Journal of Number Theory

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1 Journal of Number Theory ) Contents lists available at ScienceDirect Journal of Number Theory New analogues of Ramanujan s partition identities Heng Huat Chan, Pee Choon Toh National University of Singapore, Department of Mathematics, S17, 10 Lower Kent Ridge Road, Singapore article info abstract Article history: Received 21 June 2009 Communicated by David Goss We establish several new analogues of Ramanujan s exact partition identities using the theory of modular functions Elsevier Inc. All rights reserved. MSC: 11P83 11F03 Keywords: Modular forms Partitions Congruences Dedekind eta function 1. Introduction Let pn) denote the number of unrestricted partitions of the non-negative integer n, then p 5 j n + δ 5, j ) 0 mod 5 j ), 1.1) p 7 j n + δ 7, j ) 0 mod 7 j/2+1 ), 1.2) p 11 j n + δ 11, j ) 0 mod 11 j ), 1.3) where δ l, j = 1/2 mod l j ). These are commonly known as the Ramanujan congruences. The congruence 1.3) was proved by A.O.L. Atkin [1 while the proofs of 1.1) and 1.2) have traditionally been attributed to G.N. Watson [18, although Ramanujan had already outlined the proofs in an un- * Corresponding author. addresses: matchh@nus.edu.sg H.H. Chan), mattpc@nus.edu.sg, peechoon@gmail.com P.C. Toh) X/$ see front matter 2010 Elsevier Inc. All rights reserved. doi: /j.jnt

2 H.H. Chan, P.C. Toh / Journal of Number Theory ) published manuscript [. More details about this and other results concerning pn) can be found in [13, Chapter 5. In [15, Ramanujan proved 1.1) and 1.2) for j = 1 and 2 using the following two identities and p5n + )q n 1 q 5n ) 5 = 5 1 q n ), 6 1.) p7n + 5)q n 1 q 7n ) 3 = 7 1 q n ) + 9q 1 q 7n ) 7 1 q n ) ) Identity 1.) was described by G.H. Hardy as Ramanujan s most beautiful identity [16, p. xxxv and Ramanujan [15 provided an elegant proof using the following identity, q 1/5 where Rq) is the Rogers Ramanujan continued fraction given by 1 q n/5 ) 1 q 5n ) = 1 1 Rq), 1.6) Rq) Rq) = q1/5 q q Recently, H.-C. Chan [5, Theorem 2 proved an analogue of 1.6) involving Ramanujan s cubic continued fraction and obtained an analogue of 1.). He showed that if then an)q n 1 = 1 q n )1 q 2n ), a3n + 2)q n 1 q 3n ) 3 1 q 6n ) 3 = 3 1 q n ) 1 q 2n ). 1.7) H.-C. Chan and S. Cooper [7 subsequently obtained another analogue of 1.), namely, c2n + 1)q n 1 q 2n ) 1 q 6n ) = 2 1 q n ) 6 1 q 3n ), 6 1.8) where the generating function of cn) is given by cn)q n 1 = 1 q n ) 2 1 q 3n ). 2 Again 1.8) was derived via an identity analogous to 1.6). See also [10.) The existence of identities 1.), 1.5), 1.7) and 1.8) can all be explained by the theory of modular functions. Furthermore, they can be established without first proving identities similar to 1.6). The first author and R.P. Lewis [9 used Hecke operators to provide a uniform proof of 1.), 1.5), as well

