The Pennsylvania State University

The Graduate School

Eberly College of Science

BIJECTIVE METHODS AND COMBINATORIAL STUDIES

OF PROBLEMS IN PARTITION THEORY AND RELATED

AREAS

A Dissertation in

Mathematics

by

Shishuo Fu

c© 2011 Shishuo Fu

Submitted in Partial Fulfillmentof the Requirements

for the Degree of

Doctor of Philosophy

August 2011

ii

The dissertation of Shishuo Fu was reviewed and approved∗ by the following:

George E. AndrewsEvan Pugh Professor of MathematicsDissertation Co-AdviserCo-Chair of Committee

Ae Ja YeeAssociate Professor of MathematicsDissertation Co-AdviserCo-Chair of Committee

James A. SellersProfessor of Mathematics

Donald St. P. RichardsProfessor of Statistics

Svetlana KatokProfessor of MathematicsChair of Graduate Program

∗Signatures are on file in the Graduate School.

iii

ABSTRACT

This dissertation explores five problems that arise in the course of studying ba-sic hypergeometric series and enumerative combinatorics, partition theory in particular.Chapter 1 gives a quick introduction to each topic and states the main results. Then eachproblem is discussed separately in full detail in Chapter 2 through Chapter 6. Chapter2 starts with Bressound’s conjecture, which states that two sets of partitions under cer-tain constraints are equinumerous. The validity of the conjecture in the first two casesimplies exactly the partition-theoretical interpretation for the Rogers-Ramanujan iden-tities. We give a nearly bijective proof of the conjecture, and we provide examples todemonstrate the bijection as well. Chapter 3 preserves this combinatorial flavor andsupplies a purely combinatorial proof of one congruence that was first obtained by An-drews and Paule in one of their series papers on MacMahon’s partition analysis. Chap-ter 4 addresses an enumeration problem from graph theory and completely solves theproblem with a closed formula. Chapter 5 introduces a (q, t)-analogue of binomial co-efficient that was first studied by Reiner and Stanton. We also settles a conjecture madeby them concerning the sign of each term in this (q, t)-binomial coefficient when q ≤ −2is a negative integer. Chapter 6 focuses on two lacunary partition functions and wereproves two related identities uniformly using the orthogonality of the Little q-JacobiPolynomial. We concludes in Chapter 7 by addressing the significance of bijective andcombinatorial methods in the study of partition theory and related areas.

iv

Table of Contents

List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

Chapter 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Bressoud’s conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 A congruence for the broken 1-diamond partition . . . . . . . . . . . 21.3 An enumeration problem in graph theory . . . . . . . . . . . . . . . . 21.4 Reiner and Stanton’s conjecture on the (q,t)-binomial coefficients . . 21.5 q-Orthogonal polynomials and lacunary partition functions . . . . . 31.6 Collection of definitions and propositions . . . . . . . . . . . . . . . . 3

Chapter 2. Bressoud’s conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Bijective proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Chapter 3. Broken diamond partition . . . . . . . . . . . . . . . . . . . . . . . . 163.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Combinatorial Proof of Theorem 3.1.1 . . . . . . . . . . . . . . . . . . 183.3 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.4 More Congruence Results for Bk(n) . . . . . . . . . . . . . . . . . . . 28

Chapter 4. Enumeration of degree sequences . . . . . . . . . . . . . . . . . . . . 304.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.2 Enumeration of degree sequences with line–Hamiltonian realization 31

Chapter 5. Reiner and Stanton’s conjecture . . . . . . . . . . . . . . . . . . . . . 345.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.2 Proof of Theorem 5.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.2.1 k = 0 and k = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2.2 Cases with k ≥ 2 and n 6≡ k (mod 2) . . . . . . . . . . . . . . . 365.2.3 Cases with k ≥ 2 and n ≡ k (mod 2) . . . . . . . . . . . . . . . 38

5.3 A subspace interpretation . . . . . . . . . . . . . . . . . . . . . . . . . 39

v

Chapter 6. q-Orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . . . 426.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.2 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.3 The Main theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Chapter 7. Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

vi

List of Tables

2.1 Q4,2(11) and B4,2(11) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

vii

List of Figures

1.1 Ferrers Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Conjugate Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1 Durfee Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Footed Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Example for Case I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.4 Map φ composed of σ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Map ψ composed of τ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.1 Plane Partition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.2 Partition Diamond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3 Broken k-diamond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4 Broken 1-diamond Partition . . . . . . . . . . . . . . . . . . . . . . . . . . 193.5 Example I of 5-step process . . . . . . . . . . . . . . . . . . . . . . . . . . 203.6 Example II of 5-step process . . . . . . . . . . . . . . . . . . . . . . . . . . 213.7 Example III of 5-step process . . . . . . . . . . . . . . . . . . . . . . . . . 233.8 Infinite Bracelet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.9 Half Bracelets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

viii

Acknowledgments

I would first and foremost like to thank my advisor, Professor George E. An-drews, for suggesting interesting problems, keeping me on track, and making all kindsof effort to support me over the summers and to attend various workshops and con-ferences. His one of a kind knowledge of the subject and always encouraging attitudewere greatly appreciated during the course of writing this dissertation, as well as thewhole period of my Ph.D. study.

Hearty thanks also go to Professor James A. Sellers and Professor Ae Ja Yee(who is also my co-advisor) for serving in my thesis committee, providing invaluablecomments and suggestions, and being very supportive of my studies. Professor Wen-Ching W. Li have also provided numerous help when I had questions. They have allbeen excellent role models throughout my time at Penn State.

Thanks should also go to Barbara Baum, Becky Halpenny and all other staffmembers in the Mathmatics Department, for assisting me in various places to facilitatemy research and teaching.

Furthermore, I would like to thank my academic brothers Kagan Kursungoz andHeiko Todt, my officemates Evgeny Mayanskiy, Shih-Chang Tsao, Weisheng Wu andKun Zhou, my friends Rubing Chen, Xiao Chen, Jian Han, Han Li, Manlin Li, TianjiangLi, Ming Lu, Yanping Ma, Ruomeng Pang, Hua Qin, Ge Ruan, Huan Sun, ChenyingWang, Nana Wang, Yiqiang Wang, Xiang Xu, Yinkun Xue, Lei Zhang, Yanxiang Zhao,Yan Zhuang, the entire PSU math department, and everyone else I have known here,for making this experience a pleasant and unforgettable one.

On the personal side, I have two wonderful parents who have raised me up withall their love and have always provided me with the best educations possible. Lastbut definitely not least, I am immensely grateful to my gorgeous wife Xiaoxue for hertremendous support and sacrifices all these years.

1

Chapter 1

Introduction

My dissertation work centers on combinatorial interpretation and combinatorialanalysis of established or conjectured identities and congruences in basic hypergeomet-ric series and enumerative combinatorics, with the aim of providing new bijective orinvolutive proofs. And the new proofs will hopefully provide us with better insightsinto the problem, which consequently may lead to generalization or new related iden-tities. Several topics have been investigated under this theme. I will provide a quickoverview for each topic in this introductory chapter, and then elaborate on each of themin the following chapters. For the reader’s convenience, I conclude this chapter with acollection of definitions and propositions that will be used quite often later on.

1.1 Bressoud’s conjecture

In the theory of partitions, the most celebrated are the Rogers-Ramanujan iden-tities:

1 +∞∑n=1

qn2

(1− q)(1− q2) · · · (1− qn)=

∞∏m=1

m6≡0,±2 (mod 5)

1

1− qm, (1.1)

1 +∞∑n=1

qn2+n

(1− q)(1− q2) · · · (1− qn)=

∞∏m=1

m6≡0,±1 (mod 5)

1

1− qm. (1.2)

In 1961, B. Gordon gave a combinatorial generalization of the above Rogers-Ramanujan identities to any odd moduli, whose analytic counterpart was given byG. E. Andrews in 1974. To give a more natural partition theoretical interpretation tothe Andrews-Gordon identity that he derived, Andrews introduced the Durfee dissec-tion of partitions in 1979.

Meanwhile, D. M. Bressoud was able to extend Gordon’s interpretation to allmoduli. Later in an unpublished paper, trying to also extend Andrews’ interpretation toall moduli, Bressoud defined another set of partitions, namely footed partitions, whichis equinumerous with the partition families arising in the Rogers-Ramanujan identities.He then conjectured a generalization to all moduli.

Earlier in 1972, in a quite different direction, using sieve methods, Andrews con-nected yet another set of partitions to the above mentioned families, and Bressoud ex-tended this work to all moduli. I was able to connect Bressoud’s footed partition familywith this latter family by setting up a bijection, hence settled the conjecture.

2

1.2 A congruence for the broken 1-diamond partition

In 2007, George E. Andrews and Peter Paule continued their study of partitionfunctions via MacMahon’s Partition Analysis by considering partition functions asso-ciated with directed graphs which consist of chains of diamond shape, namely, brokenk-diamond partitions. They proved a congruence related to broken 1-diamond partitionand conjectured a number of similar congruence results. I have reproved this congru-ence by constructing an explicit way to group partitions.I have kept the essence of themethod and managed to apply it to a different kind of plane partitions to get moregeneral results and several other congruences.

1.3 An enumeration problem in graph theory

The degree sequence of any given graph, written in nonincreasing order, is natu-rally an integer partition. So if we are provided with a ”nice” (sufficient and necessary)characterization for the degree sequence of a particular set of graphs, presumably wecould try to enumerate the number of integer partitions that could be viewed as thedegree sequence of some graph from that particular set. A quite recent joint work withJames A. Sellers has solved such an enumeration problem, based on simply understand-ing the nature of the corresponding integer partitions.

1.4 Reiner and Stanton’s conjecture on the (q,t)-binomial coefficients

In a recent collaborative paper, Victor Reiner and Dennis Stanton introduced a(q, t)-generalization of a binomial coefficient.

[n

k

]q,t

:=k−1∏i=0

1− tqn−qi

1− tqk−qi

. (1.3)

By running many examples, they have found the following interesting behaviorof the (q, t)-binomial coefficients.

Theorem 1.4.1 (Conjectured by V. Reiner and D. Stanton). If q ≤ −2 is a negative inte-

ger, then[n

k

]q,t

is a Laurent polynomial in t, all of whose coefficients have the same sign as

(−1)k(n−k).

By adopting the combinatorial interpretation with weighted partitions, I havebeen led to a more accurate observation of the cancellation happened between theterms, which consequently have guided me to dissect the terms correctly so that the

positivity of the coefficients of (−1)k(n−k)

[n

k

]q,t

follows immediately.

3

1.5 q-Orthogonal polynomials and lacunary partition functions

Enlightened by Professor Andrews, this project has been an effort to uniformlytreat two lacunary partition functions (see (Lov02)) studied in (ADH88) and (CFLZ04).

Theorem 1.5.1 (Theorem 1 in (ADH88)).

∞∑n=0

q(n+12 )

(−q)n=∞∑n=0

(−1)nq

n(3n+1)2 (1− q2n+1

)n∑

j=−n(−1)

jq−j2 (1.4)

Theorem 1.5.2 (Theorem 2.3 in (CFLZ04)).

∞∑n=0

(−1)n(q)nq

(n+12 )

(−q)n=

∞∑n=0

(−1)nq2n

2+n

(1− q2n+1)

n∑j=−n

(−1)jq−j2 (1.5)

By using orthogonality of the Little q-Jacobi Polynomial (see (AA77)), I have beenable to prove two theorems which generalize theorem 1.5.1 and 1.5.2 above respectivelywith three free variables. The proof of the two generalized theorems were almost iden-tical and bypassed the use of Bailey pair. To keep the combinatorial flavor, one of thekey lemmas has been proved combinatorially using overpartitions (see (CL04)).

1.6 Collection of definitions and propositions

In this section, I collect all the definitions and propositions that are likely to beused many times in sequel. The proof of all the propositions listed here could be foundin (And76) and (GR90).

Definition 1.1. A partition of a positive integer n is a finite nonincreasing sequence ofpositive integers λ1, λ2, . . . , λr such that

∑r

i=1λi = n. The λi are called the parts of the

partition. (Note: Sometimes it will be convenient to allow 0 as part too, in which casewe will make it clear.)

