathcal{A}$ in $\mathbb{R}^n$.  Then we have that $r(\mathcal{A})=(-1)^n\chi_{\mathcal{A}}(-1)$ and $b(\mathcal{A})=(-1)^n\chi_{\mathcal{A}}(1)$.
\end{theorem}

The similarity between the two theorems above was explained by Orlik-Solomon \cite{Orlik-Solomon-Inventiones} who proved that
\begin{align*}
\chi_{\mathcal{A}}(q)=q^d\mathrm{Poin}_{\mathcal{A}}(-q^{-1}),
\end{align*}
where $d=\dim (V)$.

We now turn our focus to special classes of hyperplane arrangements that are our primary object of study. Given a positive integer $n$, let $V_{n-1}$ be the subspace in $\mathbb{R}^n$ comprising vectors $(x_1,\ldots,x_n)$ such that $\sum_{i=1}^nx_i=0$. The hyperplane arrangements we consider in this paper will all lie in $V_{n-1}$ and, in fact, will be special instances of what are called \emph{truncated affine arrangements} in \cite{Postnikov-Stanley}.
This given, since the hyperplane arrangements that we will consider are \emph{essential}, we will not be using the term relatively bounded for our results; instead we will just use the term bounded (see Page 3 in \cite{Stanley-notes}).

Fix nonegative integers $a$ and $b$  such that $a+b\geq 2$. Consider the \emph{truncated affine arrangement} $\mathcal{A}_{n-1}^{a,b}$ given by
\begin{align*}
x_i-x_j=-a+1,-a+2,\ldots, b-1, \hspace{5mm} 1\leq i< j\leq n.
\end{align*}
We will denote the characteristic polynomial of $\mathcal{A}_{n-1}^{a,b}$ by $\chi_{n-1}^{a,b}(q)$.

Instances of  truncated affine arrangements have been a focus of intense study in recent years, and have resulted in a lot of interesting combinatorics. We describe these arrangements next.
Fix a positive integer $a\geq 1$. The arrangement $\mathcal{A}^{a,a}_{n-1}$  will be called the \emph{extended Catalan arrangement}, and will be denoted by $\hypcat_{n-1,a}$. Note that the case $a=1$ corresponds to the braid arrangement, while the case $a=2$ corresponds to the classical Catalan arrangement. The arrangement $\mathcal{A}^{a,a+1}_{n-1}$ will be referred to as the \emph{extended Shi arrangement}, and will be denoted by $\hypshi_{n-1,a}$. The case $a=1$ corresponds the extremely well-studied Shi arrangement \cite{Athanasiadis-Linusson, Headley, Postnikov-Stanley,Shi-Lecture notes, Shi-JLMS,Stanley-Pnas}.
Finally the arrangement $\mathcal{A}^{a-1,a+1}_{n-1}$ will be referred to as the \emph{extended Linial arrangement}, and will be denoted by $\hyplin_{n-1,a}$. The case $a=1$ corresponds to the classical Linial arrangement mentioned in the introduction of this article. From now onwards $\hyplin_{n-1}:=\hyplin_{n-1,1}$. Finally, before proceeding to our next section, we would like to mention that various aspects of the truncated affine arrangements have been studied in great depth in \cite{Athanasiadis-survey, Athanasiadis-JACo,Athanasiadis-EuJC,Postnikov-Stanley} and the reader is referred to them for further beautiful combinatorics surrounding them.

\iffalse
A question posed in \cite{Athanasiadis-Advances} is the following: Give a combinatorial interpretation for the number of bounded regions in the Linial arrangement.
In what follows we will give multiple interpretations to the number of bounded regions of the Linial arrangement.
But, unfortunately, there will be no explicit bijection.

From Zaslavsky's formula applied to the characteristic polynomial of the Linial arrangement, we have the following.
\fi

\section{Rook placements}\label{sec:rook placements}
A \emph{board} $B$ is a finite subset of $\mathbb{P}\times \mathbb{P}$ (thought of in Cartesian coordinates) where $\mathbb{P}$ denotes the set of positive integers. A \emph{nonattacking rook placement} on $B$ is one where there is at most one rook in any row or column. A nonattacking rook placement is called \emph{maximal} if there is exactly one rook in every row, and at most one in any column. We will label the rows and columns of a board with positive integers. By row $i$, we mean the $i$-th row from the bottom, while by column $j$ we mean the $j$-th column from the left. For the purposes of this article, we will primarily be interested in \emph{skew Ferrers boards}. These are boards corresponding to Young diagrams of shape $\lambda/\mu$ for partitions $\lambda$ and $\mu$ satisfying $\mu\subseteq \lambda$.

Given a board $B$ and a nonnegative integer $k$, let $r_k(B)$ denote the number of nonattacking rook placements of $k$ rooks on $B$. Clearly if $k$ exceeds the number of rows in $B$ then $r_{k}(B)=0$. Also, we have that $r_0(B)=1$. Finally, if $m\geq $ number of rows in $B$, we denote by $R_m(x,B)$ the \emph{$m$-factorial polynomial} of $B$ \cite{GJW-iii} defined below.
\begin{align*}
R_m(x,B)=\sum_{k\geq 0}r_k(B)(x)_{m-k}
\end{align*}
Here $(x)_{j}=x(x-1)\cdots (x-j+1)$ is the usual \emph{falling factorial}. We will normally set $m$ equal to the number of rows in the board $B$. In this case we will refer to the $m$-factorial polynomial of $B$ as the \emph{factorial polynomial} of $B$, and denote it by $R(x,B)$.

\begin{example}\label{example: board b}
\em{Consider the board $B$ below which has $3$ rows.
\begin{align*}
\ytableausetup{mathmode, boxsize=0.8em}
\begin{ytableau}
\none & \none & \none & *(blue!50) & *(blue!50)& *(blue!50)\\
\none & \none & *(blue!50) & *(blue!50)& *(blue!50)\\
\none & *(blue!50) & *(blue!50)& *(blue!50)\\
\end{ytableau}
\end{align*}
 Note that from its definition it follows that $R(x,B)=(x)_3+9(x)_2+22(x)_1+14$. It is worth mentioning that the coefficients $1,9,22,$ and $14$ appear in the analysis of the `three-ply' staircase in the work of Riordan \cite{Riordan}. Of course, the above board is a special instance of the boards considered by Riordan.}
\end{example}
The theorem below, which is present in \cite{GJW-v}, will be useful for us. Let $\Pi_m$ denote the lattice of partitions of $[m]$, and let $\hat{0}$ denote the unique minimum in $\Pi_m$. Consider a board $B$ with $m$ rows. Given a subset $S \subseteq [m]$, we define a statistic $v_{B}(S)$ as follows.
\begin{align*}
v_B(S)=|\{j\;|\; (i,j)\in B \hspace{2mm}\forall i\in S\}|
\end{align*}
\begin{theorem}[Goldman-Joichi-White]\label{thm: GJW-v}
Let $B$ be a board with $m$ rows. Then
\begin{align*}
R(x,B)=\sum_{\sigma\in \Pi_m}\mu(\hat{0},\sigma)\prod_{A\in \sigma}(x+v_{B}(A))
\end{align*}
where $\mu$ denotes the M\"obius function of $\Pi_{m}$. Also, by $A\in\sigma$ we mean that $A$ is a block in $\sigma$.
\end{theorem}
\begin{example}\em{
Given below is the Hasse diagram of the partition lattice $\Pi_3$.
\begin{align*}
\begin{tikzpicture}
        \tikzstyle{every node} = [rectangle]
        \node (s) at (0,-3) {$1/2/3$};
        \node (s1) at (-2,-2) {$12/3$};
        \node (s3) at (0,-2) {$13/2$};
        \node (s2) at (2,-2) {$1/23$};
        \node (s12) at (0,-1) {$123$};
        \foreach \from/\to in {s/s1, s/s2, s1/s12, s2/s12,s/s3,s3/s12}
            \draw[-] (\from) -- (\to);
\end{tikzpicture}
\end{align*}
In writing the blocks of a set partition above, we have omitted commas and braces.

Then, by Theorem \ref{thm: GJW-v}, for the same board $B$ as in Example \ref{example: board b}, we obtain that $R(x,B)=(x+3)^3-(x+2)(x+3)-(x+1)(x+3)-(x+2)(x+3)+2(x+1)=x^3 + 6x^2 + 15x + 14$. Note that $R(1,B)=36$ and $R(-1,B)=4$. To hint at where we are going with this example, note further that 36 and 4 are the number of regions and the number of bounded regions in the Linial arrangement $\hyplin_3$ respectively. }
\end{example}
We will also be considering rook placements on rectangular boards in terms of functions as described next.
Given positive integers $n,k$, let $\mathcal{F} _{n,k}$ denote the following set.
\begin{align*}
\mathcal{F}_{n,k}=\{f:[n-1] \hookrightarrow [kn] \}
\end{align*}
That is, the elements of $\mathcal{F}_{n,k}$ are injective functions from the first $n-1$ positive integers to the first $kn$ positive integers.
We will identify an element $f\in\mathcal{F}_{n,k}$ with the rook placement on the rectangular board with $n-1$ rows and $kn$ columns obtained by placing a rook in the position $(i,f(i))$ for $1\leq i\leq n-1$.
Note that this is a maximal nonattacking rook placement.
Furthermore, statements involving rook placements on subboards of the $(n-1)\times kn$ board can be phrased in terms of elements of $\mathcal{F}_{n,k}$ satisfying certain constraints. Finally, we note here that there is a unique injective function from the empty set $\emptyset$ (corresponding to the case $n=1$)  into $[k]$ for $k$ a positive integer. This corresponds the unique rook placement of $0$ rooks on the empty board. To avoid unnecessary complications with the case $n=1$, we will assume throughout that $n\geq 2$.

Closely associated to the set $\mathcal{F}_{n,k}$ is the set $\widetilde{\Func}_{n,k}$ defined below.
\begin{align*}
\widetilde{\Func}_{n,k}=\{f: [n-1] \hookrightarrow [n] \times [k]\}
\end{align*}
The sets $\widetilde{\Func}_{n,k}$ and $\mathcal{F}_{n,k}$ are clearly equinumerous, and we will use the bijection $\Phi$ described next to associate an element of $\widetilde{\Func}_{n,k}$ with $\mathcal{F}_{n,k}$. The image of $f\in \mathcal{F}_{n,k}$ under $\Phi$ is the element $\Phi(f)\in \widetilde{\Func}_{n,k}$ defined as follows.
\begin{align*}
\Phi(f) (i) = (f(i)-n\left\lfloor \frac{f(i)-1}{n} \right\rfloor, \left\lceil \frac{f(i)}{n} \right\rceil)
\end{align*}
It is easy to see that $\Phi$ establishes a bijection between $\mathcal{F}_{n,k}$ and $\widetilde{\Func}_{n,k}$.