3 1900 H.H. Chan, P.C. Toh / Journal of Number Theory ) as an identity due to H.S. Zuckerman [19 involving the coefficients p13n + 6). In the next section, we extend their technique to prove 1.7), 1.8) and several new identities. To state these identities, let t be a positive integer and define the infinite product Et) = ) 1 q tn. 1.9) A generalized partition π is an expression of the form [ π = t t r t, 1.10) where r t are integers and only finitely many r t are nonzero. For a fixed partition π,wedefinethe associated infinite product A π j) = t 1 q jtn) rt = t E jt) r t, 1.11) and the coefficients p π n) by p π n)q n = 1 A π 1). 1.12) In this notation, an) = p [ n) and cn) = p [ n). We also adopt the convention that p π n) = 0 whenever n is negative. Theorem 1. The following identities hold. p [ n 1)qn = q E2)2 E1) 2 E1) 3 E7) 3, 1.13) p [1 5 2n 3)qn = q 2 E2) E10) E1) 8 E5) + E2)8 E10) 8 8 8q3, E1) ) E5) 12 p [ n 3)qn = 2q 2 E2)2 E22) 2 E1) E11) + 2q3 E2) E22) E1) 6 E11) 6, 1.15) p [ n 2)qn = q E3)E15) E1) 2 E5) 2 + 3q2 E3)3 E15) 3 E1) E5), 1.16) p [ n 3)qn = q 2 E2)E6) E1) 2 E23) + E2)2 E6) 2 2 q3 E1) 3 E23), ) p [ n 3)qn = 3q E5)E10) E1) 2 E2) q2 E5)3 E10) 3 E1) E2) + 125q3 E5)5 E10) 5 E1) 6 E2) 6, 1.18) p [ n )qn = 2q 2 E3)E33) E1) 2 E11) + E3)2 E33) 2 2 3q3 E1) 3 E11) + E3)3 E33) 3 3 3q E1) E11), 1.19)

4 H.H. Chan, P.C. Toh / Journal of Number Theory ) p [ n 8)qn = 16q 2 E7)E21) E1) 2 E3) + E7)2 E21) q3 E1) 3 E3) + E7)3 E21) q E1) E3) q 5 E7) E21) E1) 5 E3) + E7)5 E21) q6 E1) 6 E3) q 7 E7)6 E21) 6 E1) 7 E3) q8 E7)7 E21) 7 E1) 8 E3) ) Besides proving 1.7) Chan [6 proved a higher power analogue of the Ramanujan congruence for an). The corresponding analogue for cn) was subsequently established by Chan and Cooper [7. The result for an) is Corollary 2. See H.-C. Chan [6.) a 3 j n + c j ) 0 mod 3 2 j/2+1 ), where c j = 1/8 mod 3 j ). In Section 3, we shall prove a different higher power congruence for an), involving the prime 5, which leads to the following. Corollary 3. where d j = 1/8 mod 5 j ). a 5 j n + d j ) 0 mod 5 j 2 ), Both Corollaries 2 and 3 were first discovered implicitly by P.C. Eggan [ Proof of Theorem 1 and consequences In this section, we shall illustrate how the identities in Theorem 1 may be constructed via the theory of modular functions. Most of the results that we need can be found in [2. Let H ={τ C: Imτ > 0} be the complex upper half plane and Γ be any subgroup of SL 2 Z). Elements of Γ act on H as linear fractional transformations and Γ has a connected fundamental domain. The points of compactification of the fundamental domain on the boundary of H are called cusps. Fix an integer k for a function f τ ) on H. k is usually called the weight of f τ ), although in our treatment we are mainly interested in the case k = 0. For any matrix L = define the slash operator by aτ + b f L = ad bc) k/2 cτ + d) k f cτ + d ab cd ) GL + 2 R), we ). 2.1) We say that f τ ) is a modular function of weight 0) on Γ,if f τ ) is analytic in the interior of H, meromorphic at each cusp of Γ and satisfies f V = f, for every V Γ. 1 We thank F.G. Garvan for this reference.

