A study of the crank function with special emphasis on Ramanujan s Lost Notebook

Transcription

1 A study of the crank function with special emphasis on Ramanujan s Lost Notebook arxiv: v1 [math.ho] 26 May 2014 A Project Report Submitted in partial fulfillment of the requirements for the completion of the course MI 515: Project during the programme Integrated M.Sc. in Mathematics. Submitted by Manjil Pratim Saikia Roll No. MSI09001 Supervised by Prof. Nayandeep Deka Baruah Submitted to Department of Mathematical Sciences Tezpur University, Napaam Tezpur Assam, India May 2014

2 To my grandfather, Late Cheniram Saikia ( ) i

3 Declaration I, Manjil Pratim Saikia, do hereby declare that the subject matter in this project report, is the record of survey work, done by me during the last two semesters of my Integrated M. Sc. course with full integrity, honesty and concentration under the immaculate supervision of Prof. Nayandeep Deka Baruah, Department of Mathematical Sciences and submitted to the Department of Mathematical Sciences, Tezpur University in partial fulfilment of the requirements for completing the course MI515: Project. Signature of the Candidate ii

4 Acknowledgments It is my good fortune that I was guided by my supervisor, Prof. Nayandeep Deka Baruah with immense care and encouragement. For this, and for making me fall in love with Ramanujan s mathematics I shall be forever grateful to him. Without his support and guidance this project would not have been completed successfully. I also acknowledge the helpful comments that I recieved from Prof. Bruce C. Berndt, Dr. Atul Dixit and Mr. Zakir Ahmed related to various aspects of Ramanujan s mathematics. I am thankful to Prof. Mangesh B. Rege, Prof. Sujatha Ramdorai and Dr. Rupam Barman for encouraging me in each and every step of my mathematical life and for shoiwng me the beauties of mathematics, each in their own ways. I am thankful to my wonderful parents and sister, Aitu whose love, blessings and support has been a pillar of strength for me during my studies. I shall remain forever grateful to them. I am thankful to my friends Sirat, Madhurrya, Rupam, Parama, Kantadorshi, Dhritishna and Salik for showing such confidence in me, that it always prompted me to work harder. I am also thankful to Prof. Debajit Hazarika, Head of the Department of Mathematical Sciences and all other faculty members for their help and co-operation to make this project a successful one. Finally, I have to acknowledge the support of a DST INSPIRE Scholarship 422/2009 from the Government of India throughout my period of study at Tezpur University. iii

5 ABSTRACT In this project, we study the crank statistic defined for the partition function. In 1919 Ramanujan gave the following remarkable congruences for the partition function p(5n + 4) 0 (mod 5), and p(7n + 5) 0 (mod 7), p(11n + 6) 0 (mod 11). This motivated Freeman Dyson to combinatorially explain these congruences and in doing so, he first conjectured the existence of a crank. This crank was later found in 1987 by George E. Andrews and Frank G. Garvan. However, much earlier Ramanujan had studied the generating function for the crank and offered several beautiful results for them. In the first part of this project we systematically study these results of Ramanujan. As an example of application of cranks to some real applications, we briefly discuss the problem of the parity of p(n) with the help of some recent work by S. J. Kaavya. In the latter part of the project, we look into the spt function of Andrews and examine the Ramanujan like congrunces for this function. We study the associated crank for these congruences given by Andrews, Garvan and F. Liang. We see that many interesting problems related to crank arises in this setting and briefly mention the claimed proof of a conjecture of Andrews, Dyson and Rob Rhoades by William Chen, Kathy Q. Ji and W. J. T. Zang. iv

6 Contents 1 Introduction Motivation Notations and Preliminaries Crank for the Partition Function Cranks and Dissections in Ramanujan s Lost Notebook Other results from the Lost Notebook Cranks The final problem Crank 0 partitions Cranks and the parity of p(n) Invariance and Self-Conjugacy Crank for the spt-function Work of Andrews, Garvan and Liang on the spt-function The spt-crank for ordinary partitions Proof of the Andrews-Dyson-Rhoades conjecture Conclusion 51 1

7 Chapter 1 Introduction This thesis is divided into five chapters. This chapter is a brief introduction to the results that we discuss in the remainder of the thesis and to set the notations and preliminaries that we shall be needing in the rest of our study. In the second chapter, we formally study the crank statistic for the general partition function. We state and prove a few results of George E. Andrews and Frank G. Garvan [8] and then show conclusively that Ramanujan had also studying the crank function much earlier in another guise. We follow the works of Bruce C. Berndt and his collaboraters ([11, 12] and [13]) to study the work of Ramnaujan dealing with cranks. In the third chapter, we relate the crank function to the age old problem of determing the parity of the general partition function. Recent work by S. J. Kaavya [23] has shed much light on this. In the fourth chapter, we study another function defined by Andrews called the spt-function [2] and define the crank for the spt- function. We follow the work of Andrews, Garvan and Liang [6] as well as recent work of William Y. C. Chen Kathy Q. Ji and W. J. T. Zang [15]. In this chapter, we also briefly mention a proof of a conjecture of Andrews, Freeman Dyson and Rob Rhoades [5] which was recently proved by Chen, Ji and Zang [16]. In the final chapter, we give some concluding remarks. Although no originality is claimed with regards to the results stated or proved here, but in some cases the arguments may have been modified to give an easier justification. 2

8 1.1 Motivation The motivation for the project comes from the work of Freeman Dyson [17], in which he gave combinatorial explanations of the following famous congruences given by Ramanujan p(5n + 4) 0 (mod 5), (1.1.1) p(7n + 5) 0 (mod 7), (1.1.2) p(11n + 6) 0 (mod 11), (1.1.3) where p(n) is the ordinary partition function defined as follows. Definition (Partition Function). If n is a positive integer, let p(n) denote the number of unrestricted representations of n as a sum of positive integers, where representations with different orders of the same summands are not regarded as distinct. We call p(n) the partition function. In order to give combinatorial explanations of the above, Dyson defined the rank of a partition to be the largest part minus the number of parts. Let N(m, t, n) denote the number of partitions of n with rank congruent to m modulo t. Then Dyson conjectured that N(k, 5, 5n + 4) = p(5n + 4), 0 k 4, 5 and N(k, 7, 7n + 5) = p(7n + 5), 0 k 6, 7 which yield combinatorial interpretations of (1.1.1) and (1.1.2). These conjectures were later proven by Atkin and Swinnerton-Dyer. The generating function for N(m, n) is given by m= n=0 N(m, n)a m q n = n=0 q n2 (aq; q) n (q/a; q) n. 3

