aa r X i v : . [ m a t h . N T ] A ug ASYMPTOTICS FOR RANK PARTITION FUNCTIONS
KATHRIN BRINGMANN
Abstract.
In this paper, we obtain asymptotic formulas for an infinite class of rank generat-ing functions. As an application, we solve a conjecture of Andrews and Lewis on inequalitiesbetween certain ranks. Introduction and Statement of Results A partition of a positive integer n is any non-increasing sequence of positive integers whosesum is n . As usual, let p ( n ) denote the number of partitions of n . The partition function p ( n ) has the well known infinite product generating function(1.1) 1 + ∞ X n =1 p ( n ) q n = ∞ Y n =1 − q n = 1 + ∞ X n =1 q n (1 − q ) (1 − q ) · · · (1 − q n ) . Hardy and Ramanujan showed the following asymptotic formula for p ( n ) p ( n ) ∼ n √ · e π √ n/ ( n → ∞ ) . Using the modularity of the generating function for p ( n ), Rademacher obtained an exactformula for p ( n ). To state his result, let I s ( x ) be the usual I -Bessel function of order s , andlet e ( x ) := e πix . Furthermore, if k ≥ n are integers, then let(1.2) A k ( n ) := 12 r k X x (mod 24 k ) x ≡− n +1 (mod 24 k ) χ ( x ) · e (cid:16) x k (cid:17) , where the sum runs over the residue classes modulo 24 k , and where χ ( x ) := (cid:18) x (cid:19) . (1.3)If n is a positive integer, then Rademacher showed that(1.4) p ( n ) = 2 π (24 n − / ∞ X k =1 A k ( n ) k · I (cid:18) π √ n − k (cid:19) . Date : October 23, 2018.2000
Mathematics Subject Classification.
The partition function also satisfies some nice congruence properties; the most famous onesare the so-called Ramanujan congruences: p (5 n + 4) ≡ ,p (7 n + 5) ≡ ,p (11 n + 6) ≡ . In order to explain the congruences with modulus 5 and 7 combinatorially, Dyson [9] intro-duced the “rank” of a partition. The rank of a partition is defined to be its largest partminus the number of its parts. Dyson conjectured that the partitions of 5 n + 4 (resp. 7 n + 5)form 5 (resp. 7) groups of equal size when sorted by their ranks modulo 5 (resp. 7). Thisconjecture was proved in 1954 by Atkin and Swinnerton-Dyer [5].If N ( m, n ) denotes the number of partitions of n with rank m , then it is well known that(1.5) R ( w ; q ) := 1 + ∞ X n =1 ∞ X m = −∞ N ( m, n ) w m q n = 1 + ∞ X n =1 q n ( wq ; q ) n ( w − q ; q ) n , where ( a ; q ) n := (1 − a )(1 − aq ) · · · (1 − aq n − ) . Obviously, by letting w = 1, we obtain (1.1). Moreover, if N e ( n ) (resp. N o ( n )) denotes thenumber of partitions of n with even (resp. odd) rank, then by letting w = − ∞ X n =1 ( N e ( n ) − N o ( n )) q n = 1 + ∞ X n =1 q n (1 + q ) (1 + q ) · · · (1 + q n ) . In the following we denote this series by f ( q ) and its n -th Fourier coefficient by α ( n ). Theseries f ( q ) is one of the third order mock theta functions defined by Ramanujan in his lastletter to Hardy dated January 1920 (see pages 127-131 of [14]). There Ramanujan claimed,without including a proof, that α ( n ) = ( − n − exp (cid:16) π q n − (cid:17) q n − + O exp (cid:16) π q n − (cid:17)q n − . Dragonette [8] proved this claim in her Ph.D. thesis written 1951 under the direction ofRademacher. Andrews [2] improved this in his Ph.D. thesis 1964, also written under Rademacher,as(1.7) α ( n ) = π √ n − [ √ n ] X k =1 ( − ⌊ k +12 ⌋ A k (cid:16) n − k (1+( − k )4 (cid:17) k · I (cid:18) π √ n − k (cid:19) + O ( n ǫ ) . Moreover Andrews and Dragonette made the following conjecture.