5 1902 H.H. Chan, P.C. Toh / Journal of Number Theory ) We define { ) } a b Γ 0 m) = SL c d 2 Z): c 0 mod m), and A n = ) ) ) n ex y, S =, W e =, mz ew where W e is known as an Atkin Lehner involution and satisfies det W e = e, e m. The action of W e is independent of x, y, z and w in the definition and W e 2 = I modulo Γ 0m). W e is in the normalizer of Γ 0 m) and we denote the group generated by Γ 0 m) and W e by Γ 0 m) + W e. Finally, we define the Hecke operator Ul for some prime l by l 1 f U l = j=0 f A 1 l S j. 2.2) Lemma. Let l be prime, l m. If f τ ) is a modular function on Γ 0 m) then f Ul is a modular function on Γ 0 m). Furthermore, if l 2 mthen f Ul is a modular function on Γ 0m/l). Proof. [2, Lemma 7. Lemma 5. Let e m. If f τ ) is a modular function on Γ 0 m) then f W e is a modular function on Γ 0 m). Proof. [2, Lemma 10. Lemma 6. Let l be prime, l m, e m andl, e) = 1.If fτ ) is a modular function on Γ 0 m) then f U l ) W e = f W e ) U l. Proof. Following the proof of [2, Lemma 11, since l, e) = 1, we can choose W e ) mod l) and M j = ) l j. 0 1 By the definition 2.2), we have f U l = f M 1 j, thus it suffices to prove that each M 1 j commutes with W e. A straightforward computation reveals that M 1 W j e M j is a matrix with integral entries and is equal to some W e. Since these are involutions, f M 1 W j e M j W e 1 = f,hence f M 1 W j e = f W e M 1. j Lemma 7. Let l be prime, l 2 m, e m andl, e) = 1. If fτ ) is a modular function on Γ 0 m) + W e then f U l is a modular function on Γ 0m/l) + W e.

6 H.H. Chan, P.C. Toh / Journal of Number Theory ) Proof. By the hypothesis, f is also a modular function on Γ 0 m), hence f τ ) Ul is a modular function on Γ 0 m/l). Thus it suffices to show that f τ ) Ul is invariant under W e which follows directly from Lemma 6. From this point onwards, we only consider modular functions that vanish at all cusps. If we let q = e 2π iτ, the Fourier expansion of the modular function f τ ) is given by f τ ) = For a prime l, the Hecke operator U l is defined by f U l = an)q n. aln)q n. 2.3) Lemma 8. Let l be prime, l m. If f τ ) is a modular function of weight 0) on Γ 0 m) that vanishes at all cusps, then l f U l = f U l. Proof. [2, Lemma 1. Lemma 9. For a product of two Fourier expansions, we have an)q ln ) ) bn)q n Ul = an)q n bln)q ). n This result was known to Atkin and O Brien [3, Eq. 28), [12, p. 126 and plays a crucial role in our proof of the identities in Theorem 1. We have seen that the U l operator maps a modular function to another modular function on some Γ.LetH be the union of H and the set of cusps of Γ,ifthe genus of Γ \H is zero, then we know the function field of Γ \H over C is generated by a certain modular function h, known also as a hauptmodul. For brevity we simply say Γ has genus zero. Proof of Theorem 1. Fix a prime l and a partition π and define A j) = A π j) according to 1.11). l and π are chosen such that φ = φτ ) = q α Al2 ) A1) with α = l2 1 2 tr t, 2.) t is a modular function on Γ 0 l 2 e) + W e, where l, e) = 1. In addition, Γ 0 le) + W e should have genus zero, with hauptmodul defined by h = hτ ) = q β Al)γ l 1)γ A1) γ where β = 2 tr t. 2.5) In the above, α, β and γ are all positive integers. From Lemma 9, φ U l = q α A l 2) ) p π n)q n Ul = Al) p π ln α)q n. 2.6) t

