RANK DIFFERENCES FOR OVERPARTITIONS

RANK DIFFERENCES FOR OVERPARTITIONS JEREMY LOVEJOY AND ROBERT OSBURN Abstract In 1954 Atkin Swinnerton-Dyer proved Dyson s conjectures on the rank of a partition by establishing formulas for the generating functions for rank differences in arithmetic progressions In this paper we prove formulas for the generating functions for rank differences for overpartitions These are in terms of modular functions generalized Lambert series 1 Introduction The rank of a partition is the largest part minus the number of parts This statistic was introduced by Dyson [14] who observed empirically that it provided a combinatorial explanation for Ramanujan s congruences p(5n + 4) 0 (mod 5) p(7n + 5) 0 (mod 7) Here p(n) denotes the usual partition function Specifically Dyson conjectured that if N(s m n) denotes the number of partitions of n whose rank is congruent to s modulo m then for all 0 s 4 0 t 6 we have p(5n + 4) N(s 5 5n + 4) = 5 p(7n + 5) N(t 7 7n + 5) = 7 Atkin Swinnerton-Dyer proved these assertions in 1954 [3] In fact they proved much more establishing generating functions for every rank difference N(s l ln + d) N(t l ln + d) with l = 5 or 7 0 d s t < l Many of these turned out to be non-trivially 0 while others were infinite products still others were generalized Lambert series related to Ramanujan s third order mock theta functions Such formulas in the case l = 11 were subsequently given by Atkin Hussain [2] in a similar (though technically far more difficult) manner Dyson s rank extends in the obvious way to overpartitions Recall that an overpartition [10] is simply a partition in which the first occurrence of each distinct number may be overlined For example the 14 overpartitions of 4 are 4 4 3 + 1 3 + 1 3 + 1 3 + 1 2 + 2 2 + 2 2 + 1 + 1 2 + 1 + 1 2 + 1 + 1 2 + 1 + 1 1 + 1 + 1 + 1 1 + 1 + 1 + 1 Overpartitions (often under other names) naturally arise in diverse areas of mathematics where partitions already occur such as mathematical physics [15 16] symmetric functions [4 13] representation theory [20] algebraic number theory [21 23] In the study of basic hypergeometric series Dyson s rank for overpartitions its generalizations play an important role in combinatorial studies of Rogers-Ramanujan type identities [11 12 22] Date: May 28 2007 2000 Mathematics Subject Classification Primary: 11P81 05A17; Secondary: 33D15 The first author was partially supported by an ACI Jeunes Chercheurs et Jeunes Chercheuses 1

2 JEREMY LOVEJOY AND ROBERT OSBURN As for arithmetic properties let N(s m n) be the number of overpartitions of n with Dyson s rank congruent to s modulo m It has recently been shown [5] that if u is any natural number m is odd l 5 is a prime such that l 6m or l j = m then there are infinitely many non-nested arithmetic progressions An + B such that N(s m An + B) 0 (mod l u ) for all 0 s < m This is analogous to congruences involving Dyson s rank for partitions [6] On the other h there are no congruences of the form p(ln + d) 0 (mod l) for primes l 3 [9] Here p(n) denotes the number of overpartitions of n The generating functions for the rank differences N(s l ln + d) N(t l ln + d) then provide a measure of the extent to which the rank fails to produce a congruence p(ln + d) 0 (mod l) In this paper we find formulas for these generating functions for l = 3 5 in terms of modular functions generalized Lambert series Using the notation (11) R st (d) = n 0 ( N(s l ln + d) N(t l ln + d) ) q n where the prime l will always be clear the main results are summarized in Theorems 11 12 below Theorem 11 For l = 3 we have (12) R 01 (0) = 1 + (q3 ; q 3 ) 2 ( q; q) ( q 3 ; q 3 ) 2 (13) R 01 (1) = 2(q3 ; q 3 ) (q 6 ; q 6 ) (14) R 01 (2) = 4( q3 ; q 3 ) 2 (q 6 ; q 6 ) 2 (q 2 ; q 2 6( q3 ; q 3 ) ) (q 3 ; q 3 ) Theorem 12 For l = 5 we have ( 1) n q 3n2 +3n 1 q 3n+1 (15) R 12 (0) = 2q(q 10 ; q 10 ) (q 3 q 4 q 6 q 7 ; q 10 ) (16) R 12 (1) = 2q( q5 ; q 5 ) (q 5 ; q 5 ) ( 1) n q 5n2 +5n 1 q 5n+2 (17) R 12 (2) = 2(q10 ; q 10 ) (q q 4 ; q 5 ) (18) R 12 (3) = 2(q10 ; q 10 ) (q 2 q 3 ; q 5 )

