\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf Towards a Human Proof of Gessel's \\ \vskip .1in Conjecture } \vskip 1cm \large Arvind Ayyer\\ Institut de Physique Th\'eorique\\ IPhT, CEA Saclay, and URA 2306, CNRS\\ 91191 Gif-sur-Yvette Cedex \\ France\\ \href{mailto:arvind.ayyer@cea.fr}{\tt arvind.ayyer@cea.fr} \end{center} \vskip .2 in \begin{abstract} We interpret walks in the first quadrant with steps $\{(1,1),(1,0),(-1,0), (-1,-1)\}$ as a generalization of Dyck words with two sets of letters. Using this language, we give a formal expression for the number of walks using the steps above, beginning and ending at the origin. We give an explicit formula for a restricted class of such words using a correspondence between such words and Dyck paths. This explicit formula is exactly the same as that for the degree of the polynomial satisfied by the square of the area of cyclic $n$-gons conjectured by Robbins, although the connection is a mystery. Finally we remark on another combinatorial problem in which the same formula appears and argue for the existence of a bijection. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \newenvironment{proofof}[1]{\medskip\noindent \textbf{Proof of #1:} }{\hfill $\blacksquare$\par\medskip} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\ba}{\bar{a}} \newcommand{\bb}{\bar{b}} \newcommand{\bone}{\bar{1}} \newcommand{\btwo}{\bar{2}} \newcommand{\bc}{\bar{c}} \newcommand\cS{{\mathcal S}} \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{defn}{Definition} \newtheorem{rem}{Remark} \makeatletter \def\Ddots{\mathinner{\mkern1mu\raise\p@ \vbox{\kern7\p@\hbox{.}}\mkern2mu \raise4\p@\hbox{.}\mkern2mu\raise7\p@\hbox{.}\mkern1mu}} \makeatother \newcommand{\marginnote}[1] {\mbox{}\marginpar{\raggedright\hspace{0pt}{$\blacktriangleright$ #1}}} \section{Introduction} Ever since Gessel conjectured his formula for the number of walks using the steps $$\{(1,1),(1,0),(-1,0), (-1,-1)\}$$ (which we will call Gessel steps) starting and ending at the origin in $2n$ steps constrained to lie in the first quadrant, there has been much interest in studying lattice walks in the quarter plane. There have been conjectures for lattice walks with Gessel steps terminating at other points \cite{pw}, as well as conjectures for the number of walks ending at the origin with other sets of steps, most of which have been proven \cite{bmm}. In a remarkable {\em tour de force}, Gessel's original conjecture has been finally proven using computer algebra techniques \cite{kkz}. Even so, it is important to consider walks on the quarter plane from a human point of view because newer approaches tend to open up interesting mathematical avenues. In this article, we count a considerably restricted number of walks with Gessel steps starting and ending at the origin by rephrasing the problem using words with an alphabet consisting of four letters --- $1$, $2$, $\bone$ and $\btwo$ --- which obey certain conditions. We first show that the restatement of Gessel's conjecture in this context can be interpreted using Dyck paths. This gives a formal solution to the conjecture. Unfortunately, the solution is so formal as to be even computationally intractable!\footnote{Computing the $n$th term in the sequence involves $2n$ sums of binomial coefficients.