YMPTOTICS FOR RANK PARTITION FUNCTIONS 3
Conjecture. (Andrews-Dragonette) If n is a positive integer, then (1.8) α ( n ) = π (24 n − ∞ X k =1 ( − ⌊ k +12 ⌋ A k (cid:16) n − k (1+( − k )4 (cid:17) k · I (cid:18) π √ n − k (cid:19) . In [6] Ono and the author proved this conjecture using the theory of Maass-Poincar´e series.It turns out that q − f (cid:0) q (cid:1) is the “holomorphic part” of a weak Maass form (see [7] for thedefinition of a weak Maass form). In [7] they showed that a similar phenomenon is true forall functions R ( ζ ac ; q ) =: 1 + ∞ X n =1 A (cid:16) ac ; n (cid:17) q n , where ζ n := e πin and 0 < a < c are integers. More specificly, R ( ζ ac ; q ) is the holomorphic partof a weak Maass form of weight . Using this deeper insight, in this paper, we are able toobtain asymptotic formulas for all the coefficients A (cid:0) ac ; n (cid:1) , which in turn implies asymptoticsfor the rank partition functions.Before we state our result, we need some more notation. We let k and h be coprimeintegers, h ′ defined by hh ′ ≡ − k ) if k is odd and hh ′ ≡ − k ) if k is even, k := k gcd( k,c ) , c := c gcd( k,c ) and 0 < l < c defined by the congruence l ≡ ak (mod c ). If bc ∈ (0 , \ { , , } , then define the integer s ( b, c ) by(1.9) s ( b, c ) := < bc < , < bc < , < bc < , < bc < . In particular, set s := s ( l, c ). Let ω h,k be the multiplier occuring in the transformation lawof the partition function p ( n ). This is explicitly given by ω h,k := exp ( πit ( h, k )) , (1.10)where t ( h, k ) := X µ (mod k ) (cid:16)(cid:16) µk (cid:17)(cid:17) (cid:18)(cid:18) hµk (cid:19)(cid:19) . Here (( x )) := (cid:26) x − ⌊ x ⌋ − if x ∈ R \ Z , x ∈ Z . Moreover we define, for n, m ∈ Z , the following sums of Kloosterman type(1.11) B a,c,k ( n, m ) := ( − ak +1 sin (cid:16) πac (cid:17) X h (mod k ) ∗ ω h,k sin (cid:0) πah ′ c (cid:1) · e − πia k h ′ c · e πik ( nh + mh ′ ) KATHRIN BRINGMANN if c | k , and D a,c,k ( n, m ) := ( − ak + l X h (mod k ) ∗ ω h,k · e πik ( nh + mh ′ ) . (1.12)Here the sums run through all primitive residue classes modulo k . Moreover, for c ∤ k , let δ c,k,r := − (cid:0) + r (cid:1) lc + (cid:16) lc (cid:17) + if 0 < lc < , − l c + (cid:16) lc (cid:17) + − r (cid:16) − lc (cid:17) if < lc < , < lc < or < lc < m a,c,k,r := c (cid:0) − a k + 6 lak − ak c − l + lc − ark c + 2 lc r (cid:1) if 0 < lc < , c (cid:0) − ak c − a k + 6 lak + ak c + 6 lc if < lc < . − l − c − lc + 2 ark c + 2 c ( c − l ) r (cid:1) Remark.
It is not hard to see that m a,c,k,r ∈ Z .We obtain following asymptotic formulas for the coefficients A (cid:0) ac ; n (cid:1) . Theorem 1.1. If < a < c are coprime integers and c is odd, then for positive integers n we have that A (cid:16) ac ; n (cid:17) = 4 √ i √ n − X ≤ k ≤√ nc | k B a,c,k ( − n, √ k · sinh (cid:18) π √ n − k (cid:19) + 8 √ · sin (cid:0) πac (cid:1) √ n − X ≤ k ≤√ nc ∤ kr ≥ δc,k,r> D a,c,k ( − n, m a,c,k,r ) √ k · sinh π p δ c,k,r (24 n − √ k ! + O c ( n ǫ ) . Four remarks.
1) One can easily see that the second sum is empty for c ∈ { , } .2) For a fixed choice of a and c , we can in principle obtain exact formulas for A (cid:0) ac ; n (cid:1) bymodifying an argument given in [6] to prove (1.8).3) Similarly as in this paper, one can also prove asymptotic formulas for generalized ranks asdefined in [11]. Since the proof is basically the same, we do not give it here.4) One could also generalize our results to the case that c is even, but for simplicity we restrictus to the case that c is odd.If we denote by N ( a, c ; n ) the number of partitions of n with rank congruent a (mod c )then it is easy to conclude the following: YMPTOTICS FOR RANK PARTITION FUNCTIONS 5
Corollary 1.2.
For integers ≤ a < c , where c is an odd integer, we have N ( a, c ; n ) = 2 πc · √ n − ∞ X k =1 A k ( n ) k · I (cid:18) π √ n − k (cid:19) + 1 c c − X j =1 ζ − ajc √ i √ n − X c | k B j,c,k ( − n, √ k sinh (cid:16) π k √ n − (cid:17) + 8 √ (cid:16) πjc (cid:17) √ n − X k,rc ∤ kδc,k,r> D j,c,k ( − n, m j,c,k,r ) √ k sinh r δ c,k,r (24 n − πk !! + O c ( n ǫ ) . This corollary implies some conjectures of Andrews and Lewis. In [4, 13] they showed N (0 ,
2; 2 n ) < N (1 ,
2; 2 n ) if n ≥ ,N (0 , n ) > N (2 , n ) if 26 < n ≡ , ,N (0 , n ) < N (2 , n ) if 26 < n ≡ , . Moreover, they conjectured (see Conjecture 1 of [4]).