RANK DIFFERENCES FOR OVERPARTITIONS 3 (19) R 12 (4) = 6( q5 ; q 5 ) (q 5 ; q 5 ) ( 1) n q 5n2 +5n 1 q 5n+1 (110) R 02 (0) = 1 + ( q2 q 3 ; q 5 ) (q 5 ; q 5 ) (q 2 q 3 ; q 5 ) ( q 5 ; q 5 ) (111) R 02 (1) = 2(q 4 q 6 q 10 ; q 10 ) (q 2 q 8 ; q 10 ) 2 (q 3 q 7 ; q 10 ) + 4q( q5 ; q 5 ) (q 5 ; q 5 ) (112) R 02 (2) = 0 4(q 2 q 8 q 10 ; q 10 ) (q 4 q 6 ; q 10 ) 2 (q q 9 ; q 10 ) ( 1) n q 5n2 +5n 1 q 5n+2 (113) R 02 (3) = 2(q10 ; q 10 ) (q 2 q 3 ; q 5 ) (114) R 02 (4) = 2(q 2 q 8 q 10 ; q 10 ) (q 4 q 6 ; q 10 ) 2 (q q 9 ; q 10 ) 2( q5 ; q 5 ) (q 5 ; q 5 ) ( 1) n q 5n2 +5n 1 q 5n+1 Here we have employed the stard basic hypergeometric series notation [17] n 1 (a 1 a 2 a j ; q) n = (1 a 1 q k )(1 a 2 q k ) (1 a j q k ) k=0 following the custom of dropping the ; q unless the base is something other than q We should also remark that if the number of overpartitions of n with rank m is denoted by N(m n) then conjugating Ferrers diagrams shows that N(m n) = N( m n) [22] Hence the values of s t considered in Theorems 11 12 are sufficient to find any rank difference generating function R st (d) To prove our main theorems we adapt the method of Atkin Swinnerton-Dyer [3] This may be generally described as regarding groups of identities as equalities between polynomials of degree l 1 in q whose coefficients are power series in q l Specifically we first consider the expression (115) { } N(s l n) N(t l n) n=0 q n By (26) (27) (53) we write (115) as a polynomial in q whose coefficients are power series in q l We then alternatively express (115) in the same manner using Theorem 12 Lemma 31 Finally we use various q-series identities to show that these two resulting polynomials are the same for each pair of values of s t We should stress that this technique requires knowing all of the generating function formulas for the rank differences beforeh We cannot prove a generating function for R st (d) for some d without proving them for all d The paper is organized as follows In Section 2 we collect some basic definitions notations generating functions In Section 3 we record a number of equalities between an infinite

4 JEREMY LOVEJOY AND ROBERT OSBURN product a sum of infinite products These are ultimately required for the simplification of identities that end up being more complex than we would like principally because there is only one 0 in Theorem 12 In Section 4 we prove two key q-series identities relating generalized Lambert series to infinite products in Section 5 we give the proofs of Theorems 11 12 Before proceeding it is worth mentioning that there is a theoretical reason why some of the rank differences in the main theorems are modular others are not We do not go into great detail here but the weak Maass forms that lie behind the rank differences N(s l ln + d) N(t l ln + d) [5] can be shown in many cases (such as when d is not a square modulo l) to be weakly holomorphic modular forms This has been carried out in detail for the rank differences for ordinary partitions in [7] 2 Preliminaries We begin by introducing some notation definitions essentially following [3] With y = q l let Thus we have r s (d) := N(s l ln + d)y n n=0 r st (d) := r s (d) r t (d) l 1 N(s l n)q n = r s (d)q d n=0 To abbreviate the sums occurring in Theorems 11 12 we define Σ(z ζ q) := d=0 ( 1) n ζ n q n2 +n 1 zq n Henceforth we assume that a is not a multiple of l We write Σ(a b) := Σ(y a y b y l ) = Σ(0 b) := ( 1) n y bn+ln(n+1) 1 y ln+a ( 1) n y bn+ln(n+1) 1 y ln where the prime means that the term corresponding to n = 0 is omitted To abbreviate the products occurring in Theorems 11 12 we define P (z q) := (1 zq r 1 )(1 z 1 q r ) r=1 P (a) := P (y a y l )

RANK DIFFERENCES FOR OVERPARTITIONS 5 P (0) := (1 y lr ) Note that P (0) is not P (a) evaluated at a = 0 We also have the relations r=1 (21) P (z 1 q q) = P (z q) (22) P (zq q) = z 1 P (z q) From (21) (22) we have (23) P (l a) = P (a) (24) P ( a) = P (l + a) = y a P (a) (25) In [22] it is shown that the two-variable generating function for N(m n) is n=0 N(m n)q n = ( 1) n 1 q n2 + m n 1 qn 1 + q n From this we may easily deduce that the generating function for N(s m n) is (26) n=0 n=1 N(s m n)q n = Hence it will be beneficial to consider sums of the form (27) S(b) := We will require the relation ( 1) n q n2 +n (q sn + q (m s)n ) (1 + q n )(1 q mn ) ( 1) n q n2 +bn 1 q ln (28) S(b) = S(l b) which follows from the substitution n n in (27) We shall also exploit the fact that the functions S(l) are essentially infinite products Lemma 21 We have S(l) = + 1 2