Definition 1.2. For a positive integer n, we use p(n), called partition function, to denotethe number of partitions of n. Throughout this thesis, we will use the following nota-tions for partition functions with restrictions on the number of parts or the size of thelargest part.

• p(n,≤ m): the number of partitions of n with at most m parts;

• p(n,m): the number of partitions of n with exactly m parts;

• p(n;≤ m): the number of partitions of n with the largest part no greater than m;

• p(n;m): the number of partitions of n with the largest part being exactly m.

• p(n;≤ N,≤M) the number of partitions of n with at most M parts, each ≤ N .

4

Definition 1.3. Each partition λ = (λ1, λ2, . . . , λn) is associated with a graphical repre-sentation Gλ, namely Ferrers diagram (or Young diagram), which is formally defined tobe the set of points with integral coordinates (i, j) in the plane such that (i, j) ∈ Gλ ifand only if 0 ≥ i ≥ −n+ 1, 0 ≤ j ≤ λ|i|+1− 1. One can use either dots or unit squares toform the Ferrers diagram, as we can see in the following example. We will use the unitsquares representation throughout the rest of this thesis.

Fig. 1.1: Ferrers diagram of partition λ = (6, 5, 5, 3, 1)

Definition 1.4. For a given partition of n, say λ = {λ1, λ2, . . . , λr}. Let α1 =∣∣∣{λi|λi ≥

1}∣∣∣, α2 =

∣∣∣{λi|λi ≥ 2}∣∣∣, . . . , then λ

′= {α1, α2, . . . } (note that all but finitely many of

them will be 0) is also a partition of n, and is called the conjugate partition to λ. In termsof the Ferrers diagram, we can easily obtain the conjugate partition by reflecting thediagram with respect to its main diagonal.

Fig. 1.2: Partition λ = (6, 5, 5, 3, 1) and its conjugate λ′ = (5, 4, 4, 3, 3, 1)

Definition 1.5. The rank of a partition is the difference between the length of the largestpart and the number of parts. If we think of its Ferrers diagram, the rank is simply thenumber of boxes in the topmost row minus the number of boxes in the leftmost column.

Definition 1.6. Throughout the dissertation, we will use the q-Pochhammer symbol andGaussian Polynomial:

(a)∞ = (a; q)∞ =∞∏k=0

(1− aqk) (1.6)

(a)n = (a; q)n =(a)∞

(aqn)∞(1.7)

[n

m

]=

(q)n

(q)m(q)n−mif 0 ≤ m ≤ n,

0 otherwise.(1.8)

5

Proposition 1.7. The generating function for p(n) is given by∑n≥0

p(n)qn

=1

(q; q)∞(1.9)

.

Proposition 1.8. The generating function for p′(n) (the number of partitions of n into distinctparts) is given by ∑

n≥0p′(n)q

n= (−q; q)∞. (1.10)

Proposition 1.9.[N+MM

]generates partitions with at most M parts, each≤ N . In terms of the

generating function, [N +M

M

]=∑n≥0

p(n;≤ N,≤M)qn. (1.11)

Proposition 1.10 (Euler’s Pentagonal Number Theorem).

(q; q)∞ =∑m∈Z

(−1)mq

32m

2− 12m. (1.12)

Proposition 1.11 (q-binomial).

∞∑n=0

(A)ntn

(q)n=

(At)∞(t)∞

(1.13)

Proposition 1.12 (q-Chu-Vandermonde).

n∑j=0

(q−n

)j(b)jqj

(q)j(c)j=

(c/b)nbn

(c)n(1.14)

Proposition 1.13 (q-Gauss).

∞∑n=0

(a)n(b)n(c/ab)n

(q)n(c)n=

(c/a)∞(c/b)∞(c)∞(c/ab)∞

(1.15)

6

Chapter 2

Bressoud’s conjecture

2.1 Introduction

Let us once again bring up the Rogers-Ramanujan identities:

1 +∞∑n=1

qn2

(1− q)(1− q2) · · · (1− qn)=

∞∏m=1

1

(1− q5m−4)(1− q5m−1), (2.1)

1 +

∞∑n=1

qn2+n

(1− q)(1− q2) · · · (1− qn)=

∞∏m=1

1

(1− q5m−3)(1− q5m−2). (2.2)

In terms of partitions, the first identity can be stated as follows: the number of partitionsof n with no parts below its Durfee square (see Definition 2.1) equals the number ofpartitions of n into parts congruent to 1 or 4 modulo 5.

In (Gor61), B. Gordon gave a combinatorial generalization of the Rogers-Ramanujanidentities to any odd moduli, whose analytic counterpart was given by G. E. Andrews(And74). To give a more natural partition-theoretical interpretation to the Andrews-Gordon identity that he derived in (And74), Andrews (And79) introduced the Durfeedissection of partitions. For more details and references, see chapter 7 of (And76).

Meanwhile, D. M. Bressoud (Bre79) was able to extend Gordon’s interpretationto all moduli. Later in an unpublished paper, trying to also extend Andrews’ interpreta-tion to all moduli, Bressoud defined another set of partitions, namely footed partitions,which is equinumerous with the partition families arising in the Rogers-Ramanujanidentities. He then conjectured a generalization to all moduli. The definition of footedpartitions is rather complex. Thus we just state the conjecture of Bressoud in the fol-lowing theorem and will give the definition in Section 2.

Recall Definition 1.5 for the rank of a partition. Naturally, we can then considerthe so called successive rank (And84). The i-th successive rank is the number of boxesin the i-th row (from the top down) minus the number of boxes in the i-th column (fromleft to right). Now for any positive integer a > b with a + b ≥ 5, let Aa,b(n) denote thenumber of partitions of n into parts not congruent to 0,±b (mod a+ b), and let Ba,b(n)

denote the number of partitions of n with all the successive ranks no less than −b+ 2 invalue and with no feet with respect to a− 3 (see Definition 2.4).

Theorem 2.1.1 (Conjectured by D. M. Bressoud). For any n ≥ 0, we have Aa,b(n) =

Ba,b(n).

In a quite different direction, using sieve methods, Andrews (And72) connectedyet another set of partitions to the above mentioned families, and Bressoud (Bre80)

7

extended this work to all moduli. For any positive integer a > b with a + b ≥ 5, letQa,b(n) denote the number of partitions of n with all the successive ranks bounded in[−b+2, a−2]. Then we have the following Andrews-Bressoud Successive Rank theorem.

Theorem 2.1.2. For any n ≥ 0, Aa,b(n) = Qa,b(n).

The aim of this chapter is to prove Bressoud’s conjecture. In Section 2.3, weshow that for all n ≥ 0, Ba,b(n) = Qa,b(n) by setting up a bijection. Combining this withTheorem 2.1.2 proves the conjecture.

2.2 Definitions

Definition 2.1. If we represent a partition by its Ferrers graph as in Definition 1.3, forexample, a partition of 23, π = {7, 5, 5, 3, 2, 1} is graphically expressed as

Fig. 2.1: The Durfee square of a Ferrers diagram

then the Durfee square is the largest square (with bold sides as shown above) that can fitin the upper left corner, and the Durfee number D(π) is the length of a side of the Durfeesquare. For the example above, D(π) = 3.

Next, for our convenience, we define a new symbol, which is closely related toFrobenius Symbol (And84) and Durfee Symbol (And07).

Definition 2.2. For any given partition of n with Durfee number D(π) = r, the corre-sponding Pseudo-Durfee Symbol is a two-rowed array of r non-negative integers(

a1 a2 · · · arb1 b2 · · · br

), (2.3)

where the top row consists of the length of the rows to the right of the Durfee square,while the bottom row consists of the length of the columns below the Durfee squarewith n = r

2+∑r

i=1ai +

∑r

i=1bi. As with the Frobenius Symbol, we see that each row

is nonincreasing in order.

For the Figure 2.1 above, the corresponding Pseudo-Durfee Symbol is(4 2 23 2 1

). (2.4)

8

The following two definitions are the key to our main theorem. For both defini-tions, let π be a partition with Durfee number D(π) = r and the corresponding Pseudo-Durfee Symbol (

a1 a2 · · · arb1 b2 · · · br

).

Definition 2.3 (D. M. Bressoud). We define the following three functions on the set ofintegral j, 1 ≤ j ≤ r:

• SR(j) = aj − bj ; SR(j) is the same as the j-th successive rank (And84), also oftenreferred to as the j-th hook difference;

• k(j) is the smallest positive integer i such that ai = aj . Intuitively, in the Ferrersdiagram of π, if we view parts of the same length as one block, the k(j)-th partwill be the topmost part of the block to which the j-th part belongs;

• K(j) is the largest integer i not exceeding r and such that ai = aj . Similar to thedefinition of k(j), if we view parts of the same length as one block, the K(j)-thpart will be the bottommost part of the block to which the j-th part belongs.

For the Figure 2.1 and (2.4) above, the values of these functions are: SR(1) = 1,SR(2) = 0, SR(3) = 1; k(1) = 1, k(2) = k(3) = 2; K(1) = 1, K(2) = K(3) = 3.

Definition 2.4 (D. M. Bressoud). Take br+1 = 0 for our convenience, then π is said tohave a foot at j, 1 ≤ j ≤ r, if bj > bj+1. It is said to have a foot with respect to d at j, if ithas a foot at j and SR(k(j)) ≥ d.

The Figure 2.1 has feet at 1, 2 and 3 (see �). But it only has one foot with respectto 1, and that is at 1 (see the square with bold sides).

Fig. 2.2: Footed partition with one foot with respect to 1 at 1

Remark 2.5. In Theorem 2.1.1, if a + b = 5, then a − 3 = −b + 2, consequently allsuccessive ranks are no less than a − 3 so that there can be no feet. In particular, fora = 3, b = 2 we recover the first Rogers-Ramanujan identity (2.1), for a = 4, b = 1we get the second one (2.2). We note that in these two base cases, our interpretationsagree with Andrews’ Durfee dissection, but in general, our generalization is of a verydifferent nature from that of Andrews (And79).

9

2.3 Bijective proof

In this section we prove Theorem 2.1.1. With Theorem 2.1.2, we see that it issufficient to show thatBa,b(n) = Qa,b(n) for any n ≥ 0. We construct a bijection betweenthe two sets counted by them. Before we can describe and justify the bijection, we needa crucial lemma.

Definition 2.6. For some fixed positive a and bwith a+b ≥ 5, we define two functions sand t, as well as two maps σ and τ on a given partition represented in its Pseudo-DurfeeSymbol

π =

(a1 a2 · · · arb1 b2 · · · br

).

σ&s: If π does not have foot with respect to a − 3, then take σ(π) = π and s(π) = 0 (orwe simply do not define s in this case). Otherwise, find the greatest j such that πhas a foot with respect to a− 3 at j, let s(π) = j and take

σ(π) :=

(a1 · · · ak(j)−1 ak(j) + 1 ak(j)+1 + 1 · · · aj + 1 aj+1 · · · arb1 · · · bk(j)−1 bk(j) − 1 bk(j)+1 − 1 · · · bj − 1 bj+1 · · · br

).

τ&t: If π does not have successive rank greater than a− 2 in value, then take τ(π) = πand t(π) = 0 (or we simply do not define t in this case). Otherwise, find thesmallest j such that SR(j) > a− 2, let t(π) = j and take

τ(π) :=

(a1 · · · aj−1 aj − 1 aj+1 − 1 · · · aK(j) − 1 aK(j)+1 · · · arb1 · · · bj−1 bj + 1 bj+1 + 1 · · · bK(j) + 1 bK(j)+1 · · · br

).

Note: Due to the definitions of the functions k(j) and K(j), together with our choiceof j, it is not hard to verify that both σ and τ are well defined maps from the set of allpartitions to itself. More precisely, in order to guarantee the resulting Pseudo-DurfeeSymbols (for σ(π) and τ(π)) are legitimate, we need to show that ak(j)−1 > ak(j) andbj > bj+1 in the case of σ, and that aK(j) > aK(j)+1 and bj−1 > bj in the case of τ . All ofthem follow easily from definitions.