9 Here q < 1, q < a < 1/ q. However, the corresponding analogue of the rank doesn t hold for (1.1.3), and so Dyson conjectured the existence of another statistic which he called crank. In his doctoral dissertation F. G. Garvan defined vector partitions which became the forerunner of the true crank. Let P denote the set of partitions and D denote the set of partitions into distinct parts. Following Garvan, [19] we denote the set of vector partitions V to be defined by V = D P P. For π = (π 1, π 2, π 3 ) V, we define the weight ω( π) = ( 1) #(π 1), the crank( π) = #(π 2 ) #(π 3 ), and π = π 1 + π 2 + π 3, where π is the sum of the parts of π. The number of vector partitions of n with crank m counted according to the weight ω is denoted by We then have N V (m, n) = π V, π =n,crank( π)=m N V (m, n) = p(n). m ω( π). Let N V (m, t, n) denote the number of vector partitions of n with crank congruent to m modulo t counted according to the weights ω. Then we have Theorem (Garvan, [19]). N V (k, 5, 5n + 4) = p(5n + 4), 0 k 4, 5 p(7n + 5) N V (k, 7, 7n + 5) =, 0 k 6, 7 p(11n + 6) N V (k, 11, 11n + 6) =, 0 k On June 6, 1987 at a student dormitory at the University of Illinois, G. E. Andrews and F. G. Garvan [8] found the true crank. Definition (Crank). For a partition π, let λ(π) denote the largest part of π, let µ(π) denote the number of ones in π, and let ν(π) denote the number of parts of π larger than µ(π). The crank c(π) is then defined to be { c(π) = λ(π) if µ(π) = 0, ν(π) µ(π) if µ(π) > 0. 4

10 Let M(m, n) denote the number of partitions of n with crank m, and let M(m, t, n) denote the number of partitions of n with crank congruent to m modulo t. For n 1 we set M(0, 0) = 1, M(m, 0) = 0, otherwise M(0, 1) = 1, M(1, 1) = M( 1, 1) = 1 and M(m, 1) = 0 otherwise. The generating function for M(m, n) is given by m= n=0 M(m, n)a m q n = (q; q) (aq; q) (q/a; q). (1.1.4) The crank not only leads to combinatorial interpretations of (1.1.1) and (1.1.2), but also of (1.1.3). In fact we have the following result. Theorem (Andrews-Gravan [8]). With M(m, t, n) defined as above, M(k, 5, 5n + 4) = M(k, 7, 7n + 5) = M(k, 11, 11n + 6) = p(5n + 4), 0 k 4, 5 p(7n + 5), 0 k 6, 7 p(11n + 6), 0 k Thus, we see that an observation by Dyson lead to concrete mathematical objects almost 40 years after it was first conjectured. Following the work of Andrews and Garvan, there has been a plethora of results by various authors including Garvan himself, where cranks have been found for many different congruence relations. We see that, the crank turns out to be an interesting object to study and in this thesis, we shall devote considerable attention to some of the claims of Ramanujan in his lost notebook in the next chapter. 1.2 Notations and Preliminaries We record here some of the notations and important results that we shall be using throughout our study. For each nonnegative integer n, we set n 1 (a) n := (a; q) n := (1 aq k ), (a) := (a; q) := lim (a; q) n, q < 1. n k=0 5

11 We also set (a 1,..., a m ; q) n := (a 1 ; q) n (a m ; q) n and (a 1,..., a m ; q) := (a 1 ; q) (a m ; q). Ramanujan s general theta function f(a, b) is defined by f(a, b) := a n(n+1)/2 b n(n 1)/2, ab < 1. n= Our aim in Chapter 2 is to study the following more general function F a (q) = (q; q) (aq; q) (q/a; q). To understand the theorems mentioned above and also the results of Ramnaujan stated in his Lost Notebook and the other results to be discussed here, we need some well-known as well as lesser knowns results which we shall discuss briefly here. A more detailed discussion can be found in [3, 10] and [14]. One of the most common tools in dealing with q-series identities is the Jacobi s Triple Product Identity, given by the following. Theorem (Jacobi s Triple Product Identity). For z 0 and x < 1, we have {(1 + x 2n+2 )(1 + x 2n+1 z)(1 + x 2n+1 z 1 )} = n=0 n= x n2 z n. For the sake of completeness, we sketch the proof of this result following Andrews [1]. However, from now on most of the proofs of results that are stated in this section shall be omited. Proof. We have the following two identities by Euler which can be easily verified. and (1 + x n z) = n=0 n=0 x n(n 1)/2 z n (1 x) (1 x n ), (1.2.5) (1 + x n z) 1 = n=0 n=0 6 ( 1) n z n (1 x) (1 x n ), (1.2.6)

12 where x < 1 and in (1.2.6), z < 1. Now, it is easy to see via algebraic manipulations that the Jacobi s Identity follows from (1.2.5) and (1.2.6) for all x < z. The complete argument with no such restriction easily follows from analytic continuation. It is now easy to verify that Ramanujan s theta function f(a, b) satisfies Jacobi triple product identity f(a, b) = ( a; ab) ( b; ab) (ab; ab) (1.2.7) and also the elementary identity f(a, b) = a n(n+1)/2 b n(n 1)/2 f(a(ab) n, b(ab) n ), (1.2.8) for any integer n. We now state a few results that are used to prove some of the results discussed in the next chapter. Theorem (Ramanujan). If A n := a n + a n, then (q; q) = 1 m=1,n=0 ( 1)m q m(m+1)/2+mn (A n+1 A n ). (aq; q) (q/a; q) (q; q) It is seen that the above theorem is equivalent to the following theorem. Theorem (Kăc and Wakimoto). Let a k = ( 1) k q k(k+1)/2, then (a; a) 2 (q/x; q) (qz; q) = k= a k (1 x) 1 xq k. Several times in the sequel we shall also employ an addition theorem found in Chapter 16 of Ramanujan s second notebook [9, p. 48, Entry 31]. Lemma If U n = α n(n+1)/2 β n(n 1)/2 and V n = α n(n 1)/2 β n(n=1)/2 for each integer n, then f(u 1, V 1 ) = N 1 k=0 U k f ( UN+k U k, V ) N k. U k Also useful to us are the following famous result, called the quintuple product identity and Winquist s Identity. 7