6 JEREMY LOVEJOY AND ROBERT OSBURN Proof Using the relation (28) we have 2S(l) = 2 = ( 1) n q n2 +ln 1 q ln ( 1) n q n2 1 q ln = ( 1) n q n2 ( 1) n q n2 +ln 1 q ln The lemma now follows upon applying the case z = 1 of Jacobi s triple product identity (29) z n q n2 = ( zq q/z q 2 ; q 2 ) 3 Infinite product identities In this section we record some identities involving infinite products These will be needed later on for simplification verification of certain identities First we have a result which is the analogue of Lemma 6 in [3] Lemma 31 We have (31) (32) ( q) = ( q) = (q9 ; q 9 ) ( q 9 ; q 9 ) 2q(q 3 q 15 q 18 ; q 18 ) (q25 ; q 25 ) ( q 25 ; q 25 ) 2q(q 15 q 35 q 50 ; q 50 ) + 2q 4 (q 5 q 45 q 50 ; q 50 ) Proof This really just amounts to two special cases of [1 Theorem 12] Indeed (31) is [1 Eq (118)] We give the details for (32) Beginning with Jacobi s triple product identity (29) we have ( q) = n 0 (mod 5) ( 1) n q n2 + = ( 1) n q 25n2 + 2 = ( 1) n q 25n2 2q n ±1 (mod 5) ( 1) n q n2 + ( 1) 5n+1 q (5n+1)2 + 2 Again using (29) we obtain the right h side of (32) n ±2 (mod 5) ( 1) n q n2 ( 1) 5n+2 q (5n+2)2 ( 1) n q 25n2 +10n + 2q 4 ( 1) n q 25n2 +20n Next we quote a result of Hickerson [18 Theorem 11] along with some of its corollaries Lemma 32 P (x q)p (z q)(q) 2 = P ( xz q 2 )P ( qz/x q 2 )(q 2 ; q 2 ) 2 xp ( xzq q 2 )P ( z/x q 2 )(q 2 ; q 2 ) 2

RANK DIFFERENCES FOR OVERPARTITIONS 7 The first corollary was recorded by Hickerson [18 Theorem 12] It follows by applying Lemma 32 twice once with x replaced by x once with z replaced by z then subtracting Lemma 33 P ( x q)p (z q)(q) 2 P (x q)p ( z q)(q) 2 = 2xP (z/x q 2 )P (xzq q 2 )(q 2 ; q 2 ) 2 The second corollary follows just as the first except we add instead of subtract in the final step Lemma 34 P ( x q)p (z q)(q) 2 + P (x q)p ( z q)(q) 2 = 2P (xz q 2 )P (qz/x q 2 )(q 2 ; q 2 ) 2 For the third corollary we subtract Lemma 32 with z replaced by z from three times Lemma 32 with x replaced by x Lemma 35 3P ( x q)p (z q)(q) 2 P (x q)p ( z q)(q) 2 = 2P (xz q 2 )P (zq/x q 2 )(q 2 ; q 2 ) 2 + 4xP (xzq q 2 )P (z/x q 2 )(q 2 ; q 2 ) 2 Finally we record the addition theorem as stated in [3 Eq (37)] Lemma 36 P 2 (z q)p (ζt q)p (ζ/t q) P 2 (ζ q)p (zt q)p (z/t q) + ζ/tp 2 (t q)p (zζ q)p (z/ζ q) = 0 4 Two lemmas The proofs of Theorems 11 12 will follow from identities which relate the sums Σ(a b) to the products P (a) P (0) The key steps are Lemmas 41 42 below which are similar to Lemmas 7 8 in [3] Lemma 41 We have (41) [ ( 1) n q n2 +n ζ 2n 1 zζ 1 q n + ζ2n+2 ] 1 zζq n = ζ(ζ2 qζ 2 1 q) (ζ qζ 1 ζ qζ 1 ) + ( 1) n qn2 +n 1 zq n (ζ qζ 1 ζ 2 qζ 2 z qz 1 q q) (z qz 1 zζ qz 1 ζ 1 zζ 1 qζz 1 ζ qζ 1 ) Proof This may be deduced from [19 Eq (4) p236] by making the substitutions a 3 = a 1 q a 4 = z a 1 /q a 10 = a 1 q/z a 9 = a 1 q letting a 7 a 8 tend to infinity writing ζ = q/a 1 simplifying One might also argue as in [3 pp 94-96] But the simplest way to establish the truth of (41) kindly pointed out to us by SH Chan is to observe that it is essentially the case r = 1 s = 3 a 1 = z b 1 = z/ζ b 2 = zζ b 3 = z of [8 Theorem 21] (42) P (a 1 q) P (a r q)(q) 2 P (b 1 q) P (b s q) = P (a 1/b 1 q) P (a r /b 1 q) P (b 2 /b 1 q) P (b s /b 1 q) + idem(b 1 ; b 2 b s ) ( 1) (s r)n q (s r)n(n+1)/2 1 b 1 q n ( a1 a r b s r 1 ) n 1 b 2 b s