Conjecture. (Andrews and Lewis)
For all n > , we have N (0 , n ) < N (1 , n ) if n ≡ or ,N (0 , n ) > N (1 , n ) if n ≡ , (1.14)A careful analysis of Corollary 1.2 gives the following theorem. Theorem 1.3.
The Andrews-Lewis Conjecture is true for all n
6∈ { , , } in which case wehave equality in (1.14).Remark. From Corollary 1.2 we see that N (0 , n ) − N (1 , n ) ∼ − (cid:0) π − πn (cid:1) sinh (cid:16) π ·√ n − (cid:17) √ n − . (1.15)This directly implies the Andrews-Lewis Conjecture for n sufficiently large n . For examplefor n = 1200, we have N (0 ,
3; 1200) − N (1 ,
3; 1200) = − − . . The paper is organized as follows: In Section 2 we prove a transformation law for thefunctions R ( ζ ac ; q ). The behavior under the generators of SL ( Z ) was also studied in [12] andin [7]. However here we prove a more general result since we need the occurring roots of unityand integrals explicitly for every element in SL ( Z ). In Section 4 we prove Theorem 1.1 and KATHRIN BRINGMANN
Corollary 1.2 by using the Circle Method. For this we need some estimates shown in Section3. Section 5 is dedicated to the proof of Theorem 1.3.
Acknowledgements
The author thanks K. Ono, B. Kane, and F. Garvan for helpful comments to an earlierversion of the paper. 2.
Modular transformation formulas
In this section we prove a transformation law for the functions R ( ζ ac ; q ). For this define N (cid:16) ac ; q (cid:17) := 1( q ; q ) ∞ ∞ X n =1 ( − n (1 + q n ) (cid:0) − (cid:0) πac (cid:1)(cid:1) − q n cos (cid:0) πac (cid:1) + q n · q n (3 n +1)2 ! . (2.1)In [12] it is shown that R ( ζ ac ; q ) = N (cid:16) ac ; q (cid:17) . Moreover let N ( a, b, c ; q ) := i q ; q ) ∞ ∞ X m =0 ( − m e − πiac · q m (3 m +1)+ ms ( b,c )+ b c − e − πiac · q m + bc − ∞ X m =1 ( − m e πiac · q m (3 m +1) − ms ( b,c ) − b c − e πiac · q m − bc ! . Remark.
It is easy to see that the above definition coincides with the definition, given in [7].For each ν ∈ Z define H a,c ( x ) := cosh( x )sinh (cid:0) x + πiac (cid:1) · sinh (cid:0) x − πiac (cid:1) ,I a,c,k,ν ( z ) := Z R e − πzx k · H a,c (cid:18) πiνk − πi k − πzxk (cid:19) dx. It is easy to see that H a,c ( − x ) = H a,c ( x ) H a,c ( x ) = (1 + q ) · q (cid:0) − (cid:0) πac (cid:1) q + q (cid:1) for q = e x . We show the following transformation law for the function N (cid:0) ac ; q (cid:1) . Theorem 2.1.
We assume the same notation as in the introduction. Moreover, let z ∈ C with Re( z ) > , q := e πik ( h + iz ) , and q := e πik ( h ′ + iz ) . YMPTOTICS FOR RANK PARTITION FUNCTIONS 7 (1) If c | k , then N (cid:16) ac ; q (cid:17) = ( − ak +1 i sin (cid:0) πac (cid:1) · ω h,k sin (cid:0) πah ′ c (cid:1) z · e − πia k h ′ c · e π k ( z − − z ) · N (cid:18) ah ′ c ; q (cid:19) + 2 sin (cid:0) πac (cid:1) · ω h,k k e − πz k · z X ν (mod k ) ( − ν e − πih ′ ν k + πih ′ νk · I a,c,k,ν ( z ) . (2) If c ∤ k , then N (cid:16) ac ; q (cid:17) = 4 i ( − ak + l +1 · sin (cid:0) πac (cid:1) ω h,k z · e − πih ′ sac − πia h ′ k cc + πih ′ lacc · q slc − l c · e π k ( z − − z ) × N (cid:18) ah ′ , lcc , c ; q (cid:19) + 2 sin (cid:0) πac (cid:1) ω h,k k e − πz k · z X ν (mod k ) ( − ν e − πih ′ ν k + πih ′ νk I a,c,k,ν ( z ) . Proof.