13 Lemma (Quintuple Product Identity). Let f(a, b) be defined as above, and let f( q) := f( q, q 2 ) = (q; q), then f(p 3 Q, Q 4 /P 3 ) P 2 f(q/p 3, P 3 Q 5 ) = f( Q 2 ) f( P 2, Q 2 /P 2 ). f(p Q, Q/P ) Lemma (Winquist s Identity). Following the notations given earlier, we have a, q/a, b, q/b, ab, q/(ab), a/b, bq/a, q, q; q) = f( a 3, q 3 /a 3 ){f( b 3 q, q 2 /b 3 ) bf( b 2 q 2, q/b 3 )} ab 1 f( b 3, q 3 /b 3 ){f( a 3 q, q 2 /a 3 ) af( a 3 q 2, q/a 3 )}. In Chapter 3 and 4, we shall be dealing with a variety of combinatorial results related to the partition function. To study this function combinatorially, we shall use a set of diagrams, which are called Ferrers diagram. It is just a pictorial representation of a partition whose rows represent in descending order the parts of the partition. The conjugate of a Ferrers diagram is obtained by flipping the diagram across the diagonal. This has the effect of interchanging the rows and the columns. The Durfee square of a Ferrers diagram is the largest square shape of dots possible in the partition. To illustrate the above, we take an example. The conjugate of the partition of 58 is The Ferrers diagram of these partitions are shown in Figure 1. Fig. 1. Ferrers diagram and its conjugate. 8

14 The Durfee square of the first partition is a 5 5 square shown in Figure 2 below. Fig. 2. Durfee square. In order to prove a few results in Chapter 4, we shall make use of the following very useful lemma, whose proof can be found in [3]. Lemma (Bailey s lemma). A pair of sequences (α n (a, q), β n (a, q)) is called a Bailey pair with parameters (a, q) if n α r (a, q) β n (a, q) =. (q; q) n r (aq; q) n+r r=0 Suppose (α n (a, q), β n (a, q)) is a Bailey pair with parameters (a, q). Then (α n(a, q), β n(a, q)) is another Bailey pair with parameters (a, q), where α n(a, q) = (ρ 1; q) n (ρ 2 ; q) n (aq/ρ 1 ρ 2 ) n α n (a, q), (aq/ρ 1 ; q) n (aq/ρ 2 ; q) n β n(a, q) = n j=0 (ρ 1 ; q) j (ρ 2 ; q) j (aq/ρ 1 ρ 2 ; q) n j (aq/ρ 1 ρ 2 ) j β j (a, q). (q; q) n j (aq/ρ 1 ; q) n (aq/ρ 2 ; q) n Now, we have almost all the details that we need for a detailed study of various results related to crank in the following chapters. 9

15 Chapter 2 Crank for the Partition Function In this chapter, we shall systematically study the claims related to cranks, that are found in Ramanujan s Lost Notebook. We closely follow the exposition of Bruce C. Berndt, Heng Huat Chan, Song Heng Chan and Wen- Chin Liaw [11, 12] and [13]. We also take the help of the womderful exposition of Andrews and Berndt in [4]. This chapter is an expansion of [27]. 2.1 Cranks and Dissections in Ramanujan s Lost Notebook As mentioned in Chapter 1, the generating function for N(m, n) is given by Here q < 1, m= n=0 N(m, n)a m q n = n=0 q n2 (aq; q) n (q/a; q) n. (2.1.1) q < a < 1/ q. Ramanujan has recorded several entries about cranks, mostly about (2.1.1). At the top of page 179 in his lost notebook [26], Ramanujan defines a function F (q) and coefficient λ n, n 0 by Thus, by (2.1.1), for n > 1, F (q) := F a (q) = λ n = m= (q; q) (aq; q) (q/a; q). (2.1.2) M(m, n)a m. Then Ramanujan offers two congruences for F (q). These, like others that are to follow are 10

16 to be regarded as congruences in the ring of formal power series in the two variables a and q. Before giving these congruences, we give the following definition. Definition (Dissections). If P (q) := a n q n n=0 is any power series, then the m-dissection of P (q) is given by P (q) = m 1 j=0 n j =0 a nj m+jq njm+j. We now state the two congruences that were given by Ramanujan below. Theorem (2-dissection). We have F a ( q) f( q3 ; q 5 ) ( q 2 ; q 2 ) + ( a a ) q f( q, q 7 ) ( q 2 ; q 2 ) (mod a 2 + 1a 2 ). (2.1.3) We note that λ 2 = a 2 + a 2, which trivially implies that a 4 1 (mod λ 2 ) and a 8 1 (mod λ 2 ). Thus, in (2.1.3) a behaves like a primitive 8th root of unity modulo λ 2. If we let a = exp(2πi/8) and replace q by q 2 in the definition of a dissection, (2.1.3) will give the 2-dissection of F a (q). Theorem (3-dissection). We have F q (q 1/3 ) f( q2, q 7 )f( q 4, q 5 ) + (a 1 + 1/a) q 1/3 f( q, q8 )f( q 4, q 5 ) (q 9 ; q 9 ) (q 9 ; q 9 ) + (a 2 + 1a ) q 2/3 f( q, q8 )f( q 2, q 7 ) (mod a a ). (2.1.4) 2 (q 9 ; q 9 ) 3 Again we note that, λ 3 = a a 3, from which it follows that a 9 a 6 a 3 1 (mod λ 3 ). So in (2.1.4), a behaves like a primitive 9th root of unity. While if we let a = exp(2πi/9) and replace q by q 3 in the definition of a dissection, (2.1.4) will give the 3-dissection of F a (q). In contrast to (2.1.3) and (2.1.4), Ramanujan offererd the 5-dissection in terms of an equality. Theorem (5-dissection). We have F a (q) = f( q2, q 3 ) f 2 ( q, q 4 ) f 2 ( q 5 ) 4 cos 2 (2nπ/5)q 1/5 f 2 ( q 5 ) f( q, q 4 ) + 2 cos(4nπ/5)q 2/5 f 2 ( q 5 ) f( q 2, q 3 ) 2 cos(2nπ/5)q3/5 f( q, q4 ) f 2 ( q 2, q 3 ) f 2 ( q 5 ). 11 (2.1.5)