Lemma 2.7. For any given partition π with D(π) = r,

I if π is inQa,b(n)\Ba,b(n), andm is the smallest integer such that σm(π) ∈ Ba,b(n), then

for any integer i, 1 ≤ i ≤ m, t(σi(π)) = k(s(σi−1

(π))) and K(t(σi(π))) = s(σ

i−1(π)).

Consequently, τ ◦ σi(π) = σi−1

(π);

II if π is inBa,b(n)\Qa,b(n), andm is the smallest integer such that τm(π) ∈ Qa,b(n), then

for any integer i, 1 ≤ i ≤ m, s(τ i(π)) = K(t(τi−1

(π))) and k(s(τi(π))) = t(τ

i−1(π)).

Consequently, σ ◦ τ i(π) = τi−1

(π).

Proof of Lemma 2.7: Let us first verify the claimed results for the base case i = 1, thenexplain why they are still true in general (for any 1 ≤ i ≤ m).

10

For I: We need to show that for a given partition π in Qa,b(n)\Ba,b(n), t(σ(π)) = k(s(π))

and K(t(σ(π))) = s(π). Actually, π must have at least one foot with respect toa − 3 since it is in Qa,b(n)\Ba,b(n). And by Definition 2.6 the greatest such foot isat some j = s(π), then before we apply σ on π, ak(j)−bk(j) ≥ a−3 by the definitionof ”foot with respect to”, while after the map σ, (ak(j) + 1) − (bk(j) − 1) ≥ a − 1,and this is the leftmost column in σ(π) that has successive rank greater than a− 2.So t(σ(π)) = k(j), and K(t(σ(π))) = K(k(j)) = j since ak(j) + 1 = ak(j)+1 +

1 = · · · = aj + 1 > aj+1, therefore when we continue to apply τ on σ(π), we”undo” the changes made by σ, i.e., τ ◦ σ(π) = π (see the example below forillustration). In general, every time we apply σ, we make changes to a ”block”with subindex running from k(j) to j, for some particular j, and as we observedabove, SR(k(j)) > a − 2 for the resulting partition. Moreover, both k(j) andj are nonincreasing as we go through the sequence π, σ(π), σ

2(π), . . ., therefore

t(σi(π)) = k(s(σ

i−1(π))) and K(t(σ

i(π))) = s(σ

i−1(π)). Consequently, when we

apply τ ◦σ on σi−1(π), τ will undo what σ does on σi−1(π), i.e., τ ◦σi(π) = σi−1

(π).

For II: The key observation in this case is that every time we apply τ , we make changesto a ”block” with subindex running from j to K(j) for some particular j, and theresulting partition has a foot with respect to a − 3 at K(j). We will elaborate onthis in the proof of Theorem 2.3.1. Moreover, both j and K(j) are nondecreasingas we go through the sequence π, τ(π), τ

2(π), . . .. Then use similar argument we

can prove the results as well.(6 5 5 4 4 4 15 3 3 3 2 2 1

)σ→

(6 5 5 5 5 5 15 3 3 2 1 1 1

)τ→

(6 5 5 4 4 4 15 3 3 3 2 2 1

)π ∈ Q4,2(97)\B4,2(97) σ(π) τ ◦ σ(π) = π

Fig. 2.3: An example to illustrate Case I of Lemma 2.7�

Now we are ready to prove the main theorem.

Theorem 2.3.1. For any n ≥ 0, Ba,b(n) = Qa,b(n).

Proof of Theorem 2.3.1: We will use the Pseudo-Durfee Symbol to represent both typesof partition (to better visualize the process, see the examples in next section). The proofis mainly a description of how to bijectively convert between these two types of parti-tion.

For those partitions in Ba,b(n) ∩ Qa,b(n), no conversion is needed, or we simplytake the conversion map to be identity.

Now take a partition π in Qa,b(n)\Ba,b(n), say

π =

(a1 a2 · · · arb1 b2 · · · br

),

11

where −b+ 2 ≤ SR(i) = ai− bi ≤ a− 2 and there exists at least one foot with respect toa − 3. We repeatedly apply σ on π to get σ(π), σ2(π), . . ., until we arrive at a partitionwhere there is no foot with respect to a− 3, denote this partition as

π =

(a1 a2 · · · arb1 b2 · · · br

)∈ Ba,b(n).

Note that in π, SR(k(s(π))) ≥ a − 3 by the definition of ”the foot with respect to”. Soin σ(π), SR(k(s(π))) = ak(s(π)) − bk(s(π)) + 2 ≥ a − 1, and every time we apply σ, theaffected successive ranks will be increased by 2, in other words they will remain greaterthan a − 2 in value, which implies that π is really in Ba,b(n)\Qa,b(n). Now we have awell defined map

φ : Qa,b(n)\Ba,b(n) → Ba,b(n)\Qa,b(n)

π 7→ π

with π = φ(π) = σm

(π), for some m ≥ 1 that depends on π.Conversely, if we start with a partition π in Ba,b(n)\Qa,b(n), there must be col-

umn(s) with successive rank greater than a−2 in value. Then we repeatedly apply τ onπ to get τ(π), τ2(π), . . ., until we arrive at a partition with all successive ranks boundedin [−b+ 2, a− 2]. Denote this partition as π, we see that π ∈ Qa,b(n). Note that if

π =

(a1 a2 · · · arb1 b2 · · · br

),

then

τ(π) =

(a1 · · · at(π)−1 at(π) − 1 at(π)+1 − 1 · · · aK(t(π)) − 1 aK(t(π))+1 · · · arb1 · · · bt(π)−1 bt(π) + 1 bt(π)+1 + 1 · · · bK(t(π)) + 1 bK(t(π))+1 · · · br

).

By the definition of function t, we see that

at(π) − bt(π) ≥ a− 1,

hence (at(π) − 1)− (bt(π) + 1) ≥ a− 3,

and bK(t(π)) + 1 > bK(t(π))+1.

Therefore τ(π) has a foot with respect to a− 3 at K(t(π)). Actually every time we applyτ , there will be a foot with respect to a − 3 coming from the rightmost column that wehave changed in the Pseudo-Durfee Symbol. So π is really in Qa,b(n)\Ba,b(n). Now wehave a well defined map in the opposite direction

ψ : Ba,b(n)\Qa,b(n) → Qa,b(n)\Ba,b(n)

π 7→ π

with π = ψ(π) = τl(π), for some l ≥ 1 that depends on π.

12

Next we apply Lemma 2.7 to show that φ and ψ are inverse to each other.Take a partition π in Qa,b(n)\Ba,b(n), assume φ(π) = σ

m(π) and ψ(φ(π)) =

τl(φ(π)), for some m, l ≥ 1. If l ≥ m, then

ψ(φ(π)) = τl(σm

(π))

= τl−1

(τ ◦ σm(π))

= τl−1

(σm−1

(π)) (by Lemma 2.7 I)...= τ

l−m(π)

= π (by the definition of τ).

Otherwise we have l < m, then ψ(φ(π)) = τl(σm

(π)) = σm−l

(π). This cannot happenbecause σm−l(π) is a partition with at least one successive rank greater than a − 2 invalue, hence cannot be in Qa,b(n), while ψ(φ(π)) is in Qa,b(n)\Ba,b(n). This contradic-tion shows that we must have ψ(φ(π)) = π.

Using the same argument together with Lemma 2.7 II, we can also prove thatφ(ψ(π)) = π for any π in Ba,b(n)\Qa,b(n). So we see that Qa,b(n) and Ba,b(n) are in 1-1correspondence via the map φ and ψ, in particular, they are equinumerous.

�

2.4 Example

We use Ferrers diagram to illustrate our ”cut and paste” process, which is essen-tially what σ and τ do. We map a partition in Q4,2(25) to a partition in B4,2(25). Notethat a − 3 = 4 − 3 = 1, so we try to get rid of all feet with respect to 1. Those numbersin the boxes indicate the order we move them, and ”foot with respect to” is abbreviatedas ”fwrt”.

11

fwrt 1 at 3

σ→11

2

2

fwrt 1 at 2

σ→11 2

2

3

fwrt 1 at 1

σ→

11 2

2 3

4

fwrt 1 at 1

σ→ 11 2

2 3 4

no fwrt 1

13

Fig. 2.4: Chain of σ sending a partition in Q4,2(25) to a partition in B4,2(25)

Then we should be able to apply τ to get back. Since a − 2 = 2, we will changethose columns (hooks in the Ferrers graph) with successive rank greater than 2.

1

SR(1) = 7 > 2

τ→

2

1

SR(1) = 5 > 2

τ→

33

21

SR(1) = 3 > 2

τ→44

3

321

SR(1) = 1, SR(2) = 3 > 2

τ→ 44

3

321

SR(2) = 1, SR(3) = 2

Fig. 2.5: Chain of τ sending a partition in B4,2(25) back to a partition in Q4,2(25)

The following is a list of all partitions in Q4,2(11) and B4,2(11) using the Pseudo-Durfee Symbol. The ones in the same row are paired using our maps φ and ψ.

14

Q4,2(11) B4,2(11)(55

) (55

)(

64

) (100

)(

4 03 0

) (7 00 0

)(

3 13 0

) (3 13 0

)(

4 12 0

) (6 10 0

)(

3 22 0

) (5 20 0

)(

3 12 1

) (4 11 1

)(

2 22 1

) (2 22 1

)(

3 21 1

) (4 30 0

)(

1 0 01 0 0

) (1 0 01 0 0

)(

2 0 00 0 0

) (2 0 00 0 0

)(

1 1 00 0 0

) (1 1 00 0 0

)

Table 2.1: 1-1 correspondence between Q4,2(11) and B4,2(11)

15

2.5 Conclusion

In the end, we note that our Theorem 2.3.1 together with the Andrews-BressoudSuccessive Rank Theorem (And72), (Bre80), have established Bressoud’s conjecture (The-orem 2.1.1), which is yet another partition-theoretical generalization of the Rogers-Ramanujan Identities. While our proof gives bijection between Qa,b(n) and Ba,b(n),Andrews (And72) and Bressoud (Bre80) connected Qa,b(n) with Aa,b(n) using sievemethod. So the connection between Ba,b(n) and Aa,b(n) via these two theorems is stillnot bijective, but at least with strong combinatorial flavor in it. In addition, further ef-fort may be made to develop a generating function for the partitions in Ba,b(n), so as toderive a q-series version of this generalization.

16

Chapter 3

Broken diamond partition

3.1 Introduction

To introduce our first combinatorial object, namely a ”Broken 1-Diamond Parti-tion”, let us first get some ideas of ”plane partition”. Nowadays we usually use thisphrase to denote a two-dimensional array of nonnegative integers ni,j which are non-increasing from left to right and top to bottom. But in this chapter, we actually considerthe original definition that goes back to MacMahon (Mac60). I just quote below directlyfrom (AP07):

The ”most simple case” of classical plane partitions, treated by MacMahonin (Mac60), is the situation where the non-negative integer parts ai of thepartitions are placed at the corners of a square such that the following orderrelations are satisfied:

a1 ≥ a2, a1 ≥ a3, a2 ≥ a4, a3 ≥ a4. (3.1)

It will be convenient to use arrows as an alternative description for ≥ rela-tions; for instance, Figure 3.1 represents the relations (3.1). Here and through-out the following it will be understood that an arrow pointing from ai to ajis interpreted as ai ≥ aj .

cc

cc�

��@

@@

���@

@@

a1 a4

a3

a2

� R

�R

Fig. 3.1: The inequalities (3.1) depicted in a plane partition

Andrews and Paule (AP07) suggested one generalization of this diamond shapeas shown in Figure 3.2. Their next step was to define a ”Broken k-Diamond Partition”as a plane partition with the configuration shown in Figure 3.3 and they denoted thenumber of such partitions of n by the function4k(n).