We modify the proof of [2]. We easily see, using 1 − cos(2 x ) = 2 sin( x ) that( q ; q ) ∞ · N (cid:16) ac ; q (cid:17) = sin (cid:16) πac (cid:17) X n ∈ Z ( − n H a,c (cid:18) πink ( h + iz ) (cid:19) · e πi ( h + iz ) n k . (2.2)Writing n = km + ν with 0 ≤ ν < k , m ∈ Z and using that ( h, k ) = 1, gives that (2.2) equalssin (cid:16) πac (cid:17) k − X ν =0 ( − ν e πihν k X m ∈ Z ( − m H a,c (cid:18) πihνk − π ( km + ν ) zk (cid:19) · e − πz ( km + ν )2 k . (2.3)Using Poisson summation and substituting x kx + ν gives that the inner sum equals1 k X n ∈ Z Z R H a,c (cid:18) πihνk − πxzk (cid:19) · e πi (2 n +1)( x − ν ) k − πzx k dx. (2.4)Strictly speaking for c | k there may be a pole at x = 0. In this case we take the principalpart of the integral. Inserting (2.4) into (2.3) we see that the summation only depends on ν (mod k ). Moreover, by changing ν into − ν , x into − x , and n into − ( n + 1), we see that thesum over n with n ≤ − n ≥
0. Thus (2.3) equals(2.5) 2 sin (cid:0) πac (cid:1) k X ν (mod k ) ( − ν e πihν k X n ∈ N Z R H a,c (cid:18) πihνk − πxzk (cid:19) · e πi (2 n +1)( x − ν ) k − πzx k dx. To see where the poles of the integrant lie, we introduce the function S a,c,k ( x ) := sinh( c x )sinh (cid:0) xk + πiac (cid:1) · sinh (cid:0) xk − πiac (cid:1) which is entire as a function of x . Using that c is odd we can write the integrand in (2.5) as( − hν cosh (cid:0) πihνk − πxzk (cid:1) · e πi (2 n +1)( x − ν ) k − πzx k · S a,c,k ( πxz − πihν )sinh( πc xz ) . KATHRIN BRINGMANN
From this we see that the only poles can lie in the points x m := imc z ( m ∈ Z ) . We treat the cases with c | k or c ∤ k seperately.If c | k , then c = 1. One computes that each choice ± leads at most for one ν (mod k ) toa non-zero residue, and that this ν can for ǫ ∈ {±} be chosen as ν ǫm := − h ′ ( m − ǫak ) . We denote the corresponding residues by λ ǫn,m ( ǫ ∈ {±} ). By shifting the path of integrationthrough the points ω n := (2 n + 1) i z , we have to take those points x m into account for which n ≥ m ≥
0. Setting r := and r m := 1 for m ∈ N , we obtain by the Residue Theorem( q ; q ) ∞ · N (cid:16) ac ; q (cid:17) = X + X , where X := 4 πi sin (cid:0) πac (cid:1) k X m ≥ ǫ ∈{±} r m ( − ν ǫm e πih ( νǫm )2 k ∞ X n =3 m λ ǫn,m , X := 2 sin (cid:0) πac (cid:1) k X ν (mod k ) ( − ν e πihν k X n ∈ N Z ∞ + ω n −∞ + ω n H a,c (cid:18) πihνk − πxzk (cid:19) · e πi (2 n +1)( x − ν ) k − πx zk dx. If c ∤ k , then a pole can only occur if m ≡ ± ak (mod c ). Writing c m ± l instead of m with m ≥ m > − and l as in the introduction, we see thatto each choice there corresponds exactly one ν (mod k ) and we can choose ν for ǫ ∈ {± } as ν ǫm := − h ′ (cid:18) m ǫ c ( l − ak ) (cid:19) . As before, we denote the corresponding residues by λ ǫn,m . By shifting the path of integrationthrough the points ω n , we have to take those points x m into account for which n +16 > c mǫlc .One can see that this is equivalent to n ≥ mǫs, where s was defined in (1.9). By the Residue Theorem we obtain( q ; q ) ∞ · N (cid:16) ac ; q (cid:17) = X + X , YMPTOTICS FOR RANK PARTITION FUNCTIONS 9 where P is given as before and P is defined as4 πi sin (cid:0) πac (cid:1) k X m ≥ ( − ν + m e πih ( ν + m )2 k ∞ X n =3 m + s λ + n,m + X m ≥ ( − ν − m e πih ( ν − m )2 k ∞ X n =3 m − s λ − n,m . We first consider the sums P and P . We have λ ǫn,m = − k · cosh (cid:16) πihν ǫm k − πx m zk (cid:17) · e πi (2 n +1)( xm − νǫm ) k − πzx mk πz · cosh (cid:16) πihν ǫm k − πx m zk ǫ πiac (cid:17) · sinh (cid:16) πihν ǫm k − πx m zk − ǫ πiac (cid:17) . From this one directly sees that λ ǫn +1 ,m = exp (cid:18) πik ( x m − ν ǫm ) (cid:19) · λ ǫn,m . Thus ∞ X n =3 m + r λ ǫn,m = λ ǫ m + r,m − exp (cid:0) πik ( x m − ν ǫm ) (cid:1) ( r ∈ { , ± s } , ǫ ∈ {±} ) . A straightforward but lenghty calculation gives X = ( − ak +1 i sin (cid:0) πac (cid:1) sin (cid:0) πah ′ c (cid:1) z · e − πia k h ′ c · ( q ; q ) ∞ · N (cid:18) ah ′ c ; q (cid:19) , X = 4 i sin (cid:0) πac (cid:1) ( − ak + l +1 z e − πih ′ sac − πih ′ a k cc + πih ′ lacc q slc − “ lc ” ( q ; q ) ∞ N (cid:18) h ′ a, lcc , c ; q (cid:19) . Here we have to be a little careful since for c | k ( c ∤ k ) we only have that hν is congruent to m − ǫak ( mǫ c ( l − ak )) modulo k but not necessarily modulo 2 k if k is odd.We next turn to the computation of P . With the same argument as before we can changethe sum over N into a sum over Z . Making the translation x x + ω n and writing n = 3 p + δ with p ∈ Z and δ ∈ { , ± } gives that X = sin (cid:0) πac (cid:1) k X ν (mod k ) ( − ν e πihν k X p ∈ Z δ ∈{ , ± } e − π (6 p +2 δ +1)212 kz − πiν (6 p +2 δ +1) k Z ∞−∞ H a,c (cid:18) πihνk − πi (6 p + 2 δ + 1)6 k − πzxk (cid:19) · e − πzx k dx. Now ( h ′ , k ) = 1 implies that − h ′ ( ν + p ) runs modulo k if ν does. Thus we can change ν into − h ′ ( ν + p ) which leads to(2.6) X = sin (cid:0) πac (cid:1) k X ν (mod k ) δ ∈{ , ± } p ∈ Z ( − ν + p q p (3 p +2 δ +1)1 · e − π (2 δ +1)212 kz + πih ′ ( − ν δ +1) ν ) k Z ∞−∞ H a,c (cid:18) πiνk − πi (2 δ + 1)6 k − πzxk (cid:19) · e − πzx k dx. Now the integral is independent of p and the sum over p equals X p ∈ Z ( − p · q p (3 p +(2 δ +1))1 . (2.7)If δ = 1, then (2.7) vanishes since the p − th and the − ( p + 1)-th term cancel. Changing ν into − ν , p into − p and x into − x , we see that the terms in (2.6) corresponding to δ = − δ = 0 are equal. Moreover in this case (2.7) equals ( q ; q ) ∞ . Thus X = 2 sin (cid:0) πac (cid:1) · ( q ; q ) ∞ · e − π kz k X ν (mod k ) ( − ν e − πih ′ ν k + πih ′ νk · I a,c,k,ν ( z ) . Now the theorem follows easily using the transformation law( q ; q ) ∞ = ω h,k · z · e π k ( z − − z ) · ( q ; q ) ∞ . (cid:3) Some estimates
In this section we estimate the function I a,c,k,ν ( z ), defined in Section 2, and some Kloost-erman sums. Lemma 3.1.
Assume that n ∈ N , ν ∈ Z , z := kn − k Φ i , − k ( k + k ) ≤ Φ ≤ k ( k + k ) , where h k < hk < h k are adjacent Farey fractions in the Farey sequence of order N , with N := ⌊ n ⌋ .Then z · I a,c,k,ν ( z ) ≪ k · n · g a,c,k,ν , where g a,c,k,ν , := (cid:0) min (cid:0) kc (cid:8) νk − k + ac (cid:9) , kc (cid:8) νk − k − ac (cid:9)(cid:1)(cid:1) − , where { x } := x − ⌊ x ⌋ for x ∈ R . Here the implied constant is independent of a, k , and ν .Proof. We write πzk = Ce iA with C >
0. Then | A | < π since Re( z ) >