For any  element $g\in \widetilde{\Func}_{n,k}$ we define $1\leq i\leq n-1$ to be a \emph{$j$-excedance} (respectively \emph{$j$-subcedance}) if $g(i)$ equals $(a,j)$ for some positive integer $a$ satisfying $a>i$ (respectively $a\leq i$). This given we associate two weak compositions with $g\in \widetilde{\Func}_{n,k}$ as follows
\begin{align*}
\exc(g)&=(\alpha_1,\ldots,\alpha_{k})\\
\sub(g)&=(\beta_1,\ldots, \beta_{k}),
\end{align*}
where $\alpha_j$ (respectively $\beta_j$) counts the number of $j$-excedances (respectively $j$-subcedances) in $f$. It is easily seen that $|\exc(g)|+|\sub(g)|=n-1$.

Let $\mathsf{u} = \{u_1,u_2,\ldots\}$ and $\mathsf{v} = \{v_1,v_2,\ldots\}$ be two countably infinite sets of commuting indeterminates, which further commute with each other. This given, we associate a polynomial $G^{(1)}_{n,k}$ in the variables $\mathsf{u}$ and $\mathsf{v}$ with $\widetilde{\Func}_{n,k}$ as shown below.
\begin{align*}
G^{(1)}_{n,k}=\sum_{g\in\widetilde{\Func}_{n,k}} \mathsf{u}^{\exc(g)}\mathsf{v}^{\sub(g)}
\end{align*}
Our next two examples illustrate the various notions introduced above. The next example also serves to illustrate how we will think of elements of $\widetilde{\Func}_{n,k}$ as certain rook placements of `colored' rooks on an $(n-1)\times n$ rectangular board.
\begin{example}
\em{Let $n=6$ and $k=2$. The  maximal nonattacking rook placement below on the left corresponds to a unique element $f \in\mathcal{F}_{6,2}$. The rook placement on the right corresponds to element $g:=\Phi(f)\in\widetilde{\Func}_{6,2}$.
\begin{align*}
\ytableausetup{boxsize=1em}
\begin{ytableau}
*(red!75) & *(red!75) &*(red!75) &*(red!75) &*(red!75)\rook &*(red!40)  &*(blue!65) &*(blue!65) &*(blue!65) &*(blue!65) &*(blue!65) &*(blue!40) \\
*(red!75) & *(red!75) &*(red!75) &*(red!75) &*(red!40) &*(red!40)  &*(blue!65) &*(blue!65)\rook &*(blue!65) &*(blue!65)&*(blue!40) &*(blue!40) \\
*(red!75) & *(red!75) &*(red!75) &*(red!40) &*(red!40) &*(red!40)  &*(blue!65)\rook &*(blue!65) &*(blue!65) &*(blue!40) &*(blue!40) &*(blue!40)\\
*(red!75) & *(red!75) &*(red!40) &*(red!40) &*(red!40) &*(red!40) &*(blue!65) &*(blue!65) &*(blue!40)\rook &*(blue!40) &*(blue!40) &*(blue!40)  \\
*(red!75) & *(red!40) &*(red!40) &*(red!40) &*(red!40)\rook &*(red!40) &*(blue!65) &*(blue!40) &*(blue!40) &*(blue!40) &*(blue!40) &*(blue!40)  \\
\end{ytableau}
\hspace{10mm}
\begin{ytableau}
*(white) & *(white) &*(white) &*(white) &*(white)\textcolor{red}\rook &*(white) \\
*(white) &*(white)\textcolor{blue}{\rook} &*(white) &*(white) &*(white) &*(white) \\
*(white)\textcolor{blue}{\rook} & *(white) &*(white) &*(white) &*(white) &*(white)\\
*(white) &*(white) &*(white)\textcolor{blue!40}{\rook} &*(white) &*(white) &*(white) \\
*(white) &*(white) &*(white) &*(white) & *(white)\textcolor{red!40}{\rook} &*(white) \\
\end{ytableau}
\end{align*}
Note that the rook placement on the right has `colored' rooks, and is nonattacking in the sense that if two rooks occupy the same columns, then they must have distinct colors.
If we look in the $i$-th row for $1\leq i\leq 5$, then the color of the rook implicitly tells us about the second coordinate of $g(i)$ and the column in which it belongs gives us the first coordinate of $g(i)$.

Note that $\exc(g)=(1,1)$ and $\sub(g)=(1,2)$, and therefore, the associated monomial with $g$ is $u_1u_2v_1v_2^2$.}
\end{example}

\begin{example}
\em{
We will quickly compute $G_{2,2}^{(1)}$ next. The four maximal nonattacking rook placements corresponding to $\mathcal{F}_{2,2}$ are shown below.
\begin{align*}
\ytableausetup{boxsize=1em}
\begin{ytableau}
*(red!75)\rook & *(red!40) & *(blue!65) & *(blue!40)
\end{ytableau}\hspace{3mm}
\begin{ytableau}
*(red!75) & *(red!40)\rook & *(blue!65) & *(blue!40)
\end{ytableau}\hspace{3mm}
\begin{ytableau}
*(red!75) & *(red!40) & *(blue!65)\rook & *(blue!40)
\end{ytableau}\hspace{3mm}
\begin{ytableau}
*(red!75) & *(red!40) & *(blue!65)  & *(blue!40)\rook
\end{ytableau}\hspace{3mm}
\end{align*}
From these we obtain the four elements of $\widetilde{\Func}_{2,2}$ shown below.
\begin{align*}
\begin{ytableau}
*(white)\textcolor{red}{\rook} & *(white)
\end{ytableau}\hspace{10mm}
\begin{ytableau}
*(white) & *(white)\textcolor{red!40}{\rook}
\end{ytableau}\hspace{10mm}
\begin{ytableau}
*(white)\textcolor{blue}{\rook} & *(white)
\end{ytableau}\hspace{10mm}
\begin{ytableau}
*(white) & *(white)\textcolor{blue!40}{\rook}
\end{ytableau}
\end{align*}
Thus, we obtain that
\begin{align*}
G_{2,2}^{(1)}=v_1+u_1+v_2+u_2.
\end{align*}}
\end{example}

\subsection{Catalan, Shi, and Linial boards}
In this subsection, we consider certain special boards that are important for us.
Let $t$ be a  nonnegative integer. We define the \emph{$t$-Catalan board}, denoted by $C_{t,n}$ to be the $(n-1)\times (n-2+t)$ board. We define the \emph{$t$-Shi board}, denoted by $S_{t,n}$ to be the Ferrers board with shape $(2n-3+t,\ldots,n-1+t)$. Finally, we define the \emph{$t$-Linial board}, denoted by $L_{t,n}$, to be the skew Ferrers board with shape $(2n-3+t,\ldots,n-1+t)/(n-1,\ldots,1)$. An easy computation establishes the following.
\begin{align*}
r_{n-1}(C_{t,n})&=\prod_{j=1}^{n-1}(t+n-1-j)=\sum_{j=0}^{n-1}r_{j}(C_{0,n})(t)_{n-1-j}=R(t,C_{0,n})\\
r_{n-1}(S_{t,n})&=(t+n-1)^{n-1}=\sum_{j=0}^{n-1}r_{j}(S_{0,n})(t)_{n-1-j}=R(t,S_{0,n}).
\end{align*}
From \cite[Theorem 2.4]{Headley} (also from \cite[Corollary 9.9.1]{Postnikov-Stanley} and \cite[Theorem 3.3]{Athanasiadis-Advances}), it follows that $\chi_{n-1}^{1,2}(q)=(q-n)^{n-1}$. Similarly, from \cite[Theorem 9.8]{Postnikov-Stanley}, it follows that $\chi_{n-1}^{1,1}=(q+1-n)(q+2-n)\cdots (q+n-1-n)$. We therefore infer the following equalities.
\begin{align}\label{eqn: catshi}
R(t,C_{0,n})&=(-1)^{n-1}\chi_{n-1}^{1,1}(1-t)\nonumber\\
R(t,S_{0,n})&=(-1)^{n-1}\chi_{n-1}^{1,2}(1-t)
\end{align}
At this point, the reader is invited to compare the equalities above with the statement of Theorem \ref{thm: intro 2}.
Furthermore, given Equation \eqref{eqn: catshi}, note that Theorem \ref{thm: GJW-v} allows us to write the characteristic polynomials of the extended Catalan and extended Shi arrangements as a sum that runs over the elements of partition lattice $\Pi_{n-1}$ using the identities below.
\begin{align}\label{eqn: catshi2}
R(t,C_{0,n})&=\sum_{\sigma\in \Pi_{n-1}}\mu(\hat{0},\sigma)\prod_{A\in \sigma}(t+n-2)\nonumber\\
R(t,S_{0,n})&=\sum_{\sigma\in \Pi_{n-1}}\mu(\hat{0},\sigma)\prod_{A\in \sigma}(t+n-2+\min(A))
\end{align}

The preceding discussion naturally raises the question of finding a similar interpretation for characteristic polynomials of extended Linial arrangements. As we shall soon see, the Linial boards will be crucial to this end. The route we follow is described next: Express $r_{n-1}(L_{t,n})$ as the factorial polynomial of $L_{0,n}$ and observe that this polynomial is of degree $n-1$ in $t$. Using the bijection we describe in the next section, we compute $r_{n-1}(L_{t,n})$ for infinitely many integer specializations of $t$ and relate it to the  the characteristic polynomial of extended Linial arrangement.

