NEW ZEALAND JOURNAL OF MATEMATICS Volume 7 017, 3-56 GENERAL FAMILY OF CONGRUENCES MODULO LARGE POWERS OF 3 FOR CUBIC PARTITION PAIRS D. S. Gireesh and M. S. Mahadeva Naika Received May 5, 017 Abstract. Let bn denote the number of cubic partition pairs of n. Recently, Chern proved two of Lin s conjectures on cubic partition pairs and he asked a question about general family of congruences modulo large powers of 3 for cubic partition pairs. In this paper, we give affirmative answer to Chern s question by finding α, β, l such that b3 α n + l 0 mod 3 β. In the process, we also give alternative proofs for Lin s conjectures. 1. Introduction Let pn denote the number of unrestricted partition of n. Ramanujan s most beautiful congruences are These congruences are generalized as p5n + 0 mod 5, p7n + 5 0 mod 7, p11n + 6 0 mod 11. p5 j n + δ 5,j 0 mod 5 j, 1.1 p7 j n + δ 7,j 0 mod 7 j/ +1, 1. p11 j n + δ 11,j 0 mod 11 j, 1.3 where δ l,j 1/ mod l j. Inspired by Ramanujan s work, Chan [] studied the cubic partition function an given by anq n 1, f 1 f where f k q k ; q k 1 q km. m1 010 Mathematics Subject Classification 05A17, 11P83. Key words and phrases: Partitions; Cubic Partition Pairs; Congruences.
D. S. GIREES and M. S. MAADEVA NAIKA e found the identity a3n + q n 3 f 3 3 f 3 6 f 1 f which readily implies Ramanujan s type congruence a3n + 0 mod 3. Chan [3] also established infinite family of congruences modulo large powers of 3 for an which is analogue of Ramanujan s congruences 1.1 1.3. For each n, k 1, Chan proved that a3 k n + c k 0 mod 3 k+δk, 1. where c k is the reciprocal modulo 3 k of 8 and 1 if k is even, δk : 0 if k is odd. Chan and Toh proved a higher power analogue of the Ramanujan s congruences and 1. for an, involving the prime 5, a5 j n + d j 0 mod 5 j/, 1.5 where d j 1/8 mod 5 j. Chan and Cooper [] obtained analogue results of 1.1 1.5 for cn. For each n 0 and j 1, where cn is defined by c j n + j + 1 3 c j+1 n + 5 j + 1 3 0 mod j+1, 1.6 0 mod j+, 1.7 cnq n 1 f1 f 3. 1.8 Zhao and Zhong [13] studied the cubic partition pairs function bn given by bnq n 1 f1 f. 1.9 For each n 0, they found Ramanujan s type congruences where i, 3,, 6. b5n + 0 mod 5, b7n + i 0 mod 7, b9n + 7 0 mod 9,
CUBIC PARTITION PAIRS 5 Lin [10] also studied the cubic partition pairs and established Ramanujan s type congruences modulo 7. For example, for each n 0, he proved that e also conjectured that and b7n + 16 0 mod 7, 1.10 b7n + 5 0 mod 7, 1.11 b81n + 61 0 mod 7. 1.1 b81n + 61 0 mod 3 1.13 b81n + 7q n 9 f f 3 f 6 mod 81, 1.1 b81n + 3q n 36 f 1f 6 f 3 mod 81. 