17 We observe that (2.1.5) has no term with q 4/5, which is a reflection of (1.1.1). In fact, one can replace (2.1.5) by a congruence and in turn (2.1.3) and (2.1.4) by equalities. This is done in [11]. Ramanujan did not specifically give the 7- and 11-dissections of F a (q) in [26]. However, he vaugely gives some of the coefficients occuring in those dissections. Uniform proofs of these dissections and the others already stated earlier are given in [11]. Ramanujan gives the 5-dissection of F (q) on page 20 of his lost notebook [26]. It is interesting to note that he doesnot give the alternate form analogous to those of (2.1.3) amd (2.1.4), from which the 5-dissection will follow if we set a to be a primitive fifth root of unity. On page 59 in his lost notebook [26], Ramanujan has recorded a quotient of two power series, with the highest power of the numerator being q 21 and the highest power of the denominator being q 22. Underneath he records another power series with the highest power being q 5. Although not claimed by him, the two expressions are equal. This claim was stated in the previous chapter as Theorem In the following, we shall use the results and notations that we discussed in Chapter 1. It must be noted that some of these results have been proved by numerous authors, but we follow the exposition of Berndt, Chan, Chan and Liaw [11]. Theorem Recall that F (q) = F q (q) as defined earlier, then we have ( F a (q) f( q6, q 10 ) + a ) q f( q2, q 14 ) (mod A ( q 4 ; q 4 ) a ( q 4 ; q 4 2 ). (2.1.6) ) We note that (2.1.6) is equivalent to (2.1.3) if we replace q by q. We shall give a proof of this result using a method of rationalization. This method does not work in general, but only for those n-dissections where n is small. Proof. Throughout the proof, we assume that q < a < 1/ q and we shall also frequently use the facts that a 4 1 modulo A 2 and that a 8 1 modulo A 2. We write ( (q; q) ) ( ) = (q; q) (aq n ) k (q n /a) k. (2.1.7) (aq; q) (q/a; q) n=1 We now subdivide the series under product sign into residue classes modulo 8 and then sum the series. Using repeatedly congruences modulo 8 for the powers of a, we shall obtain from k=0 k=0 12

18 (2.1.7) the following (q; q) (aq; q) (q/a; q) (q; q) ( q 4 ; q 4 ) (1 + aq n )(1 + q n /a) (mod A 2 ), (2.1.8) n=1 upon multiplying out the polynomials in the product that we obtain and using the congruences for powers of a modulo A 2. Now, using Lemma with α = a, β = q/a amd congrunces for powers of a modulo A 2, we shall find after some simple manipulations (q; q) ( aq; q) ( q/a; q) f( q 6, q 10 ) + (A 1 1)qf( q 2, q 14 ) (mod A 2 ). Using (2.1.8) in the above, we shall get the desired result. Using similar techniques, but with more complicated manipulations, we shall be able to find the folllowing theorem. Theorem We have F a (q) f( q6, q 21 )f( q 12, q 15 ) (q 27 ; q 27 ) + (A 1 1)q f( q3, q 24 )f( q 12, q 15 ) (q 27 ; q 27 ) + A 2 q 3 f( q3, q 24 )f( q 6, q 21 ) (q 27 ; q 27 ) (mod A 3 + 1). When a = e 2πi/9, the above theorem gives us Theorem For the remainder of this section, we shall use the following notation S n (a) := We note that, when p is an odd prime then n k= n a k. (2.1.9) S (p 1)/2 (a) = a (1 p)/2 Φ p (a), where Φ n (a) is the minimal, monic polynomial for a primitive nth root of unity. We now give the 5-dissection in terms of a congruence. Theorem With f( q), S 2 and A n as defined earlier, we have F a (q) f( q10, q 15 ) f 2 ( q 5, q 20 ) f 2 ( q 25 f 2 ( q 25 ) ) + (A 1 1)q f( q 5, q 20 ) + A 2 q 2 f 2 ( q 25 ) f( q 10, q 15 ) A 1q 3 f( q5, q 20 ) f 2 ( q 10, q 15 ) f 2 ( q 25 ) (mod S 2 ). 13

19 In his lost notebook [26, p. 58, 59, 182], Ramanujan factored the coefficients of F a (q) as functions of a. In particular, he sought factors of S 2 in the coefficients. The proof of Theorem uses the famous Rogers-Ramanujan continued fraction identities, which we shall omit here. We now state and prove the following 7-dissection of F a (q). Theorem With usual notations defined earlier, we have (q; q) 1 (qa; q) (q/a; q) f( q 7 ) (A2 + (A 1 1)qAB + A 2 q 2 B 3 + (A 3 + 1)q 3 AC A 1 q 4 BC (A 2 + 1)q 6 C 2 ) (mod S 3 ), where A = f( q 21, q 28 ), B = f( q 35, q 14 ) and C = f( q 42, q 7 ). Proof. Rationalizing and using Theorem we find (q; q) (qa; q) (q/a; q) (q; q)2 (qa 2 ; q) (q/a 2 ; q) (qa 3 ; q) (q/a 3 ; q) (q 7 ; q 7 ) (2.1.10) 1 f( q 7 ) f( a 2, q/a 2 ) (1 a 2 ) f( a 3, q/a 3 (1 a 3 ) (mod S 3 ). Now using Lemma with (α, β, N) = ( a 2, q/a 2, 7) and ( a 3, q/a 3, 7) respectively, we find that and f( a 2, q/a 2 ) (1 a 2 ) f( a 3, q/a 3 ) (1 a 3 ) A q (a5 a 4 ) (1 a 2 ) B + q3 (a3 a 6 ) (1 a 2 C (mod S 3) (2.1.11) A q (a4 a 6 ) (1 a 3 ) B + q3 (a a2 ) (1 a 3 C (mod S 3). (2.1.12) Now substituting (2.1.11) and (2.1.12) in (2.1.10) and simplifying, we shall get the proof of the desired result. Although there is a result for the 11 dissection as well, but we do not discuss it here. 2.2 Other results from the Lost Notebook Apart from the results that we have discussed so far in this chapter, Ramanujan also recoreded many more entires is his lost notebook which pertains to cranks. For example, on page 58 in his lost notebook [26], Ramanujan has written out the first 21 coefficients in the power series representation of the crank F a (q), where he had incorrectly written the coefficient of q

