Andrews-Beck Type Congruences Related to the Crank of a Partition
aa r X i v : . [ m a t h . C O ] J a n ANDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF APARTITION
SHREEJIT BANDYOPADHYAY
Abstract.
In this paper, we discuss a few recent conjectures made by George Beck related to the ranksand cranks of partitions. The conjectures for the rank of a partition were proved by Andrews by usingresults due to Atkin and Swinnerton-Dyer on a suitable generating function, while the conjectures relatedto cranks were studied by Shane Chern using weighted partition moments. We revisit the conjectures onthe crank of a partition by decomposing the relevant generating function and further explore connectionswith Apple-Lerch series and tenth order mock theta functions. Introduction
The partition of a positive integer n is a weakly decreasing sequence of integers whose sum equals n , and is usually denoted by p ( n ) . For example, the number 4 has 5 partitions: 4, 3+1, 2+2, 2+1+1,1+1+1+1 and as such, p (4) = 5 . Among the identities satisfied by the partition function p ( n ) , we have the three celebrated congruencesdue to Ramanujan: p (5 n + 4) ≡ p (7 n + 5) ≡ p (11 n + 6) ≡ λ of n, if l ( λ ) is the largest part, ω ( λ ) the number of ones and µ ( λ )the number of parts in λ bigger than ω ( λ ) , then the crank of λ is given by c ( λ ) = ( l ( λ ) if ω ( λ ) = 0 µ ( λ ) − ω ( λ ) if ω ( λ ) > N ( m, k, n ) to be the num-ber of partitions of n with rank congruent to m modulo k and then establishing that N ( i, , n + 4) =15 p (5 n + 4) for 0 ≤ i ≤ N ( i, , n + 5) = 17 p (7 n + 5) for 0 ≤ i ≤ . Instead of the function N ( m, k, n ), the conjectures made by George Beck considered the number ofparts in the partitions of n with rank congruent to m modulo k, denoted by N T ( m, k, n ) and the numberof ones in the partitions of n with crank congruent to m modulo k, denoted by M ω ( m, k, n ) . We nowmake a note of these results.
Theorem 1.1. If i = 1 , , then for ∀ n ≥ , N T (1 , , n + i ) − N T (4 , , n + i ) + 2 N T (2 , , n + i ) − N T (3 , , n + i ) ≡ . Theorem 1.2. If i = 1 , , then for ∀ n ≥ , N T (1 , , n + i ) − N T (6 , , n + i ) + N T (2 , , n + i ) − N T (5 , , n + i ) − N T (3 , , n + i ) + N T (4 , , n + i ) ≡ . Penn State University, [email protected]
Theorem 1.3.
For ∀ n ≥ , M ω (1 , , n + 4) + 2 M ω (2 , , n + 4) − M ω (3 , , n + 4) − M ω (4 , , n + 4) ≡ . Theorem 1.4.
For ∀ n ≥ , M ω (1 , , n + 5) + 2 M ω (2 , , n + 5) + 3 M ω (3 , , n + 5) − M ω (4 , , n +5) − M ω (5 , , n + 5) − M ω (6 , , n + 5) ≡ . Remark 1.
As observed by Andrews in [2], Theorems 1.1 and 1.2 are trivial if we replace the function
N T by the function N of Atkin and Swinnerton-Dyer since the rank function is symmetric: N ( m, k, n ) = N ( k − m, k, n ) . For the rest of this paper, we adopt the standard q-series notations:( A ; q ) n = n − Y m =0 (1 − Aq m ), ( A ; q ) ∞ = ∞ Y m =0 (1 − Aq m ) , ( A , A , ...A r ; q ) n = r Y t =1 ( A t ; q ) n , ( A , A , ...A r ; q ) ∞ = r Y t =1 ( A t ; q ) ∞ . Preliminaries
Theorems 1.1 and 1.2 were proved by Andrews in [2] by considering the generating function whichweights partitions by the number of parts while keeping track of the rank. That function is ∂∂x (cid:12)(cid:12)(cid:12) x =1 X n ≥ x n q n ( zq ; q ) n ( xqz ; q ) n (4)where q marks the number being partitioned, z marks the rank and x keeps track of the number ofparts in the partition. As noted by Andrews in [2], this is a more complicated function compared to theuniversal mock-theta function X n ≥ q n ( zq ; q ) n ( qz ; q ) n which counts the number of partitions with a certainrank, with q marking the number being partitioned and z counting the rank itself.Theorems 1.1-1.4 were independently considered by Shane Chern in [3] by relating them to the sec-ond rank and crank moments N ( n ) and M ( n ) respectively.In this paper, our goal will be to deal with Theorems 1.3 and 1.4 by considering, instead of the functionin (4), the following function ( xq ) ∞ ( zq, xqz ; q ) ∞ (5)where the variable x keeps track of the number of ones in a partition, q marks the number being parti-tioned and z marks the crank of the partition. This function is somewhat simpler than the function in(4) used by Andrews in [2] to deal with Theorems 1.1 and 1.2.Our goal will be to decompose the function in (5) into powers of q modulo 5 or modulo 7 and thereby re-duce the proofs of Theorems 1.3 and 1.4 into proving certain equalities we obtain on comparing coefficientsfor different powers of q. Generating Functions
We will use the following theorem to decompose the crank generating function in (5) into powers of q modulo 5 or 7. Theorem 3.1. ( xq ) ∞ ( zq, xqz ; q ) ∞ = 1( q ; q ) ∞ (cid:20) X n ≥ ( − n − ( xq ; q ) n ( q ; q ) n − q ( n +12 ) (cid:18) q n (1 − zq n ) + x/z (1 − xq n z ) (cid:19)(cid:21) . NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 3
Remark 2.
We note here the similarity of this result and Theorem 3 in [2].
Corollary 3.2. If M ω ( b, k, n ) is the number of ones in the partitions λ of n with crank congruent to b modulo k , then X n ≥ M ω ( b, k, n ) q n = ∂∂x (cid:12)(cid:12)(cid:12) x =1 q ; q ) ∞ X n ≥ ( − n − ( xq ; q ) n ( q ; q ) n − q ( n +12 ) (cid:18) q n ( b − − q nk + x k − b q ( k − b − n − x k q kn (cid:19) . Corollary 3.3. X n ≥ (cid:20) M ω (1 , , n ) + 2 M ω (2 , , n ) − M ω (3 , , n ) − M ω (4 , , n ) (cid:21) q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 5) . Corollary 3.4. X n ≥ (cid:20) M ω (1 , , n ) + 2 M ω (2 , , n ) − +3 M ω (3 , , n ) − M ω (4 , , n ) − M ω (5 , , n ) − M ω (6 , , n ) (cid:21) q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 7) . Proof of Theorem 3.1.