\begin{example}
\em{
For $n=3$ and $t=2$, given below are the boards $C_{t,n}$, $S_{t,n}$ and $L_{t,n}$.
\begin{align*}
\ytableausetup{mathmode,boxsize=0.8em}
\begin{ytableau}
  *(blue!50) & *(blue!50) & *(blue!50)\\
 *(blue!50)& *(blue!50) & *(blue!50)\\
\end{ytableau}
\hspace{5mm}
\ytableausetup{mathmode,boxsize=0.8em}
\begin{ytableau}
 *(blue!50) & *(blue!50) & *(blue!50) & *(blue!50) & *(blue!50)\\
 *(blue!50)& *(blue!50) & *(blue!50) & *(blue!50) \\
\end{ytableau}
\hspace{5mm}
\ytableausetup{mathmode,boxsize=0.8em}
\begin{ytableau}
 \none & *(blue!50) & *(blue!50) & *(blue!50)\\
 *(blue!50)& *(blue!50) & *(blue!50)\\
\end{ytableau}
\end{align*}

}
\end{example}
\section{Trees}\label{sec:trees}
Given positive integers $n$ and $k$, we will denote the set of \emph{labeled rooted trees} on $n$ vertices such that any vertex has at most $k$ children by $\tree_{n,k}$. Furthermore, we will denote the set of \emph{labeled rooted plane $k$-ary trees} on $n$ vertices by $\ptree_{n,k}$. Given any vertex $v$ of a tree $T\in \ptree_{n,k}$, we will consider its children ordered from left to right, and by  the $i$-th child of $v$ we mean the $i$-th child from the left where $1\leq i\leq k$. We remark here that the trees in $\ptree_{n,k}$ are \emph{incomplete}, in that every nonleaf node need not $k$ children. Additionally, we will refer to the first child as the \emph{left child} and the $k$-th child as the \emph{right child}. We will want to distinguish between left and right children, and therefore, from this point onwards, we will work under the assumption that $k\geq 2$. For all our trees, we will implicitly assume that the edges of the tree are directed towards the root.

A tree $T\in \tree_{n,k}$ can be endowed with some additional data so as to obtain a \emph{decorated tree}. Essentially, we assign to each edge connecting any parent node to its children in $T$ a distinct label from the set $[k]$. We denote the set of decorated trees on $n$ vertices such that every node has at most $k$ children by $\dtree_{n,k}$. Note that there is an obvious natural bijection between $\dtree_{n,k}$ and $\ptree_{n,k}$. Given $T\in\dtree_{n,k}$, for an edge labeled $i$ going from $p$ to $q$, we retain the edge but designate $q$ to be the $i$-th child of $p$ (and get rid of the labels of the edge). It is easy to see that in so doing, we obtain an element of $\ptree_{n,k}$, and that the aforementioned map is indeed a bijection.

Given $T\in \ptree_{n,k}$, define the \emph{spine} of $T$, denoted by $\spine(T)$ to be the sequence of numbers encountered along the unique path from the root to the vertex labeled $n$.  Suppose $\spine(T)=a_1\cdots a_m$ where $a_1$ is the label of the root and $a_m=n$. We will call an integer $1\leq i\leq m-1$ a \emph{record} of $T$ if $a_i>a_j$ for all $1\leq j< i$. Furthermore, we will denote the number of records in $T$ by $\rec(T)$.

\begin{remark}
Note that records of a labeled tree are a straightforward generalization of records or left-to-right maxima of a permutation.
\end{remark}

 Now given a tree $T\in \ptree_{n,k}$ and $1\leq i \leq k$, call an edge from $q$ to $p$ an \emph{$i$-descent} if the following two conditions are met.
\begin{enumerate}
\item $p>q$.
\item $q$ is the $i$-th child of $p$.
\end{enumerate}
We define an \emph{$i$-ascent} similarly by reversing the inequality in the first condition above. Using these notions we define certain special subsets of $\ptree_{n,k}$.
\begin{align*}
\iptree_{n,k}&=\{T\in \ptree_{n,k}\;|\; \text{$T$ has no $i$-descents for $1\leq i\leq k$}\}\\
\riptree_{n,k}&=\{T\in \ptree_{n,k}\;|\; \text{$T$ has no $k$-descents}\}\\
\ltree_{n,k}&=\{T\in \ptree_{n,k}\;|\; \text{$T$ has no $1$-ascents or $k$-descents}\}\\
\ltree_{n,k}^{b}&=\left\{T\in \ltree_{n,k}\;|\; \begin{array}{c}\text{$1$ has no right child and}\\\text{$n$ has no left child}\end{array}\right\}
\end{align*}
Clearly, elements of $\iptree_{n,k}$ are labeled rooted \emph{increasing} plane $k$-ary trees while elements of $\riptree_{n,k}$ are labeled rooted plane $k$-ary trees that always increase to the right. Both of these classes are extremely well studied classes, and it is the other two classes defined above that are important for our purposes.

We will be interested in enumerating trees according to various constraints based on ascents and descents. To this end, it helps to introduce a multivariate generating function keeping track of these statistics. Firstly, we associate two weak compositions $\asc(T)=(\alpha_1,\ldots, \alpha_k)$ and $\dsc(T)=(\beta_1,\ldots, \beta_k)$ with $T\in \ptree_{n,k}$ as shown below.
\begin{align*}
\alpha_i &= |\{\text{$i$-descents in } T\}|\\
\beta_i &=|\{\text{$i$-ascents in } T\}|
\end{align*}
Clearly, we have that $|\alpha|+|\beta|=n-1$.
Next we associate a polynomial $G^{(2)}_{n,k}$ in the variables $\mathsf{u}$ and $\mathsf{v}$ with $\ptree_{n,k}$.
\begin{align*}
G^{(2)} _{n,k}=\sum_{T\in\ptree_{n,k}} \mathsf{u}^{\dsc(T)}\mathsf{v}^{\asc(T)}
\end{align*}

\begin{remark}
We could have easily associated analogues of the aforementioned statistics to decorated trees rather than labeled rooted plane $k$-ary trees.
\end{remark}

Our next goal is to establish that $G^{(1)}_{n,k}=G^{(2)}_{n,k}$. To this end, we will describe our all important bijection between $\widetilde{\Func}_{n,k}$ and $\ptree_{n,k}$.
We would like to emphasize that our algorithm is a mild generalization of the bijection given in \cite{Remmel-Egecioglu}. The reader should read the construction described next in tandem with Example \ref{example: main bijection}.

Let $f\in \widetilde{\Func}_{n,k}$. We construct first the decorated digraph corresponding to $f$ as follows: For a positive integer $i$ satisfying $1\leq i\leq n-1$, let $f(i)=(a,b)$. Note that $1\leq a\leq n$ and $1\leq b\leq k$. Draw a directed edge from $i$ to $a$, and assign it the label $b$. Given the definition of $\widetilde{\Func}_{n,k}$, it is clear that there is no other edge directed towards $a$ whose label is $b$. Additionally, the digraph thus obtained possesses the following features.
\begin{itemize}
\item There is a unique connected component containing $n$, and it is a tree rooted at $n$.
\item The other connected components (if any) comprise directed cycles (with length $\geq 1$) of rooted trees.
\end{itemize}
We will consider the connected components of the functional digraph thus obtained from left to right in increasing order of the largest entries in their directed cycles. Furthermore, the connected component that contains $n$ (which does not contain any directed cycles) will be the rightmost. Let these connected components be $T_1$ through $T_r$. For each $1\leq i\leq r-1$,  let $a_i$ denote the largest entry taking part in the directed cycle in $T_i$ and suppose further that the unique edge directed out of $a_i$ points towards $b_i$, and that the label of this edge is $c_i$. Furthermore, let $a_{r}=n$. Now for each $1\leq i\leq r-1$, convert each $T_i$ to a tree rooted at $a_i$  by removing the directed edge connecting $a_i$ to $b_i$ labeled $c_i$ (and remember this label). In this way, from the functional digraph, we obtain a sequence of rooted trees $T_1$ through $T_r$ with labeled edges.

Next we will assemble this sequence of rooted trees into a single tree. For $1\leq i\leq r-1$, draw a directed edge from $a_{i+1}$ to $b_{i}$ labeled $c_i$. Note that this gives us a tree rooted at $a_1$. Furthermore, note that the procedure is reversible. That is, given a rooted tree with vertices labeled from $[n]$ and edges labeled from $[k]$, we can recover a unique functional digraph as follows: Consider the path from root to the vertex labeled $n$. Let this be comprising vertices labeled $j_1,\ldots,j_p$. Let $s$ and $t$ be the indices of two successive records in the sequence $j_1,\ldots, j_m$ with $s<t$. Remove the edge connecting $j_{t-1}$ to $j_t$. Suppose further that the label of this edge was $c$. Now insert an directed edge from $j_s$ to $j_{t-1}$ with label $c$. Doing this procedure for every pair of consecutive records gives us the corresponding functional digraph.

At this point, it is easy to obtain an element of $\ptree_{n,k}$. Indeed, if the edge from $i$ to $j$ has label $l$, then we set $i$ to be the $l$-th child of $j$. Doing this gives us a tree $T\in \ptree_{n,k}$.
Let us call the tree thus obtained $\Psi(f)$. It is easy to see given the preceding paragraph that $\Psi$ is a bijection between $\widetilde{\Func}_{n,k}$ and $\ptree_{n,k}$ (and therefore induces a bijection between $\mathcal{F}_{n,k}$ and $\ptree_{n,k}$).
\begin{example}\label{example: main bijection}
\em{
Let $n=21$ and $k=3$. Consider the following element of $\widetilde{\Func}_{21,3}$.

\begin{tabular}{ |p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|}
 \hline
 \multicolumn{20}{|c|}{$f$} \\
 \hline
 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20\\
 \hline
 3,1   & 5,2    & 4,3 &  5,3 & 3,3   & 21,2    & 7,1 &  12,2 & 1,3   & 4,2    & 4,1 &  20,2 & 19,3   & 19,1    & 6,2 &  1,2 & 16,1   & 6,3    & 7,2 & 12,3 \\
 \hline
\end{tabular}

Then we obtain the following decorated digraph corresponding to $f$.