20 On the following page, beginning with the coefficient of q 13, Ramanujan listed some (but not necessarily all) of the factors of the coefficients up to q 26. He did not indicate why he recorded an incomplete list of such factors. However, it can be noted that in each case he recorded linear factors only when the leading index is 5. On pages 179 and 180 in his lost notebook [26], Ramanujan has offered ten tables of indices of coefficients λ n satisfying certain congruences. On page 61 in [26], he offers rougher drafts of nine of these ten tables, where table 6 is missing. Unlike the tables on pages 179 and 180, no explanations are given on page 61. It is clear that Ramanujan had calculated factors beyond those he has recorded on pages 58 and 59 of his lost notebook as mentioned in the ealier paragraph. In [12], the authors have verifies these claims using a computer algebra software. Although it is clear that Ramanujan believed all his tables are complete, it has not been verified yet. We explain below these tables following Berndt, Chan, Chan and Liaw [12]. Table 1. λ n 0 (mod a 2 + 1/a 2 ) Here Ramanujan indicates which coefficients of λ n have a 2 as a factor. If we replace q by q 2 in (2.1.3), we see that Table 1 contains the degree of q for those terms with zero coefficients for both f( q6, q 10 ) and q f( q2, q 14 ). There are 47 such values. ( q 4 ; q 4 ) ( q 4 ; q 4 ) Table 2. λ n 1 (mod a 2 + 1/a 2 ) Returning again to (2.1.3) and replacing q by q 2, we see that Ramanujan recorded all the indices of the coefficients that are equal to 1 in the power series expansion of f( q6, q 10 ) ( q 4 ; q 4 ). There are 27 such values. Table 3. λ n 1 (mod a 2 + 1/a 2 ) values. This table can be understood in a similar way as the previous table. There are 27 such 15

21 Table 4. λ n a a (mod a2 + 1/a 2 ) Again looking at (2.1.3), we note that a a occurs as a factor of the second expression on the right side. Thus replacing q by q 2, Ramanujan records the indices of all coefficients of q f( q2, q 14 ) which are equal to 1. There are 22 such values. ( q 4 ; q 4 ) ( Table 5. λ n a ) (mod a 2 + 1/a 2 ) a This table can also be interpreted in a manner similar to the previous one. There are 23 such values. Table 6. λ n 0 (mod a + 1/a) Ramanujan here gives those coefficients which have a 1 as a factor. There are only three such values and these values can be discerned from the table on page 59 of the lost notebook. From the calculation (q; q) (aq; q) )(q/a; q) (q; q) = f( q)f( q2 ) ( q 2 ; q 2 ) f( q 4 ) (mod a + 1/a), we see that in this table Ramanujan has recorded the degree of q for the terms with zero coefficients in the power series expansion of f( q)f( q2 ). f( q 4 ) From the next three tables, it is clear from the calculation (q; q) (aq; q) )(q/a; q) (q2 ; q 2 ) = f( q2 )f( q 3 ) ( q 3 ; q 3 ) f( q 6 ) (mod a + 1/a), that Ramanujan recorded the degree of q for the terms with coefficients 0, 1 and 1 respectively in the power series expansion of f( q2 )f( q 3 ). f( q 6 ) Table 7. λ n 0 (mod a 1 + 1/a) There are 19 such values. Table 8. λ n 1 (mod a 1 + 1/a) 16

22 There are 26 such values. Table 9. λ n 1 (mod a 1 + 1/a) There are 26 such values. Table 10. λ n 0 (mod a /a) Ramanujan has put 2 such values. From the calculation (q; q) (aq; q) )(q/a; q) (q; q)2 = f 2 ( q) ( q 3 ; q 3 ) f( q 3 ) (mod a /a), it is clear that Ramanujan had recorded the degree of q for the terms with zero coefficients in the power series expansion of f 2 ( q) f( q 3 ). The infinite products f( q6, q 10 ), q f( q2, q 14 ), f( q)f( q2 ), f( q2 )f( q 3 ) and f 2 ( q) ( q 4 ; q 4 ) ( q 4 ; q 4 ) f( q 4 ) f( q 6 ) f( q 3 ) do not appear to have monotonic coefficients for sufficiently large n. However, if these products are dissected, then we have the following conjectures by Berndt, Chan, Chan and Liaw [12]. Conjecture Each component in each of the dissections for the five products given above has monotonic coefficients for powers of q above 600. The authors have checked this conjecture for n = Conjecture For any positive integers α and β, each component of the (α + β + 1)- dissection of the product f( q α )f( q β ) f( q al+β+1 ) has monotonic coefficients for sufficiently large powers of q. It is clear that Conjecture is a special case of Conjecture for the last three infinite products given above when we set (α, β) = (1, 2), (2, 3), and (1, 1) respectively. Using the Hardy-Ramanujan circle method, these conjectures have been verified by O. -Y. Chan. On page 182 in his lost notebook [26], Ramanujan returns to the coefficients λ n in (2.1.1). He factors λ n for 1 n n as before, but singles out nine particular factors by giving them special notation. These are the factors which occured more than once. Ramanujan uses these 17

23 factors to compute p(n) which is a special case of (2.1.1) with a = 1. It is possible that through this, Ramanujan may have been searching for results through which he would have been able to give some divisibility criterion of p(n). Ramanujan has left no results related to these factors, and it is up to speculation as to his motives for doing this. Again on page 59, Ramanujan lists two factors, one of which is Theorem further below this he records two series, namely, and S 1 (a, q) := a + ( ) ( 1) n q n(n+1)/2 + ( 1)n q n(n+1)/2 1 + aq n a + q n n=1 (2.2.13) S 2 (a, q) := 1 + m=1,n=0 ( 1) m+n q m(m+1)/2+nm (a n+1 + a n ), (2.2.14) where here a 0 := 1. Although no result has been written by Ramanujan, however the authors in [12] have found the following theorem. Theorem With the notations as described above we have Proof. We multiply (2.2.13) by (1 + a) to get by Theorem (1 + a)s 1 (a, q) = S 2 (a, q) = F a (q). ( ) ( 1) n q n(n+1)/2 (1 + a)s 1 (a, q) = 1 + (1 + a) + ( 1)n q n(n+1)/2 1 + aq n a + q n n=1 ( ) ( 1) n q n(n+1)/2 = 1 + (1 + a) + ( 1)n q n(n 1)/2 1 + aq n 1 + aq n n=1 = 1 + (1 + a) n 0 = = n= ( 1) n q n(n+1)/2 1 + aq n ( 1) n q n(n+1)/2 (1 + a) 1 + aq n (q; q) 2 ( aq; q) ( q/a; q), (2.2.15) 18

