Towards a Human Proof of Gessel’s Conjecture

Arvind Ayyer Institut de Physique Théorique IPhT, CEA Saclay, and URA 2306, CNRS 91191 Gif-sur-Yvette Cedex France

                  arvind.ayyer@cea.fr

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.

1 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 [1], as well as conjectures for the number of walks ending at the origin with other sets of steps, most of which have been proven [2]. In a remarkable tour de force, Gessel’s original conjecture has been finally proven using computer algebra techniques [3]. 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, ¯1 and ¯2 — 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!^{1} 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 2 we start with the preliminaries by defining the alphabet and stating the main theorem. In Section 3, 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 4 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 5.

2 Gessel Alphabet

To rephrase the problem in the notation of formal languages, we need some definitions.

Definition 1. The Gessel alphabet consists of a set of letters S = {1, 2, … } with an order < (1 < 2 < …) along with their complements which we denote S̄ = {¯1, ¯2, … }. The order on the complement set is irrelevant.

Definition 2. Let S = [n]. Denote by N_{α}(w) the number of occurrences of the letter α ∈ S∪S̄ in the word w. (For example, N_{2}(2¯2) = N¯2(2¯2) = 1, N_{1}(2¯2) = 0.) Then a Gessel word is a word such that every prefix w of the word satisfies

∑_{i=1}^{k} (N_{n+1−i}(w) − N\overline{n+1−i}(w)) ≥ 0

for each k ∈ [1, n].

In words, this means that in each prefix, n has to occur more often than n̄, 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¯1 is a valid Gessel word but 1¯2 is not.

Definition 3. A complete Gessel word is a Gessel word w where N_{i}(w) = Nī(w) for all letters i ∈ S. In other words, the number of times the letter i appears equals the number of times ī appears for each i ∈ [n].

1Computing the nth term in the sequence involves 2n sums of binomial coefficients.

As an example, for n = 3 both 3¯221¯1¯3 and 12¯1¯21 are Gessel words but 21¯323¯2¯1 is not because the prefix consisting of three letters fails the criterion in Definition 2. Among the other two, the first one is a complete Gessel word.

Remark 1. The number G^{(d)}(n) of d dimensional lattice walks in the first 2^{d}-ant with steps

\begin{matrix}\{(1, …, 1), (1, …, 1, 0), & ⋯, (1, 0, …, 0), \\ (−1, …, −1), (−1, …, −1, 0), & ⋯, (−1, 0, …, 0)\}, \end{matrix}  (1)

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.

Starting with the complete Gessel word in d letters, one replaces the letter k by the step (1, …, 1, 0, …, 0) with k 1’s and the letter k̄ by (¯1, …, ¯1, 0, …, 0) with k ¯1’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 [4] 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

16^{n} \frac{(5/6)_{n}(1/2)_{n}}{(2)_{n}(5/3)_{n}} ,  (2)

where

(a)_{n} = a(a + 1) … (a + n − 1)  (3)

is the Pochhammer symbol or rising factorial. This is the sequence \underline{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 ¯2 appear increasing from left to right.

1 1  1 2  7  2 5  37  38  5 14  177  390  187  14  (4)

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¯1(x) (N_{2}(x) > N¯2(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 \underline{A000531}. It turns out to be exactly the one conjectured by Dave Robbins [5] to be the degree of the polynomial satisfied by 16K^{2}, where K is the area of a cyclic n-gon and proved in [6, 7]. As far as we know, this result is a coincidence without any satisfactory explanation. For a recent review of the subject, see [8]. This is also related to Simon Norton’s conjecture on the same page in the OEIS. We comment on this in Section 5.

Theorem 1. The number of complete Gessel words G_{1}(n) in two letters with n − 1 2’s and ¯2’s, and one 1 and ¯1, is given by

G_{1}(n) = \frac{(2n + 1)}{2}\binom{2n}{n} − 2^{2n−1}.  (5)

The proof uses the idea that the number of Gessel words with n_{2} 2’s and ¯2’s and n_{1} 1 and ¯1’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.

3 Complete Gessel words and Dyck paths

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 1.