\begin{tikzpicture}[every tree node/.style={fill=red!30,draw=red!60,circle,inner sep=0.5pt,minimum size=2pt},
   level distance=1.25cm,sibling distance=1cm,
   edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}]
\Tree
[.\node(5){$5$};
    \edge[<-,thick,blue] node[auto=right] {$3$};
    [.\node(4){$4$};
       \edge[<-,thick,blue] node[auto=right] {$3$};
       [.\node(3){$3$};
       \edge[<-,thick,blue] node[auto=right] {$1$};
       	[.\node(1){$1$};
       	\edge[<-,thick,blue] node[auto=right] {$2$};
       	[.\node(16){$16$};
       	\edge[<-,thick,blue] node[auto=right] {$1$};
       		[.\node(17){$17$};]]
       	\edge[<-,thick,blue] node[auto=left] {$3$};
       	[.\node(9){$9$};]
       	]
       ]
       \edge[<-,thick,blue] node[auto=right] {$1$};
       [.\node(11){$11$};]
       \edge[<-,thick,blue] node[auto=left] {$2$};
       [.\node(10){$10$};]
        ]
    \edge[<-,thick,blue] node[auto=left] {$2$};
    [.\node(2){$2$};
    ]
]
\path (5) edge[->, in=90,out=150,very thick, dashed, blue,auto=right] node {$3$} (3);
\end{tikzpicture}
\begin{tikzpicture}
[every tree node/.style={fill=red!30,draw=red!60,circle, inner sep=0.5pt, minimum size=2pt},
   level distance=1.25cm,sibling distance=1cm,
   edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}]

\Tree
[
.\node(7){$7$};
\edge[<-,thick,blue] node[auto=right] {$2$};
	[
	.\node(19){$19$};
	\edge[<-,thick,blue] node[auto=right] {$3$};
		[.\node(13){$13$};]
	\edge[<-,thick,blue] node[auto=left] {$1$};
		[.\node(14){$14$};]
	]
]
\path (7) edge [loop above, very thick, dashed, blue,auto=above] node {$1$} (7);
\end{tikzpicture}\hspace{10mm}
\begin{tikzpicture}
[every tree node/.style={ fill=red!30,draw=red!60,circle, inner sep=0.5pt, minimum size=2pt},
   level distance=1.25cm,sibling distance=1cm,
   edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}]

\Tree
[
.\node(20){$20$};
\edge[<-,thick,blue] node[auto=left] {$2$};
	[
	.\node(12){$12$};
	\edge[<-,thick,blue] node[auto=right] {$2$};
		[.\node(8){$8$};]
	]
]
\path (20) edge[->, in=120,out=210,very thick, dashed, blue,auto=right] node {$3$} (12);
\end{tikzpicture}
\hspace{10mm}
\begin{tikzpicture}
[every tree node/.style={fill=red!30, draw=red!60,circle,inner sep=0.5pt, minimum size=2pt},
   level distance=1.25cm,sibling distance=1cm,
   edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}]

\Tree
[
.\node(21){$21$};
\edge[<-,thick,blue] node[auto=right] {$2$};
	[
	.\node(6){$6$};
	\edge[<-,thick,blue] node[auto=right] {$2$};
		[.\node(15){$15$};]
	\edge[<-,thick,blue] node[auto=left] {$3$};
		[.\node(18){$18$};]
	]
]
\end{tikzpicture}

Applying the procedure outlined earlier, we obtain the following element of $\dtree_{21,3}$.
$$
\begin{tikzpicture}[every tree node/.style={fill=red!30,draw=red!60,circle,inner sep=0.5pt,minimum size=4pt},
   level distance=1.2cm,sibling distance=0.4cm,
   edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}]
\Tree
[.\node(5){$5$};
	\edge[<-,thick,blue] node[auto=right] {$2$};
	[.\node(2){$2$};
	]
	\edge[<-,thick,blue] node[auto=left] {$3$};
	[.\node(4){$4$};
		\edge[<-,thick,blue] node[auto=right]{$1$};
		[.\node(11){$11$};
		]
		\edge[<-,thick,blue] node[auto=left]{$2$};
		[.\node(10){$10$};
		]
		\edge[<-,thick,blue] node[auto=left]{$3$};
		[.\node(3){$3$};
			\edge[<-,thick,blue] node[auto=right]{$1$};
			[.\node(1){$1$};
				\edge[<-,thick,blue] node[auto=right]{$2$};
				[.\node(16){$16$};
					\edge[<-,thick,blue] node[auto=right]{$1$};
					[.\node(17){$17$};]
				]
				\edge[<-,thick,blue] node[auto=left]{$3$};
				[.\node(9){$9$};
			   ]
			]
			\edge[<-,thick,dashed,blue] node[auto=left]{$3$};
			[.\node(7){$7$};
				\edge[<-,thick,dashed,blue] node[auto=right]{$1$};
				[.\node(20){$20$};
					\edge[<-,thick,blue] node[auto=right]{$2$};
					[.\node(12){$12$};
						\edge[<-,thick,blue] node[auto=right]{$2$};
						[.\node(8){$8$};]
					\edge[<-,thick,dashed,blue] node[auto=left]{$3$};
					[.\node(21){$21$};
						\edge[<-,thick,blue] node[auto=left]{$2$};
						[.\node(6){$6$};
							\edge[<-,thick,blue] node[auto=right]{$2$};
							[.\node(15){$15$};
							]
							\edge[<-,thick,blue] node[auto=left]{$3$};
							[.\node(18){$18$};
							]
						]
					]
				]
				]
				\edge[<-,thick,blue] node[auto=left]{$2$};
				[.\node(19){$19$};
					\edge[<-,thick,blue] node[auto=right]{$1$};
					[.\node(14){$14$};
					]
					\edge[<-,thick,blue] node[auto=left]{$3$};
					[.\node(13){$13$};
					]
				]
			]
		]
	]
]
\end{tikzpicture}
$$
One can now easily obtain $\Psi(f)\in\ptree_{21,3}$ by orienting the edges according to the labels they carry.}
\end{example}
This brings us to our next theorem detailing the properties of $\Psi$.

\begin{theorem}
The map $\Psi$ is a bijection from $\widetilde{\Func}_{n,k}$ and $\ptree_{n,k}$. The number of connected components in the functional digraph corresponding to $f\in \widetilde{\Func}_{n,k}$ is $\rec(\Psi(f))$. Furthermore, we have that $\exc(f)=\dsc(\Psi(f))$ and $\sub(f)=\asc(\Psi(f))$.
\end{theorem}
\begin{proof}

We have already established earlier that $\Psi$ is a bijection, and it is implicit in the argument above that the number of connected components in the functional digraph corresponding to $f\in \widetilde{\Func}_{n,k}$ is $\rec(\Psi(f))$.
Now let $\exc(f)=(\alpha_1,\ldots,\alpha_k)$ and $\sub(f)=(\beta_1,\ldots,\beta_k)$. Additionally, let $\asc(\Psi(f))=(\gamma_1,\ldots,\gamma_k)$ and $\dsc(\Psi(f))=(\delta_1,\ldots,\delta_k)$.

For the length of this proof, let $T:=\Psi(f)$. Note that we have
\begin{align*}
&\delta_i=\text{\# edges $b\to a$ in $T$ such that $a>b$ and $b$ is the $i$-th child of $a$},\\
&\alpha_i=\text{\# edges $b\to a$ in functional digraph associated to $f$ such that $a>b$ and edge is labeled $i$}.
\end{align*}
In recovering $f$ from $T$, it suffices to recover the functional digraph of $f$. To do this, we identify the spine of $T$, and redirect certain edges leaving the rest unaltered. Observe that the redirected edges do not correspond to descents in $T$, and additionally, upon redirection they point from a larger number to a smaller number (or to itself), in the decorated functional digraph thus obtained. Thus, it follows that
$\delta_i=\alpha_i$, and therefore that $\dsc(T)=\exc(f)$.

Now we will establish that $\asc(T)=\sub(f)$ with a similar argument. Note that in going back to the decorated functional digraph from $T$, we redirect certain ascent edges $b\to a$ where $b$ is the $i$-th child of $a$ and $b>a$. But despite the redirection, they continue to point from a larger number to a smaller number and the edge label is $i$, inherited from before. This immediately implies that $\asc(T)=\sub(f)$.
\end{proof}

As a corollary of the above theorem, we obtain the following.
\begin{corollary}
We have that $
G_{n,k}^{(1)}=G_{n,k}^{(2)}$.
\end{corollary}
In view of the above corollary, we define $G_{n,k}:=G_{n,k}^{(1)}=G_{n,k}^{(2)}$ to be the \emph{Gessel polynomial} indexed by positive integers $n$ and $k$.

\begin{remark}\label{remark: work of Gessel}
The polynomials $G_{n,2}$ have been studied by Gessel \cite{Gessel-unpublished}, Kalikow \cite{Kalikow} and Drake \cite{Drake-thesis}. More precisely, the following generating function was considered by the aforementioned authors.
\begin{align*}
B(u_1,u_2,v_1,v_2;x)=\sum_{n\geq 1}G_{n,2}(u_1,u_2,v_1,v_2)\frac{x^n}{n!}
\end{align*}
It develops that $B:=B(u_1,u_2,v_1,v_2;x)$ satisfies the following functional equation.
\begin{align*}
\frac{(1+v_1B)(1+u_2B)}{(1+v_2B)(1+u_1B)}=e^{[(v_1u_2-v_2u_1)B+v_1-v_2-u_1+u_2]x}
\end{align*}
It is worthwhile to note the non-obvious symmetry in $G_{n,2}$ from the equation above, namely that  $$G_{n,2}(u_1,u_2,v_1,v_2)=G_{n,2}(u_1,v_1,u_2,v_2).$$ A recursive proof has been obtained by Kalikow \cite{Kalikow}. Note further that, from classical results of rooted labeled binary trees, we know that $G_{n,2}(1,1,1,1)=n! C_n$ where $C_n=\frac{1}{n+1}\binom{2n}{n}$ denotes the $n$-th Catalan number. Additionally, $G_{n,2}(1,1,1,0)=(n+1)^{n-1}$, $G_{n,2}(1,0,1,0)=n!$ and $G_{n,2}(u_1,u_2,0,0)=\sum_{\sigma\in \mathfrak{S}_n}u_1^{\dsc(\sigma)}u_2^{\asc(\sigma)}$ (the $n$-th homogenized Eulerian polynomial). The careful reader might recall that the number of regions in the Catalan arrangement is $n! C_n$, the number of regions in the Shi arrangement is $(n+1)^{n-1}$, and the number of regions in the braid arrangement is $n!$.
\end{remark}