0. Making thesubstitution τ = πzxk gives z · I a,c,k,ν ( z ) = kπz Z S e − kτ πz · H a,c (cid:18) πiνk − πi k − τ (cid:19) dτ, (3.1) YMPTOTICS FOR RANK PARTITION FUNCTIONS 11 where τ runs on the ray through 0 of elements with argument ± A . One can see that for0 ≤ t ≤ A (cid:12)(cid:12)(cid:12)(cid:12) e − kR e itπz · H a,c (cid:18) πiνk − πi k ± Re it (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) → R → ∞ ) . (3.2)Moreover, since c is odd, the integrant in (3.1) can only have poles in points ir with r ∈ R \{ } .Thus we can shift the path of integration to the real line and get z · I a,c,k,ν ( z ) = kπz Z R e − kt πz · H a,c (cid:18) πiνk − πi k − t (cid:19) dt. We can show the following estimates (cid:12)(cid:12)(cid:12)(cid:12) e − kt πz (cid:12)(cid:12)(cid:12)(cid:12) = e − kπ Re ( z ) t , (cid:12)(cid:12)(cid:12)(cid:12) cosh (cid:18) πiνk − πi k − t (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e t , (cid:12)(cid:12)(cid:12)(cid:12) sinh (cid:18) πiνk − πi k − t ± πiac (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ( e t √ if t ≥ , (cid:12)(cid:12) sin (cid:0) πνk − π k ± πac (cid:1)(cid:12)(cid:12) if t ≤ , (cid:12)(cid:12)(cid:12) sin (cid:16) πνk − π k + πac (cid:17)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) sin (cid:16) πνk − π k − πac (cid:17)(cid:12)(cid:12)(cid:12) ≫ (cid:18) min (cid:18)(cid:26) νk − k + ac (cid:27) , (cid:26) νk − k − ac (cid:27)(cid:19)(cid:19) . Thus z · I a,c,k,ν ( z ) ≪ k (cid:0) min (cid:0)(cid:8) νk − k + ac (cid:9) , (cid:8) νk − k − ac (cid:9)(cid:1)(cid:1) | z | Z R e − kπ t Re ( z ) dt. Making the substitution t q k Re ( z ) π · t and using the estimateRe (cid:18) z (cid:19) − · | z | − ≤ · n · k − gives the lemma. (cid:3) We next estimate certain sums of Kloosterman type.
Lemma 3.2.
Let n, m ∈ Z , ≤ σ < σ ≤ k , D ∈ Z with ( D, k ) = 1 . (1) We have X h (mod k ) ∗ σ ≤ Dh ′≤ σ ω h,k · e πik ( hn + h ′ m ) ≪ gcd(24 n + 1 , k ) · k + ǫ . (3.3) (2) If c | k , then we have (3.4)( − ak +1 sin (cid:16) πac (cid:17) X h (mod k ) ∗ σ ≤ Dh ′≤ σ ω h,k sin (cid:0) πah ′ c (cid:1) · e − πia k h ′ c · e πik ( hn + h ′ m ) ≪ gcd(24 n +1 , k ) · k + ǫ , where the implied constants are independent of a and k .Proof. Equation (3.3) is basically proven in [2], thus we only consider (3.4). We set ˜ c := c if k is odd and ˜ c := 2 c if k is even. Clearly e − πia k h ′ c sin “ πah ′ c ” only depends on the residue class of h ′ (mod ˜ c ). Thus we can write (3.4) as( − ak +1 sin (cid:16) πac (cid:17) X c j e − πia k cjc sin (cid:0) πac j c (cid:1) X h (mod k ) ∗ σ ≤ Dh ′≤ σ h ′≡ cj (mod ˜ c ) ω h,k · e πik ( hn + h ′ m ) , where c j runs through a set of primitive residues (mod ˜ c ). The inner sum can be rewrittenas 1˜ c X h (mod k ) ∗ σ ≤ Dh ′≤ σ ω h,k · e πik ( hn + h ′ m ) X r (mod ˜ c ) e πir ˜ c ( h ′ − c j ) = 1˜ c X r (mod ˜ c ) e − πircj ˜ c X h (mod k ) ∗ σ ≤ Dh ′≤ σ ω h,k · e πik ( hn + h ′ ( m + kr ˜ c )) . Using (3.3) now easily gives (3.4). (cid:3)
Remark. If a = 1 and c = 3, then it is not hard to see that the left hand side of (3.4) simplifiesto X h (mod k ) ∗ σ ≤ Dh ′≤ σ (cid:18) h (cid:19) · ω h,k · e πik ( hn + h ′ m ) . Proof of Theorem 1.1
Proof of Theorem 1.1.