17

ca1���

AAA

�

U ca2

������

-�

ca3AAAAAA

-

U

ca4

������

-�

ca5AAAAAA

-

U

ca6

. . . . . . . . . .

ca7. . . . . . . . . . ca2k−1

AAAAAA

-

U

ca2k−2

������

-�

ca2k+1

AAAU

ca2k

����

ca2k+2

Fig. 3.2: A k-elongated partition diamond of length 1

cb(2k+1)n+1���

AAAc

b(2k+1)n−1

. . . . . . .cb(2k+1)n

. . . . . . . cAAA

c

���c�

K

K

�

. . . c���AAA

c . . . . . . .�

c . . . . . . .K

cb7AAAAAA

�

�

cb6

������

�K

cb5

������

�Kc

b4

AAAAAA

�

�

cb3

cb2

b2k+2

b2k+1

b2k

ca1���

AAA

�

U ca2

. . . . . . .ca3

. . . . . . . ca2k

AAA

�

ca2k+1

���

U ca2k+2. . . c���

AAA

�

U c

. . . . . . .c

. . . . . . . ca(2k+1)n−1

AAA

�

ca(2k+1)n

���

U ca(2k+1)n+1

Fig. 3.3: A broken k-diamond of length 2n

They then note that the generating function for4k(n) is a ratio of products of etafunctions by proving that

∞∑n=0

4k(n)qn

=(−q; q)∞

(q; q)2∞

(−q2k+1; q2k+1)∞=q(k+1)/12

η(2τ)η((2k + 1)τ)

η(τ)3η((4k + 2)τ)(3.2)

where q = e2πiτ and η(τ) = q

1/24∏∞n=1

(1− qn) is Dedekind’s eta function.Next they state and prove the following Ramanujan-like congruence satisfied by

the function41. We note that their proof uses some basic generating function manipu-lations.

Theorem 3.1.1 (Theorem 5 in (AP07)). For n ≥ 0,

41(2n+ 1) ≡ 0 (mod 3). (3.3)

Proof of Theorem 3.1.1 in (AP07):

∞∑n=0

41(n)qn

=(−q; q)∞

(q; q)2∞

(−q3; q3)∞=

(q2; q

2)∞

(q; q)3∞

(−q3; q3)∞(3.4)

≡(q

2; q

2)∞

(q3; q3)∞(−q3; q3)∞(mod 3) (3.5)

(because (1−X)3 ≡ 1−X3

(mod 3))

=(q

2; q

2)∞

(q6; q6)∞(3.6)

18

We see that in (3.6), all the powers of q are even, which means after the expansion, thecoefficients of all the terms with q raised to an odd power will automatically be 0 mod3. This is exactly what the theorem states. �

3.2 Combinatorial Proof of Theorem 3.1.1

Our next goal is to show that for any broken 1-diamond partition of an oddnumber 2n+ 1, we can find a unique set (with cardinality 3 or 6) it belongs to. This willexplain the congruence in Theorem 3.1.1 combinatorially. Moreover, using our way of”grouping”, the generating function (3.6) for the even case will follow directly.

Let us first consider a preliminary lemma.

Lemma 3.1. Each broken k-diamond partition of n is in 1-1 correspondence with a partitiontriple (π1, π2, π3), where π1 and π2 are ordinary partitions, while π3 is a partition with distinctparts, no part is divisible by 2k + 1, and |π1|+ |π2|+ |π3| = n.

Remark 3.2. We note that this lemma also establishes the generating function (3.2) com-binatorially.

Proof of Lemma 3.1: For a given broken k-diamond partition π = (a1, a2, a3, . . . ; b2, b3, b4, . . .)with configuration shown in Figure 3.3, we define two ”inversion” sets

Ia := {i|ai < ai+1, i ≥ 1} and Ib := {i|bi+1 < bi+2, i ≥ 1}.

Considering all the inequalities encoded in the diamond shape configuration, we seethat

i ∈ Ia ⇒ i ≡ 2, 4, . . . , 2k (mod 2k + 1),

i ∈ Ib ⇒ i ≡ 1, 3, . . . , 2k − 1 (mod 2k + 1).

Now for a given inversion in the sequence a1a2a3 . . ., say at i, we ”peel it off” byfirst switching ai and ai+1, then subtracting 1 from each number to the left of ai.

a1a2 . . . aiai+1 . . . ⇒ a1a2 . . . ai+1ai . . . ⇒ (a1 − 1)(a2 − 1) . . . (ai+1 − 1)ai . . . .

Carrying out this ”peeling off” process for each inversion in a1a2 . . ., the se-quence we end up with will be a non-increasing sequence, naturally a partition, whichwe take to be π1. On the other hand, the numbers we have taken off each time willcompose the set Ia.

Now repeat the same process for the sequence b2b3 . . . and denote the ”core”sequence to be π2, the numbers we have peeled off for each inversion will compose theset Ib. Now combine the sets Ia and Ib to form a strictly decreasing sequence, whichwill be our π3.

Obviously (π1, π2, π3) matches the description for the partition triple mentionedin the lemma, and it’s easy to see that our steps to construct (π1, π2, π3) are reversible.So we have set up the 1-1 correspondence as claimed. �

19

Now we proceed to describe the steps to assign a unique set for a given broken1-diamond partition of 2n+ 1 (see Figure 3.4 below).

b6

��

b3

��

a3

��

a6

��· · · b7

__

��

b4

^^

��

a1

??

��

a4

??

��

a7

??

��

· · ·

b5

^^

b2

^^

a2

??

a5

??

Fig. 3.4: A broken 1-diamond partition

Step 0: Find the corresponding partition triple (π1, π2, π3). As stated in Lemma 3.1, π1 andπ2 are unrestricted partitions while π3 is a partition with distinct parts ≡ 1 or 2(mod 3).

Step 1: For any integer k ≥ 0, consider the subpartition in π1 (resp. π2) that is composedof parts ≡ 3

k or 5 · 3k (mod 6 · 3k), denote it as πk1

(resp. πk2

). We note that for

i 6= j, πi1∩ πj

1= ∅, and

⋃k≥0 π

k

1= {all odd parts in π1}. Similar results hold for

πk

2.

Step 2: Keep splitting even parts in π3 into a pair of odd parts until there are no evenparts left, like (4)→ (2

2)→ (1

4). Denote the new partition as π0

3. Then π0

3will be

composed of parts ≡ 1 or 5 (mod 6), possibly repeated.

Step 3: Compare π01, π0

2and π

0

3. By definition, they are all partitions with parts ≡ 1 or 5

(mod 6). If π01

= π0

2= π

0

3, go to Step 4. Otherwise, by permutation we get a group

of cardinality 3 (if two of them are the same,(31

)= 3) or 6 (if they are different

pairwise, 3! = 6). Our procedure ends here.

Step 4: If in Step 3 we get π01

= π0

2= π

0

3, we combine them to get π1

3, like (5 1

2) + (5 1

2) +

(5 12)→ (15 3

2). Then π1

3will be composed of parts ≡ 3 or 15 (mod18).

Step 5: Compare π11, π1

2and π1

3like we did in Step 3 and Step 4. We probably will need to

further combine (Step 4) and get π23, π3

3, etc., but in any case, this process must ter-

minate after finitely many steps because 2n+1 is finite and after each ”combining”we get bigger parts.

We see that for any given broken 1-diamond partition, there exists one uniquegroup of 3 or 6 partitions to which it belongs, via the above steps. For example, takethis broken 1-diamond partition of 25:

20

4

��

6

��

1

1 6

��

@@

2

��

@@

2

^^

3

@@

0

↓Step 0

(π1, π2, π3) =((4

22 1), (3 2 1), (5 2 1)

)↓Step 1, 2

(π0

1, π

0

2, π

0

3) =

((1), (1), (5 1

3))

↓Step 3 (Permutation){((4

22 1), (3 2 1), (5 1

3)),((4

22 1), (5 3 2 1

3), (1)

),((5 4

22 1

3), (3 2 1), (1)

).}

Fig. 3.5: An example of finding the group (3 elements)using only permutation

It is quite easy to recover the corresponding visual presentations for the othertwo partition triples in this group.

For((4

22 1), (5 3 2 1

3), (1)

):

1

��

6

��

2

��1 2

^^

��

4

��

@@

1

1

^^

3

^^

4

@@

For((5 4

22 1

3), (3 2 1), (1)

):

4

��

4

��

1

��1 5

@@

��

2

��

@@

1

2

^^

4

@@

1

@@

21

We include one more example where the unique group we find has 6 elements.We begin with this broken 1-diamond partition of 14:

0 1

��

3

1

��

^^

4

��

@@

1 3

^^

1

↓Step 0

(π1, π2, π3) =((3 2 1), (3 1

3), (2)

)↓Step 1, 2

(π0

1, π

0

2, π

0

3) =

((1), (1

3), (1

2))

↓Step 3 (Permutation){ ((3 2 1), (3 1

3), (1

2)),((3 2 1), (3 1

2), (1

3)),((3 2 1

2), (3 1

3), (1)

),(

(3 2 12), (3 1), (1

3)),((3 2 1

3), (3 1

2), (1)

),((3 2 1

3), (3 1), (1

2)).

}

Fig. 3.6: Another example of finding the group (6 elements)using only permutation

We can similarly recover the corresponding visual presentations for the otherfive partition triples in this group.

For((3 2 1), (3 1

2), (1

3))↔((3 2 1), (3 1

2), (2 1)

):

4

��

3

1 4

@@

��1

^^

1

22

For((3 2 1

2), (3 1

3), (1)

):

0 4

��

1

��1

^^

��

3

@@

��

1

1 1

^^

2

@@

For((3 2 1

2), (3 1), (1

3))↔((3 2 1

2), (3 1), (2 1)

):

4

��

3

��0 4

@@

��

1

1

^^

1

@@

For((3 2 1

3), (3 1

2), (1)

):

4

��

1

��

0

1 3

@@

��

1

@@

��1

^^

2

@@

1

For((3 2 1

3), (3 1), (1

2))↔((3 2 1

3), (3 1), (2)

):

1

��

3

��

0

0 4

@@

��

1

��

@@

3

^^

1

@@

1

Here is another example involving ”combining”. We follow the 5-step process tolook for the unique group for this broken 1-diamond partition of 11:

23

4

��

0

1 3

��

@@

2

^^

1

↓Step 0& Step 2

(π1, π2, π3 = π0

3) =

((3 1), (3 2 1), (1)

)↓Step 4 (Combining)

(π1

1, π

1

2, π

1

3) =

((3), (3), (3)

)↓Step 4 (Combining)

(π2

1, π

2

2, π

2

3) =

(∅, ∅, (9)

)↓Step 3 (Permutation)

{(∅, (2), (9)

),(∅, (9 2), ∅

),((9), (2), ∅

).}

Fig. 3.7: An example of finding the group usingboth permutation and combining

And these are the corresponding broken 1-diamond partitions for the other twoin this group:

2

��0

9

^^

0

��

0

0 9

@@

��2

^^

0

24

Now the reader might be asking, why doesn’t this process work for even num-bers 2n? There seems to be nothing special about 2n+1 during our process. But we mustkeep in mind that in order for our machinery (especially Step 3 and Step 4) to even makea start, there must be some odd parts in any one of the triple (π1, π2, π

0

3). In other words,

if for one particular broken 1-diamond partition π = (a1, a2, a3, . . . ; b2, b3, b4, . . .), wehave that π3 = ∅ and π0

1= π

0

2= π

1

1= π

1

2= . . . = π

k

1= π

k

2= . . . = ∅, then there is noth-

ing to be ”permuted” or ”combined”. Since by definition⋃k≥0 π

k

1= {all odd parts in π1}

and⋃k≥0 π

k

2= {all odd parts in π2}, we note that in this case all the ai’s and bi’s are

even and there are no inversions. Therefore for 2n + 1, we will not have this case, be-cause there will be at least one odd part. But for 2n, there will be partitions with onlyeven parts and without any inversions, which will cause this method to fail. To be moreprecise, for the 2n case, we could still ”group” those partitions with at least one oddpart in one of the triples (π1, π2, π

0

3), so if we count partitions modulo 3, it will suffice

to count those ”failed” cases only (in which π1 and π2 contain only even parts and π3 isempty). Therefore,

∞∑n=0

41(n)qn ≡ 1

(q2; q2)∞(q2; q2)∞(mod 3) (3.7)

=(q

2; q

2)∞

(q2; q2)3∞

(3.8)

≡(q

2; q

2)∞

(q6; q6)∞(mod 3) (3.9)

3.3 Generalization

In order to properly generalize our method so as to give similar congruence forhigher dimensional diamonds, we need to take a closer look at our ”grouping” process.Any decent grouping should satisfy at least two major principles, namely, no repeatingnor missing of the objects that we are trying to group.