7 190 H.H. Chan, P.C. Toh / Journal of Number Theory ) On the other hand, by Lemma 7, φ U l is a modular function on Γ 0 le) + W e which has genus zero. Thus, it must be generated by the hauptmodul h. For each of the identities in Theorem 1, as well as 1.7) and 1.8), a computation reveals that φ U l are all polynomials in h with integral coefficients, i.e. p π ln α)q n = 1 Al) N a j h j, a j Z. 2.7) We shall illustrate by proving identity 1.20) in detail. The Dedekind eta function is defined as j=1 ητ ) = ) q 1/2 1 q n = q 1/2 E1), where q = e 2π iτ. 2.8) For a generalized partition π, we define the eta-quotient associated with π by η π τ ) = t ηtτ ) r t. 2.9) Many examples of modular functions of weight 0 on Γ 0 m) + W e can be given in terms of etaquotients. In particular, Γ 0 21) + W 3 has genus zero, and is generated by the hauptmodul h = η7τ )η21τ ) ητ )η3τ ) = q A7) A1), where A j) = E j)e3 j) and π =[ Now using the results in [8, we can check that φ = η9τ )η17τ ) ητ )η3τ ) is a modular function on Γ 0 17) + W 3.Thus = q 8 A72 ) A1) = A 7 2) φ U 7 = A7) p [11,3 1 7n 8)qn p [1 1,3 1 n 8)qn is a modular function on Γ 0 21) + W 3. A direct calculation shows that it is equal to the following polynomial ph) = 16h h h h h h h 8, 2.10) which proves identity 1.20). The proofs of the other identities are similar, and can be deduced from Table 1. The list of identities given in Theorem 1 is by no means exhaustive. In theory, there should be many modular functions where one can apply the U l operator to lower the level by l. Iftheresulting congruence subgroup has genus zero, an identity in terms of the hauptmodul exists and can be easily calculated. But having genus zero is not a necessary condition. M.L. Lang has classified all congruence subgroups of level less than 300 that have genus zero and a list of these groups is given as an appendix in [8. There are a total of 15 groups that explicitly meet the conditions set out in the proof of Theorem 1. The remaining 5 groups share the property that e in Γ 0 m) + W e is composite. Their associated identities can be deduced from Table 2.

8 H.H. Chan, P.C. Toh / Journal of Number Theory ) Table 1 Identity Γ 0 l 2 e) + W e l φτ ) = η π π h = η π Ph) 1.7) Γ 0 18) + W 2 3 [ [ [ h 1.8) Γ 0 12) + W 3 2 [ [ [ h 1.13) Γ 0 28) + W 7 2 [ [ [ h 1.1) Γ 0 20) + W 5 2 [ [1 5 [ h 2 + 8h ) Γ 0 ) + W 11 2 [ [ [ h 2 + 2h ) Γ 0 5) + W 5 3 [ [ [ h + 3h ) Γ 0 92) + W 23 2 [ [ [ h 2 + h ) Γ 0 50) + W 2 5 [ [ [ h + 25h h ) Γ 0 99) + W 11 3 [ [ [ h 2 + 3h 3 + 3h 1.20) Γ 0 17) + W 3 7 [ [ [ ph) defined in 2.10) Table 2 Γ 0 l 2 e) + W e l φτ ) = η π π h = η π Ph) Γ 0 36) + W 3 [ [ [ h Γ 0 36) + W 9 2 [ [ [ h 2 + 2h 3 Γ 0 60) + W 15 2 [ [ [ h 2 + h 3 Γ 0 72) + W 8 3 [ [ [ h 2 + 6h 3 + 3h Γ 0 100) + W 5 [ [ [ h h h + 250h h From the above identities, we see that only an), cn), p [1 5 n) and p [ n) satisfy p π ln α) 0 mod l). 2.11) J. Sinick [17 has obtained some necessary conditions for such congruences. We further note that identities 1.13), 1.16), 1.18), 1.19) and 1.20) can be written as the following congruences: E2)E1) p [ n 1)qn η 3 τ )η 3 7τ ) mod 2), 2.12) E3)E15) p [ n 2)qn η τ )η 5τ ) mod 3), 2.13) E5)E10) p [ n 3)qn 3η 8 τ )η 8 2τ ) mod 5), 2.1) E3)E33) p [ n )qn 2η τ )η 11τ ) mod 3), 2.15) E7)E21) p [ n 8)qn 2η 12 τ )η 12 3τ ) mod 7). 2.16)