} We give a closed-form expression for the restricted problem and hope a generalization of this method will give a better understanding of Gessel's conjecture. Admittedly this result is a long way from a solution of the problem, but one hopes that this technique can be generalized to obtain a complete proof of Gessel's conjecture. In Section \ref{sec:alph} we start with the preliminaries by defining the alphabet and stating the main theorem. In Section \ref{sec:dyck}, we make the connection to Dyck paths and give a formal expression for the number of walks beginning and ending at the origin using Gessel steps. Section \ref{sec:pf} contains the proof which involves summations of hypergeometric type. In principle, such sums can be tackled by computer packages, but a certain amount of manipulation is needed before they are summable. Lastly, we comment on related problems in Section \ref{sec:open}. \section{Gessel Alphabet} \label{sec:alph} To rephrase the problem in the notation of formal languages, we need some definitions. \begin{defn} \label{defn:galph} The {\em Gessel alphabet} consists of a set of letters $S = \{1,2,\dots\}$ with an order $<$ ($1<2<\dots$) along with their complements which we denote $\bar{S}=\{\bone,\btwo,\dots\}$. The order on the complement set is irrelevant. \end{defn} \begin{defn} \label{defn:gword} Let $S = [n]$. Denote by $N_\alpha(w)$ the number of occurrences of the letter $\alpha \in S \cup \bar S$ in the word $w$. (For example, $N_2(2 \btwo) = N_{\btwo}(2 \btwo) = 1, N_1(2 \btwo) = 0$.) Then a {\em Gessel word} is a word such that every prefix $w$ of the word satisfies $$\sum_{i=1}^k \left( N_{n+1-i}(w) - N_{\overline{n+1-i}}(w) \right) \geq 0$$ for each $k \in [1,n]$. \end{defn} In words, this means that in each prefix, $n$ has to occur more often than $\bar{n}$, %the sum of $n$ and $n-1$ occurrences must be larger %than or equal to the sum of their barred counterparts and so on. the number of occurrences of $n$ and $n$-1 must be at least equal to the number of occurrences of their barred counterparts and so on. For example, $2 \bone$ is a valid Gessel word but $1 \btwo$ is not. \begin{defn} \label{defn:cgword} A {\em complete Gessel word} is a Gessel word $w$ where $N_i(w) = N_{\bar i}(w)$ for all letters $i \in S$. In other words, the number of times the letter $i$ appears equals the number of times $\bar i$ appears for each $i \in [n]$. \end{defn} As an example, for $n=3$ both $3 \btwo 2 1 \bone \bar 3$ and $1 2 \bone \btwo 1$ are Gessel words but $2 1 \bar 3 2 3 \btwo \bone$ is not because the prefix consisting of three letters fails the criterion in Definition \ref{defn:gword}. Among the other two, the first one is a complete Gessel word. \begin{rem} The number $G^{(d)}(n)$ of $d$ dimensional lattice walks in the first $2^d$-ant with steps \be \begin{split} \{(1,\dots,1),(1,\dots,1,0),&\cdots,(1,0,\dots,0), \\ (-1,\dots,-1),(-1,\dots,-1,0),&\cdots,(-1,0,\dots,0)\}, \end{split} \ee starting at the origin and returning in $2n$ steps is the same as the number of complete Gessel words of length $2n$ in $d$ letters. \end{rem} Starting with the complete Gessel word in $d$ letters, one replaces the letter $k$ by the step $(1,\dots,1,0,\dots,0)$ with $k$ 1's and the letter $\bar{k}$ by $(\bone,\dots,\bone,0,\dots,0)$ with $k$ $\bone$'s to obtain the bijection. The condition of being a Gessel word is the same as that of the walk being in the first $2^d$-ant. None of these sequences seem to be present in \cite{sloane} for dimensions higher than two and it would be interesting to see if they are holonomic. Furthermore, none of these higher dimensional sequences seem to have the property of small factors which is present for the Gessel case. $G^{(2)}(n)$ is conjectured by Gessel to be given by the closed form expression \be \label{gconj} 16^n \frac{(5/6)_n (1/2)_n}{ (2)_n (5/3)_n}, \ee where \be (a)_n = a (a+1) \dots (a+n-1) \ee is the Pochhammer symbol or rising factorial. This is the sequence \seqnum{A135404}. For the remainder of the paper, we implicitly assume $d=2$ and omit the superscripts in defining various constrained Gessel numbers. We express the number of walks of length $2n$ as a number triangle based on the number of times $2$ and $\btwo$ appear increasing from left to right. \be \begin{array}{ccccccccc} & & & &1 & & & & \\ & & &1 & &1 & & & \\ & &2 & &7 & &2 & & \\ &5 & &37 & &38 & &5 & \\ 14 & &177 & &390 & &187 & &14 \end{array} \ee One immediately notices that the leftmost and rightmost entries are the Catalan numbers. This is because the number of complete Gessel words with $N_2(w) = 0$ ($N_1(w)= 0$) in a word $w$ of length $2n$ is in immediate bijection with the number of Dyck paths ending at $(2n,0)$ because $N_1(x) > N_{\bone}(x)$ ($N_2(x) > N_{\btwo}(x)$) for each prefix $x$ of the word $w$. What is more interesting, and the main result of the paper is the next-to-rightmost sequence beginning $1,7,38,187$. Strangely enough, this sequence is already present in the OEIS as \seqnum{A000531}. It turns out to be exactly the one conjectured by Dave Robbins \cite{robb1} to be the degree of the polynomial satisfied by $16K^2$, where $K$ is the area of a cyclic $n$-gon and proved in \cite{fp,mrr}. As far as we know, this result is a coincidence without any satisfactory explanation. For a recent review of the subject, see \cite{pak}. This is also related to Simon Norton's conjecture on the same page in the OEIS. We comment on this in Section \ref{sec:open}. \begin{thm} \label{thm:gws} The number of complete Gessel words $G_1(n)$ in two letters with $n-1$ $2$'s and $\btwo$'s, and one $1$ and $\bone$, is given by \be \label{cn} G_1(n) = \frac{(2n+1)}{2}\binom{2n}{n} - 2^{2n-1}. \ee \end{thm} The proof uses the idea that the number of Gessel words with $n_2$ $2$'s and $\btwo$'s and $n_1$ $1$ and $\bone$'s can be calculated using a bijection with Dyck paths. The answer can be written as a sum of products of expressions counting the number of Dyck paths between two different heights. The summation can be done explicitly when $n_1=1$. \section{Complete Gessel words and Dyck paths} \label{sec:dyck} We consider Dyck paths to be paths using steps $\{(1,1),(1,-1)\}$ starting at the origin, staying on or above the $x$-axis and ending on the $x$-axis. In this section we exhibit a bijection between complete Gessel words (the counting of which is stated by the conjecture of Gessel) and a set of restricted Dyck paths which will be useful in the proof of Theorem~\ref{thm:gws}. \begin{defn} Let $P=(P_1,\dots,P_{m})$ be a nondecreasing list of positive integers and $H=(H_1,\dots,H_{m})$ be a list of nonnegative integers of the same length. We define a {\em $(P,H)$-Dyck path} to be a Dyck path of length $\geq P_m$ which satisfies the constraint that between positions $P_i$ and $P_{i+1}$ (both inclusive), the ordinate of the path is greater than or equal to $H_i$ for $i=1,\dots,m-1$. \end{defn} Notice that this forces the ordinates of the path at positions $P_i$ to be greater than or equal to the height $\max\{H_{i-1},H_i\}$. We now associate lists $P$ and $H$ with every complete Gessel word $w$ in two letters, using the following algorithm. \begin{enumerate} \item Construct the list $S$ of length $2n_1$ of letters $1$ or $\bone$ as they occur in the word. \item From the list $S$, construct the list $T$ by replacing $1$ by $1$ and $\bone$ by $-1$. \item Construct the list $\tilde P$ whose elements are positions of the letter $S_i$ in $w$. Similarly, construct the elements of the list $P$ as $\tilde P_i -i$. \item Finally, each element of the list $H$ is given by \be \label{hts} H_i = \max\Big\{-\sum_{k=1}^i T_k,0\Big\}. \ee \end{enumerate} As an example, consider the Gessel word $w=2 \bone 2 1 \btwo \btwo$, which is the only complete Gessel word of length six with $\bone$ at position two at $1$ at position four. For this word, $S=(\bone,1), T=(-1,1), \tilde P = (2,4), P=(1,2)$ and $H=(1,0)$. There is exactly one such $((1,2),(1,0))$-Dyck path of length four, namely $(\nearrow,\nearrow,\searrow,\searrow)$. Clearly, $P$ and $\tilde P$ determine each other. However, $S$ and $H$ do not. In particular, $H$ is the zero list if $S$ is a valid