We have the following identity φ " a, q √ a, − q √ a, b, c, d √ a, −√ a, qab , qac , qad ; q, qabcd = ( aq, aqbc , aqbd , aqcd ; q ) ∞ ( aqb , aqc , aqd , aqbcd ; q ) ∞ (6)Set a = x, b = xz , c = z, d → ∞ in (6) whence the right side reduces to ( xq, q ; q ) ∞ ( zq, xqz ; q ) ∞ . Under the samesubstitution, the left side becomes X n ≥ ( − n ( x ; q ) n (1 − xq n )(1 − x )( q ; q ) n (1 − xz )(1 − z )(1 − zq n )(1 − xz q n ) q ( n +12 ) . (7)Next we replace (1 − z )(1 − xz )(1 − xq n )(1 − zq n )(1 − xz q n ) by − (1 − q n )(1 − xq n )( 1 q n (1 − zq n ) + x/z (1 − xq n z ) ) + ( 1 − xq n q n ) , so that (7) becomes X n ≥ ( − n − ( x ; q ) n (1 − x )( q ; q ) n q ( n +12 )( 1 q n (1 − zq n ) + x/z (1 − xq n z ) )(1 − xq n )(1 − q n )+ X n ≥ ( − n ( x ; q ) n (1 − x )( q ; q ) n q ( n +12 ) (1 − xq n ) q n . (8)We next note that the second sum in (8) vanishes. Indeed, if we set a = x, bc = qx, d → ∞ in (6), we getthat X n ≥ ( − n ( x ; q ) n (1 − x )( q ; q ) n q ( n − )(1 − xq n ) = ( xq, q ) ∞ ( xqb , b ; q ) ∞ = 0 . SHREEJIT BANDYOPADHYAY
Thus, by (6), (7) and (8), we have that X n ≥ ( − n − ( x ; q ) n (1 − x )( q ; q ) n q ( n +12 )(1 − xq n )(1 − q n )( 1 q n (1 − zq n ) + x/z (1 − xq n z ) ) = ( xq, q ; q ) ∞ ( zq, xqz ; q ) ∞ . Theorem 3.1 now follows on dividing both sides by ( q ; q ) ∞ . (cid:3) Proof of Corollary 3.2.
We write 1 q n (1 − zq n ) = q − n (1+ zq n + z q n + ... ) = X j ≥ z j q n ( j − and x/z (1 − xq n z ) = X j ≥ x j +1 z − j − q nj . Taking exponents of the form kt + b for z (since we count the partitions with crank congruent to b modulo k ) , we thus get that X n ≥ M ω ( b, k, n ) q n = ∂∂x (cid:12)(cid:12)(cid:12) x =1 q ; q ) ∞ X n ≥ ( − n − q ( n +12 ) ( xq ; q ) n ( q ; q ) n [ q n ( b − − q nk + x k − b q ( k − − b ) n − x k q nk ]which is Corollary 3.2. (cid:3) Proof of Corollary 3.3.
Set k = 5 , b = 1 , , , X n ≥ [ M ω (1 , , n ) + 2 M ω (2 , , n ) − M ω (3 , , n ) − M ω (4 , , n )] q n = ∂∂x (cid:12)(cid:12)(cid:12) x =1 q ; q ) ∞ X n ≥ ( − n − ( xq ; q ) n ( q ; q ) n − q ( n +12 ) × " − q n + x q n − x q n + 2 q n − q n + 2 x q n − x q n − q n − q n − x q n − x q n − q n − q n − x − x q n = ∂∂x (cid:12)(cid:12)(cid:12) x =1 q ; q ) ∞ X n ≥ ( − n − ( xq ; q ) n ( q ; q ) n − q ( n +12 ) × (1 − x )(1 − q n )(1 − x q n ) " q n (1 + x ) − q n (1 + x + x ) − q n (1 + x )(1 + x ) + q n x (1 + x )(1 + x ) + 2 q n x (1 + x + x ) − q n x (1 + x ) − q n x = 1( q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n )(1 − q n ) " q n − q n − q n + 4 q n + 6 q n − q n − q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 5)where we have used ∂∂x (cid:12)(cid:12)(cid:12) x =1 (1 − x ) F ( x, q, z ) = − F (1 , q, z ) NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 5 in the penultimate step. This proves Corollary 3.3. (cid:3)
Proof of Corollary 3.4.
Exactly as in the proof of Corollary 3.3, we get using Corollary 3.2, X n ≥ [ M ω (1 , , n ) + 2 M ω (2 , , n ) + 3 M ω (3 , , n ) − M ω (4 , , n ) − M ω (5 , , n ) − M ω (6 , , n )] q n = ∂∂x (cid:12)(cid:12)(cid:12) x =1 q ; q ) ∞ X n ≥ ( − n − ( xq ; q ) n ( q ; q ) n − q ( n +12 ) × " − q n + x q n − x q n + 2 q n − q n + 2 x q n − x q n + 3 q n − q n + 3 x q n − x q n − q n − q n − x q n − x q n − q n − q n − x q n − x q n − q n − q n − x − x q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 7)where we have simplified as in the proof of Corollary 3.3 and again used ∂∂x (cid:12)(cid:12)(cid:12) x =1 (1 − x ) F ( x, q, z ) = − F (1 , q, z ) . This proves Corollary 3.4. (cid:3) The Decomposition modulo 5
In this section, we attempt to prove certain equalities which should imply Theorem 1.3. We start withthe following conjecture.
Conjecture 4.1. q ; q ) ∞ X n ≥ ( − n q ( n +12 )(1 − q n ) (1 + q n )1 − q n = − q ( q ; q ) ∞ G ( q ) + q ( q ; q ) ∞ H ( q )+ q ( q ; q ) ∞ H ( q ) G ( q ) where G ( q ) = 1( q ; q ) ∞ ( q ; q ) ∞ and H ( q ) = 1( q ; q ) ∞ ( q ; q ) ∞ represent the familiar infinite productsarising in the context of Rogers Ramanujan identities. We will next show that Conjecture 4.1 follows from five different identities we get on comparingcoefficients of powers of q after properly expanding both sides of 4.1. We also give proofs of four of thesefive identities. We have not been able to yet prove the fifth identity, but we give different formulationsof it and various ways we have attacked it. SHREEJIT BANDYOPADHYAY
Remark 3.
We note that, Corollary 3.3 and Conjecture 4.1 would together imply that X n ≥ (cid:20) M ω (1 , , n ) + 2 M ω (2 , , n ) − M ω (3 , , n ) − M ω (4 , , n ) (cid:21) q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 5) ≡ − q ( q ; q ) ∞ G ( q ) + q ( q ; q ) ∞ H ( q ) + q ( q ; q ) ∞ H ( q ) G ( q ) (mod 5) (9) so that X n ≥ (cid:20) M ω (1 , , n + 4) + 2 M ω (2 , , n + 4) − M ω (3 , , n + 4) − M ω (4 , , n + 4) (cid:21) ≡ since there’s no exponent of q which is 4 modulo 5 on the right side of Conjecture 4.1. This means that,in conjunction with Corollary 3.3, a proof of Conjecture 4.1 will yield Theorem 1.3 Theorem 4.2.
Conjecture 4.1 follows from the following five equalities. ( i ) X m ( − m q m m − q m +1 − X m ( − m q m − m − q m − = ( q ; q ) ∞ G ( q ) H ( q ) (10)( ii ) X m ( − m q m m − q m +1 = ( q ; q ) ∞ G ( q ) (11)( iii ) X m ( − m q m m − q m +2 = ( q ; q ) ∞ H ( q ) (12)( iv ) X m ( − m q m m − − q m − − X m ( − m q m − m − − q m − = ( q ; q ) ∞ H ( q ) G ( q ) (13)( v ) − ′ X m ( − m q m m − − q m + X m ( − m q m − m − − q m − + 2 ′ X m ( − m q m m − − q m +2 X m ( − m q m m − q m +2 = ( q ; q ) ∞ H ( q ) G ( q ) (14)In this theorem and subsequently, we adopt the notation that X n means ∞ X n = −∞ while ′ X n stands for ∞ X n = −∞ n =0 . Proof.