24 Secondly by Theorem 1.2.2, we have S 2 (a, q) = 1 + = m=1,n=0 ( 1) m q m(m+1)/2+nm ( ( a) n+1 ( a) n 1 + ( a) n + ( a) n ) (q; q) 2 ( aq; q) ( q/a; q). (2.2.16) Thus, (2.2.15) and (2.2.16) completes the proof. From the preceeding discussion, it is clear that Ramanujan was very interested in finding some general results with the possible intention of determining arithmetical properties of p(n) from them by setting a = 1. Although he found many beautiful results, but his goal eluded him. The kind of general theorems on the divisibility of λ n by sums of powers of a appear to be very difficult. Also, a challenging problem is to show that Ramanujan s Table 6 is complete. 2.3 Cranks The final problem In his last letter to G. H. Hardy, Ramanujan announced a new class of functions which he called mock theta functions, and gave several examples and theorems related to them. The latter was dated 20th January, 1920, a little more than three months before his death. Thus for a long time, it was widely believed that the last problem on which Ramanujan worked on was mock theta functions. But the wide range of topics that are covered in his lost notebook, [26] suggests that he had worked on several problems in his death bed. Of course, it is only speculation and some educated guess that we can make. But, the work of Berndt, Chan, Chan and Liaw [13] has provided conclusive evidence that the last problem on which Ramanujan worked on was cranks, although he would not have used this terminology. We have already seen, the various dissections of the crank generating function that Ramanujan had provided. In the preceeding section, we saw various other results of Ramanujan related to the coefficients λ n. In order to calculate those tables, Ramanujan had to calculate with hand various series upto hundreds of coefficients. Even for Ramanujan, this is a tremendous task and we can only wonder what might have led him to such a task. Clearly, there was very 19

25 little chance that he might have found some nice congruences for these values like (1.1.1) for example. In the foregoing discussion of the crank, pages 20, 59-59, 61, 53-64, 70-71, and are cited from [26]. Ramanujan s 5-dissection for the crank is given on page 20. The remaining ten pages are devoted to the crank. Infact in pages 58-89, there is some scratch work from which it is very difficult to see where Ramanujan was aiming them at. But it is likely that all these pages were related to the crank. In [13], the authors have remarked that pages 65 (same as page 73), 66, 72, 77, 80-81, and are almost surely related to the crank, while they were unable to determine conclusively if the remaining pages pertain to cranks. In 1983, Ramanujan s widow Janaki told Berndt that there were more pages of Ramanujan s work than the 138 pages of the lost notebook. She claimed that during her husband s funeral service, some gentlemen came and took away some of her husband s papers. The ramining papers of Ramanujan were donated to the University of Madras. It is possible that Ramanujan had two stacks of paper, one for scratch work or work which he did not think complete, and the other where he put down the results in a more complete form. The pages that we have analysed most certainly belonged to the first stack and it was Ramanujan s intention to return to them later. In the time before his death, it is certainly clear that most of Ramanujan s mathematical thoughts had been only on one topic - cranks. However, one thing is clear that Ramanujan probably didn t think of the crank or rank as we think of it now. We do not know with certainty whether or not Ramanujan thought combinatorially about the crank. Since his notebooks contain very little words and also since he was in his death bed so he didn t waste his time on definitions and observations which might have been onvious to him; so we are not certain about the extend to which Ramanujan thought about these objects. It is clear from some of his published papers that Ramanujan was an excellent combinatorial thinker and it would not be surprising if he had many combinatorial insights about the crank. But, since there is not much recorded history from this period of his life, we can at best only speculate. 20

26 Chapter 3 Crank 0 partitions One of the most well-known problems in partition theory is the parity problem, that is how often is p(n) even or odd? Motivated by this problem, S. J. Kaavya [23] had obtained the generating function for the number of crank 0 partitions of n and an involution on these partitions whose fixed points ae called invariant partitions. Also the generating function for the number of self-conjugate rank 0 partitions was obtained. In this chapter, we shall discuss these advances briefly, which shall also give us an application of the crank to the well-known problem of determining the parity of p(n). 3.1 Cranks and the parity of p(n) The Ramanujan partition functions (1.1.1), (1.1.2) and (1.1.3) have been a long standing subject of study. Various people have tried to find generalizations of these beautiful congruences. Recently, the work of Ken Ono has revealed that similar congruences for p(n) exists for any modulus coprime to 6. But this leaves open the characterizations of those values of n for which p(n) is divisible by moduli not coprime to 6, in particular moduli 2 and 3. Recent work by Kaavya [23] has shared some light on new means of attacking the problem for determining when p(n) 0 mod 2. It was shown by Berkovich and Garvan that there is always an involution between the crank k and crank k partitions of n for all values of k. Thus p k (n) = p k (n). 21

27 Hence it follows that the crank 0 partitions of n have the same parity as p(n) itself. Since such partitions are less in number than p(n) so it may be eaier to investigate the parity of p 0 (n) than p(n). Now the question comes how to make a crank 0 partition? A crank 0 partition is made by building a Ferres diagram on something called a k-root. This is a k by k + 1 rectangle of dots, with a column of k dots underneath, for a total of k 2 + 2k dots. We observe that this is the smallest number for which there is a crank 0 partition with exactly k parts equal to 1. We call the k by k + 1 rectangle the body and the column of k dots the tail. The Ferres diagram of crank 0 partitions can be built as follows, first in between the body and the tail new rows of length greater than 1 and less than k + 1 may be inserted in order of non-increasing length. Second, to the right of the body any number of columns of length k may be inserted again in order of non-decreasing length. Fig. 3. Crank 0 k-root. We denote the number of crank m partitions of n which are built on k-roots as p k m(n). This is also the number of partitions of n such that the crank equals m and the number of parts equal to 1 is exactly k. We now have the following result. 22