9 1906 H.H. Chan, P.C. Toh / Journal of Number Theory ) Recall that in each of the above, the expression on the left-hand side is a modular function on Γ 0 le) + W e which is now congruent modulo l) to a modular form on Γ 0 e), where the weight of this modular form is divisible by l 1. To state our results involving higher powers of the prime l, we fix a partition π =[ t tr t,the associated infinite product A π j), and the modular functions φτ ) and hτ ). See 1.11), 2.) and 2.5) for the respective definitions.) For positive integers j, define where See 2.) for a definition of α.) ) L 2 j 1 = A π l) p π l 2 j 1 n δ 2 j 1 q n, ) L 2 j = A π 1) p π l 2 j n δ 2 j q n, 2.17) δ 2 j 1 = δ 2 j = 1 l2 j 1 l 2 α. Theorem 10. In each of the fifteen cases listed in Tables 1 and 2, L j is a polynomial in h, the hauptmodul, with integral coefficients. In terms of L j, we can reformulate the results of [6 and [7 as the next two theorems. Theorem 11. Let π =[ , l = 3 and α = 1,thenforeach j, L j c j 3 2 j 2 +1 η 8 τ )η 8 2τ ) mod 3 2 j 2 +2), where c j is some integer relatively prime to 3. Theorem 12. Let π =[ , l = 2 and α = 1,thenforeach j, L j d j 2 j 2 +1 η 6 τ )η 6 3τ ) mod 2 j 2 +2), where d j is some integer relatively prime to 2. We also have the following result. Theorem 13. Let π =[ , l = 5 and α = 1,thenforeach j, L j e j 5 j 2 η 8 τ )η 8 2τ ) mod 5 j 2 +1), where e j is some integer relatively prime to 5. Corollaries 2 and 3 in Section 1 are immediate consequences of Theorems 11 and 13. Remark 1. Theorems are motivated by an observation of the first author. He found that for prime numbers l, ifl n is the smallest positive integral solution to 2x 1 mod l n)

10 H.H. Chan, P.C. Toh / Journal of Number Theory ) and L n,l = { k=1 1 qlk ) k=0 pln k + l n )q k+1, if n is odd, k=1 1 qk ) k=0 pln k + l n )q k+1, if n is even, then L n,5 5 n a n mod 5 n+1), 2.18) L n,7 7 n/2+1 b n mod 7 n/2+2) 2.19) and L n,11 11 n c n E 8 mod 11 n+1 ), 2.20) where 5, a n ) = 7, b n ) = 11, c n ) = 1, = ) q 1 q k 2 k=1 and k 7 q k E 8 = q. k k=1 Note that 2.18) and 2.19) follow immediately from Watson s work see [12). The congruence 2.20), on the other hand, does not follow immediately from existing methods. Y.F. Yang, at the request of the first author, verified 2.20) for the first few values of n. It is very likely that one can obtain a rigorous proof of 2.20) using Atkin s method given in [1. It is our hope that by writing the congruences in the form as in 2.18) 2.20), new proofs which are independent of Atkin s identities and explicit use of modular equations such as 3.6) can be found. 3. Identities involving higher prime powers We shall illustrate how Theorem 10 may be proved by induction on j, with the identities of Theorem 1 forming the respective base cases, L 1. This method is due to Atkin [1, [12, Chapters 7 and 8. We first establish the following general lemmas. Lemma 1. For a positive integer j, we have L 2 j 1 U l = L 2 j, φl 2 j ) U l = L 2 j+1. Proof. We shall only prove the second equation.