Recall that Conjecture 4.1 is the following:1( q ; q ) ∞ X n ≥ ( − n q ( n +12 )(1 − q n ) (1 + q n )1 − q n = − q ( q ; q ) ∞ G ( q ) + q ( q ; q ) ∞ H ( q )+ q ( q ; q ) ∞ H ( q ) G ( q ) (15) NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 7
By Lemma 6 in [1], ( q ; q ) ∞ = − qP (0) " − q − P (2) P (1) + q P (4) P (2) with P (0) = ( q ; q ) ∞ , P (1) = P (4) = ( q , q ; q ) ∞ , P (2) = ( q , q ; q ) ∞ . Thus, ( q ; q ) ∞ = ( q ; q ) ∞ G ( q ) H ( q ) − q − q H ( q ) G ( q ) ! . We get that( q ; q ) ∞ " − q ( q ; q ) ∞ G ( q ) + q ( q ; q ) ∞ H ( q ) + q ( q ; q ) ∞ H ( q ) G ( q ) =( q ; q ) ∞ − q G ( q ) H ( q ) + 2 q G ( q ) + q H ( q ) − q H ( q ) G ( q ) − q H ( q ) G ( q ) ! (16)On the other hand, X n ≥ ( − n q ( n +12 )(1 − q n ) (1 + q n )1 − q n = X n ≥ ( − n q n ( n +1) / (1 − q n + 2 q n − q n )1 − q n . Since on replacing n with − n , X n ≥ ( − n q n ( n +1) / q n − q n = − X n ≤− ( − n q n ( n +1) / q n − q n and X n ≥ ( − n q n ( n +1) / q n − q n = − X n ≤− ( − n q n ( n +1) / − q n , we get that, X n ≥ ( − n q ( n +12 )(1 − q n ) (1 + q n )1 − q n = ′ X n ( − n q n ( n +1) / − q n − ′ X n ( − n q n ( n +3) / − q n (17)Set n = 5 m + t. Then ′ X n ( − n q n ( n +1) / − q n = X t = − ( − t q t ( t +1) / X m ( − m q ((5 m + t )(5 m + t +1) − t ( t +1)) / − q m +5 t = X t = − ( − t q t ( t +1) / X m ( − m q (25 m +10 mt +5 m ) / − q m +5 t (with the term for m = t = 0 omitted)= ′ X m ( − m q (25 m +5 m ) / − q m − X m ( − m q (25 m +15 m +2) / − q m +5 − X m ( − m q (25 m − m ) / − q m − − X m ( − m q (25 m +25 m +6) / − q m +10 + X m ( − m q (25 m − m +2) / − q m − . (18) SHREEJIT BANDYOPADHYAY
Similarly, ′ X n ( − n q n ( n +3) / − q n = ′ X m ( − m q (25 m +15 m ) / − q m − X m ( − m q (25 m +25 m +4) / − q m +5 − X m ( − m q (25 m +5 m − / − q m − + X m ( − m q (25 m +35 m +10) / − q m +10 + X m ( − m q (25 m − m − / − q m − . (19)Using (17),(18) and (19), we get that X n ≥ ( − n q ( n +12 )(1 − q n ) (1 + q n )1 − q n = q " − X m ( − m q (25 m +15 m ) / − q m +5 + X m ( − m q (25 m − m ) / − q m − + 2 q " X m ( − m q (25 m +25 m ) / − q m +5 + q " X m ( − m q (25 m +25 m ) / − q m +10 + 2 q " − X m ( − m q (25 m +5 m − / − q m − + X m ( − m q (25 m − m − / − q m − + q " ′ X m ( − m q (25 m +5 m − / − q m − X m ( − m q (25 m − m − / − q m − − ′ X m ( − m q (25 m +15 m − / − q m − X m ( − m q (25 m +35 m ) / − q m +10 (20)Comparing coefficients of q, q , q , q , q between (16) and (20) and replacing q by q / , the proof iscomplete. (cid:3) Theorem 4.3.
Equations (11) and (12) in Theorem 4.2 are valid.Proof.
By Theorem 3.2 in [4], we have φ " z, q √ z, − q √ z, a , a , a √ z, −√ z, zqa , zqa , zqa ; q, zqa a a = Y " zq, zqa a , zqa a , zqa a zqa , zqa , zqa , zqa a a (21)The left side equals X m ≥ ( z ; q ) m (1 − zq m )(1 − z )( q ; q ) m ( a , a , a ; q ) m ( zqa , zqa , zqa ; q ) m zqa a a ! m = X m ≥ ( − m (1 + q m ) q m ( m +1) / ( q, q ; q ) m ( q , q ; q ) m (put z = 1 , a → ∞ , a = q, a = 1 q , q = q )= 1 + X m ≥ ( − m (1 + q m ) q m ( m +1) / (1 − q )(1 − q )(1 − q m +1 )(1 − q m − )= 1 + X m ≥ ( − m q m ( m +1) / (1 − q )(1 − q − ) − q )(1 − q m − ) + 1(1 − q − )(1 − q m +1 ) ! NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 9 = 1 + X m ≤− ( − m q (5 m +5 m ) / (1 − q )1 − q m +1 + X m ≥ ( − m q (5 m +5 m ) / (1 − q )1 − q m +1 = (1 − q ) X m ( − m q (5 m +5 m ) / − q m +1 . On the other hand, setting z = 1 , a → ∞ , a = q, a = 1 q , q = q on the right side of (21), we get( q ; q ) ∞ ( q , q ; q ) ∞ = (1 − q )( q ; q ) ∞ G ( q ) and thus equation (11) is proved.Again, if we set z = 1 , a → ∞ , a = q , a = 1 q , q = q on the left side of (21), we similarly get X m ≥ ( q ; q ) m (1 − zq m )(1 − z )( q ; q ) m ( a , a , a ; q ) m ( zqa , zqa , zqa ; q ) m zqa a a ! m = X m ≥ ( − m (1 + q m ) q m ( m +1) / ( q , q ; q ) m ( q , q ; q ) m = 1 + X m ≥ ( − m (1 + q m ) q m ( m +1) / (1 − q )(1 − q )(1 − q m +2 )(1 − q m − )= (1 − q ) X m ( − m q (5 m +5 m ) / − q m +2 exactly as before, whereas under the same substitutions, the right side of (21) equals ( q ; q ) ∞ ( q , q ; q ) ∞ =(1 − q )( q ; q ) ∞ H ( q ), proving (12). (cid:3) Theorem 4.4.
Equations (10) and (13) in Theorem 4.2 are valid.Proof.