1.15 Recently, Chern confirmed Lin s conjectures and he proved stronger result of 1.13 that b81n + 61 0 mod 79. In concluding remarks, Chern asked a question about general family of congruences modulo large powers of 3 for cubic partition pairs. In this paper, we give affirmative answer to Chern s question by finding α, β, l such that b3 α n + l 0 mod 3 β, which is analogue of 1.1 1.7. In the process, we also give alternative proofs for Lin s conjectures. We list our main results in the following theorem. Theorem 1.1. For each α 0 and n 0, b 3 α+1 n + 3α+1 + 1 0 mod 3 α, 1.16 b 3 α+ n + 3α+5 + 1 0 mod 3 α+5, 1.17 b 3 α+5 n + 7 3α+ + 1 0 mod 3 α+6, 1.18 b 3 α+5 n + 11 3α+ + 1 0 mod 3 α+6. 1.19
6 D. S. GIREES and M. S. MAADEVA NAIKA. Preliminaries In this section, we list identities which are useful in proving our main results. The Ramanujan s cubic continued fraction denoted by r c q is defined by r c q 1 + q 1/3 q + q 1 + q + q 1 + q3 + q 6 1 + Lemma.1. [7, Theorem 3.1] Let xq q 1/3 r c q. Then 1 f 1 f f 9 f 18 xq 3 q q xq 3..1 Lemma.. [7, Theorem 3.1] We have Now let f3 f6 f9 f 18 1 xq 3 3 7q3 8q 6 xq 3 3.. ζ f 1f, ρ 1 qf 9 f 18 qxq 3, T f 3 f6 q 3 f9 f 18..3 Then, from.1,. and.3,. and ζ f 1f qf 9 f 18 ρ 1 ρ. T ρ 3 7 8 ρ 3..5 From. and.5, we have It follows from.6 that We can write.7 so ζ 3 ρ 3 3ρ 3ρ + 11 + 6 ρ 1 ρ 8 ρ 3 T + 18 3ρ 3ρ + 6 ρ 1 ρ T + 9 3ζ 9ρ + 18 ρ T 9ζ 3ζ..6 ζ 3 + 3ζ + 9ζ T..7 1 ζ 1 T 9 + 3ζ + ζ,.8 1 ζ i 1 9 T ζ i 1 + 3 ζ i + 1 ζ i 3..9
CUBIC PARTITION PAIRS 7 Now let be the huffing operator modulo 3, that is, an q n a 3n q 3n. If we apply to.9, we find 1 ζ i 1 1 1 1 9 T ζ i 1 + 3 ζ i + ζ i 3..10 Now, From.10.13, we find ζ ρ ρ 3 + ρ + ρ 3,.11 ζ ρ 1 1,.1 ρ 1 1..13 1 3 ζ T,.1 T 33 ζ + T,.15 1 1 ζ 3 and so on. Indeed, for i 1 we can write 1 ζ i 1T + 33 T + 35 T 3,.16 i m i,j T j,.17 where the m i,j are defined in the following matrix. The m i,j form a matrix M, the first ten rows of which are 3 0 0 0 0 0 3 3 0 0 0 0 1 3 3 3 5 0 0 0 0 3 3 3 7 0 0 0 5 3 3 5 3 6 5 3 9 0 M 0 1 3 7 3 6 5 3 9 3 11.18 0 0 3 7 3 7 3 8 7 3 10 7 0 0 3 3 3 19 3 7 3 9 7 0 0 1 3 3 9 3 9 5 0 0 0 3 5 3 5 17 3 8 5...... and for i, m i,1 0, and for j, m i,j 9m i 1,j 1 + 3m i,j 1 + m i 3,j 1..19
8 D. S. GIREES and M. S. MAADEVA NAIKA In fact m i,j 0 for j i, so we can write where 1 ζ i i ji+1 m i,j T j 3i Similarly, m i+,j 0 if j i, so we can write where 1 ζ i+ We can write.0 i+ ji+1 and this can be rearranged to m i,i+j T i+j 3i a i,j,.0 T i+j a i,j m i,i+j..1 3i+ m i+,j m i+,i+j T j T i+j 3i+ b i,j,. T i+j b i,j m i+,i+j..3 q f i 9f 18 3i a i,j q 3 f 9 f i+j 18 f 1 f f3 f 6,. i q i f3 f 6 f 1 f The equation. can be written as q f i+ 9f 18 f 1 f and this can be rearranged to i+ q i 1 f3 f 6 f 1 f 3i It follows from the binomial theorem that 3i+ 3i+ j a i,j q 3j f9 f 18..5 f 3 f 6 b i,j q 3 f 9 f18 i+j f3 f 6,.6 j b i,j q 3j 3 f9 f 18..7 f 3 f 6 f 3 1 f 3 mod 3,.8 f 9 1 f 3 3 mod 9..9 To prove our main results, we use these preliminaries without explicitly mentioning.