Dyck word in the letters $1$ and $\bone$. However, if we fix $S$, then we can associate $(P,H)$-Dyck paths and Gessel words, in a bijective way. \begin{lem} \label{lem:bij} Fix the list $S$ of length $2n_1$ in letters $1$ and $\bone$ beforehand. Complete Gessel words of length $2(n_1+n_2)$ in two letters with the positions of the letters in $S$ given by the list $\tilde P$ are in bijection with $(P,H)$-Dyck paths of length $2n_2$ where the pairs of lists $(P,H)$ are constructed by the algorithm described above. \end{lem} \begin{proof} Starting with the complete Gessel word, one replaces each occurrence of the letter $2$ by the step $(1,1)$ and that of $\btwo$ by the step $(1,-1)$. The constraint defining the $(P,H)$-Dyck path is simply another way of expressing the inequality in Definition~\ref{defn:gword}. \end{proof} One could generalize this bijection to include paths not ending on the $x$-axis and Gessel words which are not complete, but this is sufficient for our purposes. One of the main tools in the proof of Theorem~\ref{thm:gws} is an expression for the number of Dyck paths between two different heights, which can be readily obtained from the reflection principle \cite{co}. \begin{lem} \label{lem:dyck} The number of Dyck paths $a_{i,j}(k)$ that stay above the $x$-axis starting at the position $(0,i)$ and end at position $(k,j)$ is given by \be \label{aij} a_{i,j}(k) = \begin{cases} \displaystyle \binom{k}{(k+i-j)/2} -\binom{k}{(k+i+j)/2+1}, & \\ & \hspace{-2cm} \text{if } (k+i+j) \equiv 0 \ ({\rm mod}\ 2); \\ 0, & \hspace{-2cm} \text{if } (k+i+j) \equiv 1 \ ({\rm mod}\ 2). \end{cases} \ee \end{lem} We now use the bijection in Lemma~\ref{lem:bij} and the formula in Lemma~\ref{lem:dyck} to write an expression for the number of complete Gessel words of length $2n$ for fixed positions of $1,\bone$. \begin{lem} \label{lem:gwn2} Let us fix the positions of $n_1$ $1,\bone$ by the lists $S,\tilde P$. Calculate the lists $T$ and $H$ by the algorithm above and let $G_{n_1}(\tilde P,S;2n)$ denote the number of such complete Gessel words. Then \be \label{cpt} \begin{split} &G_{n_1}( \tilde P,S;2n) = \displaystyle \sum_{k_1=H_1}^{\tilde P_1-1} \sum_{k_2=H_2}^{\tilde P_2-1} \dots \sum_{k_i=H_i}^{\tilde P_i-1} \dots \sum_{k_{2n_1}=H_{2n_1}}^{\tilde P_{2n_1}-1} \\ &\displaystyle a_{0,k_1}(\tilde P_1-1) \; a_{k_{2n_1},0}(2n-\tilde P_{2n_1}) \prod_{i=2}^{2n_1} a_{k_{i-1}-H_{i-1},k_i-H_{i-1}}(\tilde P_i-\tilde P_{i-1}-1). \end{split} \ee \end{lem} \begin{proof} The proof is straightforward, using the bijection of Lemma~\ref{lem:bij} to rewrite each Gessel word with the positions of $1,\bone$ given by the lists $S, \tilde P$ as a Dyck path with heights at the points $P_i$ (given by $k_i$) being not less than $H_i$ and then the reflection principle in Lemma~\ref{lem:dyck} to count the number of paths between position $P_{i-1}$ and $P_i$ for each $i$. \end{proof} \begin{cor} \label{cor:cat} For a given configuration of $1,\bone$, replace each $+1$ in $T$ by an upward Dyck step and each $-1$ by a downward Dyck step. If the whole of $T$ forms a legal Dyck path, then $G_{n_1}( \tilde P,S;2n) = C_{n-n_1}$, the $(n-n_1)$th Catalan number independent of the list $P$. \end{cor} \begin{proof} Whenever the above condition is satisfied, $H_i=0$ for all $i$, which means we simply count the number of Dyck paths of length $2n_1$ in \eqref{cpt} by definition. \end{proof} Now we obtain a formula for the number of complete Gessel words with $n_1$ $1,\bone$'s using Lemma~\ref{lem:gwn2} and writing down all possibilities for $\tilde P$ and $S$. The number of ways of writing all possible $\tilde P$'s is simply $\binom{2n}{2n_1}$ because one has to choose $2n_1$ positions out of $2n$ positions. For each $\tilde P$, one has to choose $n_1$ positions for $1$ and $\bone$ each and therefore the number of such ways is $\binom{2n_1}{n_1}$. Let us form the set \be \cS = \left\{(\tilde P,S) \Biggl\vert \begin{array}{l} S \text{ is a list of $n_1$ $1$'s and $n_1$ $\bone$'s.