We start with the Apple-Lerch series m ( x, q, z ) = − zj ( z ; q ) X r ( − r q r ( r +1) / z r − xzq r with j ( z ; q ) =( z, q/z, q ; q ) ∞ . Then m ( q ; q ; 1 /q ) = − qj (1 /q ; q ) X m ( − m q m ( m +1) / q − m − q m +1 = − qj (1 /q ; q ) S where S = X m ( − m q (5 m +3 m ) / − q m +1 . Similarly, m ( q ; q ; 1 /q ) = 1 q j (1 /q ; q ) S where S = − X m ( − m q (5 m − m ) / − q m − . The left side of (10) is therefore S + S = − qj (1 /q ; q ) m ( q , q , /q ) + q j (1 /q ; q ) m ( q , q , /q )= ( q, q , q ; q ) ∞ " m ( q , q , /q ) − m ( q , q , /q ) (22) as j (1 /q ; q ) = (1 /q, q , q ; q ) ∞ = − q ( q, q , q ; q ) ∞ and j (1 /q ; q ) = (1 /q , q , q ; q ) ∞ = − q ( q, q , q ; q ) ∞ . By lemma 11.3.4 in [5], m ( x, q, z ) − m ( x, q, z ) = z ( q ; q ) ∞ j ( z /z ; q ) j ( xz z ; q ) j ( z ; q ) j ( z ; q ) j ( xz ; q ) j ( xz ; q )whence, setting x = q , z = 1 q , z = 1 q , q = q ,m ( q , q , /q ) − m ( q , q , /q ) = 1 q ( q ; q ) ∞ j ( q ; q ) j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j ( q ; q )= ( q , q , q ; q ) ∞ ( q, q, q, q , q , q ; q ) ∞ By (22), the proof of equation (10) is now finished.For equation (13), we note similarly that m ( q ; q ; 1 /q ) = − q j (1 /q ; q ) X m ( − m q (5 m − m ) / − q m − and m ( q ; q ; 1 /q ) = − q j (1 /q ; q ) X m ( − m q (5 m + m ) / − q m − whence the left side of (13) reduces to − q j (1 /q ; q ) m ( q, q , /q ) + qj (1 /q ; q ) m ( q, q , /q )= 1 q ( q , q ; q ) ∞ " m ( q, q , /q ) − m ( q, q , /q ) (23)as j (1 /q ; q ) = − q ( q , q , q ; q ) ∞ , j (1 /q ; q ) = − q ( q , q , q ; q ) ∞ . By 11.3.4 in [5], we check as before that (setting x = q, z = 1 q , z = 1 q , q = q ), m ( q, q , /q ) − m ( q, q , /q ) = q ( q, q , q ; q ) ∞ ( q , q , q , q , q , q ; q ) ∞ whence by (23), equation (13) is proved. (cid:3) The only equation left to be proven to establish the validity of Conjecture 4.1 and hence of Theorem 1.3is equation (14) in the statement of Theorem 4.2. We haven’t been able to prove this, but the followingtheorem gives three equivalent versions of this equation.
Theorem 4.5.
If (A) is the statement − ′ X m ( − m q m m − − q m + X m ( − m q m − m − − q m − + 2 ′ X m ( − m q m m − − q m +2 X m ( − m q m m − q m +2 = ( q ; q ) ∞ H ( q ) G ( q ) , NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 11 (B) is the statement q ∞ X j =0 q j +2 − q j +1 − q j +5 − q j +4 − q j +3 − q j +2 + 3 q j +4 − q j +3 ! = ( q ; q ) ∞ H ( q ) G ( q ) and (C) is the statement X m q m (1 − q m +1 ) − q m +5 = ( q ; q ) ∞ H ( q ) G ( q ) then (A), (B) and (C) are equivalent.Proof. We will prove Theorem 4.5 by reducing the left side of (A) to the left side of (B) and the left sideof (B) to the left side of (C).Let T = ′ X m ( − m q m m − − q m , T = X m ( − m q m − m − − q m − , T = ′ X m ( − m q m m − − q m T = X m ( − m q m m − q m +2 . Then since m ( q , q , q ) = − q j (1 /q ; q ) X m ( − m q (5 m − m ) / − q m − , we get T = − q j (1 /q ; q ) m ( q , q , q ) . Similarly, T = − j ( q ; q ) q m ( q, q , q ) . Unfortunately, T and T are not proper Apple-Lerch sums sincethey skip the m = 0 term.Let m ′ ( x, q, z ) = − zj ( z ; q ) ′ X r ( − r q r ( r +1) / z r − xzq r , so m ′ ( q , q , q ) = − q j (1 /q ; q ) ′ X m ( − m q (5 m + m ) / − q m and hence, T = − qj (1 /q ; q ) m ′ ( q , q , q ) = − qj (1 /q ; q ) lim z → q − m ( q , q , z ) + zj ( z ; q ) . − q z ! . In a similar fashion, we may verify that T = − j (1 /q ; q ) lim z → q − m ( q, q , z ) + zj ( z ; q ) . − qz ! . Thus the left side of (A)= ( − T + T ) + 2( T + T )= " qj (1 /q ; q ) lim z → q − m ( q , q , z ) + zj ( z ; q ) . − q z ! − q j (1 /q ; q ) m ( q , q , q ) +2 " − j (1 /q ; q ) lim z → q − m ( q, q , z ) + zj ( z ; q ) . − qz ! − j ( q ; q ) q m ( q, q , q ) = X + 2 Y ,say.Thus, X = q ( 1 q , q , q ; q ) ∞ lim z → q − m ( q , q , z ) + zj ( z ; q ) . − q z ! − q ( 1 q , q , q ; q ) ∞ m ( q , q , q )= 1 q ( q , q , q ; q ) ∞ lim z → q − " m ( q , q , q ) − m ( q , q , z ) − zj ( z ; q ) 11 − q z But, by 11.3.4 of [5], we can check that m ( q , q , q ) − m ( q , q , z ) = z ( q ; q ) ∞ j ( 1 q z ; q ) j ( zq ; q ) j ( z ; q ) j ( 1 q ; q ) j ( q z ; q ) j ( 1 q ; q ) = z ( q , q z , q z, zq , q z ; q ) ∞ ( z q z , q , q , q z, q z , q , q ; q ) ∞ so that X = 1 q ( q , q , q ; q ) ∞ lim z → q − " z ( q , q z , q z, zq , q z ; q ) ∞ ( z, q z , q , q , q z, q z , q , q ; q ) ∞ − zj ( z ; q )(1 − q z ) = 1 q ( q , q , q ; q ) ∞ lim z → q − " z ( z, q z , q , q , q z, q z , q , q , q ; q ) ∞ × ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − q z ) = 1 q ( q , q , q ; q ) ∞ /q ( 1 q , q , q , q , q , q , q , q , q ; q ) ∞ × lim z → q − ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − qz )= − q q, q , q , q , q , q ; q ) ∞ × lim z → q − ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − qz )Suppose, f ( z ) = ( q , q , q z , q z, zq , q z ; q ) ∞ . Then f ′ ( z ) = ( q ; q ) ∞ " ( q z, zq , q z ; q ) ∞ ∞ X j =0 − q j q z ! ′ ∞ Y n =0 n = j − q n q z ! + ( 1 q z , zq , q z ; q ) ∞ ∞ X j =0 − ( q z ) q j ! ′ ∞ Y n =0 n = j − ( q n .q z ) ! + ( 1 q z , q z, q z ; q ) ∞ ∞ X j =0 − zq .q j ! ′ ∞ Y n =0 n = j − q n . zq ! + ( 1 q z , q z, zq ; q ) ∞ ∞ X j =0 − q z .q j ! ′ ∞ Y n =0 n = j − q n . q z ! NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 13