For our purpose, we actually do not have to worry about the issue of ”no miss-ing”. As we can clearly see from the above proof, for an odd number 2n+1, our processcan always start for any partition, so no one is missed. In contrast, for an even number2n, we do miss some partitions, so we simply count those missed ones since we countpartitions modulo 3.

Now in the view of ”no repeating”, more precisely, we need to make sure thatunder our process, no partition is ever assigned into more than one group. Actually,Step 0 and Step 2 are naturally bijections, but Step 4 (Combining) is in danger of in-ducing double counting, namely, after ”combining” for one partition we might arriveat another partition that does not go through any ”combining”. Fortunately, the natureof the partition πk

3prevents this from happening. We see that π0

3is composed of parts

≡ 1 or 5 (mod 6), which are not divisible by 3. After one ”combining”, we get π13,

25

which will be composed of parts ≡ 3 or 15 (mod 18), divisible by 3 but not by 9. Aftertwo ”combinings”, we get π2

3, which will be composed of parts ≡ 9 or 45 (mod 54),

divisible by 9 but not by 27, and so on. Therefore there is no chance of repeats betweenthem.

Based on the above observation, we will inevitably run into trouble if we insiston generalizing this method to n-diamond partitions. Guided and encouraged by Pro-fessor George Andrews, the author decided to preserve the nature of this method andapply it to a slightly different kind of partition in seeking a generalization. For the mo-ment, we will call it ”the n dots bracelet partition” for the sake of simplicity. A precisedefinition is as follows.

Definition 3.3. For n ≥ 3, an ”Infinite Bracelet with n Dots” is the configuration shownin Figure 3.8. It is composed of repeating ”diamonds” and ”dots”, with n − 2 ”dots”between two consecutive ”diamonds”.

am−1

""

am+n−1

%%· · · // am−3

;;

##

am// · · · // am+n−3

88

&&

am+n// · · ·

am−2

<<

am+n−2

99

Fig. 3.8: Infinite Bracelet with n dots

We see that there are essentially n − 1 different ways to cut an Infinite Braceletwith n dots in half. For each different ”cut”, we get a half bracelet (the right half) indifferent configuration. Graphically:

am+n−1

%%am

// · · · // am+n−3

88

&&

am+n// · · ·

am+n−2

99

26

am+n−1

%%am+1

// · · · // am+n−3

88

&&

am+n// · · ·

am+n−2

99

...

am+n−1

%%

am+2n−1

%%am+n−3

88

&&

am+n// · · · // am+2n−3

88

&&

am+2n// · · ·

am+n−2

99

am+2n−2

99

am+n−1

%%

am+2n−1

%%am+n

// · · · // am+2n−3

88

&&

am+2n// · · ·

am+n−2

99

am+2n−2

99

Fig. 3.9: n− 1 different half bracelets

Definition 3.4. A ”k Dots Bracelet Partition” is a partition which consists of the k − 1different half bracelets (in the sense of different configuration) as shown in Figure 3.9above. We use Bk(n) to denote the number of such ”k Dots Bracelet Partitions” of n.

We are now led to a natural generalization of Theorem 3.1.1.

Theorem 3.3.1. For n ≥ 0, k ≥ 3, if k = pr is a prime power, we have

Bk(2n+ 1) ≡ 0 (mod p). (3.10)

Remark. Note that Theorem 3.1.1 is just the special case k = 3 of Theorem 3.3.1.

Proof of Theorem 3.3.1: We could just adopt the machinery we have developed in theproof of Theorem 3.1.1 to group all partitions into subsets by permutation, and all ofthem will have cardinality divisible by p because

p

∣∣∣∣( prm )for 1 ≤ m ≤ pr − 1.

This explains the congruence combinatorially.

27

Actually, when p 6= 2, k will be an odd number and the whole machinery willgo through smoothly with minor changes, just like what we have explained for the casewhen k = 3. More precisely, in Step 0, we ”smooth” (peel off inversions) all k − 1different half bracelets to get k − 1 ”core” sequences as our unrestricted partitions π1,π2, . . ., πk−1, then use those numbers that we have taken off during the ”smoothing” toform the partition πk, which is a partition with distinct parts ≡ 1, 2, . . ., k − 1 (mod k).Then after Step 2, we get π0

k, which is a partition with parts ≡ 1, 3, . . ., k − 2, k + 2,

. . ., 2k − 1 (mod 2k), possibly repeated. Next by applying Step 3 and 4 (permutationand combination), we could manage to get every single partition assigned to a uniquesubset. Again we observe that there is no chance of ”double counting” for the samereason as when k = 3. And similarly, for an even number 2n, to count Bk(2n) modulop, it will suffice to examine those partitions with π1, π2, . . ., πk−1 all composed of evenparts, and πk empty. Thus,

∞∑n=0

Bk(n)qn ≡ 1

(q2; q2)k−1∞

(mod p) (3.11)

=(q

2; q

2)∞

(q2; q2)k∞

(3.12)

≡(q

2; q

2)∞

(q2k; q2k)∞(mod p). (3.13)

Next for the case of k = 2r even, note that after Step 2, we get π0

k, which is a

partition with parts≡ 1, 3, . . ., k−3, k−1 (mod k), all odd, possibly repeated. But unlikethe case of k being odd, we actually will not run into any ”combining” process. For ifin Step 3 we have π0

1= π

0

2= . . . = π

0

k−1= π

0

k(note that in this case, for 1 ≤ i ≤ k − 1,

π0

i:= {all odd parts in πi}), then there will be k identical copies of π0

k, with the rest

parts of π1, π2, . . ., πk−1 all being even. So their sum can never be an odd number2n + 1. Therefore we see in this case, the ”grouping” is even more straightforward for2n+1 (without Step 4). And we actually have a much stronger result for the k even case(see Theorem 3.4.2). Moreover, for the even case 2n, we could still start our procedureto group partitions (probably with several times ”combining”) with one exception, i.e.,when π1, π2, . . ., πk−1 are all composed of even parts and πk is empty. So we can get thesimilar q-series formula as above. �

Remark 3.5. If we view our k Dots Bracelet Partition equivalently as the k-tuple parti-tions we get after Step 0 (this can be justified in a similar manner as Lemma 3.1), namely,k − 1 unrestricted partitions π1, π2, . . ., πk−1, and a partition πk with distinct parts ≡ 1,2, . . ., k − 1 (mod k), then we get the following generating function proof.

28

∞∑n=0

Bk(n)qn

=(−q; qk)∞(−q2; qk)∞ · · · (−q

k−1; qk)∞

(q; q)k−1∞

(3.14)

=(−q; q)∞

(q; q)k−1∞

(−qk; qk)∞(3.15)

=(q

2; q

2)∞

(q; q)k∞

(−qk; qk)∞(3.16)

≡(q

2; q

2)∞

(qk; qk)∞(−qk; qk)∞(mod p) (3.17)

(because (1−X)pr

≡ 1−Xpr

(mod p), for p prime)

=(q

2; q

2)∞

(q2k; q2k)∞. (3.18)

3.4 More Congruence Results for Bk(n)

By personal communication with James A. Sellers, I have been then led to somefurther congruence results, which might motivate more work on the function Bk(n) inthe future.

With a standard technique that traces back to (AR97) and (Sel03) among others,we have the following result.

Theorem 3.4.1. For any k ≥ 3, s an integer between 1 and p−1 such that 12s+1 is a quadraticnonresidue modulo p, and any n ≥ 0, if p

∣∣k for some prime p ≥ 5, say k = pm, then we have

Bk(pn+ s) ≡ 0 (mod p). (3.19)

Proof of Theorem 3.4.1: Starting with the generating function (3.16), we have

∞∑n=0

Bk(n)qn

=(q

2; q

2)∞

(q; q)k∞

(−qk; qk)∞

≡(q

2; q

2)∞(q

k; qk)∞

(qp; qp)m∞

(q2k; q2k)∞(mod p) (3.20)

=(∑i1∈Z

(−1)i1q

3i12−i1)(∑

i2∈Z(−1)

i2qk3i2

2−i22)( ∞∑

j1=0

p(j1)qpj1)m

·( ∞∑j2=0

p(j2)q2kj2)

(3.21)

(using Proposition 1.7 and Proposition 1.10)

29

So if there exists i1, i2, j1 and j2 such that the exponent of q on the right hand sidesum to pn+ s, then we will have

s ≡ 3i12 − i1 (mod p) and (3.22)

12s+ 1 ≡ 36i12 − 12i1 + 1 (3.23)

≡ (6i1 − 1)2

(mod p). (3.24)

This contradicts the fact that 12s+1 has been chosen to be a quadratic nonresiduemodulo p. So we get (3.19). �

Theorem 3.4.2 (Conjectured by J.A. Sellers). For n ≥ 0, k ≥ 3 even, say k = 2ml, where l

is odd, we have

Bk(2n+ 1) ≡ 0 (mod 2m

). (3.25)

Proof of Theorem 3.4.2: Actually in the proof of Theorem 3.3.1, we have already noticedthat for the case of k being even, we do not have to worry about ”combining”. More-over, we observe that in Step 3, when we compare π0

kwith π

0

1, π0

2, . . ., π0

k−1(note that

they are all odd parts in this case), there must exist one partition that appears odd times.Otherwise each partition appears an even number of times, summing together to get aneven number, and with the remaining even parts in π1, π2, . . ., πk−1, the total cannotbe 2n + 1. So we can always assume that there is a partition which appears r times, rbeing odd. And note that the cardinality of the subset generated by permutation (Step3) will be

(kr

)·N , with N ≥ 1. Therefore, our proof reduces to the following lemma for

binomial coefficients. �

Lemma 3.6. Let k = 2ml, where l is odd. Then for any r odd, we have(

k

r

)≡ 0 (mod 2

m). (3.26)

Proof of Lemma 3.6: Just note that(k

r

)=k

r

(k − 1

r − 1

)= 2

m ·l

(k − 1

r − 1

)r

, and that both(k

r

)and

(k − 1

r − 1

)are integers, with r being odd we see that 2

m

∣∣∣∣(kr)

as desired. �

30

Chapter 4

Enumeration of degree sequences

4.1 Introduction

In this chapter, all graphs G = (V,E) under consideration will be finite, undi-rected, and loopless but may contain multiple edges. We denote the degree sequence ofthe vertices v1, v2, . . . , vm by d1, d2, . . . , dm with the convention that d1 ≥ d2 ≥ · · · ≥ dm.We say that a sequence d1, d2, . . . , dm with d1 ≥ d2 ≥ · · · ≥ dm is multigraphic if thereexists a multigraphGwith this degree sequence. This multigraphG is then called a real-ization of this degree sequence. Lastly, we say that a degree sequence is line–Hamiltonianif it has a multigraphic realizationG such that L(G), the line graph ofG, is Hamiltonian.

In 2008, Fan, Lai, Shao, Zhang, and Zhou (FLS+08) characterized those degreesequences for which there exists a simple line–Hamiltonian graph realization. This wasfollowed immediately in 2009 by Lai and Liang (LL09) who provided a characterizationof those degree sequences for which there exists a multigraphic line–Hamiltonian real-ization. The main result of Lai and Liang is the following:

Theorem 4.1.1. Let d1 ≥ d2 ≥ · · · ≥ dn ≥ 1 be integers with n ≥ 2. There exists a multi-graphic line–Hamiltonian realization of d1, d2, . . . , dn if and only if

1) d1 + d2 + · · ·+ dn is even, and2) d1 ≤ d2 + d3 + · · ·+ dn, and3a)

∑di=1

di ≤∑dj>1

(dj − 2) or 3b) d1 = n− 1.

In a joint work with James A. Sellers (FS11), we have been able to enumerate all degreesequences of sum 2m for which there exists a multigraphic line–Hamiltonian realiza-tion. We will denote the number of degree sequences of sum 2m with a multigraphicline–Hamiltonian realization by dlh(2m). Then we are going to prove the followingtheorem in next section:

Theorem 4.1.2. For all m ≥ 2,

dlh(2m) = p(2m)− 2(m−1∑j=0

p(j)) + 1.