} \\ \tilde P \text{ is an increasing list of} \\ \; \text{ $2n_1$ positions between $1$ and $2n$,} \end{array} \right\}, \ee whose cardinality is \be \binom{2n}{2n_1} \binom{2n_1}{n_1} = \frac{(2n)!}{(n_1)!^2 (2n-2n_1)!}. \ee Therefore the number of complete Gessel words with exactly $n_1$ $1,\bone$'s is given by \be \label{cgess} G_{n_1}(n) = \sum_{(\tilde P,S) \in \cS} G_{n_1}(\tilde P,S;2n), \ee and the number of complete Gessel words in $2n$ letters is \be G(n) = \sum_{n_1=0}^n G_{n_1}(n). \ee Showing that $G(n)$ is equal to the expression \eqref{gconj} would be the ultimate (and possibly hopeless) aim of this line of approach. We now have all the ingredients necessary to prove Theorem~\ref{thm:gws} which corresponds to the special case $n_1=1$. Before we go on to the proof, however, we make some observations about complete Gessel words with exactly one $1$ and $\bone$. Let $d_{i,j}$ be the number of possibilities where a $1$ or a $\bone$ is at position $i$ and its counterpart at position $j$. Then we draw the following triangle for a specific $n$, \be \begin{array}{ccccccccc} & & & &d_{1,2n} & & & & \\ & & &d_{1,2n-1} & &d_{2,2n} & & & \\ & &d_{1,2n-2} & &d_{2,2n-1} & &d_{3,2n} & & \\ &\Ddots& & & & & &\ddots & \\ d_{1,2} & &d_{2,3} & &\cdots & &d_{2n-2,2n-1} & &d_{2n-1,2n}. \end{array} \ee For $n=3$, the triangle is \be \begin{array}{ccccccccc} & & & &2 & & & & \\ & & &2 & &2 & & & \\ & &2 & &3 & &2 & & \\ &2 & &3 & &3 & &2 & \\ 2 & &4 & &3 & &4 & &2, \end{array} \ee and for $n=4$, the triangle is \be \begin{array}{ccccccccccccc} & & & & & &5 & & & & & &\\ & & & & &5 & &5 & & & & &\\ & & & &5 & &7 & &5 & & & &\\ & & &5 & &7 & &7 & &5 & & &\\ & &5 & &8 & &7 & &8 & &5 & & \\ &5 & &8 & &8 & &8 & &8 & &5 &\\ 5 & &10 & &8 & &10 & &8 & &10 & &5. \end{array} \ee It is clear that if one stacks these triangles on top of one another, one gets a kind of Pascal's pyramid, where each layer $n$ fits in the vacancies of the layer $n-1$ above it. \begin{remark} We note some properties of these triangles. \begin{enumerate} \item The sum of the entries in the triangle are precisely what we claim are given by \eqref{cn}. \item One notices immediately that the extremal columns are Catalan numbers $C_{n-1}$. This follows immediately from Corollary~\ref{cor:cat} and the fact that a Gessel word cannot begin with $\bone$ or $\btwo$, or end with a $1$ or $2$. The entries at even positions in the last row are $2 C_{n-1}$. This is clear since if $\bone$ and $1$ are at positions $2i$ and $2i+1$ respectively, then any legal Dyck word in the letters $2$ and $\btwo$ is a Gessel word after this insertion. %We calculate the sum of these entries as $S_1$ in Section~\ref{sec:pf}. \item Every number in the interior of the triangle occurs $4k$ times for $k$ a positive integer. Furthermore, they are organized as rhombus-shaped blocks of size four. %There are diamond-shaped blocks of numbers %in the interior of these triangles. This turns out to be true for all $n$. We will need this fact in the proof later and we state it as Lemma~\ref{lem:diamond}. %We calculate the sum of these diamonds as $4(S_2 - S_3)$ in %Section~\ref{sec:pf}. \end{enumerate} \end{remark} \section{Proof of Theorem~\ref{thm:gws}} \label{sec:pf} %Before we go on to the proof, we remark on the pattern of the number %of Gessel words with a single $1,\bone$ by illustrating %cross-sections of three-dimensional number triangles, or number pyramids. %\proofof{Theorem~\ref{thm:gws}} One simply has to analyze all possibilities of occurrences of $1$ and $\bone$ case by case. Suppose $1$ occurs at position $i$ and $\bone$ occurs at position $j$ in a word of length $2n$ and $i