Thus f ′ ( 1 q ) = ( q ; q ) ∞ ( 1 q , q , q , q ; q ) ∞ ∞ X j =0 q j − z − q j − z + − q j +7 − q j +7 z + − q j − − q j − z + q j +7 z − q j +7 z ! z =1 /q = ( q ; q ) ∞ q ( q, q , q , q ; q ) ∞ ∞ X j =0 q j − q j − − q j +7 − q j +6 − q j − − q j − + q j +9 − q j +8 ! . And if g ( z ) = ( 1 q , q , q , q ; q ) ∞ ( q z, q z ; q ) ∞ , then we can similarly check that g ′ ( 1 q ) = 0 . Thus, by L’Hospitals rule, X = 1 q ∞ X j =0 q j − q j − − q j +7 − q j +6 − q j − − q j − + q j +9 − q j +8 ! . Similarly, Y = − ( 1 q , q , q ; q ) ∞ lim z → q − m ( q, q , z ) + zj ( z ; q ) . − qz ! − ( q, q , q ; q ) ∞ q m ( q, q , q )= − q ( q, q , q ; q ) ∞ lim z → q − " m ( q, q , q ) − m ( q , q , z ) − zj ( z ; q ) 11 − qz Exactly as before, we can check that m ( q, q ; q ) − m ( q, q , z ) = z ( qz , q z, q z, q z , q ; q ) ∞ ( z, q z , q, q , q z , qz, q , q ; q ) ∞ so that Y = − q ( q, q , q ; q ) ∞ lim z → q − " z ( qz , q z, q z, q z , q ; q ) ∞ ( z, q z , q, q , q z , qz, q , q ; q ) ∞ − zj ( z ; q )(1 − qz ) = − q ( q, q , q ; q ) ∞ lim z → q − " z ( z, q z , q, q , q z , q z, q , q , q ; q ) ∞ × ( qz , q z, q z, q z , q , q ; q ) ∞ − ( q, q , q z , q z, q , q ; q ) ∞ (1 − qz ) = 1 q ( q, q , q , q , q , q ; q ) ∞ × lim z → q − ( qz , q z, q z, q z , q , q ; q ) ∞ − ( q, q , q z , q z, q , q ; q ) ∞ (1 − qz )after simplification as before.As before, if h ( z ) = ( qz , q z, q z, q z , q , q ; q ) ∞ , then h ′ ( 1 q ) = ( q ; q ) ∞ ( q, q , q , q ; q ) ∞ ∞ X j =0 q j +3 − q j +2 − q j +4 − q j +3 − q j +2 − q j +1 + q j +5 − q j +4 ! and if p ( z ) = ( q, q , q z , q z, q , q ; q ) ∞ , then p ′ ( 1 q ) = 0so that Y = − q ∞ X j =0 q j +3 − q j +2 − q j +4 − q j +3 − q j +2 − q j +1 + q j +5 − q j +4 ! . Thus, the left side of (A)= X + 2 Y = 1 q ∞ X j =0 q j − q j − − q j +7 − q j +6 − q j − − q j − + q j +9 − q j +8 ! − q ∞ X j =0 q j +3 − q j +2 − q j +4 − q j +3 − q j +2 − q j +1 + q j +5 − q j +4 ! = 1 q ∞ X j =0 q j +5 − q j +4 − q j +2 − q j +1 − q j +3 − q j +2 + q j +4 − q j +3 ! + 1 q − /q + q − q − /q − /q − q − q ! − q ∞ X j =0 q j +3 − q j +2 − q j +4 − q j +3 − q j +2 − q j +1 + q j +5 − q j +4 ! = 1 q ∞ X j =0 q j +2 − q j +1 − q j +5 − q j +4 − q j +3 − q j +2 + 3 q j +4 − q j +3 ! which is the left side of (B). This shows that (B) implies (A).Next we note that ∞ X j =0 q j − q j + A = X j,m ≥ q j +5 jm + Am = ∞ X m =0 q Am − q m +5 . Thus the left side of (B) equals X m ≥ q m − q .q m − q.q m + 3 q .q m − q m +5 = X m ≥ q m (1 − q m +1 ) − q m +5 which shows that (B) implies (C) and the proof of Theorem 4.5 is complete. (cid:3) The Decomposition modulo 7
In this section, we will study Theorem 1.4 by again connecting it to the proof of certain equalities weobtain as in the last section. We start with the following conjecture.
Conjecture 5.1. q ; q ) ∞ X n ≥ ( − n q n ( n +1) / (1 − q n ) (1 + q n )1 − q n = − qA + 3 q B − q C + q D − q E where A = ( q ; q ) ∞ ( q , q ; q ) ∞ , B = ( q , q , q ; q ) ∞ ( q , q , q , q ; q ) ∞ , C = ( q ; q ) ∞ ( q , q ; q ) ∞ , D = ( q ; q ) ∞ ( q , q ; q ) ∞ ,E = ( q , q , q ; q ) ∞ ( q , q , q , q ; q ) ∞ . NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 15
We set L ( q ) = 1( q, q ; q ) ∞ , N ( q ) = 1( q , q ; q ) ∞ , Q ( q ) = 1( q , q ; q ) ∞ so that Conjecture 5.1 becomes1( q ; q ) ∞ X n ≥ ( − n q n ( n +1) / (1 − q n ) (1 + q n )1 − q n = ( q ; q ) ∞ " − qL ( q ) + 3 q L ( q ) N ( q ) Q ( q ) − q N ( q ) + q Q ( q ) − q N ( q ) Q ( q ) L ( q ) Remark 4.
By Corollary 3.4, Conjecture 5.1 would imply that X n ≥ (cid:20) M ω (1 , , n ) + 2 M ω (2 , , n ) − +3 M ω (3 , , n ) − M ω (4 , , n ) − M ω (5 , , n ) − M ω (6 , , n ) (cid:21) q n ≡ q ; q ) ∞ X n ≥ ( − n q ( n +12 ) (1 − q n ) (1 + q n )(1 − q n ) (mod 7)= ( q ; q ) ∞ " − qL ( q ) + 3 q L ( q ) Q ( q ) N ( q ) − q N ( q ) + q Q ( q ) − q N ( q ) Q ( q ) L ( q ) (mod 7) which would mean X n ≥ (cid:20) M ω (1 , , n + 5) + 2 M ω (2 , , n + 5) + 3 M ω (3 , , n + 5) − M ω (4 , , n + 5) − M ω (5 , , n + 5) − M ω (6 , , n + 5) (cid:21) ≡ which is Theorem 1.4. We will now show that Conjecture 5.1 follows from seven equalities we obtain by comparing coefficientsof powers of q by properly expanding both sides of the conjecture. As in the modulo 5 case, we also giveproofs of six of these seven equalities, and give various forms we have obtained for the seventh. Theorem 5.2.