\subsection{Drake's results}
In this subsection we describe a beautiful result of Drake \cite{Drake-thesis} which allows us to compute the generating function for the ascent-descent statistics on $\ptree_{n,k}$. First, consider the following generating function.
\begin{align*}
P_k(\mathsf{u},\mathsf{v};x)=\sum_{n\geq 1}\sum_{T\in \ptree_{n,k}}\mathsf{u}^{\dsc(T)}\mathsf{v}^{\asc(T)}\frac{x^n}{n!}=\sum_{n\geq 1}G_{n,k}\frac{x^n}{n!}
\end{align*}
The following theorem of Drake \cite[Theorem 1.8.4]{Drake-thesis} expresses $P_k:=P_k(\mathsf{u},\mathsf{v};x)$ as a compositional inverse of another function.
\begin{theorem}[\cite {Drake-thesis}]\label{thm: drake}
Consider an auxiliary function $Z(a,b,c,d)=(ad-bc)x+(a-c)-(b-d)$. Then we have that
\begin{align*}
P_k=\left(\sum_{i=1}^{k}(v_i-u_i)^{k-2}\log{\left(\frac{1+v_ix}{1+u_ix}\right)}\prod_{j\in [k], i\neq j}\frac{1}{Z(v_i,u_i,v_j,u_j)}\right)^{\langle -1 \rangle}
\end{align*}
\end{theorem}
The reader is invited to compare the above functional equation to the one obtained for $P_2$ in Remark \ref{remark: work of Gessel}.
We will now use the above result to compute the cardinality of $\ltree_{n,k}$ where $k\geq 2$. Thus, we are counting the number of trees in $\ptree_{n,k}$ that constrained to have no $1$-ascents and no $k$-descents. Let $l_{n,k}$ denote this number, that is, $l_{n,k}=|\ltree_{n,k}|$. We can obtain the exponential generating function for these number by evaluating $P$ at the values $u_k=0,v_1=0$ and all the other variables set to $1$.
Consider
\begin{align*}
\widetilde{P}_k=\sum_{n\geq 1}l_{n,k}\frac{x^n}{n!}.
\end{align*}
Then from Theorem \ref{thm: drake}, we get the following functional equation for $\widetilde{P}_k$ (the details are similar to the many examples discussed in \cite[Chapter 1.9]{Drake-thesis}).
\begin{align*}
\widetilde{P}_k=\left(\frac{2\log(1+x)}{(1+x)^{k-2}(2+x)}\right)^{\langle -1\rangle}
\end{align*}
Substituting $f=1+\widetilde{P}_k$, we get that $f$ satisfies the following relation.
\begin{align*}
f^2=e^{x(f^{k-2}+f^{k-1})}
\end{align*}
We see that the functional equation above is precisely the one in the statement of \cite[Theorem 9.3]{Postnikov-Stanley} with $b=k$ and $a=k-2$. Thus we see that the coefficient of $\frac{x^n}{n!}$ in $f$ for $n\geq 1$, that is, $l_{n,k}$,
is given by the following expression.
\begin{align}\label{eqn: local k-ary search trees}
l_{n,k}=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(1+(k-2)n+j)^{n-1}
\end{align}
\begin{remark}
In the case $k=2$, we are in fact counting local binary search trees, and the formula obtained above coincides with the formula obtained by Postnikov \cite{Postnikov-JCTA} and Athanasiadis \cite[Theorem 4.2]{Athanasiadis-Advances}.
\end{remark}

Now we return to our question of counting rook placements on skew Ferrers boards. Consider the sequence of Linial boards $L_{t,n}$ for $t=(a-1)n+2$ where $a$ ranges over all positive integers. For a fixed positive integer $a$, maximal nonattacking rook placements on $L_{(a-1)n+2,n}$ are clearly in bijection with the subset of $\mathcal{F}_{n,a+1}$ that comprises injective functions $f:[n-1] \to [(a+1)n]$ satisfying the criterion $i+1\leq f(i)\leq an+i$ for $1\leq i\leq n-1$. But by our bijection $\Psi$ from before, the number of such rook placements is the number of trees $T\in \ptree_{n,a+1}$ such that there are no $1$-ascents and no $a+1$-descents. The condition $i+1\leq f(i)$ implies that $T$ can have no $1$-ascents while the condition $f(i)\leq an+i$ implies that $T$ can have no $a+1$-descents. This motivates our next theorem that gives a closed form expression for the factorial polynomial of the Linial board $L_{0,n}$.

\begin{theorem}\label{thm: factorial rook polynomial Linial board}
We have that
\begin{align*}
R(t,L_{0,n})=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(t-1+j)^{n-1},
\end{align*}
\begin{proof}
Note that since $L_{0,n}$ has $n-1$ rows, the factorial polynomial $R(t,L_{0,n})$ is a polynomial of degree $n-1$ by its definition. Furthermore, its definition also implies that the evaluations of $R(t,L_{0,n})$ at $t=m$ for $m$ a nonnegative integer agree with $r_{n-1}(L_{m,n})$. By comparison with Equation \eqref{eqn: local k-ary search trees}, we know that
\begin{align*}
r_{n-1}(L_{m,n})=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(1+(m-2)+j)^{n-1}
\end{align*}
when $m$ is of the form $(a-1)n+2$ for a positive integer $a\geq 1$. Thus the polynomial of degree $n-1$ below:
\begin{align*}
\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(1+(t-2)+j)^{n-1},
\end{align*}
agrees with $R(t,L_{0,n})$ for infinitely many integer values $t$. Therefore the two must be equal.
\end{proof}
\end{theorem}

Of course, given the similarity between the expression obtained in the theorem above and \cite[Equation 9.11]{Postnikov-Stanley}\cite[Theorem 4.2]{Athanasiadis-Advances}, the next corollary is no surprise.
\begin{corollary}\label{cor: relation factorial and characteristic}
Given a positive integer $a$, the following hold.
\begin{align*}
(-1)^{n-1}\chi_{n-1}^{a-1,a+1}(q)&=R(1+(a-1)n-q, L_{0,n})\\
(-1)^{n-1}\chi_{n-1}^{a-1,a+1}(q)&=\sum_{\sigma\in \Pi_{n-1}}\mu(\hat{0},\sigma)\prod_{A\in\sigma}(an-q-1+\min(A)-\max(A))
\end{align*}
\end{corollary}
\begin{proof}
The first equality follows from comparing the expression for $\chi_{n-1}^{a-1,a+1}$ in \cite[Equation 9.11]{Postnikov-Stanley} to the expression for $R(t,L_{0,n})$ in the  statement of Theorem \ref{thm: factorial rook polynomial Linial board}.

For the second equality, we apply Theorem \ref{thm: GJW-v} to obtain an expression for the factorial polynomial of $L_{0,n}$, and use the fact that the statistic $v_{L_{0,n}}(A)=n-2-(\max(A)-\min(A))$ for $A\subseteq [n-1]$.
\end{proof}
Note that in view of the above corollary and Equation \eqref{eqn: catshi2}, we have expressed characteristic polynomials of the extended Catalan, extended Shi and extended Linial arrangements as sums over the lattice of set partitions. It is interesting to note that similar results were achieved in \cite{Forge-and-gang} from a very different perspective, and the equivalence with our results is not obvious. Furthermore, the case $a=1$ in the above corollary implies Theorem \ref{thm: intro 2} from the introduction. Finally, the case $a=1$ and $q=-1$ gives us the following alternate formula for the number of local binary search trees on $n$ nodes.
\begin{align}\label{eqn: alternate lbs}
|\ltree_{n,2}|=\sum_{\sigma\in \Pi_{n-1}}\mu(\hat{0},\sigma)\prod_{A\in\sigma}(n+\min(A)-\max(A))
\end{align}
Admittedly, this is not as nice as the more well-known closed form $\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(1+j)^{n-1}$, but it would be interesting to relate the said closed form to the expression in Equation \eqref{eqn: alternate lbs}.

Now we are ready to state our main enumerative result, a special case of which was present in the introduction in the form of Theorem \ref{thm: intro 1}.
\begin{corollary}\label{cor: enumerative stuff}
For a fixed positive integer $a$,  let $\mathcal{A}:=\hyplin_{n-1,a}=\mathcal{A}^{a-1,a+1}_{n-1}$. Then the following hold.
\begin{enumerate}
\item $r(\mathcal{A})=R((a-1)n+2,L_{0,n})$ and $b(\mathcal{A})=R((a-1)n,L_{0,n})$.
\item $r(\mathcal{A})=r_{n-1}(L_{(a-1)n+2,n})$ and $b(\mathcal{A})=r_{n-1}(L_{(a-1)n,n})$.
\item $r(\mathcal{A})=|\ltree_{n,a+1}|$ and $b(\mathcal{A})=|\ltree_{n,a+1}^{b}|$.
\item $r(\mathcal{A})$ equals the number of sequences $(x_1,\ldots,x_{n-1})$ such that $1\leq x_i\leq an$ for $1\leq i\leq n-1$  and  $x_i+i\neq x_j+j$ for $1\leq i\neq j\leq n-1$; and $b(\mathcal{A})$ equals the number of sequences $(x_1,\ldots,x_{n-1})$ such that $1\leq x_i\leq an-2$ for $1\leq i\leq n-1$  and  $x_i+i\neq x_j+j$ for $1\leq i\neq j\leq n-1$.
\end{enumerate}
\end{corollary}
\begin{proof}
The first claim follows from plugging $q=\pm 1$ in Corollary \ref{cor: relation factorial and characteristic} and recalling Zaslavsky's result, that is, Theorem \ref{thm: Zaslav}. The second claim follows from the first by utilizing the relation between factorial polynomial and maximal nonattacking rook placements.

The discussion preceding Theorem \ref{thm: factorial rook polynomial Linial board} establishes that $r_{n-1}(L_{(a-1)n+2,n})=|\ltree_{n,a+1}|$ from which the first part of the third claim follows. To see how the second part follows, note that maximal nonattacking rook placements on $L_{(a-1)n,n}$ can be thought of maximal nonattacking rook placements on $L_{(a-1)n+2,n}$ where we forbid the rooks to belong to columns $n$ and $an+1$ (one can forbid any two columns of length $n-1$ in fact, but our choice leads to a nicer combinatorial interpretation for the number of bounded regions in extended Linial arrangements). Under the map $\Psi$, maximal nonattacking rook placements on $L_{(a-1)n+2,n}$ map to elements of $\ltree_{n,a+1}$.
This given, the condition that we are forbidding rooks to belong to columns $n$ and $an+1$ translates to trees in $\ltree_{n,a+1}$ which satisfy the criterion that the node labeled $1$ does not have a right child and the node labeled $n$ does not have a left child. But these are precisely trees that belong to $\ltree_{n,a+1}^{b}$. This establishes the third claim.