Definition 4. Let P = (P_{1}, …, P_{m}) be a nondecreasing list of positive integers and H = (H_{1}, …, H_{m}) be a list of nonnegative integers of the same length. We define a (P, H)-Dyck path to be a Dyck path of length ≥ 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, …, m − 1.

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.

1. Construct the list S of length 2n_{1} of letters 1 or ¯1 as they occur in the word.

2. From the list S, construct the list T by replacing 1 by 1 and ¯1 by −1.

3. Construct the list P̃ whose elements are positions of the letter S_{i} in w. Similarly, construct the elements of the list P as P̃_{i} − i.

4. Finally, each element of the list H is given by

H_{i} = max { − ∑_{k=1}^{i} T_{k}, 0}.  (6)

As an example, consider the Gessel word w = 2¯121¯2¯2, which is the only complete Gessel word of length six with ¯1 at position two at 1 at position four. For this word, S = (¯1, 1), T = (−1, 1), P̃ = (2, 4), P = (1, 2) and H = (1, 0). There is exactly one such ((1, 2), (1, 0))-Dyck path of length four, namely (↗, ↗, ↘, ↘).

Clearly, P and P̃ 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 ¯1. However, if we fix S, then we can associate (P, H)-Dyck paths and Gessel words, in a bijective way.

Lemma 2. Fix the list S of length 2n_{1} in letters 1 and ¯1 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 P̃ 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.

Proof. Starting with the complete Gessel word, one replaces each occurrence of the letter 2 by the step (1, 1) and that of ¯2 by the step (1, −1). The constraint defining the (P, H)-Dyck path is simply another way of expressing the inequality in Definition 2.

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 1 is an expression for the number of Dyck paths between two different heights, which can be readily obtained from the reflection principle [9].

Lemma 3. 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

a_{i,j}(k) = \begin{cases}\binom{k}{(k + i − j)/2} − \binom{k}{(k + i + j)/2 + 1}, \\ if (k + i + j) ≡ 0 (mod 2); \\ 0, & if (k + i + j) ≡ 1 (mod 2). \end{cases}  (7)

We now use the bijection in Lemma 2 and the formula in Lemma 3 to write an expression for the number of complete Gessel words of length 2n for fixed positions of 1, ¯1.

Lemma 4. Let us fix the positions of n_{1} 1, ¯1 by the lists S, P̃. Calculate the lists T and H by the algorithm above and let G_{n_{1}}(P̃, S; 2n) denote the number of such complete Gessel words. Then

\begin{matrix}G_{n_{1}}(P̃, S; 2n) = ∑_{k_{1}=H_{1}}^{P̃_{1}−1} ∑_{k_{2}=H_{2}}^{P̃_{2}−1} ⋯ ∑_{k_{i}=H_{i}}^{P̃_{i}−1} ⋯ ∑_{k_{2n_{1}}=H_{2n_{1}}}^{P̃_{2n_{1}}−1} \\ a_{0,k_{1}}(P̃_{1} − 1) a_{k_{2n_{1}},0}(2n − P̃_{2n_{1}}) ∏_{i=2}^{2n_{1}} a_{k_{i−1}−H_{i−1},k_{i}−H_{i−1}}(P̃_{i} − P̃_{i−1} − 1). \end{matrix}  (8)

Proof. The proof is straightforward, using the bijection of Lemma 2 to rewrite each Gessel word with the positions of 1, ¯1 given by the lists S, P̃ 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 3 to count the number of paths between position P_{i−1} and P_{i} for each i.

Corollary 5. For a given configuration of 1, ¯1, 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}}(P̃, S; 2n) = C_{n−n_{1}}, the (n − n_{1})th Catalan number independent of the list P.

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 (8) by definition.

Now we obtain a formula for the number of complete Gessel words with n_{1} 1, ¯1’s using Lemma 4 and writing down all possibilities for P̃ and S. The number of ways of writing all possible P̃’s is simply \binom{2n}{2n_{1}} because one has to choose 2n_{1} positions out of 2n positions. For each P̃, one has to choose n_{1} positions for 1 and ¯1 each and therefore the number of such ways is \binom{2n_{1}}{n_{1}}.