CUBIC PARTITION PAIRS 9 3. Generating Functions In this section, we found some generating functions which are useful in proving our main results. Theorem 3.1. For each α 0, b 3 α+1 n + 3α+1 + 1 q n 1 f3 f 6 and b 3 α+ n + 3α+3 + 1 q n 1 f1 f where the coefficient vectors x α x α,1, x α,,... are given by and i x α,i q i 1 f3 f 6 3.1 f 1 f i x α+1,i q i 1 f3 f 6, 3. f 1 f x 0 x 0,1, x 0,, x 0,3,..., 7, 0,..., 3.3 where A a i,j i,j 1 and B b i,j i,j 1. Proof. Using., 1.9 can expressed as bnq n 1 1 q 3 f9 f 18 Taking the operator on both sides b3n + 1q 3n 1 q 3 f 9 f 18 1 q 3 f9 f 18 1 q 3 f9 f 18 x α+1 x α A if α 0, 3. x α x α 1 B if α 1, 3.5 1 ζ T + 33 T After simplification, we arrive at b3n + 1q n f 3 f6 f1 f 1 ζ. 3.6 q 3 f 9 f18 f3 f 6 + 7q 6 f 9 8 f18 8 f3 8f 6 8. 3.7 + 7q f 3 6 f6 6 f1 8f 8, 3.8 The identity 3.8 is the α 0 case of 3.1. Suppose 3.1 holds for some α 0. Then b 3 α+1 n + 3α+1 + 1 q n 1 i f3 f 6 x α,i q i 1 f3 f 6, 3.9 f 1 f
50 D. S. GIREES and M. S. MAADEVA NAIKA which is equivalent to b 3 α+1 n + 3α+1 + 1 q n 1 f3 f 6 i x α,i q i 3 f3 f 6. 3.10 f 1 f Applying the operator to 3.10, we find that b 3 α+1 3n + + 3α+1 + 1 q 3n i x α,i q 3 f 3 f 6 q 3 f 3 f 6 1 f 3 f 6 1 f 3 f 6 3i x α,i q i f3 f 6 f 1 f a i,j q 3j f9 f 18 f 3 f 6 x α,i a i,j j 1 j 1 j q 3j 3 f9 f 18 f 3 f 6 j x α+1,j q 3j 3 f9 f 18. f 3 f 6 j Which implies that b 3 α+ n + 3α+3 + 1 q n 1 f1 f j 1 j x α+1,j q j 1 f3 f 6, 3.11 f 1 f which is 3.. Now suppose 3. holds for some α 0. Then b 3 α+ n + 3α+3 + 1 q n 1 f1 f i x α+1,i q i 1 f3 f 6, 3.1 f 1 f which is same as b 3 α+ n + 3α+3 + 1 q n 1 f3 f 6 i+ x α+1,i q i 1 f3 f 6. 3.13 f 1 f
CUBIC PARTITION PAIRS 51 Applying the operator to 3.13, we find that b 3 α+ 3n + 3α+3 + 1 q 3n 1 i+ f3 f 6 x α+1,i q i 1 f3 f 6 f 1 f 1 f 3 f 6 1 f 3 f 6 1 f 3 f 6 j 1 j 3i+ x α+1,i b i,j q 3j 3 f9 f 18 f 3 f 6 x α+1,i b i,j q 3j 3 f9 f 18 f 3 f 6 j 1 j x α+,j q 3j 3 f9 f 18. f 3 f 6 After simplification, we obtain b 3 α+3 n + 3α+3 + 1 q n 1 f1 f 1 f 3 f 6 j 1 j 1 j j x α+,j q j 1 f3 f 6 f 1 f j x α+,j q j 1 f3 f 6, f 1 f which is 3.1 with α + 1 in place of α. This completes the proof of 3.1 and 3. by induction.. Congruences Let νn be the largest power of 3 that divides N. Note that ν0 +. Proof of the Theorem 1.1. It follows from.18 and.19 that and then follows from.1,.3 and.1 that and But note that and It not hard to show that νm i.j 3j i 1,.1 νa i.j 3i + j i 1 3j i 1. νb i.j 3i + j i + 1 3j i 3..3 νa 1,j νm,j+1 j. νb 1,j νm 6,j+1 j 1.5 νx α,j α + j 1.6