The final claim follows easily once one left-justifies our Linial boards. The columns occupied by the rooks in the left-justified board give us our sequence $(x_1,\ldots, x_{n-1})$ (reading the rows from bottom to top) and the condition $x_i+i\neq x_j+j$ is tantamount to the nonattacking condition.\end{proof}
\section{Chromatic polynomials and Linial graphs}\label{sec: chromatic poly}
We will now associate a bipartite graph $G_{t,n}$ with the $t$-Linial board $L_{t,n}$. For fixed values of $t$ and $n$, consider two sets of vertices $A_1=\{v_1,\ldots,v_{2n-4+t}\}$ and $A_2=\{v_{2n-3+t},\ldots,v_{3n-5+t}\}$. Note that $|A_1|=2n-4+t$ and $|A_2|=n-1$. Now using these two sets we will construct our bipartite graph $G_{t,n}$ as follows. For $1\leq i\leq n-1$, we connect vertex $v_{2n-4+t+i}\in A_2$ to the vertices $v_i,v_{i+1},\ldots,v_{n-3+t+i}$ (all belonging to $A_1$).

Note that all we are doing is labeling the rows of $L_{t,n}$ by elements of $A_2$ bottom to top and the columns of $L_{t,n}$ by elements of $A_1$ left to right and finally, drawing an edge for every square belonging to $L_{t,n}$. From this viewpoint, it is immediate that a nonattacking rook placement on $L_{t,n}$ corresponds to a matching on $G_{t,n}$. Recall that a \emph{matching} in a graph is a set of pairwise non-adjacent edges. Furthermore, a matching is said to be \emph{maximum} if it contains the largest possible number of edges.

The obvious relation between nonattacking rook placements and matchings gives us our next theorem.
\begin{theorem}\label{thm: rooks and matchings}
For a fixed positive integer $a$, let $\mathcal{A}:=\hyplin_{n-1,a}$. Then we have that
\begin{align*}
r(\mathcal{A})&=\text{ number of maximum matchings in $G_{(a-1)n+2,n}$}\\
b(\mathcal{A})&= \text{ number of maximum matchings in $G_{(a-1)n,n}$}.
\end{align*}
\end{theorem}
While the above theorem is not surprising given Corollary \ref{cor: enumerative stuff}, it is interesting to note that the complement of the graph $G_{t,n}$, denoted by $\overline{G_{t,n}}$, has some combinatorial value in the following sense: $$\text{The chromatic polynomial of $\overline{G_{t,n}}$ coincides with the $(3n-5+t)$-factorial polynomial of $L_{t,n}$.}$$
We explain this briefly next. Observe that $3n-5+t$ equals the sum of the number of rows and the number of columns in $L_{t,n}$, and is also equal to the total number of vertices in $\overline{G_{t,n}}$. Furthermore, it develops that, upto an interchange of rows and columns, the graph $G_{t,n}$ is obtained from a proper $(3n-5+t)$-board.  The result follows from invoking \cite[Theorem 2]{GJW-iii}. We refer the reader to \cite{GJW-iii} for the definition of proper boards and more details on the relation between factorial polynomials of proper boards and chromatic polynomials of associated graphs.

\begin{theorem}\label{thm: relation chromatic and characteristic polynomial}
Denote the chromatic polynomial of $\overline{G_{t,n}}$ by $c(x,\overline{G_{t,n}})$. Let $a$ be a fixed positive integer. Then
\begin{align*}
\frac{1}{(x)_{2n-4+t}}c(x,\overline{G_{t,n}})=(-1)^{n-1}\,\chi_{n-1}^{a-1,a+1}((a+1)n-3-x)
\end{align*}
\end{theorem}
\begin{proof}
By \cite[Theorem 2]{GJW-iii} we know that
\begin{align*}
c(x,\overline{G_{t,n}})=R_{3n-5+t}(x,L_{t,n})
\end{align*}
This given we have the following equalities.
\begin{align*}
c(x,\overline{G_{t,n}})&=
(x)_{2n-4+t}\left(\sum_{k=0}^{n-1}r_k(L_{t,n})(x-2n+4-t)_{n-1-k}\right)\\
&=
(x)_{2n-4+t}R(x-2n+4-t,L_{t,n})
\\&=
(x)_{2n-4+t}R(x-2n+4,L_{0,n})
\end{align*}
In the last step above, we have made use of the fact that $R(x,L_{t,n})=R(x+t,L_{0,n})$. Now from Corollary \ref{cor: relation factorial and characteristic}, we know that $R(x-2n+4,L_{0,n})=(-1)^{n-1}\chi_{n-1}^{a-1,a+1}((a+1)n-3-x)$, thereby establishing the claim.
\end{proof}
Thus, the content of Theorem \ref{thm: relation chromatic and characteristic polynomial} is that upto a certain normalizing factor, the chromatic polynomial of the graphs $\overline{G_{t,n}}$ coincides with the characteristic polynomial of extended Linial arrangements. In fact, the normalizing factor $(x)_{2n-4+t}$ can be thought of the chromatic polynomial of the complete graph on $2n-4+t$ vertices, that is, the complete graph on the vertices from the set $A_1$. Note that this complete graph is a subgraph of $\overline{G_{t,n}}$.

\begin{example}
\em{
Consider the case where $n=3$, $t=1$ and $a=1$. Drawn below is the Linial board $L_{1,3}$.
\begin{align*}
\ytableausetup{mathmode,boxsize=0.8em}
\begin{ytableau}
 \none & *(blue!50) & *(blue!50)\\
 *(blue!50)& *(blue!50)\\
\end{ytableau}
\end{align*}
The orange edges in the graph below on 5 vertices compose $G_{1,3}$, while the blue edges on the same set of vertices correspond to $\overline{G_{1,3}}$.
$$
\begin{tikzpicture}
\node[draw, circle,fill=red!30] (v1) at ({360/5 * (1 - 1)}:1.6cm) {$v_1$};
\node[draw, circle,fill=red!30] (v2) at ({360/5 * (2 - 1)}:1.6cm) {$v_2$};
\node[draw, circle,fill=red!30] (v3) at ({360/5 * (3- 1)}:1.6cm) {$v_3$};
\node[draw, circle,fill=green!40] (v4) at ({360/5 * (4 - 1)}:1.6cm) {$v_4$};
\node[draw, circle,fill=green!40] (v5) at ({360/5 * (5 - 1)}:1.6cm) {$v_5$};
\draw[blue,very thick] (v1)--(v2);
\draw[blue,very thick] (v1)--(v3);
\draw[blue,very thick] (v2)--(v3);
\draw[blue,very thick] (v4)--(v5);
\draw[blue,very thick] (v1)--(v5);
\draw[blue,very thick] (v4)--(v3);
\draw[orange, very thick] (v1)--(v4);
\draw[orange, very thick] (v2)--(v4);
\draw[orange, very thick] (v2)--(v5);
\draw[orange, very thick] (v3)--(v5);
\end{tikzpicture}
$$
Now, a computation reveals that $c(x,\overline{G_{1,3}})=x(x-1)(x-2)(x^2-3x+3)$. Additionally, we have that $\chi_{2}^{0,2}(x)=x^2-3x+3=\chi_{2}^{0,2}(3-x)$. We can now immediately verify that $\frac{1}{(x)_3}c(x,\overline{G_{1,3}})=\chi_{2}^{0,2}(3-x)$.}
\end{example}

\section{Final remarks}\label{sec:final}
We close with some remarks about future avenues and related work.
\begin{enumerate}
\item It is natural to ask if we can extend our results to the entire class of truncated affine arrangements; or even to subarrangements thereof.
The Shi($G$) arrangements \cite{Athanasiadis-Advances, Athanasiadis-Linusson} and the $G$-Shi arrangements \cite{Hopkins-Perkinson} have garnered plenty of attention in the recent past, and remain an active area of research. A rook-theoretic perspective on them might be of interest and lend new insight. Furthermore, it would be interesting to obtain $q$-analogs of our formulae and extend the results of Haglund-Remmel \cite{Haglund-Remmel} to our skew boards. Finally, it might be worthwhile to explore the relation to work of Ardila on Tutte polynomials associated with hyperplane arrangements \cite{Ardila}. In work in progress \cite{Griffin-Tewari}, we aim to answer some of the aforementioned  questions.

\item We will focus on the case $k=2$ of our bijection. We know that the number of maximal nonattacking rook placements on the skew Ferrers board $(2n-1,\ldots,n+1)/(n-1,\ldots,1)$ equals the number of local binary search trees  on $[n]$. Clearly, the set of such rook placements is equinumerous with the set of bijections $\{f: \mathbb{Z}\mapsto \mathbb{Z}\}$ satisfying the following conditions:
\begin{enumerate}
\item $f(i+n-1)=f(i)+(n-1),$
\item $i\leq f(i)\leq i+n-1$.
\end{enumerate}
It is worth remarking that this is a subset of conditions used to define \emph{bounded affine permutations} \cite{Knutson-Lam-Speyer} and we wonder if there is a stronger connection. Also, nonattacking rook placements on skew Ferrers boards have also been studied in \cite{Postnikov, Sjostrand} and it might be interesting to put our results in the context of the aforementioned articles.
\item If we consider the rook placements corresponding to increasing $k$-ary labeled rooted trees on $n$ nodes, we see that we are essentially counting maximal nonattacking rook placements on the Ferrers board of shape $(k(n-1),k(n-2),\ldots,k)$. Then it is easy to obtain the equality
\begin{align*}
|\iptree_{n,k}|=\prod_{i=1}^{n-1}(1+i(k-1)).
\end{align*}
Josuat-Verg\`es \cite{Josuat-Verges} studied the case $k=2$ using stammering tableaux and the language of growth diagrams. It would be interesting to explore the analogue of stammering tableaux for $k>2$.
\end{enumerate}
Finally, we would be remiss to not mention other work inspired by Gessel's original question, namely \cite{Corteel-Forge-Ventos,Forge}, and more recently \cite{Bernardi} which answers Gessel's questions from a different perspective than ours. Connections with the aforementioned work, especially \cite{Bernardi}, and ours remain to be explored.