For the proof of Theorem 1.1 we use the Hardy-Ramanujan method.By Cauchy’s Theorem we have for n > A (cid:16) ac ; n (cid:17) = 12 πi Z C N (cid:0) ac ; q (cid:1) q n +1 dq, where C is an arbitrary path inside the unit circle surrounding 0 counterclockwise. Choosingthe circle with radius e − πn and as a parametrisation q = e − πn +2 πit with 0 ≤ t ≤
1, gives A (cid:16) ac ; n (cid:17) = Z N (cid:16) ac ; e − πn +2 πit (cid:17) · e π − πint dt. YMPTOTICS FOR RANK PARTITION FUNCTIONS 13
Define ϑ ′ h,k := 1 k ( k + k ) , ϑ ′′ h,k := 1 k ( k + k ) , where h k < hk < h k are adjacent Farey fractions in the Farey sequence of order N := (cid:4) n / (cid:5) .From the theory of Farey fractions it is known that1 k + k j ≤ N + 1 ( j = 1 , . (4.1)We decompose the path of integration in paths along the Farey arcs − ϑ ′ h,k ≤ Φ ≤ ϑ ′′ h,k , whereΦ = t − hk and 0 ≤ h ≤ k ≤ N with ( h, k ) = 1. Thus A (cid:16) ac ; n (cid:17) = X h,k e − πihnk Z ϑ ′′ h,k − ϑ ′ h,k N (cid:16) ac ; e πik ( h + iz ) (cid:17) · e πnzk d Φ , where z = kn − k Φ i . Applying Theorem 2.1 gives A (cid:16) ac ; n (cid:17) = i sin (cid:16) πac (cid:17) X h,kc | k ω h,k ( − ak +1 sin (cid:0) πah ′ c (cid:1) · e − πia k h ′ c − πihnk Z ϑ ′′ h,k − ϑ ′ h,k z − · e πzk ( n − ) + π kz × N (cid:18) ah ′ c ; q (cid:19) d Φ − i sin (cid:16) πac (cid:17) X h,kc ∤ k ω h,k ( − ak + l e − πih ′ sac − πih ′ a k cc + πih ′ lacc − πihnk Z ϑ ′′ h,k − ϑ ′ h,k z − · e πzk ( n − ) + π kz · q slc − l c · N (cid:18) ah ′ , lcc , c ; q (cid:19) d Φ + 2 sin (cid:16) πac (cid:17) X h,k ω h,k k · e − πihnk X ν (mod k ) ( − ν e − πih ′ ν k + πih ′ νk Z ϑ ′′ h,k − ϑ ′ h,k e πzk ( n − ) · z · I a,c,k,ν ( z ) d Φ =: X + X + X . To estimate P , we write N (cid:18) ah ′ c ; q (cid:19) =: 1 + X r ∈ N a ( r ) · e πimrh ′ k · e − πrkz , where m r is a sequence in Z and the coefficients a ( r ) are independent of a, c, k , and h . Wetreat the constant term and the term coming from from r ≥ S and S , respectively and first estimate S . Throughout we need the easily verified fact thatRe( z ) = kn , Re (cid:0) z (cid:1) > k , | z | − ≤ n · k − , and ϑ ′ h,k + ϑ ′′ h,k ≤ k ( N +1) . Since k , k ≤ N , wecan write Z ϑ ′′ h,k − ϑ ′ h,k = Z k ( N + k ) − k ( N + k ) + Z − k ( N + k ) − k ( k k ) + Z k ( k k )1 k ( N + k ) (4.2) and denote the associated sums by S , S , and S , respectively.We first consider S . Using Lemma 3.2 gives S ≪ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X r =1 a ( r ) X c | k ( − ak +1 sin (cid:16) πac (cid:17) X h ω h,k sin (cid:0) πah ′ c (cid:1) · e − πia k h ′ c − πihnk + πimrh ′ k Z k ( N + k ) − k ( N + k ) z − · e − πkz ( r − ) + πzk ( n − ) d Φ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ ∞ X r =1 | a ( r ) | · e − πr X k k − ǫ · (24 n − , k ) ≪ X d | (24 n − d ≤ N d X k ≤ Nd ( dk ) − ǫ ≪ n ǫ X d | (24 n − d ≤ N d − ≪ n ǫ . Since S and S are treated in exactly the same way we only consider S . Writing Z − k ( N + k ) − k ( k + k = N + k − X l = k + k Z − k ( l +1) − kl , we see(4.3) S ≪ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X r =1 a ( r ) X c | k N + k − X l = N +1 Z − k ( l +1) − kl z − · e − πkz ( r − ) + πzk ( n − ) d Φ( − ak +1 sin (cid:16) πac (cid:17) X hN
0, and the number is bounded in terms of c . Moreover the coefficient b ( r )is independent of k and a in a fixed class. Now the terms with negative exponents can beestimated as before, thus P equals2 sin (cid:16) πac (cid:17) X k,rc ∤ kδc,k,r> ( − ak + l X h ω h,k e πik ( − nh + m a,c,k,r h ′ ) Z ϑ ′′ h,k − ϑ ′ h,k z − e πzk ( n − ) + πkz δ c,k,r d Φ+ O ( n ǫ ) . We turn back to the estimation of P . We can write Z ϑ ′′ h,k − ϑ ′ h,k = Z kN − kN − Z − k ( k + k − kN − Z kN k ( k + k and denote the associated sums by S , S , and S , respectively. The sums S and S contribute to the error terms. Since they have the same shape, we only consider S . Writing Z − k ( k + k − kN = k + k − X l = N Z − k ( l +1) − kl gives S ≪ X c | k N + k − X l = N Z − k ( l +1) − kl z − e π kz + πzk ( n − ) d Φ( − ak +1 sin (cid:16) πac (cid:17) X hl
1) sinh r r (24 n − πk ! + O (cid:16) n − (cid:17) . YMPTOTICS FOR RANK PARTITION FUNCTIONS 17
Thus we obtain, estimating the Kloosterman sums with the same arguments as before andusing Lemma 3.1, X + X = 4 √ i √ n − X c | k B a,c,k ( − n, √ k · sinh (cid:16) π k √ n − (cid:17) + 8 √ (cid:0) πac (cid:1) √ n − X k,rc ∤ kδc,k,r> D a,c,k ( − n, m a,c,k,r ) √ k · sinh r δ c,k,r (24 n − πk ! + O ( n ǫ ) . To estimate P , we split the path of integration as in (4.2) and estimate the sums in thesame way as before, using that X ≤ ν ≤ k g a,c,k,ν ≪ X ≤ ν ≤ ck ν ≪ k ǫ . (cid:3) Proof of Corollary 1.2.