5 D. S. GIREES and M. S. MAADEVA NAIKA and νx α+1,j α + j..7 The identity.6 is true for α 0, by 3.3. Suppose.6 is true for some α 0. Then νx α+1,j min νx α,i + νa i,j νx α,1 + νa 1,j α + j, which is.7. Now suppose.7 is true for all α 0. Then νx α+,j min νx α+1,i + νb i,j νx α+1,1 + νb 1,j α + + j 1 α + j, which is.6 with α + 1 in place of α. This completes the proof of.6 and.7 by induction. The congruence 1.16 follows from 3.1 together with.6. It follows from 3.1 and.6 that b 3 α+1 n + 3α+1 + 1 q n 3 α f 3 f 6 f 1 f + 3 α+ q f 6 3 f 6 6 f 8 1 f 8 + 3 α+ q f 3 10 f6 10 f1 1f 1 mod 3 α+6..8 Using.9 and.3,.8 can be expressed as b 3 α+1 n + 3α+1 + 1 3α q 3 f 3 f 6 q n T ζ + 3 T ζ 8 + 3 1 ζ 3 mod 3 α+6..9
CUBIC PARTITION PAIRS 53 Taking the operator on both sides of.9, we obtain b 3 α+ n + 3α+3 + 1 3α q 3 f3 f 6 3α q 3 f3 f 6 3α q 3 f3 f 6 3α+ q 3 f 3 f 6 which implies that 1 T ζ q 3n + 3 T 1 ζ 8 1 + 3 3 T T + 3 T 3 + 3 T 3 19 3 118 3 + T T 19q 3 f 9 f18 f3 f 6 + 118 3 q 6 f 9 8 f18 8 f3 8f 6 8 b 3 α+ n + 3α+3 + 1 3α+ f 1 f 3α+ q 3 f 9 f 18 19 f 3 f6 f1 f q n ζ 3 T 3 + 33 19 T + 3 1 T mod 3 α+6,.10 + 3 q f 3 8 f6 8 f1 8f 8.11 mod 3 α+6..1 Tζ T 19 + 3 6 ζ 10 Taking the operator on both sides, we obtain b 3 α+3 n + 3α+3 + 1 q 3n 3α+ q 3 f9 f 18 3α+ q 3 f9 f 18 3α+ q 3 f9 f 18 3α+ q 3 f9 f 18 19T 19T 19 1 ζ 6 1 + 3 T 1 1 3 + T T 3 T + 8 3 T ζ 10 19q 3 f 9 f18 f3 f 6 + 8 3 q 6 f 9 8 f18 8 f3 8f 6 8 + 3 T 5 3 T mod 3 α+6,.13 which implies that b 3 α+3 n + 3α+3 + 1 q n 3 α+ 19 f 3 f6 f1 f + 8 3 q f 3 6 f6 6 f1 8f 8.1 3α+ q 1 Tζ T 19 f3 f 6 + 8 3 ζ 8 mod 3 α+6,.15
5 D. S. GIREES and M. S. MAADEVA NAIKA which same as b 3 α+3 n + 3α+3 + 1 q n 3α+ q 3 Tζ T 19 f3 f 6 + 8 3 ζ 8 Taking the operator on both sides, we obtain b 3 α+ n + 3α+5 + 1 q 3n 1 19T ζ 3α+ q 3 f3 f 6 3α+ q 3 f3 f 6 3α+5 q 3 f3 f 6 3α+5 q 3 f3 f 6 1 + 8 3 T 19T 3 T + 8 3 T 3 T 3 3 T q3 f 9 f 18 f 3 f 6 Replacing q 3 by q, we obtain that b 3 α+ n + 3α+5 + 1 q n 3 α+5 f 3 f6 f1 6f 6 mod 3 α+6..16 ζ 8 mod 3 α+6..17 3 α+5 f 3 f 6 mod 3 α+6..18 Congruence 1.17 is easily follows from.18. Extracting the terms in which powers of q are congruent to q 3n+1 and q 3n+ from.18, we obtain 1.18 and 1.19, respectively. 5. Proofs of Lin s conjectures To prove Lin s conjectures, we need the following lemmas. Lemma 5.1. We have f 5 f1 f 6 6 f9 10 f3 10f 18 5 + q f 6 5 f9 7 f3 9f 18 + 9q f 6 f9 f 18 f3 8 + 10q 3 f 6 3 f 9 f18 f3 7 Proof. From [1, p. 9, Corollary], we have + q f 6 f18 7 f3 6f 9. 5.1 irschhorn [9] proved that f f 1 f 6f 9 f 3 f 18 + q f 18 f 9. 5. f f1 f 6 f9 6 f3 8f 18 3 + q f 6 3 f9 3 f3 7 + q f 6 f18 3 f3 6. 5.3 Squaring 5. and then multiplying the resulting equation with 5.3, we arrive at 5.1.