Let us form the set

S = \{(P̃, S)| \begin{matrix}S is a list of n_{1} 1’s and n_{1} ¯1’s. \\ P̃ & is an increasing list of \\ 2n_{1} positions between 1 and 2n, \end{matrix} \} ,  (9)

whose cardinality is

\binom{2n}{2n_{1}}\binom{2n_{1}}{n_{1}} = \frac{(2n)!}{(n_{1})!^{2}(2n − 2n_{1})!}.  (10)

Therefore the number of complete Gessel words with exactly n_{1} 1, ¯1’s is given by

G_{n_{1}}(n) = ∑_{(P̃,S)∈S} G_{n_{1}}(P̃, S; 2n),  (11)

and the number of complete Gessel words in 2n letters is

G(n) = ∑_{n_{1}=0}^{n} G_{n_{1}}(n).  (12)

Showing that G(n) is equal to the expression (2) would be the ultimate (and possibly hopeless) aim of this line of approach.

We now have all the ingredients necessary to prove Theorem 1 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 ¯1. Let d_{i,j} be the number of possibilities where a 1 or a ¯1 is at position i and its counterpart at position j. Then we draw the following triangle for a specific n,

\begin{aligned}& & & & d_{1,2n} \\ & & & d_{1,2n−1} & & d_{2,2n} \\ & & d_{1,2n−2} & & d_{2,2n−1} & & d_{3,2n} \\ & ... & & & & & & ... \\ d_{1,2} & & d_{2,3} & & ⋯ & & d_{2n−2,2n−1} & & d_{2n−1,2n}. \end{aligned}  (13)

For n = 3, the triangle is

\begin{aligned}& & & & 2 \\ & & & 2 & & 2 \\ & & 2 & & 3 & & 2 \\ & 2 & & 3 & & 3 & & 2 \\ 2 & & 4 & & 3 & & 4 & & 2, \end{aligned}  (14)

and for n = 4, the triangle is

\begin{aligned}& & & & & & 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{aligned}  (15)

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.

Remark 1. We note some properties of these triangles.

1. The sum of the entries in the triangle are precisely what we claim are given by (5).

2. One notices immediately that the extremal columns are Catalan numbers C_{n−1}. This follows immediately from Corollary 5 and the fact that a Gessel word cannot begin with ¯1 or ¯2, or end with a 1 or 2. The entries at even positions in the last row are 2C_{n−1}. This is clear since if ¯1 and 1 are at positions 2i and 2i + 1 respectively, then any legal Dyck word in the letters 2 and ¯2 is a Gessel word after this insertion.

3. 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. This turns out to be true for all n. We will need this fact in the proof later and we state it as Lemma 6.

4 Proof of Theorem 1

One simply has to analyze all possibilities of occurrences of 1 and ¯1 case by case. Suppose 1 occurs at position i and ¯1 occurs at position j in a word of length 2n and i < j. Then by Corollary 5, the number of such Gessel words is C_{n−1}. The number of possibilities of i, j such that 1 ≤ i < j ≤ 2n is n(2n − 1). Therefore, the number of Gessel words where the 1 occurs before the ¯1 is

(2n − 1)\binom{2n − 2}{n − 1}.  (16)

We now use Lemma 4 to count the number of words where ¯1 occurs at site i before 1 at site j,

\begin{matrix}G_{1}([i, j], & [¯1, 1]; 2n) \\ = & ∑_{k_{1}=1}^{i−1} ∑_{k_{2}=0}^{j−1} a_{0,k_{1}}(i − 1) a_{k_{1}−1,k_{2}−1}(j − i − 1) a_{k_{2},0}(2n − j), \end{matrix}  (17)

which, using (7) gives

\begin{matrix}G_{1}([i, j], & [¯1, 1]; 2n) = ∑_{k_{1}=1}^{i−1} ∑_{k_{2}=0}^{j−1} C_{(i−1−k_{1})/2}^{(i−1+k_{1})/2} C_{(2n−j−k_{2})/2}^{(2n−j+k_{2})/2} \\ [\binom{j − i − 1}{(j − i − 1 + k_{1} − k_{2})/2} − \binom{j − i − 1}{(j − i − 1 + k_{1} + k_{2})/2}] \end{matrix}  (18)

where C_{n}^{m} is the Catalan triangle number given by \frac{(m−n+1)}{(m+1)} \binom{m+n}{n} for 0 ≤ n ≤ m and is zero when either m or n are not integers.