8 JEREMY LOVEJOY AND ROBERT OSBURN Here we use the notation F (b 1 b 2 b m ) + idem(b 1 ; b 2 b m ) := F (b 1 b 2 b m ) + F (b 2 b 1 b 3 b m ) + + F (b m b 2 b m 1 b 1 ) We shall use the following specialization of Lemma 41 which is the case ζ = y a z = y b q = y l : (43) We now define y 2a Σ(a + b 2a) + Σ(b a 2a) y a P (2a)P ( 1 yl ) P (a)p ( y a y l Σ(b 0) ) P (a)p (2a)P ( yb y l )P (0) 2 P (b + a)p (b a)p (b)p ( y a y l ) = 0 g(z q) := z P (z2 q)p ( 1 q) P (z q)p ( z q) Σ(z 1 q) z2 Σ(z 2 z 2 q) ( 1) n z 2n q n(n+1) 1 q n (44) g(a) := g(y a y l ) = y a P (2a)P ( 1 yl ) P (a)p ( y a y l ) Σ(a 0) y2a Σ(2a 2a) Σ(0 2a) The second key lemma is the following Lemma 42 We have (45) 2g(z q) g(z 2 q) + 1 2 = (q)2 P ( z 4 q) P (z 4 q)p ( 1 q) + z P ( 1 q)2 (q) 2 P (z 2 q) P (z q) 2 P ( z q) 2 (46) g(z q) + g(z 1 q q) = 1 Proof We first require a short computation involving Σ(z ζ q) Note that (47) z 2 Σ(z ζ q) + ζσ(zq ζ q) = = = ( 1) n z2 ζ n q n(n+1) 1 zq n + ζ n q n(n 1)( z 2 q 2n 1 ) 1 zq n ( 1) n ζ n q n(n 1) (1 + zq n ) ( 1) n ζn+1 q n(n+1) 1 zq n+1

RANK DIFFERENCES FOR OVERPARTITIONS 9 upon writing n 1 for n in the second sum of the first equation Taking ζ = 1 yields (48) z 2 Σ(z 1 q) + Σ(zq 1 q) = = z = z r=1 ( 1) n q n(n 1) (1 + zq n ) ( 1) n q n2 1 q r 1 + q r where (29) was used for the last step Now write g(z q) in the form where g(z q) = f 1 (z) f 2 (z) f 3 (z) f 1 (z) := z P (z2 q)p ( 1 q) Σ(z 1 q) P (z q)p ( z q) By (21) (22) (48) f 3 (z) := f 2 (z) := z 2 Σ(z 2 z 2 q) ( 1) n z 2n q n(n+1) 1 q n (49) A similar argument as in (47) yields (410) f 2 (zq) f 2 (z) = (411) f 3 (zq) f 3 (z) = 2 + f 1 (zq) f 1 (z) = P (z2 q)p ( 1 q) 1 q r P (z q)p ( z q) 1 + q r r=1 = 2 ( 1) n z 2n q n2 ( 1) n z 2n 2 q n(n 1) + ( 1) n z 2n q n(n 1) + Adding (410) (411) then subtracting from (49) gives ( 1) n z 2n q n2 ( 1) n z 2n q n2

10 JEREMY LOVEJOY AND ROBERT OSBURN (412) g(z q) g(zq q) = 2 If we now define f(z) := 2g(z q) g(z 2 q) + 1 2 (q)2 P ( z 4 q) P (z 4 q)p ( 1 q) z P ( 1 q)2 (q) 2 P (z 2 q) P (z q) 2 P ( z q) 2 then from (21) (22) (412) one can verify that (413) f(zq) f(z) = 0 Now the only possible poles of f(z) are simple ones at points equivalent (under z zq) to those given by z 4 = 1 q q 2 or q 3 For each such point w it is easy to calculate lim z w (z w)f(z) see that there are in fact no poles Hence f(z) is analytic except at z = 0 applying (413) to its Laurent expansion around this point shows that f(z) is constant Next let us show that f( 1) = 0 Since the non-constant terms in f(z) have simple poles at z = 1 we must d consider lim z 1 dz (z + 1)f(z) We omit the computation but mention that the term g(z q) gives 1/4 g(z 2 q) gives 7/8 the last two terms give 1/8 1 respectively Then 2( 1/4) 7/8 + 1/2 1/8 + 1 = 0 we conclude that f(z) is identically 0 This proves (45) To prove (46) it suffices to show after (412) (414) g(z 1 q) + g(z q) = 1 Note that (415) Σ(z 1 q) + z 2 Σ(z 1 1 q) = = z 1 ( 1) n qn(n+1) 1 zq n z 1 ( 1) n q n2 q n2 ( 1) n 1 zq n where we have written n for n in the second sum in the first equation Thus by (21) (22) (415) we have (416) f 1 (z) + f 1 (z 1 ) = z 1 Again a similar argument as in (415) gives (417) f 2 (z) + f 2 (z 1 ) = ( 1) n q P n2 (z2 q)p ( 1 q) P (z q)p ( z q) ( 1) n z 2n q n2