28 Theorem (Kaavya [23]). The generating function for p 0 (n) is given by Proof. We have p 0 (n)q n q k2 +2k = (1 q). (q; q) 2 k n=1 k=1 p 0 (n) = p 1 0(n) + p 2 0(n) + p 3 0(n) +. For each k we note that the terms in the generating function of p k 0(n) has one-one correspondence with the Ferres diagram for crank 0 partitions built up from the k-root. So, we can write the generating function one term at a time, by listing the corresponding Ferrers diagrams. We observe that the unique crank 0 partition built up from the 1-root is of the form (n 1)+1 for n 3. Since the Ferrers diagram consists of three dots so each term in the generating function for p 1 0(n) contains a factor of q 3. We can then add i dots to the largest part of the root, and that will contribute successive factors of q i. Thus we have p 1 0(n) = q 3 (1 + q + q 2 + ) = q3 1 q. For the 2-root (which is ) there are new possibilities. We xannot only lengthen the first part, but we may also lengthen both size-3 parts and we can insert parts of size 2 in between the body and the tail. Lengthening both size-3 parts by the same amount is equivalent to adding some number of columns of length 2 to the right of the body. If we add i columns of size 2 to the right of the body, then we have the following possibilites (1) all i columns are to the right of the body, (2) o 1 columns are next to the body and a single row of a pair of dots is in between the body and the tail, (3) i 2 columns are to the right of the body and two rows of pairs of dots are in between the body and the tail,. (i+1) all i pairs of dots are in between the body and the tail in i rows. Thus we have p 2 0(n) = q 8 (1 + q )(1 + 2q 2 + 3q 4 + 4q 6 + ) = q 8 (1 q)(1 q 2 ) 2 q 8 = (1 q) (q; q) 2 2 We can extend this argument and get the following from which the result follows. p k +2k 0(n) = (1 q) qk2, (q; q) 2 k 23

29 We also notice that the conjugate of any rank m partition is a rank m partition. Therefore, there is a bijection between the rank m and rank m partitons of n, which means that the parity of rank 0 partitions gives us the parity of p(n). Let r m (n) denote the number of rank m partitions of n. Then Dyson [17] had obtained the following generating function for r 0 (n) 1 ( 1) k 1 q (3k2 k)/2 (1 q k ). (q; q) k>0 3.2 Invariance and Self-Conjugacy In this section we shall describe an involution on crank 0 partitions of m demonstrated by Figure 4. The partition on the left is built using a 7-root represented by plain dots. The dots to the right of the body are of two kinds. The excess of the largest part over the second largest is represented by decorated dots and is called the nose while the rest shown in the dotted box A are hatched dots. The black dots shown in the dotted box B represent the stuff in between the body and the tail. To apply the involution, we first conjugate the hatched dots in box A and the black dots in box B. Then we swap their positions. This yields a partition as shown in the right of the figure. We note that the nose remains inert in the process. The fixed points under this involution will be called invariant crank 0 partitions. Also, the number of such partitions will be denoted by p 0(n). The involution defined above will pair up the crank 0 partitions except for the invariant ones. Thus the parity of p 0(n) will determine the parity of p(n). 24

30 Fig. 4. Crank 0 involution. By another clever combinatorial technique, similar to the proof of Theorem 3.1.1, we have the following result. Theorem The generating function for p 0(n) is given by p 0(n)q n q k2 +2k = (1 + q). (q 2 ; q 2 ) k n=1 The j-root rank 0 partition is defined simply to be a j j square of dots. In general, the root of any rank 0 partition will just be the Durfee square of the partition. We observe that the conjugate of any rank 0 partition is also a rank 0 partition with the same Durfee square. Further, conjugacy is an involution on the set of rank 0 partitions. The number of self-conjugate rank 0 partiotns of n will be denoted by p 0 (n). For similar reasons like above, the parity of p 0 (n) determines the parity of p(n). Once we have the root, the dots next to the Durfee square and underneath it will have to be conjugates of each other for the entire partition to be self-conjugate. After, we use j 2 dots to make a j-root, we shall use a similar argument like in Theorem to get the following result. Theorem The generating function for p 0 (n) is given by p 0 (n)q n q j2 =. (q 2 ; q 2 ) j n=1 j=1 k=1 25

31 Thus, so far there have been several results stated in this chapter which gives us an alternate approach to a classical problem. We shall close this chapter with the following conjecture of Kaavya [23]. Conjecture r 0 (n) p 0 (n) for all n. Recent results of R. Mao [25] has asymptotically proved this conjecture. In the next chapter, we shall give a brief description of another conjecture which implies Conjecture for a special case. 26

32 Chapter 4 Crank for the spt-function In the study of partition identities, it has often been of interest to consider weighted counts of partitions. For example we have the following beautiful result. Theorem 4.1 (Fokkink, Fokkink and Wang, [18]). If D n denotes the set of partitions π of n into distinct parts, then π D n ( 1) #(π) σ(π) = d(n), where #(π) is the number of parts of π, σ(π) is the smallest part of π and d(n) is the number of divisors of n. The aim of this chapter is to study the spt-function of Andrews introduced in 2008, [2]. We denote by spt(n) the total number of appearences of the smallest part in each integer partition of n. For example, we have spt(4) = 10. In view of Theorem 4.1, it is not unreasonable to think that this spt function will also have some interesting properties. In [2], G. E. Andrews found some remarkable properties of the spt-function which reminds us of the Ramanujan congruence of partitions. This chapter is devoted to understanding these congruences. In the first section we shall follow the work of Andrews [2] and Andrews, Garvan and Jie Liang [6] and [7] on the spt-crank. In the next section, we shall study the work of Andrews, Dyson and Rhoades [5] and that of Chen, Ji and Zang [15] and [16]. In the last section of this chapter, we shall sketch a proof of a conjecture made in [5]. 27

33 4.1 Work of Andrews, Garvan and Liang on the sptfunction In [2], Andrews found the following remarkable congruence properties of the spt-function which reminds one of the famous Ramanujan congruences for p(n). spt(5n + 4) 0 (mod 5) (4.1.1) spt(7n + 5) 0 (mod 7) (4.1.2) Andrews also proved the following two theorems in [2]. spt(13n + 6) 0 (mod 13) (4.1.3) Theorem spt(n)q n = n 1 Theorem (q; q) n=1 nq n 1 q + 1 n (q; q) n=1 spt(n) = np(n) 1 2 N 2(n), ( 1) n q n(3n+1)/2 (1 + q n ) (1 q n ) 2. where N 2 (n) is the second Atkin-Garvan moment given by N j (n) = m j N(m, n). m= The generating function for spt(n) is not very difficult to find, once we have the right tools at our disposal. Here, we shall prove Theorem below. Proof. We know that Thus we see that n=0 p(n)q n = n=0 np(n)q n = q d 1 = dq (q; q) by logarithmic differentiation. Next we have the following result from [2]. 1 (q; q). 1 (q; q) n=1 nq n 1 q n, (4.1.4) 28