We now use the following result to simplify calculations. The proof of this assertion is easily verified by expanding (18) and noting that the answer is the same when i is replaced by either 2i or 2i + 1 and similarly for j.

Lemma 6. For 1 ≤ i < j ≤ n − 1,

\begin{matrix}G_{1}([2i, 2j], [¯1, 1]; 2n) = G_{1}([2i, 2j + 1], [¯1, 1]; 2n) \\ = G_{1}([2i + 1, 2j], [¯1, 1]; 2n) = G_{1}([2i + 1, 2j + 1], [¯1, 1]; 2n). \end{matrix}  (19)

Then the total number of Gessel words with a ¯1 preceding a 1 is given by

\begin{matrix}∑_{i=1}^{2n−1} ∑_{j=i+1}^{2n} G_{1}([i, j], [¯1, 1]; 2n) = & 4 ∑_{i=1}^{n−2} ∑_{j=i+1}^{n−1} G_{1}([2i, 2j], [¯1, 1]; 2n) \\ + ∑_{i=1}^{n−1} G_{1}([2i, 2i + 1], [¯1, 1]; 2n) \\ = 4(S_{2} − S_{3}) + S_{1}. \end{matrix}  (20)

where we have split the sum in three parts, with

S_{1} = ∑_{i=1}^{n−1} G_{1}([2i, 2i + 1], [¯1, 1]; 2n).  (21)

The remainder in (20) we split using (18), and using the variables r = (2i − k_{1} − 1)/2, s = (2n − 2j − k_{2})/2,

∑_{i=1}^{n−2} ∑_{j=i+1}^{n−1} G_{1}([2i, 2j], [¯1, 1]; 2n) = S_{2} − S_{3}  (22)

where we have extended the upper and lower index of the inner sum, instead of from n − 2j to n − j, to range from 0 to n − j − 1 for later convenience, as

\begin{aligned}S_{2} & = ∑_{i=1}^{n−2} ∑_{j=i+1}^{n−1} ∑_{r=0}^{i−1} ∑_{s=0}^{n−j−1} C_{r}^{2i−r−1}C_{s}^{2n−2j−s}\binom{2j − 2i − 1}{2j + s − n − r − 1}, \\ S_{3} & = ∑_{i=1}^{n−2} ∑_{j=i+1}^{n−1} ∑_{r=0}^{i−1} ∑_{s=0}^{n−j−1} C_{r}^{2i−r−1}C_{s}^{2n−2j−s}\binom{2j − 2i − 1}{n − s − r − 1}. \end{aligned}  (23)

This change of summation index does not matter because S_{2} − S_{3} is zero for s = n − j and s between 0 and n − 2j − 1 inclusive. We now estimate these three sums in turn.

4.1 The sum S_{2}

Replacing r → i − 1 − r and s → n − j − s − 1, substituting k = j − i and rearranging the variables, we get

S_{2} = ∑_{r=0}^{n−3} ∑_{k=1}^{n−r−2} ∑_{i=r+1}^{n−1−k} ∑_{s=0}^{n−i−k−1} C_{i−r−1}^{i+r}C_{n−k−i−s−1}^{n−k−i+s+1}\binom{2k − 1}{k − s + r − 1}.  (24)

Now replace k → k − 1, i → i − r − 1 to get

S_{2} = ∑_{r=0}^{n−3} ∑_{k=0}^{n−r−3} ∑_{i=0}^{n−r−k−3} ∑_{s=0}^{n−r−k−i−3} C_{i}^{i+2r+1}C_{n−k−i−r−s−3}^{n−k−i−r+s−1}\binom{2k + 1}{k − s + r}.  (25)

We now replace the variable r by u = k+r. Notice that the binomial coefficient is independent of i for which we use the identity