It should be noted that similar enumeration results were discovered recently byRødseth, Sellers, and Tverberg (RST09) when they proved that the number of degree se-quences with multigraphic connected (respectively, non–separable) realizations equalssimilar linear combinations of the partition function p(k). In (RST09), the proofs utilizedincluded generating function manipulations and bijections. In this new joint work, the

31

proof techniques are even more elementary and involve simply understanding the com-binatorial nature of the integer partitions in question. In other words, the techniqueshave even stronger bijective flavor and should not be discounted for their simplicity.

4.2 Enumeration of degree sequences with line–Hamiltonian realization

We begin our proof of Theorem 4.1.2 by considering the enumeration of thosedegree sequences satisfying the criteria 1, 2, and 3a of Theorem 4.1.1 above. To this end,assume d1 + d2 + · · ·+ dn = 2m (this will automatically account for criteria 1) for somefixed value of m, so that the graphs in question have exactly m edges. Then we see thatcriterion 2 is equivalent to 2d1 ≤ d1 + d2 + d3 + · · · + dn or 2d1 ≤ 2m or d1 ≤ m. Froma partition–theoretic point of view, this means that the largest part in our partition is atmost m. Note also that criterion 3a is equivalent then to saying that n ≤ m because ofthe following:

∑di=1

di ≤∑dj>1

(dj − 2) ⇔

∑di=1

di ≤∑dj>1

dj − 2∑dj>1

1 ⇔

2∑di=1

di ≤∑

all dj

dj − 2∑dj>1

1 ⇔

2∑di=1

1 ≤∑

all dj

dj − 2∑dj>1

1 ⇔

2∑

all di

1 ≤∑

all dj

dj ⇔

2n ≤ 2m ⇔n ≤ m

This gives us an upper bound on the number of parts in each partition in ques-tion. Taking all of this information into account, and also recall the partition–theoreticinterpretation of the q–binomial coefficient (see Proposition 1.9), we see that the num-ber of partitions satisfying criteria 1, 2, and 3a is given by the coefficient of q2m in theq–binomial coefficient[

2m

m

]=

(1− qm+1)(1− qm+2

)(1− qm+3) . . . (1− q2m)

(1− q)(1− q2)(1− q3) . . . (1− qm).

This generating function is equivalent to

(1− qm+1)(1− qm+2

)(1− qm+3) . . . (1− q2m)

∞∑N=0

p(N ;≤ m)qN

32

where as in Definition 1.2, p(N ;≤ m) is the number of partitions of N where the largestpart is at most m. A closer consideration reveals that the coefficient of q2m in this gen-erating function is given by

p(2m;≤ m)− p(2m− (m+ 1);≤ m)− p(2m− (m+ 2);≤ m)− · · · − p(2m− 2m;≤ m)

orp(2m;≤ m)− p(m− 1;≤ m)− p(m− 2;≤ m)− · · · − p(0;≤ m)

upon simplification. Next, note that whenever a ≤ m, p(a;≤ m) = p(a). Thus, we nowknow that the coefficient of q2m in the generating function in question is actually equalto

p(2m;≤ m)− p(m− 1)− p(m− 2)− · · · − p(0).

Lastly, note that

p(2m;≤ m) = p(2m)− p(2m;m+ 1)− p(2m;m+ 2)− · · · − p(2m; 2m)

where p(a; b) equals the number of partitions of a with the largest part exactly equal tob. But, of course, when a ≥ b, we have p(a; b) = p(a − b;≤ b) by simply removing thislargest part b from every partition counted by p(a; b). Therefore,

p(2m;≤ m) = p(2m)− p(m− 1;≤ m+ 1)− p(m− 2;≤ m+ 2)− · · · − p(0;≤ 2m)

= p(2m)− p(m− 1)− p(m− 2)− · · · − p(0).

Combining all the results above, we know that the number of partitions satisfying cri-teria 1, 2, and 3a equals

p(2m)− 2m−1∑j=0

p(j). (4.1)

Notice that (4.1) gives almost all of the formula for dlh(2m) as given in Theorem 4.1.2. Asan aside, it should be noted that the values generated by (4.1), namely 1, 3, 8, 18, 39, . . . ,appear as sequence A128552 in Sloane’s Online Encyclopedia of Integer Sequences(Han07).

To close the proof, we must now consider those partitions of 2m which satisfycriteria 1, 2, 3b and the negation of 3a of Theorem 4.1.1 above. Combining criteria 2 and3b we know that n − 1 ≤ m. We also know from the negation of criterion 3a, and thework completed above, that n > m. Clearly, this implies m = n− 1. This means d1 = msince criterion 3b states that d1 = n−1. So we know that the partitions in question musthave largest part equal to m and have exactly n = m+ 1 parts. Lastly, from criterion 1,using the same assumption that d1 + d2 + · · · + dn = 2m we know that the sum of allthe parts in the partition must be 2m. Combining all these facts implies that there canbe only one partition satisfying all these requirements, the partition

2m = m+ 1 + 1 + · · ·+ 1︸ ︷︷ ︸m ones

.

33

(Notice that the graph realization of this degree sequence is a star graph. The line graphof such a star graph is complete, and therefore is clearly Hamiltonian.) This meansthat the contribution to the formula for dlh(2m) which arises from this set of criteria(1+2+3b-3a) is simply 1 (accounting for this one partition above). Combining this with(4.1) yields

dlh(2m) = p(2m)− 2m−1∑j=0

p(j) + 1.

This completes the proof of Theorem 4.1.2.�

34

Chapter 5

Reiner and Stanton’s conjecture

5.1 Introduction

Definition 5.1. The (q, t)-binomial coefficient we are going to discuss in this chapter wasfirst studied in a recent joint paper of Reiner and Stanton (RS10), and it was defined as

[n

k

]q,t

:=k−1∏i=0

1− tqn−qi

1− tqk−qi

. (5.1)

Then we see that

limt→1

[n

k

]q,t

= limt→1

k−1∏i=0

1− tqn−qi

1− tqk−qi

(5.2)

=k−1∏i=0

qn − qi

qk − qi=

[n

k

]q

.

Just like binomial coefficients, there is also a Pascal relation for the (q, t)-binomialcoefficients.

Proposition 5.2 (Reiner and Stanton (RS10)). The (q, t)-binomial coefficients satisfy the fol-lowing:

[n

k

]q,t

=

[n− 1

k − 1

]q,tq

+ tqk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

[n− 1

k

]q,tq. (5.3)

It appears that q being a negative integer is interesting. Reiner and Stanton havethe following partition-theoretical interpretation.

Theorem 5.1.1 (Reiner and Stanton (RS10)). If q ≤ −2 is an integer, then[n

k

]q,t

=∑

λ∈Pk×(n−k)

(−1)number of odd parts of λ ∑

µ compatible with λ

t|µ|, (5.4)

35

where Pk×(n−k) is the collection of all partitions whose Ferrers diagram could fit inside a k ×(n− k) rectangle, and we say µ = p

m1

kpm2

k−1· · · pmk

1is compatible with λ = λ1λ2 · · ·λk if

qλi+1 + · · ·+ q

λi−1 ≤ mi ≤ qλi+1 + · · ·+ q

λi − 1 when λi is even,

qλi+1 + · · ·+ q

λi ≤ mi ≤ qλi+1 + · · ·+ q

λi−1 − 1 when λi is odd.

By running many examples, they (FRS11) have found the following interestingbehavior of the (q, t)-binomial coefficients.

Theorem 5.1.2 (Conjectured by Reiner and Stanton). If q ≤ −2 is a negative integer, then[n

k

]q,t

is a Laurent polynomial in t, all of whose coefficients have the same sign as (−1)k(n−k).

The main goal of this chapter is to provide a proof of this theorem. It shouldbe noted that the proof uses only induction and iteration of the forementioned Pascalrelation (5.3). One key step however, is due to understanding the combinatorial in-terpretation of the (q, t)-binomial coefficient. We give full details of the proof in nextsection and we will elaborate on a subspace interpretation in Section 5.3.

5.2 Proof of Theorem 5.1.2

5.2.1 k = 0 and k = 1

We can directly use (5.1) to compute[n

k

]q,t

for the cases k = 0 and 1. Note that

we assume q ≤ −2 is a negative integer and we will use the following two identitiesthroughout the proof.

For an integer N > 0,

1− tN

1− t= 1 + t+ t

2+ · · ·+ t

N−1; (5.5)

1− t−N

1− t= −(t

−1+ t−2

+ · · ·+ t−N

). (5.6)

Now it is routine to check the cases k = 0 and k = 1.

k = 0: [n

0

]q,t

= 1,

which has the same sign as (−1)0

= 1.

k = 1: [2n

1

]q,t

=1− tq

2n−1

1− tq−1=

1− t(q−1)(q2n−1

+q2n−2

+···+q+1)

1− tq−1

= −(t(−1)(q−1)

+ t(−2)(q−1)

+ · · ·+ t(q

2n−1+q

2n−2+···+q+1)(q−1)

),

36

which has the same sign as (−1)1·(2n−1)

= −1;

[2n+ 1

1

]q,t

=1− tq

2n+1−1

1− tq−1=

1− t(q−1)(q2n

+q2n−1

+···+q+1)

1− tq−1

= 1 + tq−1

+ t2(q−1)

+ · · ·+ t(q

2n+q

2n−1+···+q+1−1)(q−1)

,

which has the same sign as (−1)1·(2n+1−1)

= 1.

5.2.2 Cases with k ≥ 2 and n 6≡ k (mod 2)

We will use induction and the Pascal relation (5.3). Note that just like binomial

coefficient,[n

k

]q,t

= 0 when n < k.

We assume the theorem holds true for[m

k − 1

]q,t

and[m

k − 2

]q,t

with any m, and

also true for[m

k

]q,t

when m 6≡ k (mod 2), m < n. Now we are going to prove the

conjecture for[n

k

]q,t

. We will first use the Pascal relation on[n

k

]q,t

, and then iterate this

relation on[n− 1

k − 1

]q,t

and[n− 1

k

]q,t

.

[n

k

]q,t

=

[n− 1

k − 1

]q,tq

+ tqk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

[n− 1

k

]q,tq

=

[n− 2

k − 2

]q,tq

2+ t

qk−q

k−2∏i=0

1− tqk+1−qi+2

1− tqk−qi+1

[n− 2

k − 1

]q,tq

2+

tqk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

·([n− 2

k − 1

]q,tq

2+ t

qk+1−q

k−1∏i=0

1− tqk+2−qi+2

1− tqk+1−qi+1

[n− 2

k

]q,tq

2

)

=

[n− 2

k − 2

]q,tq

2+ t

qk−q

k−2∏i=0

1− tqk+1−qi+2

1− tqk−qi+1

[n− 2

k − 1

]q,tq

2+ t

qk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2

+tqk+1

+qk−q−1

k−1∏i=0

1− tqk+2−qi+2

1− tqk−qi

[n− 2

k

]q,tq

2

= A(n, k; q)(t) +B(n, k; q)(t) + C(n, k; q)(t) +D(n, k; q)(t),

37

where

A(n, k; q)(t) =

[n− 2

k − 2

]q,tq

2, (5.7)

B(n, k; q)(t) = tqk−q

k−2∏i=0

1− tqk+1−qi+2

1− tqk−qi+1

[n− 2

k − 1

]q,tq

2, (5.8)

C(n, k; q)(t) = tqk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2, (5.9)

D(n, k; q)(t) = tqk+1

+qk−q−1

k−1∏i=0

1− tqk+2−qi+2

1− tqk−qi

[n− 2

k

]q,tq

2(5.10)

are all Laurent polynomials in t. Now by our induction hypothesis,[n− 2

k

]q,tq

2has

the same sign as (−1)k(n−k−2)

= (−1)k(n−k), and note that since q2 > 0, every sin-

gle1− tq

k+2−qi+2

1− tqk−qi

=1− t(q

k−qi)q2

1− tqk−qi

takes plus sign. Consequently, D(n, k; q)(t) has the

same sign as (−1)k(n−k). Therefore we only need to work on the sign of A(n, k; q)(t) +

B(n, k; q)(t) + C(n, k; q)(t).