\section*{Acknowledgements}
I owe a lot to Ira Gessel for his generosity in sharing his ideas, wisdom and notes, as well as his prompt and detailed responses to all my queries. I would also like to thank Joel Lewis and Alejandro Morales for an invaluable push in the right direction and helping me view my results in the right context; further thanks go to the former for helpful suggestions on improving exposition. I  have benefitted enormously from the many suggestions of Christos Athanasiadis, Sara Billey and Richard Stanley regarding avenues to pursue and pointers to relevant literature.
Finally, I would like to thank Sean Griffin, Brendon Rhoades, Jos\'e Samper, Andy Wilson, and Alex Woo for many enlightening discussions and numerous helpful comments.

redpajama_set_name: RedPajamaArXiv


My first real piece with acrylics. I'm extremely happy with the sky and slightly less happy with each element going down. I will be giving this to a friend as a birthday gift. It is loosely based on a photo I took from Kelly Butte last month. Despite its flaws, I am overall quite proud of the piece. I won't lie and say I'm not. PS- Those little bars at the bottom are part of the easel. I actually have no idea what their purpose is yet.
I'm thinking about maybe narrating a pixel art project some time in the future. I'm a bit shy and a bit lazy, which means it'll be awhile before I set everything up, but I think it could be a fun sort of tutorial. I definitely have a lot to say on the subject. I figure it would be fun to get it all out in a video.
Oh man!!! That art is Bob-Ross worthy. I don't know HOW you or other incredible artists do this kinda stuff. My mind only works in L I N E A R and S T R A I G H T L I N E S. Those kind of organic shapes make sense to me but when I try to recreate them they look like trash. I don't know how shading works, and everything I draw looks flat. I make video games and stuff, and my code and music is ok, but sprites and graphics really limit me. Please teach me the way!!!


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text:
This Chicago surgeon is training citizens to be first responders
by VICE News
This segment originally aired Feb. 3, 2016, on VICE News Tonight on HBO.
In 2016, there were more than 760 homicides in Chicago, the highest figure the city had seen in 30 years.
Chicago has four level one adult trauma centers, but not one of them is within 30 minutes of the city's south side, where most of the murders occur. In an emergency, that distance could mean the difference between life and death.
After a decades-long battle, a trauma center is scheduled to open at the University of Chicago next year. Until then, some south side residents are trying to address the problem themselves.
"I had patient after patient bleeding out, coming from the south side of Chicago," Northwestern Memorial Hospital surgeon Dr. Mamta Swaroop told VICE News. "And I wasn't able to save them. They would die, they would either be dead on arrival or die on my trauma bay."
So Dr Swaroop decided to activate local Chicagoans by offering first responder courses to citizens interested in helping. "We are recognizing that having people right there who are able to help their fellow citizens, turning people from citizens to immediate responders, is a way of reducing morbidity and mortality," Dr Swaroop said.
Tagged:videoVICE NewsChicagosouth sideHBOchicago violencehomocideDr. Mamta Swarooptrauma centers

redpajama_set_name: RedPajamaCommonCrawl


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The second part of the conference will be dedicated to the stories of successful deployment. A representative of Azerbaijan Express Bank will share his experience of implementing the WSS Docs e-document workflow system and will tell about the economic effect of business process automation in the bank.


text:
Bad Witch Series
"[The Importance of Being Emily] is a wonderfully imaginative tale that begs a re-read just so that every detail and nuance can be savored." RT Book Reviews
"[Poison in the Blood] brings fans right back into the dark, imaginative world of a London filled with magic and the fascinating characters who control it. … Bachar keeps the plot moving along nicely, and her inventiveness makes the world of this story come to life." RT Book Reviews
"Sigh I'm in love. To me, [Blood, Smoke and Mirrors] is classic urban fantasy done very well." Bitten By Books
The Chronicles of Emily: Books 1-2
Bad Witch: The Chronicles of Emily
The Importance of Being Emily
Magic, matchmaking and murder…
Lord Willowbrook's spring ball is supposed to be a magical celebration, but Miss Emily Wright is bored. The only outlet allowed for her magic is matchmaking—for others, not herself. Why bother? The only man she wants, Michael Black, is a man she can never have.
Suddenly the guests are abuzz with news of a young sorceress found drained of blood in the parlor. The mystery calls to her, and since she is the only available seer in all England, she jumps at the chance to prove herself.
Michael has spent his life preparing for his ritual death, when he will join the Order of St. Jerome as an immortal chronicler. Now that dream hangs in the balance, his mentor accused of the murder. Worse, gentle Emily, the woman he silently loves, is walking into a world of horrors beyond her imagination.
Torn between duty to the order and desire to keep her safe, Michael fights his growing need for a love that can never be his. All the while the real killer stalks the shadows of Willowbrook Hall, homing in on the next victim.
Poison in the Blood
She is honey to his throat…but one sweet taste could part them forever…
Banished from home while her husband adjusts to life as an immortal, blood drinking chronicler, Emily Black is homesick, heartsick and struggling under the constant sensory barrage of a city drenched in ancient magic. When an old friend asks for her aid in solving a string of murders, she welcomes the distraction, despite the danger.
Justine Dubois is grateful for a seer's help, and more understanding than anyone of Emily's plight. As a guardian, Justine commands respect; as a woman, her magic is considered inferior. Together, they are determined to prove their worth to London's magicians, starting with solving these murders—with maybe a bit of matchmaking on the side.
Long before he met his soul mate, Michael Black made a commitment to join the Order of St. Jerome. He will live forever, forced to watch the woman he loves age and die. As Emily hunts the murderer, Michael struggles protect her. But if he loses control of his hunger, the greatest threat to her safety could be Michael himself.
Blood, Smoke and Mirrors
Even a bad witch deserves a second chance.
Catherine Baker stopped caring about magician politics after the witches threw her out four years ago, convicted of the "crime" of using harmful magic in self-defense. Life as an outcast was nothing new, thanks to her father's wicked reputation, but the injustice left her heart scarred. Especially since she'd been turned in by the man she loved.
Alexander Duquesne has never faltered in his duties as a guardian—until now. He still suffers lingering guilt over his role in Cat's exile, and when the old Titania is murdered, Cat is in line to be the next victim. Lex will do anything to keep Cat alive, even if she hates him for it.
Hunted by a powerful stranger intent on drawing her into an evil web, Cat reluctantly accepts Lex's protection and the resurrected desire that comes along with it. Lex faces the fight of his life to keep her safe…and win her back. If they both survive.
Bloodlines and Broomsticks
Liars and tigers and weres, oh my!
Riley O'Driscoll's perfect life cracked thanks to her nasty divorce, and then her world shattered when her beloved grandmother's murder left her with a surprise inheritance of magic and mayhem. Riley has less than a year to learn a lifetime's worth of witchcraft, and after a lifetime of academic excellence Riley has only failed in two subjects: love and magic.
When witch hunters break down Riley's door, a miscast spell sends her through the looking glass and into the path of sexy shapeshifter Jeremiah Galestrom. Jere was an agent of the Wild Hunt—an elite team of sorcerers sworn to protect magiciankind by hunting dangerous shapeshifters–who was infected and left for dead after his team was massacred. His faerie family saved his life, but his rescue left him trapped in Faerie with no hope of escape—until Riley arrives.
They strike a bargain that trades a year of Jeremiah's protection for Riley's help, and Riley finds herself with a new roommate who's a real tiger. With the hunters closing in, can Riley trust her life—and her heart—with a man who is counting the days to his freedom? And while Jeremiah knows he can protect Riley from the hunters, can he protect her from himself, and the wild magic within him?
Book 3.5
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Bewitched, Blooded and Bewildered
It's better to be a bad witch than a dead one.
One thing stands between Catherine Duquesne and happily-ever-after with her soul mate Lex—her unwanted bond with billionaire vampire Zachary Harrison. Married life is hard enough without adding an uninvited vampire to the mix, and being tied to Zach endangers more than just her relationship. If Zach dies, she dies, and with his new seat on the necromancer council being challenged by older, more powerful masters, his life expectancy is dwindling fast.
Vampire politics is only the beginning of Cat's problems. Hunters are stalking magicians, abducting entire families and killing anyone who resists. As the current Titania and Oberon, Cat and Lex must protect their people, but Cat's dark past keeps the victims from seeking their help. Worse, the former Titania's granddaughter wants to oust Cat from the job, believing her vampire ties make her unfit.
To break the bond Cat must broker deals with devils on a journey that will take her to hell and back—literally. Freedom comes at a high price, but Cat will risk anything to save her people and her marriage.
Blood, Toil and Trouble
Why settle for bad, when you can be wicked?
Marie is the wild child of the venerable Duquesne family, born to break her mama's heart. After failing at her guardian duties on two previous assignments, protecting Chicago's magician community is Marie's last chance at redemption. But with hunters preying on the people she's supposed to protect, Marie must join forces with sexy, tempting Dr. Brian Dannaher to combat them—and nothing is more dangerous than falling for a vampire.
Brian lives in the shadow of the other area chroniclers, often overlooked but grudgingly content to live a quiet, unremarkable unlife—until bold, vivacious Marie stirs desires he has long ignored. Chroniclers are meant to investigate and record magician history, and never to interfere with it. They most certainly do not charge into the fray and battle the rising darkness, but Brian was a soldier in life and hates the idea of watching from the sidelines. Brian would do anything to fight at Marie's side, even if that means violating the oaths of his order.
When the hunters strike too close to home, Marie has the chance to prove herself worthy of her family's upstanding guardian reputation. But being a good guardian means ignoring the desire drawing her into Brian's arms, and it's so much more fun being bad…
Fire in the Blood
It's good to be bad…
Patience Roberts is the last summoner standing between magiciankind and certain demon invasion. After banishing two or three demons a day for too long, gods know she'd like nothing better than a little down time with her number one distraction—Faust.
But with vampires, hunters and assassins lined up to take her out, who has the time? Still, she has to admit her resistance to the amorous faerie is wearing thin. Not that she'll ever let on—after all, faeries are notorious for their short romantic attention spans.
Faust, a Shadowspawn faerie, watched as his outcast clan dwindled to nothing. Determined to hold on to the woman he loves, he'll do whatever it takes to protect Patience. And one day build a life with her.
When an old demon enemy punches through the barrier between the worlds, Patience must draw on every ounce of her reputation as a cast-iron bitch to temporarily banish him. To get rid of him for good, she'll have to sacrifice one too many pieces of her soul to leave room for love…
How do you convince a skeptic to reverse your curse?
Natividad Houlihan travels the world searching for stunning historical sites with juicy secrets…and ghosts. Nati doesn't believe in the supernatural, but as a location scout for the Spirit Seekers paranormal investigation show she works for true believers. The team is thrilled at the opportunity to investigate an infamous Spanish haunted mansion—but the price of admission requires Nati to spend two weeks stumbling around in the dark searching for EVPs and light anomalies. Nati's only company is Cris, the estate's surly, sexy caretaker, who intrigues her during the day and stars in her vivid erotic fantasies at night.
Don Cristobal Mendoza loved unwisely and paid the price for his indiscretion, suffering a vicious curse that trapped him within the bounds of his ancestral estate. By day he is the manor's mysterious caretaker, and by night he is the shadow demon who stalks its halls. For centuries he has hidden his condition from the world as he despaired of ever being freed from his curse.
Nati is determined to unearth the secrets Cris has kept for centuries and bring his past to light, but Cris can't resist the chance to indulge his dark desires with a bold, beautiful American skeptic. As the balance of magical power shifts, they must work together to break the curse before Cris's time runs out.
Blood, Book and Candle
A love worth dying for…twice?
Anne Williams is a seer—the rarest, most dangerous kind of magician—gifted with the ability to divine the past, present and future. Anne spent her life hiding her true nature to protect her from those who would abuse her prophetic powers, but with an impending apocalypse on the horizon Anne can't afford to stay in the shadows. Anne steps up the plate to help magiciankind, but someone tries to take her out of the game—permanently.
Wounded and with nowhere else to go, Anne finds shelter with chronicler Simon St. Jerome, but with one touch he plunges Anne into a tangled web of secrets and lies. Simon spent centuries concealing his past while recording the history of magician society, but nothing can hide the truth from a seer. As his soul mate, Anne is determined to discover every secret he's buried—including the tale of how she died once before in a failed attempt to spend eternity with her immortal lover.
As danger closes in around them, Anne and Simon must overcome the scars of their pasts, or risk never seeing their futures.
The Bloody End
This Apocalypse Bites
It sucks to be a vampire at the end of the world.
When an out-of-control spell triggers the zombie apocalypse, Lizzy Addams is left in the ruins of Chicago with only the unhinged commentary—and endless erotic appetites—of her inner demon for company. Her blood supply dwindling, Lizzy must find survivors to feed from, or die trying.
Officer Angela Kinney was on duty when hungry corpses overwhelmed the city. Angie's group of survivors look to her for leadership, but nothing prepared her for a beautiful monster who offers safety in exchange for blood.
Sean MacMillan never expected to see Lizzy again after she rejected his attempts to lure her back to the vampire fold. But with his flock threatened by the horde and his murderous vampire brother, Lizzy is the only one he can trust to keep them safe.
The veil of secrecy shrouding the supernatural world is shattered, and humanity's only hope is to forge an alliance with vampires, werewolves, and things that go bump in the night. Though accepting their aid could be a devil's bargain that puts humans at the bottom of the food chain.