RANK DIFFERENCES FOR OVERPARTITIONS 11 (418) f 3 (z) + f 3 (z 1 ) = 1 ( 1) n z 2n q n2 Adding (417) (418) then subtracting from (416) yields (414) Letting z = y a q = y l in Lemma 42 we get (419) 2g(a) g(2a) + 1 2 = P ( y4a y l )P (0) 2 P (4a)P ( 1 y l ) + P ( 1 ya yl ) 2 P (0) 2 P (2a) P (a) 2 P ( y a y l ) 2 (420) g(a) + g(q a) = 1 These two identities will be of key importance in the proofs of Theorems 11 12 5 Proofs of Theorems 11 12 We now compute the sums S(l 2m) The reason for this choice is twofold First we would like to obtain as simple an expression as possible in the final formulation (53) Secondly to prove Theorem 12 we will need to compute S(1) S(3) For l = 5 we can then choose m = 2 m = 1 respectively As this point we follow the idea of Section 6 in [3] Namely we write (51) n = lr + m + b where < r < The idea is to simplify the exponent of q in S(l 2m) Thus ln 2mn + n 2 = l 2 r(r + 1) + 2blr + (b + m)(b m + l) We now substitute (51) into (27) let b take the values 0 ±a ±m Here a runs through 1 2 l 1 2 where the value a ±m mod l is omitted As in [3] we use the notation to denote the sum over these values of a We thus obtain S(l 2m) = ( 1) n q(l 2m)n+n2 1 y n = ( 1) r+m+b (b+m)(b m+l) ylr(r+1)+2br q 1 y lr+m+b b r= where b takes values 0 ±a ±m the term corresponding to r = 0 b = m is omitted Thus a

12 JEREMY LOVEJOY AND ROBERT OSBURN (52) S(l 2m) = ( 1) m q m(l m) Σ(m 0) + Σ(0 2m) + y 2m Σ(2m 2m) + ( 1) m+a q (a+m)(a m+l){ } Σ(m + a 2a) + y 2a Σ(m a 2a) a Here the first three terms arise from taking b = 0 m m respectively We now can use (43) to simplify this expression By taking b = m dividing by y 2a in (43) the sum of the two terms inside the curly brackets becomes y a P (2a)P ( 1 yl ) P (a)p ( y a y l ) Σ(m 0) + P (a)p (2a)P ( y m y l )P (0) 2 y 2a P (m)p (m + a)p (m a)p ( y a y l ) Similarly upon taking a = m in (44) then the sum of the second third terms in (52) is In total we have y m P (2m)P ( 1 yl ) P (m)p ( y m y l Σ(m 0) g(m) ) (53) S(l 2m) = g(m) + ( 1) m+a q (a+m)(a m+l) y 2a P (a)p (2a)P ( y m y l )P (0) 2 P (m)p (m + a)p (m a)p ( y a y l ) a { + Σ(m 0) ( 1) m q m(l m) + y m P (2m)P ( 1 yl ) P (m)p ( y m y l ) + } ( 1) m+a q (a+m)(a m+l) y a P (2a)P ( 1 yl ) P (a)p ( y a y l ) a We can simplify some of the terms appearing in (53) as we are interested in certain values of l m a To this end we prove the following result Let { } denote the coefficient of Σ(m 0) in (53) Proposition 51 If l = 3 m = 1 then If l = 5 m = 2 a = 1 then If l = 5 m = 1 a = 2 then { } = q 2 ( q 9 ; q 9 ) ( q) (q 9 ; q 9 ) { } = q 6 ( q 25 ; q 25 ) ( q) (q 25 ; q 25 ) { } = q 4 ( q 25 ; q 25 ) ( q) (q 25 ; q 25 ) Proof This is a straightforward application of Lemma 31 We are now in a position to prove Theorems 11 12 We begin with Theorem 11