A(n, k; q)(t) +B(n, k; q)(t) + C(n, k; q)(t)

=

[n− 2

k − 2

]q,tq

2+k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2·(tqk−q

+ tqk−1 1− tq

k+1−q

1− tqk−1

)

=

[n− 2

k − 2

]q,tq

2+

k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2

·(tqk−q − tq

k−1(t(−1)(qk−1)

+ t(−2)(qk−1)

+ · · ·+ tq(q

k−1)))=

[n− 2

k − 2

]q,tq

2+

k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2·(tqk−q − 1−

(t(−1)(qk−1)

+ · · ·+ t(q+1)(q

k−1)))

=

k−3∏i=0

1− tqn−qi+2

1− tqk−qi+2 +

k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

k−2∏i=0

1− tqn−qi+2

1− tqk+1−qi+2

(tqk−q − 1

)+

k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2

(− t(−1)(q

k−1) − · · · − t(q+1)(qk−1)

)

38

=k−3∏i=0

1− tqn−qi+2

1− tqk−qi+2

(1 +

1− tqn−qk

1− tqk−q

(tqk−q − 1

))+

k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2

(− t(−1)(q

k−1) − · · · − t(q+1)(qk−1)

)

= tqn−qk

[n− 2

k − 2

]q,tq

2+k−1∏i=1

1− tqk+1−qi+1

1− tqk−qi

[n− 2

k − 1

]q,tq

2

(− t(−1)(q

k−1) − · · · − t(q+1)(qk−1)

)

By induction,[n− 2

k − 2

]q,tq

2has the same sign as (−1)

(k−2)(n−k)= (−1)

k(n−k). In

addition, since q is negative, every single1− tq

k+1−qi+1

1− tqk−qi

=1− t(q

k−qi)q

1− tqk−qi

takes minus

sign, so the second summand has the same sign as (−1)k−1

(−1)(k−1)(n−k−1)

(−1) =

(−1)(k−1)(n−k)+1

= (−1)k(n−k), note here we need the assumption that n 6≡ k (mod 2).

Because both summands in the last expression above has the desired sign, together with

the correct sign for D(n, k; q)(t), we have proved the theorem for[n

k

]q,t

when n 6≡ k

(mod 2).

5.2.3 Cases with k ≥ 2 and n ≡ k (mod 2)

This case is actually quite easy with what we have established in the first case.

Now we assume the theorem holds true for[m

k − 1

]q,t

with any m, and also true for[m

k

]q,t

with any m such that m 6≡ k (mod 2). We only need to apply the Pascal relation

once in this case.[n

k

]q,t

=

[n− 1

k − 1

]q,tq

+ tqk−1

k−1∏i=0

1− tqk+1−qi+1

1− tqk−qi

[n− 1

k

]q,tq

Now by induction,[n− 1

k − 1

]q,tq

has the same sign as (−1)(k−1)(n−k)

= (−1)k(n−k), here

we need the fact that n ≡ k (mod 2). On the other hand, every single1− tq

k+1−qi+1

1− tqk−qi

has a minus sign, so the second summand has the same sign as (−1)k(−1)

k(n−k−1)=

(−1)k(n−k).

�

39

5.3 A subspace interpretation

Now that we have established Theorem 5.1.2, if we let[n

k

]′q,t

:= (−1)k(n−k)

[n

k

]−q,t

,

it is consequently a Laurent polynomial in t, all of whose coefficients are positive. Soa natural question is to find out what these coefficients count. We provide a possibleanswer in this section. Note that Reiner and Stanton’s original interpretation (Theo-rem 5.1.1) only gives a signed sum, hence will not easily extend to similar results likewe present here.

Fix a basis for the n-dimensional space V over Fq. Relative to this basis, any kdimensional subspace U of V is the row-space of a unique k × n matrix A over Fq inrow-reduced echelon form. This matrix A will have exactly k pivot columns, and the spar-sity pattern for the (possibly) nonzero entries in its nonpivot columns have an obviousbijection to the cells x in a partition λ inside a k× (n− k) rectangle; call these |λ| entriesaij of A the parametrization entries for U . We call a k-dimensional subspace U admis-sible if the corresponding partition λ that fits inside a k × (n − k) rectangle satisfy thefollowing conditions:

• If n 6≡ k (mod 2), each even part (including parts of size 0) has even multiplicity,and the parametrization entry that corresponds to the rightmost cell of the lastcopy of any odd part does not have the value of 0 in Fq;

• if n ≡ k (mod 2), each odd part has even multiplicity, and the parametrizationentry that corresponds to the rightmost cell of the last copy of any even part doesnot have the value of 0 in Fq.

To define the statistics s(U) and s′(U), first fix four bijections φ0, φ1, φ′0

and φ′1:

Fqφ0→ {0, 1, . . . , q − 1},

Fqφ1→ {1, 2, . . . , q},

Fqφ′0→ {−1,−2, . . . ,−q},

Fqφ′1→ {0,−1, . . . ,−q + 1}.

40

For each parametrization entry aij ofA, define two integral functions vU (aij) and v′U

(aij)

as following:

vU (aij) :=

{φ1(aij) if (i, j) is the lowest parametrization entry in its column of Aφ0(aij) otherwise

v′U

(aij) :=

{φ′1(aij) if (i, j) is the lowest parametrization entry in its column of A

φ′0(aij) otherwise

Then define

dU (i, j) := k − i+ j − |{pivot columns left of j}| − 1

s(U) :=∑(i,j)

vU (aij)(qi+dU (i,j) − qdU (i,j)

)

s′(U) :=

∑(i,j)

v′U

(aij)((−q)i+dU (i,j) − (−q)dU (i,j)

)

where both the summation has (i, j) ranging over all parametrization positions in therow-reduced echelon form A for U .

Due to the extra tqn−qk sitting in the last line of the long equation we analyzed on

page 4, we need to define another statistic r(U) to account for that factor. Let mi denotethe multiplicity of the part i in a partition λ.

• If n 6≡ k (mod 2),

r(U) =

(n−k−1)/2∑i=0

m2i

2((−q)n − (−q)k+2i

);

• if n ≡ k (mod 2),

r(U) =

(n−k)/2∑i=1

m2i−12

((−q)n − (−q)k+2i−1).

Theorem 5.3.1. If q is a prime power then[n

k

]q,t

=∑U

ts(U)

,

in which the summation runs over all k-dimensional Fq subspaces U of the n-dimensional Fq-vector space V .

41

Theorem 5.3.2. If q is a prime power then[n

k

]′q,t

:= (−1)k(n−k)

[n

k

]−q,t

=∑U

ts′(U)+r(U)

,

in which the summation runs over all admissible k-dimensional Fq subspaces U of the n-dimensional Fq-vector space V .

42

Chapter 6

q-Orthogonal polynomials

6.1 Introduction

A function on N which is almost always 0 is called lacunary (see (Lov02)). Forq-series or partition functions, say

∑∞n=0

a(n)qn, we also call it lacunary if the coefficient

function a(n) is lacunary. A simple example of lacunary partition function could beEuler’s pentagonal number theorem (Theorem 1.10):

Theorem 6.1.1.∞∏n=1

(1− qn) = 1 +

∞∑m=1

q12m(3m−1)

(1 + qm

) =

∞∑m=−∞

(−1)mq

12m(3m−1) (6.1)

The goal in this chapter is to uniformly treat two lacunary partition functionsstudied in (ADH88) and (CFLZ04).

Theorem 6.1.2 (Theorem 1 in (ADH88)).

∞∑n=0

q(n+12 )

(−q)n=

∞∑n=0

(−1)nq

n(3n+1)2 (1− q2n+1

)

n∑j=−n

(−1)jq−j2 (6.2)

Theorem 6.1.3 (Theorem 2.3 in (CFLZ04)).

∞∑n=0

(−1)n(q)nq

(n+12 )

(−q)n=∞∑n=0

(−1)nq2n

2+n

(1− q2n+1)

n∑j=−n

(−1)jq−j2 (6.3)

It is not that obvious to see the lacunarity for these two partition functions asin the case of Euler’s pentagonal number theorem, but if we use σ(q) to denote theleft-hand side of identity (6.2), then it was first noted in (And86) that

σ(q) =∞∑n=0

S(n)qn

= 1 + q − q2 + 2q3

+ . . .+ 4q45

+ . . .+ 6q1609

+ . . .

Let us try to understand the left-hand side of identity (6.2) combinatorially using

integer partition. Obviously q(n+12 ) could be viewed as the collection of the first n con-

secutive natural numbers, i.e. q(n+12 ) = q

1+2+...+n. On the other hand, 1/(−q)n generatespartitions with the largest part no greater than n in value, and weighted by (−1)

m with

43

m being the number of parts. Now if we union these two pieces of partitions in a properway and recall the definition for the rank of a partition (Definition 1.5), we see that interms of integer partition, S(n) is equal to the number of partitions of n into distinctparts with even rank minus the number with odd rank. By relating the right-hand sideof identity (6.2) to the arithmetic of Q[

√6], Andrews et al. (ADH88) showed that S(n)

assumes all integer values and does so infinitely often. This result trivially implies thelacunary property for σ(q). Theorem 6.1.3 was another successful application of themethod developed in (ADH88), but this time the interaction was with the arithmeticof Q[

√2]. Owing to such strong connection, one would naturally expect some sort of

uniformity in proving these two theorems. Unfortunately, despite the similarity in theprocedures, both the original proofs used seemingly unrelated and mysterious Baileypairs, which obscured the connection.

In Section 6.3, by using orthogonality of the Little q-Jacobi Polynomial (see (AA77))and several other lemmas we are about to list out in the next section, we ”expand” twosimilar-looking q-series in terms of the Little q-Jacobi Polynomial and consequentlyprove two theorems which generalize theorem 6.1.2 and 6.1.3 respectively with threefree variables. We note that the proofs of the two generalized theorems are almost iden-tical and bypass the use of Bailey pair. Despite that a decent amount of work has beendone by q-series manipulation, one of the key lemmas has been proved combinatoriallyusing overpartitions (see (CL04)).

6.2 Lemmas

Let us first prepare ourselves with all the lemmas we need to prove the maintheorems. We will give the proofs for two of these lemmas, while omitting the othersand give reference instead.

Throughout this chapter, we will use the q-Pochhammer symbol and Gaussian Poly-nomial (Definition 1.6).

Definition 6.1 (Little q-Jacobi Polynomial).

Pn(y;A,B; q) =n∑j=0

(q−n

)j(ABqn+1

)j(yq)j

(q)j(Aq)j(6.4)

Lemma 6.2 (Lemma 20 in (And10)).

Pn(−1;−1,−1; q) = (−1)nq(n+1

2 )n∑

j=−n(−1)

jq−j2 (6.5)

We should note that the bilateral series∑n

j=−n(−1)

jq−j2 also appears in the

right-hand side of (6.2) and (6.3), and this is actually the key observation that moti-vated this whole study. We will view both (6.2) and (6.3) as expansions in terms ofLittle q-Jacobi polynomials. Then proving the identities is a matter of verifying thecoefficients in the expansions. To this end, we multiply both sides by a Little q-Jacobi

44

polynomial each time, with n running through all possible values, then both right-handsides will become the corresponding coefficient multiply with ”something” we alreadyknow. Meanwhile, the next two lemmas will take care of the left-hand side of (6.2) and(6.3) respectively. Finally, we simply divide that ”something” from both sides to recoverthe desired coefficient.

Lemma 6.3 (Proved in Theorem 9 from (AA77)).

∞∑i=0

αiqi(qi+1

)∞(βqi+1)∞

Pm(qi;α, β; q)q

in=

(q)∞(αβqm+n+2

)∞(q−n

)mαmqm(n+1)

(βqm+1)∞(αqn+1)∞(αq)m(6.6)

Lemma 6.4.