CUBIC PARTITION PAIRS 55 Lemma 5.. [1, Eq. 5.1] We have f f 8 3 f 8 1 f 6 Now we are in position to prove Lin s conjectures. Put α 0 in 1.17, we find that which is 1.13. Put α 0 in.1, we deduce that + q f f6 5 f1 5f 1 mod 9. 5. 3 b81n + 61 0 mod 3, 5.5 b7n + 7q n 9 f 3 f 6 f 1 f 9 f 3 f 5 f 1 f 6 mod 81. 5.6 Invoking 5.1 and 5.6, b7n + 7q n 9 f 3 f 6 From which we obtain f 6 6 f9 10 f3 10f 18 5 + q f 6 5 f9 7 f3 9f 18 + q 3 f 6 3 f 9 f18 f3 7 + q f 6 f18 7 f3 6f 9 b81n + 7q n 9 f 1 f 9 f f 3 f 6 f 6 f 10 3 f1 10f 6 5 f f 8 3 f 8 1 f 6 + q f 3 f 3 f6 f 7 1 + q f f6 5 f1 5f 3 mod 81. 5.7 mod 81. 5.8 Using 5. in 5.8, we arrive at 1.1. Extracting the terms of 5.7 in which powers of q are congruent to 1 modulo 3, dividing by q, and then replacing q 3 by q, we obtain b81n + 3q n 36 f 1 f 5 f3 7 f f1 9f 6 36 f 1f 6 f 3 Using 5. in 5.9, we arrive at 1.15. f f 8 3 f 8 1 f 6 + q f f6 7 f1 6f 3 + q f f 5 6 f 5 1 f 3 mod 81. 5.9 Acknowledgments The authors would like to thank the anonymous referee for helpful comments and suggestions.
56 D. S. GIREES and M. S. MAADEVA NAIKA References [1] B.C. Berndt, Ramanujan s Notebooks, Part III, Springer-Verlag, New York, 1991. [].-C. Chan, Ramanujan s cubic continued fraction and a generalization of his most beautiful identity, Int. J. Number Theory, 6 010, 673 680. [3].-C. Chan, Ramanujan s cubic continued fraction and Ramanujan type congruences for a certain partition function, Int. J. Number Theory, 6 010, 819 83. [].-C. Chan and S. Cooper, Congruences modulo powers of for a certain partition function, Ramanujan J., 010, 101 117. [5].. Chan and P.C. Toh, New analogues of Ramanujan s partition identities, J. Number Theory, 130 010, 1898 1913. [6] S. Chern, Arithmetic properties for cubic partition pairs modulo powers of 3, arxiv:1609.08578v1. [7] S. Cooper, Series and iterations for 1/π, Acta Arithmetica, 11 010, 33 58. [8] M.D. irschhorn and D.C. unt, A simple proof of the Ramanujan conjecture for powers of 5, J. Reine Angew. Math., 36 1981, 1 17. [9] M.D. irschhorn and J.A. Sellers, Infinitely many congruences modulo 5 for -colored Frobenius partitions, Ramanujan J., 0 1 016, 193 00. [10] B.L.S. Lin, Congruences modulo 7 for cubic partition pairs, J. Number Theory, 171 017, 31. [11] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 1998. [1] S. Son, Cubic identities of theta functions, Ramanujan J., 1998 303 316. [13]. Zhao and Z. Zhong, Ramanujan type congruences for a partition function, Electron. J. Combin., 18 011. D. S. Gireesh Department of Mathematics, Dayananda Sagar College of Engineering, Shavige Malleshwara ills, Kumaraswamy Layout, Bengaluru-560 078, Karnataka, India. gireeshdap@gmail.com M. S. Mahadeva Naika Department of Mathematics, Bangalore University, Central College Campus, Bengaluru-560 001, Karnataka, India. msmnaika@rediffmail.com