11 1908 H.H. Chan, P.C. Toh / Journal of Number Theory ) φl 2 j = q α Al2 ) A1) A1) ) p π l 2 j n δ 2 j q n = A l 2) p π l 2 j n α) 1 + l 2 + +l 2 j 2) α ) q n = A l 2) ) p π l 2 j n δ 2 j+1 q n. The result then follows from Lemma 9. We need to study the effect of applying the Hecke operator U l, on the modular functions φ and h. For a positive integer r, define l 1 ) τ + j S r = φ r l j=0 l 1 = φ r τ ) A 1 l S j j=0 = φ r τ ) U l = lφ r τ ) Ul. 3.1) Lemma 15. For a positive integer p, we have Proof. h p Ul = S pγ lh p, φh p ) Ul = S pγ +1 lh p. h p pβ Al)pγ = q A1) pγ = qpβ Al) pγ q pβl+1) Al 2 ) pγ qαpγ Al2 ) pγ A1) pγ = 1 Al) pγ q pβl Al 2 ) pγ φ pγ. 3.2) In the above, we used the fact that αγ = βl + 1). By applying U l and Lemma 9, h p Ul = 1 q pβ A1) pγ Al) pγ φ pγ ) Ul 3.3) = 1 h p 1 l S pγ. The proof of the second equation is similar.

12 H.H. Chan, P.C. Toh / Journal of Number Theory ) Next, we have to show that φ τ + j l ) are distinct roots of a polynomial with coefficients in terms of h. By Newton s formula for symmetric power sum of roots [12, p. 118, each S r satisfies a recurrence relation and can be written as a polynomial in terms of h. ThusifL j is a polynomial in h, multiplying L j by φ if necessary and applying U l, we see that L j+1 can be written in terms of a sum of S r which in turn can be written as a polynomial in h. At this stage, the proof depends crucially on a modular equation satisfied by the corresponding φ and h and we shall provide the details for l = 5 and the partition π =[ HenceA π j) = A j) = E j)e2 j) and p π n) = an). Wehave and φτ ) = η25τ )η50τ ) ητ )η2τ ) hτ ) = η2 5τ )η 2 10τ ) η 2 τ )η 2 2τ ) = q 3 A25) A1), 3.) = q A5)2 A1) ) φτ ) is a modular function on Γ 0 50) + W 2 and hτ ) is the hauptmodul of Γ 0 10) + W 2.Furthermore h5τ ) is also a modular function on Γ 0 50) + W 2 and thus φτ ) and h 5 τ ) = h5τ ) belong to the same congruence subgroup and as a result, satisfy a polynomial equation of degree 5 and degree 3 respectively [1, p. 110, i.e. φ 5 τ ) = 15h 5 τ ) h 2 τ 5 ) + 5 h 3 τ 5 )) φ τ ) 5 2 h 2 τ 5 ) + 53 h 3 τ 5 )) φ 3 τ ) + 5h 2 τ 5 ) + 52 h 3 τ 5 )) φ 2 τ ) 5h 3 τ )φτ 5 ) + h3 5 τ ). 3.6) The above equation holds with h 5 τ ) = h5τ ) invariant when we replace τ by τ + j/5 for j from 1 to, thus we conclude that the following polynomial Q u) = u 5 15h h h 3) u h h 3) u 3 5h h 3) u 2 + 5h 3 u h 3 3.7) has five distinct roots φ τ + j 5 ),for j from 0 to. Now by Newton s theorem, S r for r 5, satisfies We also have S r = 15h h h 3) S r h h 3) S r 2 + 5h h 3) S r 3 5h 3 S r + h 3 S r ) S 1 = 15 h h h 3, S 2 = 175 h h h h h 6, S 3 = 15 h h h h h h h h 9, S = 280 h h h h h h h h h h )