34 1 2 N 2(n)q n = 1 (q; q) n 0 n=1 ( 1) n q n(3n+1)/2 (1 + q n ) (1 q n ) 2. Finally substituting (4.1.4) in the above and comparing coefficients we shall get the desired result. The first two congruences (4.1.1) and (4.1.2) follow very easily from results of Atkin and Swinnerton Dyer, once we have Theorem The third congruence is a little tricky and uses much more machinary and uses some results of J. N. O Brien. Like the crank for the ordinary partition function, it was believed that there exists a crank for the spt-function, in order to explain (4.1.1), (4.1.2) and (4.1.3). In [6], Andrews, Garvan and J. Liang found a vector crank, like the one found by Garvan which explained the first two congruences for the spt-function. We give a brief description of the same below. For a partition π, we define by s(π), the smallest part that appears in π. For the empty partition, we have s( ) =. We define the following subset of the set of vector partitions S := { π = (π 1, π 2, π 3 ) V : 1 s(π 1 ) < and s(π 1 ) min(s(π 2 ), s(π 3 ))}. For π S, we define the weight ω 1 by ω 1 ( π) = ( 1) #(π1) 1. The number of vector partitions of n in S with crank m counted according to the weight ω 1 is denoted by N S (m, n). We can prove that N S (m, n) = N S ( m, n) and using a similar notation we have N S (m, t, n) = N S (t m, t, n). The main result in [6] is the following Theorem (Andrews, Garvan and Liang, [6]). N S (k, 5, 5n + 4) = N S (k, 7, 7n + 5) = spt(5n + 4), 0 k 4, 5 spt(7n + 5), 0 k 6. 7 This explained the first two congruences of the spt-function. It remains an open question to find a result like the above for the third congruence. In [20], it was proven that N V (m, n) 0. We similarly have the following remarkable result. 29

35 Theorem (Andrews, Garvan and Liang, [6]). for all (m, n). N S (m, n) 0 To prove Theorem 4.1.3, the techniques that are employed in [6] are q-series manipulation and Lemma The main idea is to write the generating function for the spt-function in an alternate form and then use Bailey s lemma. The proof of Theorem uses many different techniques of manipulating basic hypergeometric series and as such we shall omit it here. The authors in [6] also posed a few problems, some of which we shall partially answer in the remainder. They posed the following two main problems. (1) Find a statistic on partitions that explains Theorem combinatorially. More precisely find a statistic s crank : P Z and a weight function φ : P N such that φ(π) = spt(n) π P, π =n and π P, π =n,s rank(π)=m φ(π) = N S (m, n). (2) Find a crank type result that explains (4.1.3). Unfortunately the spt-crank does not work for (4.1.3), in fact it does not even hold for the first case. We now come to the work of Andrews, Garvan and Liang in [7], where they relate a few results on the parity of spt(n) aming other things. We define the map ι : S S given by ι( π) = ι(π 1, π 2, π 3 ) = ι(π 1, π 3, π 2 ), where S and π are as defined earlier in this section. This map is a natural involution. An S-partition is a fixed point of ι if and only if π 2 = π 3. We call these fixed-points self-conjugate 30

36 S-partitions. The number of self-conjugate S-partitions counted according to the weight ω 1 is denoted by N SC (n), so that N SC (n) = ω 1 ( π). π S, π =n,ι( π)= π SInce ι is an involution that preserves the weight ω 1, so we have N SC (n) spt(n) (mod 2), for all n. One of the main results in [7] is the following theorem. Theorem SC(q) := N SC (n)q n = n=1 n=1 ( 1) n 1 q n2 (q; q 2 ). The authors have found many more results related to q-series and mock theta functions in [7], but they lie outside the scope of this report. It is also a natural question to ask about the largest parts partition function or the lpt-function and its properties. It turns out that this function is not as interesting as the spt-function due to the generating function of lpt-function. It is highly unlikely that nifty congruences like (4.1.1), (4.1.2) and (4.1.3) are true for the lpt-function. For the sake of completeness, we mention that the generating function for the lpt-function is given by lpt(n)q n = n=1 n=1 q 2 (q; q) n 1 (1 q n ) The spt-crank for ordinary partitions In [5], the authors defined what are called marked partitions. Definition (Marked Partition). A marked partition is a pair (λ, k) where λ is a partition and k is an integer identifying one of its smallest parts. If there are s smallest parts then k = 1, 2,..., s. They also proposed a challenge to define the spt-crank in terms of ordinary partitions or marked partitions. This challenge was accepted by Chen, Qi and Zang [15] and they found a neat combinatorial way to define the crank in terms of marked and doubly marked partitions. 31

37 In this section, we introduce the structure of doubly marked partitions and define the sptcrank of a doubly marked partition. We show that N S (m, n) can be interpreted as the number of doubly marked partitions of n with spt-crank m. Moreover, we establish a bijection between marked partitions of n and doubly marked partitions of n. A marked partition is defined by Andrews, Dyson and Rhoades as a partition with exactly one of the smallest parts marked. They consider it a challenge to find a definition of the spt-crank of a marked partition so that the set of marked partitions of 5n+4 and 7n+5 can be divided into five and seven equinumerous classes. The definition of spt-crank for doubly marked partitions and the bijection between the marked partitions and doubly marked partitions leads to a solution to the problem of Andrews, Dyson and Rhoades. We closely follow the work of Chen, Ji and Zang [15]. In the following definition, we assume that a partition λ of n is represented by its Ferrers diagram, and we use D(λ) to denote size of the Durfee square of λ. Definition (Doubly Marked Partition). A doubly marked partition of n is an ordinary partition λ of n along with two distinguished columns indexed by s and t, denoted (λ, s, t) where 1. 1 s D(λ), 2. s t λ t, 3. λ s = λ t. For example ((3, 2, 2), 1, 2) is a doubly marked partition, but ((3, 2, 1), 1, 2) and ((3, 2, 2), 2, 1) are not. Fig. 5. Illustration for doubly marked partition. To define the spt-crank of a doubly marked partition (λ, s, t), we let g(λ, s, t) = λ s s + 1, 32