∑_{i=0}^{C} C_{i}^{i+A}C_{C−i}^{B−i} = C_{C}^{A+B+1},  (26)

which means we are left with

\begin{aligned}S_{2} & = ∑_{u=0}^{n−3} ∑_{k=0}^{u} ∑_{s=0}^{n−u−3} C_{n−u−s−3}^{u+n+s−2k+1}\binom{2k + 1}{u − s} \\ & = ∑_{k=0}^{n−3} ∑_{u=0}^{n−k−3} ∑_{s=0}^{n−u−k−3} C_{n−u−s−k−3}^{u+n+s−k+1}\binom{2k + 1}{u + 1 − s}. \end{aligned}  (27)

Let A = n − k − 3, v = u − s and v′ = s − u. Then one easily verifies that

∑_{u=0}^{A} ∑_{s=0}^{A−u} = ∑_{v=0}^{A} ∑_{\begin{matrix}u=v \\ (s=u−v) \end{matrix}}^{A/2+v/2} + ∑_{v′=0}^{A} ∑_{\begin{matrix}s=v′ \\ (u=s−v′) \end{matrix}}^{A/2+v′/2} − ∑_{\begin{matrix}s=0 \\ (u=s) \end{matrix}}^{A/2}  (28)

The binomial coefficient is independent of u in the first sum and of s in the remaining two and hence the innermost sum can be done using the identity

∑_{s=v}^{n/2} C_{n−2s}^{B+2s} = \binom{B + n − 1}{n − 2v}.  (29)

This reduces the sum (after a change of variables) to

\begin{matrix}S_{2} = ∑_{k=0}^{n−3} ∑_{v=0}^{k} & \binom{2k + 3}{k − v}\binom{2n − 2k − 4}{n − k − v − 2} \\ − ∑_{k=0}^{n−3} \binom{2k + 1}{k + 1}\binom{2n − 2k − 3}{n − k − 3} \end{matrix}  (30)

These sums are handled as special cases of the Chu-Vandermonde identity to yield

S_{2} = \frac{n + 2}{4}\binom{2n}{n} − 3 · 2^{2n−3},  (31)

which appears as sequence \underline{A045720} because it is the threefold convolution of the sequence a_{n} = \binom{2n+1}{n+1}.

4.2 The sum S_{3}

S_{3} = ∑_{i=1}^{n−2} ∑_{j=i+1}^{n−1} ∑_{r=0}^{i−1} ∑_{s=0}^{n−j−1} C_{r}^{2i−r−1}C_{s}^{2n−2j−s}\binom{2j − 2i − 1}{n − s − r − 1},  (32)

which after replacing i → i − r − 1 and subsequently j → j − r − i − 2 and rearranging becomes

S_{3} = ∑_{j=0}^{n−3} ∑_{i=0}^{n−j−3} ∑_{s=0}^{n−j−i−3} ∑_{r=0}^{n−j−i−s−3} C_{r}^{2i+r+1}C_{n−j−i−r−s−3}^{n−j−r−i+s−1}\binom{2j + 1}{s + j + i + 2}.  (33)

We now use (26) to do the summation with respect to r and get

S_{3} = ∑_{j=0}^{n−3} ∑_{i=0}^{n−j−3} ∑_{s=0}^{n−j−i−3} C_{n−j−i−s−3}^{n−j+i+s+1}\binom{2j + 1}{s + j + i + 2}.  (34)

Now, replacing s by k = i + s, we get

\begin{aligned}S_{3} & = ∑_{j=0}^{n−3} ∑_{k=0}^{n−j−3} ∑_{i=0}^{k} C_{n−j−k−3}^{n−j+k+1}\binom{2j + 1}{k + j + 2} \\ & = ∑_{j=0}^{n−3} ∑_{k=0}^{n−j−3}(k + 1)C_{n−j−k−3}^{n−j+k+1}\binom{2j + 1}{k + j + 2} \\ & = ∑_{k=0}^{n−3} ∑_{j=0}^{n−k−3}(k + 1)C_{n−j−k−3}^{n−j+k+1}\binom{2j + 1}{k + j + 2}. \end{aligned}  (35)