13 1910 H.H. Chan, P.C. Toh / Journal of Number Theory ) Hence if we write S r = a r,p h p, 3.10) we obtain the recurrence relation a r,p = 15a r 1,p a r 1,p a r 1,p 3 25a r 2,p 2 125a r 2,p 3 + 5a r 3,p a r 3,p 3 5a r,p 3 + a r 5,p 3, 3.11) for r 5. Moreover, an induction argument shows that a r,p 0 ifand only if 3r + 5 p 3r. 3.12) If we now define νa r,p ) as the highest power of 5 that divides a r,p and ν0) =,wecanshowby induction that 5p 3r + νa r,p ). 3.13) We are now ready to prove Theorem 10 for φ and h defined in 3.) and 3.5). Eq. 1.18) is our base case and can be rewritten as L 1 = 3h + 25h h 3. Suppose that we have L 2 j 1 = b 2 j 1,r h r, 3.1) where only finitely many b 2 j 1,r are nonzero and each is an integer. Applying U 5, L 2 j 1 U 5 = L 2 j = b 2 j 1,r h r U5 = 6r b 2 j 1,r It follows from 3.12) that for every p in the second sum, 6r + p μ = r + 5 p=μ r + 5 a 2r,p h p 5h r. 3.15) r ) Furthermore, 5p 6r + νa 2r,p ) 6r + 1 6r ) Thus L 2 j is also a polynomial in h with integral coefficients. The inductive case from L 2 j to L 2 j+1 is similar. The other fourteen cases for Theorem 10 can be dealt with in the same way. We will proceed to prove Theorem 13. We first note that hτ ) = η2 5τ )η 2 10τ ) η 2 τ )η 2 2τ ) η 8 τ )η 8 2τ ) mod 5). 3.18)

14 H.H. Chan, P.C. Toh / Journal of Number Theory ) If we set L j = b j,r h r, 3.19) then Theorem 13 follows from the next lemma. Lemma 16. νb 2 j 1,1 ) = j 1, 5r νb 2 j 1,r ) j 1 + νb 2 j,1 ) = j, 5r 5 νb 2 j,r ) j +, if r > 1,, if r > 1. Proof. We shall prove this by induction. Since L 1 = 3h + 25h h 3, νb 1,1 ) = 0, νb 1,2 ) = 2 and νb 1,3 ) = 3. Now from 3.15), L 2 j = b 2 j,s h s = 6r ) a 2r,p h p r b 2 j 1,r. 3.20) 5 s 1 r 1 p=μ Equating coefficients, we have b 2 j,s = r 1 ) a2r,s+r b 2 j 1,r. 3.21) 5 Hence, for s > 1, we have { } 5r 5s + r) 32r) + νb 2 j,s ) 1 + min j r 1 { } 5r 5s r j 1 + min + r 1 5s 1 j 1 + 5s 5 j +. We made use of the fact that the minimum occurs at r = 1. Next for the case s = 1, we have 5b 2 j,1 = r 1 b 2 j 1,r a 2r,1+r ) = b 2 j 1,1 a 2,2 ) + r 2 b 2 j 1,r a 2r,1+r ) = 175b 2 j 1,1 + r 2 b 2 j 1,r a 2r,1+r ). 3.22)

15 1912 H.H. Chan, P.C. Toh / Journal of Number Theory ) Thus, νb 2 j,1 ) = 1 + min { = 1 + j + 1) = j. 2 + j 1, min r 2 { 5r j }} 5 r + Next, we have L 2 j+1 = b 2 j+1,s h s = φl 2 j ) U 5 = a 2r+1,p h p r b 2 j,r. 3.23) 5 p Equating coefficients, we have b 2 j+1,s = r 1 ) a2r+1,s+r b 2 j,r. 3.2) 5 Thus, for s > 1, we have When s = 1, { 5r 5 j + { 5r 5 j 1 + min r 1 5s j 1 + 5s j +. νb 2 j+1,s ) 1 + min r } 5s + r) 32r + 1) + } 5s r + 1 5b 2 j+1,1 = b 2 j,1 a 3,2 ) + r 2 b 2 j,r a 2r+1,1+r ) = 15b 2 j,1 + r 2 b 2 j,r a 2r+1,1+r ). Finally, νb 2 j+1,1 ) = 1 + min { 1 + j, min r 2 = j) = j, { 5r 5 j + + }} 5 r + 1 which completes the proof of Lemma 16.

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