Corollary 1.2 follows directly from the identity ∞ X n =0 N ( a, c ; n ) q n = 1 c ∞ X n =0 p ( n ) q n + 1 c c − X j =1 ζ − ajc · R ( ζ jc ; q )(4.6)and (1.4). (cid:3) Proof of Theorem 1.3
Here we prove the Andrews-Lewis conjecture. Using equation (4.6) and N (cid:18)
13 ; q (cid:19) = N (cid:18)
23 ; q (cid:19) easily gives that N (0 , n ) < N (1 , n ) ⇔ A (cid:18)
13 ; n (cid:19) < . Thus Theorem 2 follows from the following.
Proposition 5.1.
We have for all n
6∈ { , , } A (cid:18)
13 ; 3 n (cid:19) < , (5.1) A (cid:18)
13 ; 3 n + 1 (cid:19) > , (5.2) A (cid:18)
13 ; 3 n + 2 (cid:19) < . (5.3) Remark.
For n ∈ { , , } , equation (5.2) and (5.3) still hold and in (5.1) we have equality. Proof.
With MAPLE we compute that the proposition holds for n ≤ n > n ≥ √ i (24 n − X | k B , ,k ( − n, √ k · sinh (cid:16) π k √ n − (cid:17) . (5.4)Using definition (1.10) it is easy to see that B , , ( − n,
0) = 2 i sin (cid:18) π − πn (cid:19) . Thus the term corresponding to k = 3 is given by − (cid:0) π − πn (cid:1) · sinh (cid:0) π √ n − (cid:1) (24 n − . Using the trivial bound for Kloosterman sums gives that the remaining terms can be estimatedagainst 12(24 n − X ≤ k ≤ N k · sinh (cid:16) π k √ n − (cid:17) . Next we make the O -term in Theorem 1.1 explicit. For this we use the same notation as inthe proof of Theorem 1.1. Instead of using Lemma 3.2, we estimate the Kloosterman sumstrivially. The contribution coming from N (cid:16) h ′ ; q (cid:17) can be estimated against2 · e π + π √ ∞ X r =1 | a ( r ) | · e − πr X ≤ k ≤ N k − . Computing the first few coefficients and then using that for n > p ( n ) < exp π r n ! easily gives that ∞ X r =1 | a ( r ) | · e − πr ≤ . . In the same way we can estimate the contribution coming from N ( h ′ , l, q ) by4 √ · e π ∞ X r =1 | b ( r ) | · e − πr X ≤ k ≤ N ∤ k k − . Moreover, distinguishing the cases l = 1 and l = 2, we get ∞ X r =1 | b ( r ) | · e − πr ≤ . . YMPTOTICS FOR RANK PARTITION FUNCTIONS 19
By making the path of integration symmetric, we introduce an error that can be estimatedagainst 2 √ · e π + π · n − X ≤ k ≤ N k . Integrating along the smaller arc of Γ gives an error that can be estimated against8 π · e π + π · n − X ≤ k ≤ N k. Moreover the estimate in Lemma 3.1 can be made explicit as2 · e + e − · n (cid:18) min (cid:18)(cid:26) νk − k + 13 (cid:27) , (cid:26) νk − k + 13 (cid:27)(cid:19)(cid:19) − . This easily gives X ≤ · ( e + e − ) · e π · n − X k k k X ν =1 (cid:18) min (cid:18)(cid:26) νk − k + 13 (cid:27) , (cid:26) νk − k − (cid:27)(cid:19)(cid:19) − . Combining all errors one can show that the term in (5.4) is dominant for n ≥ (cid:3) References [1] G. E. Andrews,
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School of Mathematics, University of Minnesota, Minneapolis, MN 55455, U.S.A.
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