We now use the identity

∑_{j=C}^{B} C_{B−j}^{A−j}\binom{2j + 1}{j − C} = \binom{A + B + 2}{B − C},  (36)

for the inner sum to get

\begin{aligned}S_{3} & = ∑_{k=0}^{n−3}(k + 1)\binom{2n}{n − 4 − 2k}, \\ & = ∑_{k=0}^{(n−4)/2}(k + 1)\binom{2n}{n − 4 − 2k}, \\ & = \frac{1}{2} ∑_{k=0}^{(n−4)/2}(2k + 4)\binom{2n}{n − 4 − 2k} − ∑_{k=0}^{(n−4)/2} \binom{2n}{n − 4 − 2k} \\ & = \frac{n}{2}\binom{2n − 2}{n − 4} − ∑_{k=0}^{(n−4)/2} \binom{2n}{n − 4 − 2k} \\ & = \frac{n}{2}\binom{2n − 2}{n − 4} − 2^{2n−2} + \frac{(2n)!(3n^{2} + n + 2)}{2n!(n + 2)!}. \end{aligned}  (37)

4.3 The sum S_{1}

S_{1} = ∑_{i=1}^{n−1} ∑_{r=1}^{i} ∑_{s=1}^{i} C_{i−r}^{i−1+r}C_{n−i−s}^{n−i+s−1} [\binom{0}{r − s} − \binom{0}{r + s − 1}] .  (38)

The first term forces r = s and the second term is identically zero because r + s ≥ 2. This means we are left with

\begin{aligned}S_{1} & = ∑_{i=1}^{n−1} ∑_{r=1}^{i} C_{i−r}^{i−1+r}C_{n−i−r}^{n−i+r−1} \\ & = (n − 1)C_{n−1}. \end{aligned}  (39)

Thus the total number of Gessel words where a ¯1 occurs before a 1 defined in (20) is given, using (31),(37) and (39), by

4(S_{2} − S_{3}) + S_{1} = \frac{(n^{3} + 4n^{2} + 5n + 2)(2n)!}{2n!(n + 2)!} − 2^{(2n−1)},  (40)

and therefore, the total number of complete Gessel words is

G_{1}(n) = 4(S_{2} − S_{3}) + S_{1} + (2n − 1)\binom{2n − 2}{n − 1} = \frac{(2n + 1)}{2}\binom{2n}{n} − 2^{2n−1},  (41)

which is exactly the same expression as (5). ■

5 Remarks

This section is intended to be speculative in nature and consequently, the statements are unproven as far as we know, though not necessarily very deep. In 2001, Simon Norton made the following conjecture in \underline{A000531}.

A conjectured definition: Let 0 < a_{1} < a_{2} < ⋯ < a_{2n} < 1. Then how many ways are there in which one can add or subtract all the a_{i} to get an odd number. For example, take n = 2. Then the options are a_{1} + a_{2} + a_{3} + a_{4} = 1 or 3; one can change the sign of any of the a_{i}’s and get 1; or −a_{1} − a_{2} + a_{3} + a_{4} = 1. That’s a total of 7, which is the 2nd number of this

sequence.

We want to connect this conjecture to Theorem 1. Before that, we need some preliminaries. One can represent every equation of the form ±a_{1} ⋯ ± a_{2n} = 1 as a 2n-tuple of +, − symbols. Let us replace every − by a 0 and every + by a 1. Then, one can represent all possible ways of ordering the +’s and −’s by binary words of length 2n.

To state the conjecture precisely, we need a different notion of subword than that existing in the literature, which is probably best defined recursively.

Definition 5. Suppose the word w is taken from the set S which is ordered. We say that the motif μ of length k occurs in the word w of length n if there is an increasing sequence of length k of elements ∈ [n] such that the subword w_{1} of w formed by taking the letters of w from this sequence has the same order relation among its letters as do the letters of μ.

We then say that the number of occurrences of μ is zero if no such sequence exists, and is one more than the number of motifs of the new word formed by removing the letters of the subword w_{1} from w.