Proof By (26) (27) (28) we have (54) RANK DIFFERENCES FOR OVERPARTITIONS 13 { } N(0 3 n) N(1 3 n) n=0 By (23) (24) (53) Proposition 51 we have q n = 3S(1) + S(3) (55) S(1) = g(1) + q 2 Σ(1 0) ( q 9 ; q 9 ) ( q) (q 9 ; q 9 ) By Lemma 21 we have (56) S(3) = + 1 2 Thus we have that 3g(1) 3q 2 Σ(1 0) ( q 9 ; q 9 ) ( q) (q 9 ; q 9 + 1 { ) 2 = r 01 (0)q 0 + r 01 (1)q + r 01 (2)q 2} We now multiply the right h side of the above expression using Lemma 31 the r 01 (d) from Theorem 12 (recall that r 01 (d) is just R 01 (d) with q replaced by q 3 ) We then equate coefficients of powers of q verify the resulting identities The only power of q for which the resulting equation does not follow easily upon cancelling factors in infinite products is the constant term We obtain 3g(1) + 1 2 = (q9 ; q 9 ) 3 ( q 3 ; q 3 ) 2(q 3 ; q 3 ) ( q 9 ; q 9 ) 3 4q 3 ( q9 ; q 9 ) 3 (q 18 ; q 18 ) 3 (q 6 ; q 6 ) ( q 3 ; q 3 ) But this follows from (419) some simplification This then completes the proof of Theorem 11 We now turn to Theorem 12 Proof We begin with the rank differences R 12 (d) By (26) (27) (28) we have (57) { } N(1 5 n) N(2 5 n) n=0 by (23) (24) (53) Proposition 51 q n = S(1) 3S(3) (58) S(1) = g(2) + qyσ(2 0) ( q 25 ; q 25 ) ( q) (q 25 ; q 25 q 2 (q 25 ; q 25 ) 2 ( q 10 q 15 ; q 25 ) ) (q 10 q 15 ; q 25 ) ( q 5 q 20 ; q 25 ) (59) S(3) = g(1) q 4 Σ(1 0) ( q 25 ; q 25 ) ( q) (q 25 ; q 25 + q 3 (q 25 ; q 25 ) 2 ( q 5 q 20 ; q 25 ) ) (q 5 q 20 ; q 25 ) ( q 10 q 15 ; q 25 ) By (57) (58) (59) we have

14 JEREMY LOVEJOY AND ROBERT OSBURN g(2) qyσ(2 0) ( q 25 ; q 25 ) ( q) (q 25 ; q 25 + q 2 (q 25 ; q 25 ) 2 ( q 10 q 15 ; q 25 ) ) (q 10 q 15 ; q 25 ) ( q 5 q 20 ; q 25 ) + 3g(1) + 3q 4 Σ(1 0) ( q 25 ; q 25 ) ( q) (q 25 ; q 25 3q 3 (q 25 ; q 25 ) 2 ( q 5 q 20 ; q 25 ) ) (q 5 q 20 ; q 25 ) ( q 10 q 15 ; q 25 ) { = r 12 (0)q 0 + r 12 (1)q + r 12 (2)q 2 + r 12 (3)q 3 + r 12 (4)q 4} We now multiply the right h side of the above expression using Lemma 31 the R 12 (d) from Theorem 12 equating coefficients of powers of q The coefficients of q 0 q 1 q 2 q 3 q 4 give us respectively (q 25 ; q 25 ) 2 (510) g(2) + 3g(1) = y (q 15 q 20 q 30 q 35 ; q 50 + 4y (q10 q 15 q 35 q 40 ; q 50 ) (q 50 ; q 50 ) 2 ) (q 20 q 30 ; q 50 ) (q 2 5 q 45 ; q 50 ) (511) y (q50 ; q 50 ) (q 15 q 35 q 50 ; q 50 ) (q 15 q 20 q 30 q 35 ; q 50 ) = y (q50 ; q 50 ) (q 5 q 45 q 50 ; q 50 ) (q 5 q 20 ; q 25 ) (512) (513) (q 25 ; q 25 ) 2 ( q 10 q 15 ; q 25 ) (q 10 q 15 ; q 25 ) ( q 5 q 20 ; q 25 ) = (q25 ; q 25 ) 2 (q 5 q 20 ; q 25 ) 2y (q50 ; q 50 ) 2 (q 5 q 45 ; q 50 ) (q 10 q 15 ; q 25 ) 3(q 25 ; q 25 ) 2 ( q 5 q 20 ; q 25 ) (q 5 q 20 ; q 25 ) ( q 10 q 15 ; q 25 ) = (q25 ; q 25 ) 2 (q 10 q 15 ; q 25 ) + 2(q50 ; q 50 ) 2 (q 15 q 35 ; q 50 ) (q 5 q 20 ; q 25 ) + 4y (q10 q 40 ; q 50 ) (q 50 ; q 50 ) 2 (q 20 q 30 ; q 50 ) 2 (514) (q 10 q 40 q 50 ; q 50 ) (q 25 ; q 25 ) (q 20 q 30 ; q 50 ) 2 (q 5 q 45 ; q 50 ) ( q 25 ; q 25 ) = (q50 ; q 50 ) 2 (q 15 q 35 ; q 50 ) (q 10 q 15 ; q 25 ) + y (q50 ; q 50 ) 2 (q 5 q 45 ; q 50 ) (q 15 q 20 q 30 q 35 ; q 50 ) Equation (511) is immediate Upon clearing denominators in (512)-(514) simplifying we see that (512) is equivalent to the case (x z q) = ( q 5 q 10 q 25 ) of Lemma 33 (513) is the case (x z q) = (q 5 q 10 q 25 ) of Lemma 35 (514) follows from the case (z ζ t q) = (q 20 q 10 q 5 q 50 ) of Lemma 36 As for (510) let us take a = 1 a = 2 in (419) then replace g(4) by 1 g(1) using (420) This gives 3g(1) + g(2) = ( q5 q 20 q 25 q 25 ; q 25 ) 2(q 5 q 20 q 25 q 25 ; q 25 ) 4y(q15 q 35 q 50 q 50 ; q 50 ) (q 10 q 25 q 25 q 40 ; q 50 ) ( q10 q 15 q 25 q 25 ; q 25 ) 2(q 10 q 15 q 25 q 25 ; q 25 ) + 4y2 (q 5 q 45 q 50 q 50 ; q 50 ) (q 20 q 25 q 25 q 30 ; q 50 ) Now after making a common denominator in the first third terms we may apply the case (x z q) = (q 5 q 10 q 25 ) of Lemma 33 to these two terms the result being precisely the first term