Conjecture 5.1 follows from the following seven equalities. ( i ) − X m ( − m q m m − q m +1 + X m ( − m q m − m − q m − = ( q ; q ) ∞ − L ( q ) N ( q ) + q Q ( q ) N ( q ) L ( q ) ! (24)( ii ) 4 X m ( − m q m m − q m +1 + 4 X m ( − m q m m +22 − q m +3 = ( q ; q ) ∞ L ( q ) N ( q ) Q ( q ) + 3 L ( q ) Q ( q ) N ( q ) + q Q ( q ) L ( q ) ! (25)( iii ) X m ( − m q m m − q m +2 − X m ( − m q m − m − q m − − X m ( − m q m m − q m +1 = − q ; q ) ∞ L ( q ) (26) ( iv ) 5 X m ( − m q m m − − q m − − X m ( − m q m − m − − q m − = ( q ; q ) ∞ N ( q ) Q ( q ) − L ( q ) Q ( q ) N ( q ) − q Q ( q ) N ( q ) L ( q ) ! (27)( v ) − X m ( − m q m m − q m +2 − X m ( − m q m m − − q m − − X m ( − m q m m +22 − q m +3 = ( q ; q ) ∞ N ( q )(28)( vi ) − X m ( − m q m m − q m +3 + 4 X m ( − m q m m − − q m − − X m ( − m q m − m − − q m − = − q ; q ) ∞ Q ( q )(29)( vii ) ′ X m ( − m q m m − − q m − X m ( − m q m − m − − q m − − ′ X m ( − m q m m − − q m + 4 X m ( − m q m − m − − q m − + 5 ′ X m ( − m q m m − − q m + 5 X m ( − m q m m − q m +2 = 3( q ; q ) ∞ Q ( q ) N ( q ) + N ( q ) L ( q ) ! (30)First we prove the following lemma. Lemma 5.3. ( q ; q ) ∞ = ( q ; q ) ∞ " L ( q ) N ( q ) − q N ( q ) Q ( q ) − q + q Q ( q ) L ( q ) Proof.
By Lemma 6 in [1],( q ; q ) ∞ = − q P (0) " − q − P (2) P (1) + q − P (4) P (2) − q P (6) P (3) with P (0) = ( q ; q ) ∞ , P (1) = P (6) = ( q , q ; q ) ∞ , P (2) = ( q , q ; q ) ∞ , P (3) = P (4) = ( q , q ; q ) ∞ . Thus,( q ; q ) ∞ = P (0) " P (2) P (1) − q P (4) P (2) − q + q P (6) P (3) = ( q ; q ) ∞ " L ( q ) N ( q ) − q N ( q ) Q ( q ) − q + q Q ( q ) L ( q ) as desired. (cid:3) NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 17
Proof of Theorem 5.2.
Using the expansion of ( q ; q ) ∞ from Lemma 5.3, Conjecture 5.1 becomes X n ≥ ( − n q n ( n +1) / (1 − q n ) (1 + q n )1 − q n = ( q ; q ) ∞ " − qL ( q ) + 3 q L ( q ) Q ( q ) N ( q ) − q N ( q ) + q Q ( q ) − q N ( q ) Q ( q ) L ( q ) × " L ( q ) N ( q ) − q N ( q ) Q ( q ) − q + q Q ( q ) L ( q ) = ( q ; q ) ∞ " − q L ( q ) N ( q ) + q L ( q ) N ( q ) Q ( q ) + 3 L ( q ) Q ( q ) N ( q ) ! − q L ( q )+ 2 q N ( q ) Q ( q ) − L ( q ) Q ( q ) N ( q ) ! + q N ( q ) − q Q ( q ) + 3 q Q ( q ) N ( q ) + N ( q ) L ( q ) ! + q Q ( q ) N ( q ) L ( q ) + q Q ( q ) L ( q ) − q Q ( q ) N ( q ) L ( q ) (31)The left side in (31) is X n ≥ ( − n q n ( n +1) / − q n − q n + 5 q n − q n + 4 q n − q n ! = ′ X n ( − n q n ( n +1) / − q n − ′ X n ( − n q n ( n +3) / − q n + 5 ′ X n ( − n q n ( n +5) / − q n (32)since − X n ≥ ( − n q n ( n +1) / − q n .q n = X n ≤− ( − n q n ( n +1) / − q n , X n ≥ ( − n q n ( n +1) / − q n .q n = − X n ≤− ( − n q n ( n +3) / − q n and − X n ≥ ( − n q n ( n +1) / − q n .q n = X n ≤− ( − n q n ( n +5) / − q n . Now, ′ X n ( − n q n ( n +1) / − q n = X t = − ( − t q t ( t +1) / X m ( − m q ((7 m + t )(7 m + t +1) − t ( t +1)) / − q m +7 t , (with the m = t = 0 term omitted)= ′ X m ( − m q (49 m +7 m ) / − q m − X m ( − m q (49 m +21 m +2) / − q m +7 − X m ( − m q (49 m − m ) / − q m − + X m ( − m q (49 m +35 m +6) / − q m +14 + X m ( − m q (49 m − m +2) / − q m − − X m ( − m q (49 m +49 m +12) / − q m +21 − X m ( − m q (49 m − m +6) / − q m − (33)Similarly, we get, ′ X n ( − n q n ( n +3) / − q n = ′ X m ( − m q (49 m +21 m ) / − q m − X m ( − m q (49 m +35 m +4) / − q m +7 − X m ( − m q (49 m +7 m − / − q m − + X m ( − m q (49 m +49 m +10) / − q m +14 + X m ( − m q (49 m − m − / − q m − − X m ( − m q (49 m +63 m +18) / − q m +21 − X m ( − m q (49 m − m ) / − q m − (34)and ′ X n ( − n q n ( n +5) / − q n = ′ X m ( − m q (49 m +35 m ) / − q m − X m ( − m q (49 m +49 m +6) / − q m +7 − X m ( − m q (49 m +21 m − / − q m − + X m ( − m q (49 m +63 m +14) / − q m +14 + X m ( − m q (49 m +7 m − / − q m − − X m ( − m q (49 m +77 m +24) / − q m +21 − X m ( − m q (49 m − m − / − q m − (35)Thus, by (32) through (35) the left side in (31) equals " ′ X m ( − m q m m − q m − X m ( − m q m − m − q m − − ′ X m ( − m q m m − q m + 4 X m ( − m q m − m − q m − + 5 ′ X m ( − m q m m − q m + 5 X m ( − m q m m +142 − q m +14 NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 19 + q " − X m ( − m q m m − q m +7 + X m ( − m q m − m − q m − + 4 q " X m ( − m q m m − q m +7 + X m ( − m q m m +142 − q m +21 + q " X m ( − m q m m − q m +14 − X m ( − m q m − m − q m − − X m ( − m q m m − q m +7 + 5 q (cid:20) X m ( − m q m m − − q m − − X m ( − m q m − m − − q m − + q " − X m ( − m q m m − q m +14 − X m ( − m q m m − − q m − − X m ( − m q m m +142 − q m +21 + q " − X m ( − m q m m − q m +21 + 4 X m ( − m q m m − − q m − − X m ( − m q m − m − − q m − . (36)Comparing the coefficients of q, q , q , q , q , q , q in (36) and the right side of (31), and replacing q by q / , we get the seven equalities in Theorem 5.2. (cid:3) Theorem 5.4.
Equations (26), (28) and (29) in Theorem 5.2 are true.
Lemma 5.5.
For i = 1 , , , X n ( − n q n ( n +1) / − q n + i ! (1 − q i ) = ( q ; q ) ∞ ( q − i , q i ; q ) ∞ Proof.