Let w be such a binary word. Then define n_{1}(w) to be number of occurrences of the symbol 1 in w, which trivially is the number of occurrences of the motif 1 in w. Also define n_{10}(w) to be the number of occurrences of the motif 10 in w. For example, n_{10}(1110) = 1 + n_{10}(11) = 1 and n_{10}(0110000) = 1 + n_{10}(01000) = 2 + n_{10}(000) = 2. We now form the multiset S, where each word w occurs

m(w) = ⌊\frac{n_{1}(w) − n_{10}(w)}{2}⌋_{+}  (42)

times. Note that if m(w) is zero or negative, it never appears. Then, it seems that the cardinality of S is the same as the conjecture in the sequence. Moreover there is a bijection from the ± notation to the binary notation. This means that the number of times a binary word appears in S seems to be the same as the number of positive odd integers in the right hand side of the equation corresponding to the same binary word which admit solutions. We give a concrete example in Table 1.

The connection between the two problems is as follows. For each fixed number n_{1} of + signs from 2 to 2n, count only those sums in which all possible \binom{2n}{n_{1}} combinations give rise to that sum and add them up. This number is precisely the same as the number of Gessel words stated in Theorem 1 in which the 1 precedes the ¯1. The formula for the number of

±  Odd integer Binary n_{1}(w) n_{10}(w) m(w)
word  sums  word
+ + ++  1,3  1111  4  0  2
+ + +−  1  1110  3  1  1
+ + −+  1  1101  3  1  1
+ − ++  1  1011  3  1  1
− + ++  1  0111  3  0  1
− − ++  1  0011  2  0  1

Table 1: All allowed possibilities for n = 2.

such Gessel words is given by (16). If one considers the set of only those ± words for fixed n_{1} such that a number strictly smaller than \binom{2n}{n_{1}} contribute, then this set is equinumerous with the Gessel words stated above in which the ¯1 precedes the 1 and is given by (40). This leads us to conjecture the presence of a bijection between the multiset S and the number of complete Gessel words with exactly one 1 and ¯1.

Number of + and Sum=1 Sum=3 Sum=5 Sum=7

and − signs

8+  1  1  1  1
7+, 1−  8  8  8
6+, 2−  28  28  1
5+, 3−  56  8
4+, 4−  28  1
3+, 5−  8
2+, 6−  1

Table 2: The number of words for a fixed number of + and − signs and fixed sum in the case n = 4.

For example, there are 6 complete Gessel words for n = 2 where the 1 precedes the ¯1. From Table 1, one sees that all possible terms contribute when we have either 4+ or 3+, 1− signs. There are two possibilities for the former (when the sums are 1 and 3) and four for the latter (when the sum is 1). Similarly, there is only one complete Gessel word for n = 2 where ¯1 precedes 1, which is given by 2¯121¯2¯2 and for 2+, 2− signs, there are 6 possible words, but only one contributes.

For any fixed n_{1} and any fixed odd integer sum, the number of words which allow this seem to be of the form \binom{2n}{k} where k varies from 0 to n − 1. We illustrate this via another concrete example in Table 2. Notice that the only integers appearing in the table are the binomial coefficients \binom{8}{k} with k = 0, 1, 2 or 3. Another observation is that if one draws lines of 45^{∘} starting from the first column in Table 2 and looks at the diagonal columns, one finds the pattern,

1 1 1 1 1 1 1 8 8 8 8 8 28 28 28 56  (43)

from which it is clear that each of these diagonal columns in (43) starts with \binom{2n}{0} with subsequent values of the lower index increasing by 1. The first four columns above correspond exactly to the Gessel words where 1 precedes ¯1 and the sum of the entries is precisely (2n − 1)\binom{2n−2}{n−1} with n = 4. This pattern has been observed until n = 6 and we conjecture that it holds for all n.

6 Acknowledgements

The author would like to thank Doron Zeilberger and Tewodros Amdeberhan for useful discussions, Lara Pudwell and Eric Rowland for many helpful comments on an earlier version of the draft, and Simon Norton for discussing his conjecture. Special thanks go the anonymous referee for an exceptionally careful reading of the manuscript.

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2000 Mathematics Subject Classification: Primary 05A15. Keywords: Gessel’s conjecture, quarter plane walks, Dyck paths.

(Concerned with sequences \underline{A000531}, \underline{A045720}, and \underline{A135404}.)

Received March 2 2009; revised version received April 30 2009. Published in Journal of Integer Sequences, May 12 2009.

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