RANK DIFFERENCES FOR OVERPARTITIONS 15 in (510) For the second fourth terms we make a common denominator multiply top bottom by (q 5 q 20 ; q 25 ) Then the case (z ζ t q) = (q 20 q 10 q 5 q 50 ) of Lemma 36 applies we obtain the second term in (510) We now turn to the rank differences R 02 (d) proceeding as above Again by (26) (27) (28) we have (515) { } N(0 5 n) N(2 5 n) n=0 By Lemma 21 (with l = 5) (515) (58) (59) we have q n = S(5) + 2S(1) + S(3) + 1 2 2g(2) + 2qyΣ(2 0)( q 25 ; q 25 ) ( q) (q 25 ; q 25 2q 2 (q 25 ; q 25 ) 2 ( q 10 q 15 ; q 25 ) ) (q 10 q 15 ; q 25 ) ( q 5 q 20 ; q 25 ) g(1) q 4 Σ(1 0) ( q 25 ; q 25 ) ( q) (q 25 ; q 25 + q 3 (q 25 ; q 25 ) 2 ( q 5 q 20 ; q 25 ) ) (q 5 q 20 ; q 25 ) ( q 10 q 15 ; q 25 ) { = r 02 (0)q 0 + r 02 (1)q + r 02 (2)q 2 + r 02 (3)q 3 + r 02 (4)q 4} Again substituting the claim in Theorem 12 equating coefficients of powers of q yields the following identities (516) 1 2 2g(2) g(1) = 1 ( q 10 q 15 ; q 25 ) (q 25 ; q 25 ) 2 2 (q 10 q 15 ; q 25 ) ( q 25 ; q 25 ) 2 2y (q10 q 40 ; q 50 ) (q 15 q 35 ; q 50 ) (q 50 ; q 50 ) 2 (q 20 q 30 ; q 50 ) 2 (q 5 q 45 ; q 50 ) + 2y (q20 q 30 ; q 50 ) (q 5 q 45 ; q 50 ) (q 50 ; q 50 ) 2 (q 10 q 40 ; q 50 ) 2 (q 15 q 35 ; q 50 ) (517) (q 20 q 30 q 50 ; q 50 ) (q 25 ; q 25 ) (q 10 q 40 ; q 50 ) 2 (q 15 q 35 ; q 50 ) ( q 25 ; q 25 = ( q10 q 15 ; q 25 ) (q 25 ; q 25 ) (q 15 q 35 q 50 ; q 50 ) ) (q 10 q 15 ; q 25 ) ( q 25 ; q 25 ) (518) (519) (q 25 ; q 25 ) 2 ( q 10 q 15 ; q 25 ) ( q 5 q 20 ; q 25 ) (q 10 q 15 ; q 25 ) = (q50 ; q 50 ) 2 (q 20 q 30 ; q 50 ) (q 10 q 40 ; q 50 ) 2 (q 25 ; q 25 ) 2 ( q 5 q 20 ; q 25 ) ( q 10 q 15 ; q 25 ) (q 5 q 20 ; q 25 ) = y (q50 ; q 50 ) 2 (q 5 q 45 ; q 50 ) (q 10 q 15 ; q 25 ) (q25 ; q 25 ) 2 (q 10 q 15 ; q 25 + 2y (q50 ; q 50 ) 2 (q 10 q 40 ; q 50 ) ) (q 20 q 30 ; q 50 ) 2 (520) (q 10 q 40 q 50 ; q 50 ) (q 25 ; q 25 ) (q 20 q 30 ; q 50 ) 2 (q 5 q 45 ; q 50 ) ( q 25 ; q 25 + ( q10 q 15 ; q 25 ) (q 25 ; q 25 ) (q 5 q 45 q 50 ; q 50 ) ) (q 10 q 15 ; q 25 ) ( q 25 ; q 25 ) = 2 (q50 ; q 50 ) 2 (q 15 q 35 ; q 50 ) (q 10 q 15 ; q 25 )