Recall by Theorem 3.2 in [4] that6 φ " z, q √ z, − q √ z, a , a , a √ z, −√ z, zqa , zqa , zqa ; q, zqa a a = Y " zq, zqa a , zqa a , zqa a zqa , zqa , zqa , zqa a a Setting z = 1 , a → ∞ , a = q i , a = q − i , q = q , the right side equals the right side of Lemma 5.5. Underthe same substitutions, the left side becomes1 + ∞ X n =1 ( − n q n ( n +1) / (1 + q n ) (1 − q i )(1 − q − i )(1 − q n + i )(1 − q n − i ) ! = 1 + ∞ X n =1 ( − n q n ( n +1) / (1 − q i )(1 − q − i ) − q i )(1 − q n − i ) + 1(1 − q − i )(1 − q n + i ) ! = 1 + X n ≥ ( − n q n ( n +1) / (1 − q i )1 − q n + i + X n ≤− ( − n q n ( n +1) / (1 − q i )1 − q n + i = (1 − q i ) X n ( − n q n ( n +1) / − q n + i , proving the lemma. (cid:3) Setting i = 1 , , X n ( − n q n ( n +1) / − q n +1 = ( q ; q ) ∞ L ( q ) (37) X n ( − n q n ( n +1) / − q n +2 = ( q ; q ) ∞ N ( q ) (38) X n ( − n q n ( n +1) / − q n +3 = ( q ; q ) ∞ Q ( q ) (39) Proof of Theorem 5.4.
We have that m ( q , q , q − ) = − qj (1 /q ; q ) X m ( − m q (7 m +5 m ) / − q m +2 and m ( q , q , q − ) = − q j (1 /q ; q ) X m ( − m q (7 m − m ) / − q m − . It thereby follows that X m ( − m q (7 m +5 m ) / − q m +2 − X m ( − m q (7 m − m ) / − q m − = − qj (1 /q ; q ) m ( q , q , q − )+ q j (1 /q ; q ) m ( q , q , q − ) = ( q, q , q ; q ) ∞ " m ( q , q , q − ) − m ( q , q , q − ) Using Lemma 11.3.4 of [5], we can check as before that m ( q , q , q − ) − m ( q , q , q − ) = 1 q ( q ; q ) ∞ j ( q ; q ) j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j ( q ; q ) = ( q ; q ) ∞ ( q, q , q, q ; q ) ∞ so that X m ( − m q (7 m +5 m ) / − q m +2 − X m ( − m q (7 m − m ) / − q m − = ( q ; q ) ∞ ( q, q ; q ) ∞ = ( q ; q ) ∞ L ( q )whence by (37), equation (26) follows.The proof of equation (28) is similar. Indeed, m ( q, q , q − ) = − q j (1 /q ; q ) X m ( − m q (7 m +3 m ) / − q m − and m ( q, q , q ) = − q j ( q ; q ) X m ( − m q (7 m +11 m ) / − q m +3 whence similarly as before, we can check by Lemma11.3.4 of [5] that X m ( − m q (7 m +3 m − / − q m − + X m ( − m q (7 m +11 m +2) / − q m +3 = − ( q ; q ) ∞ N ( q ) . By (38), equation (28) now follows.Equation (29) follows similarly from (39). We omit the details. (cid:3)
Theorem 5.6.
Equations (24), (25) and (27) in the statement of Theorem 5.2 are true.
First we prove the following lemma.
Lemma 5.7. ( I ) − L ( q ) N ( q ) + q Q ( q ) N ( q ) L ( q ) = − L ( q ) N ( q ) Q ( q )( II ) L ( q ) N ( q ) Q ( q ) + q Q ( q ) L ( q ) = L ( q ) Q ( q ) N ( q )( III ) N ( q ) Q ( q ) − L ( q ) Q ( q ) N ( q ) = − q Q ( q ) N ( q ) L ( q ) NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 21
Proof. (I) is equivalent to − ( q , q ; q ) ∞ ( q, q , q, q ; q ) ∞ + q ( q, q ; q ) ∞ ( q , q , q , q ; q ) ∞ + ( q , q , q , q ; q ) ∞ ( q, q , q , q , q , q ; q ) ∞ = 0 ⇐⇒ q ; q ) ∞ " − ( q , q , q , q , q , q , q , q , q , q , q , q ; q ) ∞ + q ( q, q , q, q , q, q , q , q , q , q , q , q ; q ) ∞ + ( q , q , q , q , q , q , q , q , q, q , q , q ; q ) ∞ = 0 (40)which is proved by − ( q , q , q , q , q , q , q , q ; q ) ∞ + q ( q, q , q, q , q, q , q , q ; q ) ∞ + ( q , q , q , q , q , q , q, q ; q ) ∞ = 0 . Similarly (II) is equivalent to1( q ; q ) ∞ " ( q , q , q , q , q , q , q, q , q , q , q , q ; q ) ∞ + q ( q, q , q, q , q, q , q , q , q , q , q , q ; q ) ∞ − ( q , q , q , q , q , q , q , q , q , q , q , q ; q ) ∞ = 0 (41)while (III) is equivalent to1( q ; q ) ∞ " ( q , q , q , q , q , q , q, q , q, q , q , q ; q ) ∞ − ( q , q , q , q , q , q , q, q , q , q , q , q ; q ) ∞ + q ( q, q , q, q , q, q , q, q , q , q , q , q ; q ) ∞ = 0 (42)Dividing (40), (41) and (42) by ( q , q , q , q ; q ) ∞ , ( q , q , q , q ; q ) ∞ and ( q, q , q , q ; q ) ∞ respectively,we see that all three equations in Lemma 5.7 are proved by( q , q , q , q , q , q , q, q ; q ) ∞ − ( q , q , q , q , q , q , q , q ; q ) ∞ + q ( q, q , q, q , q, q , q , q ; q ) ∞ = 0(43)Lemma 7.4.4 in [6] is( aqb , ba , aqef , efa , aqdf , dfa , aqbde , bdea ; q ) ∞ − ( aqf , fa , aqbe , bea , aqbd , bda , aqdef , defa ; q ) ∞ + ba ( d, qd , e, qe , bqf , fb , a qbdef , bdefa ; q ) ∞ = 0Setting ba = q, fa = q , e = q , q = q so that bf = 1 q , bdefa = q , (43) and hence Lemma 5.7 areproved. (cid:3) Proof of Theorem 5.6.
We have m ( q , q , q − ) = − q j (1 /q ; q ) X m ( − m q (7 m +3 m ) / − q m +1 and m ( q , q , q − ) = − q j (1 /q ; q ) X m ( − m q (7 m − m ) / − q m − so that the left side of (24) equals q j (1 /q ; q ) m ( q , q , q − ) − q j (1 /q ; q ) m ( q , q , q − ) = ( q , q , q ; q ) ∞ " m ( q , q , q − ) − m ( q , q , q − ) . By Lemma 11.3.4 of [5], we check as before that m ( q , q , q − ) − m ( q , q , q − ) = 1 q ( q ; q ) ∞ j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j (1 /q ; q ) j ( q ; q ) j (1 /q ; q ) = − ( q , q , q , q , q ; q ) ∞ ( q, q , q , q , q , q , q , q ; q ) ∞ so that the left side of (24) equals − ( q ; q ) ∞ L ( q ) N ( q ) Q ( q ) whence (24) follows from (I) of Lemma 5.7.For (25), we start with m ( q , q , q − ) = − qj (1 /q ; q ) X m ( − m q (7 m +5 m ) / − q m +1 and m ( q , q , q ) = − qj ( q ; q ) X m ( − m q (7 m +9 m ) / − q m +3 so that the left side of (25) equals4( q, q , q ; q ) ∞ " m ( q , q , q − ) − m ( q , q , q ) . By Lemma 11.3.4 of [5], m ( q , q , q − ) − m ( q , q , q ) = ( q , q , q , q , q ; q ) ∞ ( q, q , q, q , q, q , q , q ; q ) ∞ so that the left sideof (25) equals 4( q ; q ) ∞ L ( q ) Q ( q ) N ( q ) whence (25) follows from (II) of Lemma 5.7.Equation (27) follows in exactly the same way from (III) of Lemma 5.7. We again omit the details. (cid:3) The only equation left to be proven to finish the proof of Conjecture 5.1 and hence of Theorem 1.4is equation (30) in the statement of Theorem 5.2. We haven’t been able to show this, but the followingtheorem gives different expressions for the left side of this equation.