∞∑i=0

αiqi(qi+1

)∞(βqi+1)∞

Pm(qi;α, β; q)(βq

i+1)n =

(q)∞(αβqm+n+2

)∞(qn−m+1

)m(αβqm+1

)m

(βqn+1)∞(αq)∞(6.7)

Proof of Lemma 6.4:

∞∑i=0

αiqi(qi+1

)∞(βqi+1)∞

Pm(qi;α, β; q)(βq

i+1)n

=∞∑i=0

αiqi(qi+1

)∞(βqi+n+1)∞

m∑j=0

(q−m

)j(αβqm+1

)j(qi+1

)j

(q)j(αq)j

=(q)∞

(βqn+1)∞

m∑j=0

(q−m

)j(αβqm+1

)j(q)j

(q)j(αq)j

∞∑i=0

αi(qj+1

)i(βq

n+1)i

(q)i

(use Proposition 1.11) =(q)∞

(βqn+1)∞

m∑j=0

(q−m

)j(αβqm+1

)j(q)j

(q)j(αq)j·

(αβqn+j+2

)∞(αqj+1)∞

=(q)∞(αβq

n+2)∞

(βqn+1)∞(αq)∞

m∑j=0

(q−m

)j(αβqm+1

)j(q)j

(q)j(αβqn+2)j

(use Proposition 1.12) =(q)∞(αβq

n+2)∞

(βqn+1)∞(αq)∞·

(qn−m+1

)m(αβqm+1

)m

(αβqn+2)m

=(q)∞(αβq

m+n+2)∞(q

n−m+1)m(αβq

m+1)m

(βqn+1)∞(αq)∞

�

Next lemma reveals the orthogonality of the Little q-Jacobi polynomial.

Lemma 6.5 (Theorem 9 in (AA77)).

∞∑i=0

αiqi(qi+1

)∞(βqi+1)∞

Pn(qi;α, β; q)Pm(q

i;α, β; q) =

0 m 6= n

αnqn(q)∞(αβq

n+1)∞(q)n

(βqn+1)∞(αq)∞(αq)n(1− αβq2n+1)m = n

(6.8)

45

The last lemma of this section is of simplificative purpose. We provide twoproofs here. One uses q-Gauss summation (Proposition 1.13), the other does it com-binatorially using overpartition.

Lemma 6.6. For any integer m,n ≥ 0,

∞∑j=0

(−1)jq(

j+12 )q

jn

(qn+1)j+m+1

(j +m

j

)q

= 1 (6.9)

Proof of Lemma 6.6:

Proof I:

∞∑j=0

(−1)jq(

j+12 )q

jn

(qn+1)j+m+1

(j +m

j

)q

= limτ→0

1

(qn+1)m+1

∞∑j=0

(qm+1

)j(q/τ)jqjnτj

(q)j(qn+m+2)j

(use Proposition 1.13) = limτ→0

1

(qn+1)m+1

·(qn+1

)∞(τqn+m+1

)∞(qn+m+2)∞(τqn)∞

=(qn+1

)∞(qn+1)∞

= 1

Proof II: The idea is to interpret the infinite series on the left as generating function forcertain type of partition, then by constructing an involution, we see a lot of cancelation,and eventually we end up with only the term 1 on the right. First note that(

j + 1

2

)+ jn = (n+ 1 + n+ j)j/2 = (n+ 1) + (n+ 2) + . . .+ (n+ j),

and(j+mj

)q

generates partitions with at most j parts, each ≤ m (Proposition 1.9), to-

gether we see q(j+12 )q

jn(j+mj

)q

generates strict partitions with exactly j parts, each part

is from the set {n + 1, n + 2, . . . , n + j + m}. On the other hand, the denominator gen-erates partitions with parts from Sn,m,j = {n+ 1, n+ 2, . . . , n+ j +m+ 1}. Now if weuse the notion of overpartition (CL04), which is a partition wherein the first occurrenceof a part can be overlined, then we can interpret

(−1)jq(

j+12 )q

jn

(qn+1)j+m+1

(j +m

j

)q

46

as overpartitions with parts from Sn,m,j , and part n + j + m + 1 cannot be overlined,

each overpartition is weighted by (−1)j in the infinite series, where j is the number of

overlined parts. We call the set of all these overpartitions P .Next, let us construct an involution φ on P . φ acts on an overpartition π by

switching the status (i.e. overlined or not) of the largest part in π. For example, ifπ = (6, 6, 6, 5, 5, 4), then φ(π) = (6, 6, 6, 5, 5, 4). φ is well defined, actually the onlydoubtful situation is when the largest part of π happens to be n+m+j+1 or n+m+ j,and the associated set for π is Sn,m,j . We just need to note that by adding on a baror taking off a bar for one part, we have also changed Sn,m,j to Sn,m,j+1 or Sn,m,j−1accordingly, so the image φ(π) is still an overpartition in P . And it’s also obvious thatφ is an involution, i.e., φ2 = 1. Since π and φ(π) are weighted with opposite sign in theseries, they cancel out with each other, and we are left with only the empty partition,which explain the term 1 on the right. �

6.3 The Main theorems

Theorem 6.3.1.

∞∑n=0

αnq(

n+12 )y

n

(αq)n=

∞∑n=0

Cn(α, β; q)Pn(y;α, β; q), (6.10)

where

Cn(α, β; q) = (−α)nqn2

(1− αβq2n+1)∞∑j=0

αjq(

j+12 )q

jn

(αβqn+1)j+n+1

(j + n

j

)q

. (6.11)

Proof of Theorem 6.3.1: For a fixedm ≥ 0, we use the orthogonality of the Little q-Jacobi

Polynomial to findCm(α, β; q). First multiply both sides of (6.10) byαiqi(qi+1

)∞(βqi+1)∞

Pm(qi;α, β; q),

then let y = qi and sum over i. Due to Lemma 6.5, we get

RHS = Cm(α, β; q) ·αmqm

(q)∞(αβqm+1

)∞(q)m(βqm+1)∞(αq)∞(αq)m(1− αβq2m+1)

47

While for the left hand side, since y appears as yn, we can use Lemma 6.3 toevaluate.

LHS =

∞∑n=0

αnq(

n+12 )

(αq)n

∞∑i=0

αiqi(qi+1

)∞(βqi+1)∞

Pm(qi;α, β; q)q

in

=

∞∑n=0

αnq(

n+12 )

(αq)n·

(q)∞(αβqm+n+2

)∞(q−n

)mαmqm(n+1)

(βqm+1)∞(αqn+1)∞(αq)m

=∞∑n=m

αn+m

q(n+12 )(q)∞(αβq

m+n+2)∞(q

−n)mq

m(n+1)

(βqm+1)∞(αq)∞(αq)m(Note that (q

−n)m = 0 for n < m)

=

∞∑n=0

αn+2m

q(n+m+1

2 )(q)∞(αβq2m+n+2

)∞(q−n−m

)mqm(n+m+1)

(βqm+1)∞(αq)∞(αq)m

So we get:

Cm(α, β; q) =∞∑n=0

αn+m

q(n+m+1

2 )(q−n−m

)mqm(n+m)

(1− αβq2m+1)

(αβqm+1)n+m+1(q)m

=∞∑n=0

αn+m

q(n+m+1

2 )(−1)mq(

m2 )(q

n+1)m(1− αβq2m+1

)

(αβqm+1)n+m+1(q)m

= (−α)mqm

2

(1− αβq2m+1)

∞∑n=0

αnq(

n+12 )q

nm

(αβqm+1)n+m+1

(n+m

n

)q

�

Corollary 6.7 (Theorem 6.1.2).

∞∑n=0

q(n+12 )

(−q)n=∞∑n=0

(−1)nq

n(3n+1)2 (1− q2n+1

)n∑

j=−n(−1)

jq−j2 (6.12)

Proof: Just set y = α = β = −1 in Theorem 6.3.1, then apply Lemma 6.2 and Lemma 6.6.�

Theorem 6.3.2.

∞∑n=0

αnq(

n+12 )(βyq)n

(αq)n=

∞∑n=0

Cn(α, β; q)Pn(y;α, β; q), (6.13)

where

Cn(α, β; q) = (αβ)nq

3n2+n2 (1− αβq2n+1

)∞∑j=0

αjq(

j+12 )q

jn(βq

n+1)j

(αβqn+1)j+n+1(αqn+1)j

(j + n

j

)q

. (6.14)

48

Proof of Theorem 6.3.2: Almost identical as Theorem 6.3.1, except that instead of Lemma 6.3,we use Lemma 6.4 to evaluate the left hand side, since now y

n appears as (βyq)n. �

Corollary 6.8 (Theorem 6.1.3).

∞∑n=0

(−1)n(q)nq

(n+12 )

(−q)n=∞∑n=0

(−1)nq2n

2+n

(1− q2n+1)

n∑j=−n

(−1)jq−j2 (6.15)

Proof: Just set y = α = β = −1 in Theorem 6.3.2, then apply Lemma 6.2 and Lemma 6.6.�

49

Chapter 7

Conclusion

To conclude this dissertation, I would like to look back on all five topics thatwe have examined in previous chapters and to emphasize on the strong combinatorialflavor they all share. For Bressoud’s conjecture (Theorem 2.1.1) and the congruencefor broken 1-diamond partition (Theorem 3.1.1), direct bijections have been constructedto combinatorially tackle the crucial part (if not the whole bulk) of the problem. No-tably, in the case of broken diamond partition, further generalization naturally presentitself from our in-depth analysis of the original combinatorial construction. Chapter 4and Chapter 5 become a little lighter in the sense of combinatorial flavor. In there, thecombinatorial interpretation as well as the combinatorial analysis are sort of ”behindthe scene” but are still indispensable ingredient for the proof. Moreover, both topicsexemplified how easily and naturally partition theory could be intertwined with otherareas of mathematics, and often elegant results will be discovered from such interplay.I should admit that the alternative proofs using orthogonal polynomials we give in thelast chapter are by no means combinatorial. To compensate that, I have intentionallykept there a combinatorial proof to one of the key lemmas, so that hopefully it will fitwith the theme of the thesis a little better.

The problems we have showcased here only reflect a very small aspect of ap-plying bijective or combinatorial methods in the study of partition theory and relatedareas. Historically, most partition identities were first proved analytically, and onlymuch later combinatorially. Therefore combinatorial proofs are genuinely hard to find,and in a sense are more desirable, but once found, it’s my feeling that they are oftennot difficult to understand and are more likely to be qualified as ”the proof from thebook”. While a large number of combinatorial proofs have been generated by manypeople in the ”golden age” (see Pak’s survey (Pak06)), not much have happened sincethe Garsia-Milne paper (GM81). With the wide acceptance of the usefulness of newtools like modular form, the importance of the classical combinatorial methods seemsto have plunged once again. It is my hope that this study will help to stop this course,if not to reverse it.

50

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VitaShishuo Fu

EducationThe Pennsylvania State University State College, Pennsylvania, USA 2005.8–Present

Ph.D. in Mathematics, expected in August 2011Area of Specialization: Partition Theory and Enumerative Combinatorics

Tsinghua University Beijing, China 2001.8–2005.7

B.S. in Mathematics

Awards and HonorsAward Name yearMathematics Research Communities program 2011 2011.6Grad Student Travel Grant to the Joint Mathematics Meetings 2011.1Graduate Teaching Assistantship, The Pennsylvania State University 2005–2011Graduate Research Assistantship, The Pennsylvania State University 2009–2011

Journal Publications

• Fu, Shishuo. A proof of Bressoud’s conjecture related to the Rogers-Ramanujanidentities. European J. Combin. 31 (2010), No. 8, 2141–2148.

• Fu, Shishuo. Combinatorial proof of one congruence for the Broken 1-diamondPartition and a generalization. International Journal of Number Theory Vol. 7, No. 1(2011) 133-144.

Teaching ExperienceTeaching Associate The Pennsylvania State University 2010.9-2011.5

Teaching Assistant The Pennsylvania State University 2006.1–2010.8Courses taught:

• Math 21 (Pilot Mastery Program): College Algebra I

• Math 22: College Algebra II

• Math 26: Plane Trigonometry

• Math 110: Techniques of Calculus I

• Math 220: Matrices

• Math 231: Calculus of Several Variables

• Math 251: Ordinary and Partial Differential Equations