16 JEREMY LOVEJOY AND ROBERT OSBURN Now (517) is immediate After clearing denominators simplifying (518) follows from the case (z ζ t q) = (q 20 q 15 q 10 q 50 ) of Lemma 36 (519) is the case (x z q) = (q 5 q 10 q 25 ) of Lemma 33 For (520) we simplify the first term apply the case (x z q) = (q 5 q 10 q 25 ) of Lemma 34 As for (516) taking the case a = 2 of (419) together with an application of (420) gives 1 2 2g(2) g(1) = ( q10 q 15 ; q 25 ) (q 25 ; q 25 ) 2 2(q 10 q 15 ; q 25 ) ( q 25 ; q 25 ) 2 4y2 (q 5 q 45 ; q 50 ) (q 50 ; q 50 ) 2 (q 20 q 30 ; q 50 ) (q 25 ; q 50 ) 2 Now the first terms of the above equation (516) match up while after some simplification of the final two terms of (516) we may apply the case (x z q) = (q 5 q 10 q 25 ) of Lemma 33 to obtain the final term above Acknowlegements The second author would like to thank the Institut des Hautes Études Scientifiques the Max-Planck-Institut für Mathematik for their hospitality support during the preparation of this paper References [1] GE Andrews D Hickerson Ramanujan s lost notebook VII: The sixth order mock theta functions Adv Math 89 (1991) 60-105 [2] AOL Atkin SM Hussain Some properties of partitions II Trans Amer Math Soc 89 (1958) 184 200 [3] AOL Atkin H Swinnerton-Dyer Some properties of partitions Proc London Math Soc 66 (1954) 84 106 [4] F Brenti Determinants of super-schur functions lattice paths dotted plane partitions Adv Math 98 (1993) 27-64 [5] K Bringmann J Lovejoy Dyson s ranks overpartitions weak Maass forms Int Math Res Not to appear [6] K Bringmann K Ono Dyson s ranks Maass forms Ann Math to appear [7] K Bringmann K Ono R Rhoades Eulerian series as modular forms J Amer Math Soc to appear [8] SH Chan Generalized Lambert series identities Proc London Math Soc 91 (2005) 598-622 [9] D Choi Weakly holomorphic modular forms of half-integral weight with non-vanishing constant terms mod l preprint [10] S Corteel J Lovejoy Overpartitions Trans Amer Math Soc 356 (2004) 1623-1635 [11] S Corteel J Lovejoy O Mallet An extension to overpartitions of the Rogers-Ramanujan identities for even moduli J Number Theory to appear [12] S Corteel O Mallet Overpartitions lattice paths Rogers-Ramanujan identities J Combin Theory Ser A to appear [13] P Desrosiers L Lapointe P Mathieu Jack polynomials in superspace Comm Math Phys 242 (2003) 331-360 [14] FJ Dyson Some guesses in the theory of partitions Eureka (Cambridge) 8 (1944) 10 15 [15] J-F Fortin P Jacob P Mathieu Jagged partitions Ramanujan J 10 (2005) 215 235 [16] J-F Fortin P Jacob P Mathieu Generating function for K-restricted jagged partitions Electron J Combin 12 (2005) Research paper R12 [17] G Gasper M Rahman Basic Hypergeometric Series Cambridge Univ Press Cambridge 1990 [18] D Hickerson A proof of the mock theta conjectures Invent Math 94 (1988) 639-660 [19] M Jackson On some formulae in partition theory bilateral basic hypergeometric series J London Math Soc 24 (1949) 233 237

RANK DIFFERENCES FOR OVERPARTITIONS 17 [20] S-J Kang J-H Kwon Crystal bases of the Fock space representations string functions J Algebra 280 (2004) 313-349 [21] J Lovejoy Overpartitions real quadratic fields J Number Theory 106 (2004) 178 186 [22] J Lovejoy Rank conjugation for the Frobenius representation of an overpartition Ann Comb 9 (2005) 321-334 [23] J Lovejoy Overpartition pairs Ann Inst Fourier 56 (2006) 781-794 CNRS LIAFA Université Denis Diderot 2 Place Jussieu Case 7014 F-75251 Paris Cedex 05 FRANCE School of Mathematical Sciences University College Dublin Belfield Dublin 4 Irel IHÉS Le Bois-Marie 35 route de Chartres F-91440 Bures-sur-Yvette FRANCE E-mail address: lovejoy@liafajussieufr E-mail address: robertosburn@ucdie osburn@ihesfr