Theorem 5.8.
If (D) is the statement ′ X m ( − m q m m − − q m − X m ( − m q m − m − − q m − − ′ X m ( − m q m m − − q m + 4 X m ( − m q m − m − − q m − + 5 ′ X m ( − m q m m − − q m + 5 X m ( − m q m m − q m +2 = 3( q ; q ) ∞ Q ( q ) N ( q ) + N ( q ) L ( q ) ! , NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 23 (E) is the statement q ∞ X j =0 q j +5 − q j +4 − q j +4 − q j +3 + 2 q j +2 − q j +1 − q j +7 − q j +6 + 3 q j +6 − q j +5 − q j +3 − q j +2 ! = 3( q ; q ) ∞ Q ( q ) N ( q ) + N ( q ) L ( q ) ! and (F) is the statement X m q m (1 − q m +1 ) (2 q m +2 + 3 q m +1 + 2)1 − q m +7 = 3( q ; q ) ∞ Q ( q ) N ( q ) + N ( q ) L ( q ) ! , then (D), (E) and (F) are equivalent.Proof. We reduce the left side of (D) to the left side of (E) and the left side of (E) to the left side of (F).Let U = ′ X m ( − m q m m − − q m , U = X m ( − m q m − m − − q m − , U = ′ X m ( − m q m m − − q m ,U = X m ( − m q m − m − − q m − , U = ′ X m ( − m q m m − − q m , U = X m ( − m q m m − q m +2 , and let φ = U − U , ψ = U − U , ρ = U + U . The left side of (D) is thus U − U − U + 4 U + 5 U + 5 U = φ − ψ + 5 ρ. Since m ( q , q , /q ) = − q j (1 /q ; q ) X m ( − m q (7 m − m ) / − q m − , we get U = 1 q ( q , q , q ; q ) ∞ m ( q , q , /q ) . Similarly, U = 1 q ( q , q , q ; q ) ∞ m ( q , q , /q ) and U = − q ( q, q , q ; q ) ∞ m ( q, q , q ) . If m ′ ( x, q, z ) = − zj ( z ; q ) ′ X r ( − r q r ( r +1) / − xzq r , then m ′ ( q , q , /q ) = − q j (1 /q ; q ) ′ X m ( − m q (7 m + m ) / − q m so that U = − q j (1 /q ; q ) m ′ ( q , q , /q ) = 1 q ( q , q , q ; q ) ∞ lim z → q − m ( q , q , z ) + zj ( z ; q ) . − q z ! . Similarly, U = 1 q ( q , q , q ; q ) ∞ lim z → q − m ( q , q , z ) + zj ( z ; q ) . − q z ! and U = 1 q ( q, q , q ; q ) ∞ lim z → q − m ( q, q , z ) + zj ( z ; q ) . − qz ! . Now, φ = U − U = − q ( q , q , q ; q ) ∞ lim z → q − m ( q , q , q − ) − m ( q , q , z ) − zj ( z ; q ) . − q z ! . By Lemma 11.3.4 in [5], we verify that m ( q , q , q − ) − m ( q , q , z ) = z ( q ; q ) ∞ j (1 /q z ; q ) j ( z/q ; q ) j ( z ; q ) j (1 /q ; q ) j ( q z ; q ) j (1 /q ; q )= z ( q , q z , q z, zq , q z ; q ) ∞ ( z, q z , q , q , q z, q z , q , q ; q ) ∞ so that φ = − q ( q , q , q ; q ) ∞ lim z → q − " z ( q , q z , q z, zq , q z ; q ) ∞ ( z, q z , q , q , q z, q z , q , q ; q ) ∞ − zj ( z ; q ) . − q z = − q ( q , q , q ; q ) ∞ lim z → q − " z ( z, q z , q , q , q z, q z , q , q , q ; q ) ∞ × ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − q z ) = − q ( q , q , q ; q ) ∞ q ( 1 q , q , q , q , q , q , q , q , q ; q ) ∞ × lim z → q − ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − qz )= q ( q , q , q , q, q , q ; q ) ∞ lim z → q − ( q , q , q z , q z, zq , q z ; q ) ∞ − ( 1 q , q , q z, q z , q , q ; q ) ∞ (1 − qz )Now, if F ( z ) = ( q , q , q z , q z, zq , q z ; q ) ∞ , then we can check that by differentiation, F ′ ( 1 q ) = ( q ; q ) ∞ ( q, q , q , q ; q ) ∞ q ∞ X j =0 q j − q j − − q j +9 − q j +8 − q j − − q j − + q j +12 − q j +11 ! . Similarly, if F ( z ) = ( 1 q , q , q z, q z , q , q ; q ) ∞ , then F ′ ( 1 q ) = 0 . This implies that φ = 1 q ∞ X j =0 − q j − q j − + q j +9 − q j +8 + q j − − q j − − q j +12 − q j +11 ! by L’Hospitalsrule. NDREWS-BECK TYPE CONGRUENCES RELATED TO THE CRANK OF A PARTITION 25
In an exactly similar fashion, we get ψ = 1 q ∞ X j =0 − q j − − q j − + q j +11 − q j +10 + q j − − q j − − q j +13 − q j +12 ! and ρ = − q ∞ X j =0 q j +3 − q j +2 − q j +6 − q j +5 − q j +2 − q j +1 + q j +7 − q j +6 ! . Collecting like powers of q modulo 7, this gives after simplification that the left side of (D) equalsthe left side of (E).Further, noting that X j ≥ q j − q j + A = X j,m ≥ q j +7 jm +7 Am = X m ≥ q Am − q m +7 , the left side of (E) reduces to3 q X m ≥ q m +5 − q m +4 + 2 q m +2 − q m +7 + 3 q m +6 − q m +3 − q m +7 = 3 X m ≥ q m (1 − q m +1 ) (2 q m +2 + 3 q m +1 + 2)1 − q m +7 on factorization, which is the left side of (F). The proof of theorem 5.8 is now complete. (cid:3) Conclusion and Further Work
In this paper, we have thus given a new approach to proving George-Beck’s conjectures on the crankof a partition modulo 5 or 7. The Apple-Lerch sums we come across our decompositions are intricatelyrelated to tenth-order mock theta functions as discussed in [5] Chapter 11. This points to the possibilitythat George Beck’s conjectures, which appear striking and unique at first glance, may actually be comingfrom identities in tenth order mock-theta functions. We note however, that by our approach, we havenot managed to prove completely the crank conjectures, missing an identity each in the cases modulo 5and 7. These unproven identities seem to be a little different that the ones we prove in this paper, andwe cannot rule out the possibility that they are actually special cases of a wider range of results, againintricately related to mock-theta functions.
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