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Tonight, Dr Glenn meets Philip, a patient who averages 90 minutes of sleep per night and can't remember the last time he had a good night's rest.
Traditional treatments for sleeping disorders have failed Philip, so it's up to Dr Glenn to find a way to help a patient who wakes up every day feeling fatigued.
Then, Dr Frances determines whether stress is the cause of painful skin blisters for one fed-up patient.


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You are here: When You Eat, Not Just What, May Impact Health
When You Eat, Not Just What, May Impact Health
time restriction, time-restricted eating, when you eat, blood sugar, diabetes
May 7, 2015 — Watching when you eat, without necessarily changing what or how much, may yield big health benefits, including weight loss.
Although more research is needed, intriguing new findings in people and mice suggest that eating within a strict 8- to 12-hour time frame each day changes metabolism at the genetic level, lowering blood sugar and body weight, even without cutting calories.
Scientists think the changes may be powerful enough to lower the risks for cancer, heart disease, dementia, and diabetes.
The latest study on this phenomenon, called time-restricted eating, looked at the link between meal timing and blood sugar control in more than 2,200 women. The average age of women in the study was 47, and the average body mass index (BMI) was 28, making them overweight.
Poor blood sugar control is a risk factor for diabetes and cancer, among other things. Blood sugar that swings wildly before and after eating indicates that the body isn't very sensitive to insulin, the hormone that signals cells to take in calories from food. That means more insulin has to be released from the pancreas to get the blood sugar into cells. The trouble is that extra insulin doesn't just impact blood sugar. It also promotes the growth of cells — including cancer cells. And over time, the body can't keep up with the demand for more and more insulin. When that happens, blood sugar levels climb dangerously high, leading to diabetes.
The women in the study reported what and when they ate and gave blood samples. Researchers could see how high their blood sugar climbed after meals and how steady their blood sugar had stayed over the previous 2 to 3 months. About half the women reported not eating or drinking anything for at least 12 hours, from 6 p.m. to 6 a.m., for example. The other half fasted for less than 12 hours, eating both early and late.
"What we found, in general, [is] that women who fasted for longer nightly intervals had better blood sugar control than those who didn't fast as long, and that was independent of other eating behaviors such as how many calories women were eating," says study researcher Catherine Marinac, a doctoral candidate in public health at the University of California at San Diego.
"That's similar to what was seen in animal models," she says.
Interesting Animal Studies
Satchin Panda, PhD, an associate professor at the Salk Institute in La Jolla, CA, studies time restriction in mice and flies.
Panda took mice that share the same genes — the mouse version of identical twins. He fed some of them normal mouse chow while others got high-fat food. In each group, some of the mice were allowed to eat whenever they wanted, while others only had access to food for 8 hours each day.
The mice that were fed normal chow, which was low in fat, ate at night, which is when mice are naturally awake. But the mice on the high-fat diet started eating all the time and seemed to lose their natural ability to feel satisfied. They quickly got fat and seemed to be on the road to diabetes.
But the mice that were time restricted didn't become obese. That was remarkable, Panda says, because they were still eating the same number of calories as the others and getting those from a high-fat diet. He has since repeated those results with different kinds of high-fat and high-sugar diets. Each time, it turns out the same way. No matter how poor the diets of the time-restricted mice are, they don't gain weight.
What's more, Panda says, time restriction seems to act like a switch that improves blood sugar control and turns down inflammation, two major drivers of chronic disease. They also have better endurance and better motor coordination.
That makes sense to Mark Mattson, PhD, chief of the laboratory of neurosciences at the National Institute on Aging. He's found evidence that going for long periods of time without eating benefits the brain. In mice bred to have diseases like Alzheimer's and Parkinson's, fasting seems to protect their brain cells. And fasting animals perform better when he's tested their learning and memory in mazes.
Mattson says time restriction mimics the way our ancestors had to eat. Back in the hunter-gatherer days, food was scarce. Early humans had to hunt for hours before scoring their next meal. He says it wouldn't make sense for their brains and bodies to become more and more sluggish the hungrier they got.
"If your brain isn't working well when you haven't eaten for a long time, then you're going to be at a competitive disadvantage in being able to acquire food," he says.
Mattson is getting ready to test another kind of time-restriction, one that alternates 2 days of fasting with 5 days of normal eating, in people with early signs of Alzheimer's disease.
He's so convinced that time restriction works, he's doing it himself.
"I go sometimes 18 hours without eating," he says.
Panda says most of the technicians who work in his lab are trying time restriction, too.
"Everybody that looks at these mice realizes what a profound impact this has on metabolism, body weight, etc.," he says, "and that spurs them to want to do something. So I think almost everybody in the lab is affected by this."
More Research Needed
Sadly, mice are not humans. And it's harder than you might think to tease out the potential benefits of time restriction in people.
For one thing, the mice are eating the same food at every meal. That's not the way people eat.
"In humans, the diet is very different at different times of the day," Panda says.
In 2012, researchers at Brigham Young University in Utah asked 29 healthy, college-aged men not to change their diets, but to time restrict their eating. They stopped eating between 7 p.m. and 6 a.m. every day for 2 weeks, then ate as usual for 2 more.
During the time-restricted period, the men lost about a pound of body weight, but they also ate about 200 fewer calories than they did on the days when they could eat whenever they wanted. So it's not clear whether fewer calories or less late-night snacking whittled their waistlines.
Another flaw of studying time restriction in people has to do with memory. The most practical way to study dietary habits in large groups of people is just to ask them to write down everything they ate and when they ate it over the last day or so. The trouble is that people can be bad at remembering that stuff, especially little things like coffee creamer and mayonnaise on a sandwich.
Even though it's not perfect, sometimes it is the only way to test a question to see if it merits bigger, more expensive studies.
That was the case for Marinac's study. She's quick to point out that her study can only show links. She can't yet say whether an extended overnight fast will lead to better blood sugar control.
But she'd like to find out. She says she's writing a grant now in the hopes of funding a larger study to test time restriction in people. Researchers at Northwestern University in Chicago are also testing time restriction in people to see if it changes body weight and certain indicators of diabetes and cancer risk.
Experts are hopeful these future studies will connect a few more dots. They'd like to know, among other things, if it matters when someone starts their fast each day, and they hope to learn exactly how long a person should go without eating to see benefits.
"We're kind of using 12 hours as a sweet spot. We think a 12- to 14-hour fast would be attainable, feasible," and would improve measures of metabolism, she says.
Satchin Panda, PhD, associate professor, Regulatory Biology Laboratory, Salk Institute, La Jolla, CA.
Mark Mattson, PhD, Chief, Laboratory of Neurosciences; Chief, Cellular and Molecular Neurosciences Section, National Institute on Aging, Bethesda, MD.
Catherine Marinac, doctoral candidate in public health, The University of California at San Diego, San Diego.
Cell Metabolism, June 6, 2012.
Cell Metabolism, Dec. 2, 2014.
Proceedings of the National Academy of Sciences, Nov. 25, 2014.
British Journal of Nutrition, December 2013.
Cancer Epidemiology, Biomarkers & Prevention, April 20, 2015.
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