Research Article Some Congruence Properties of a Restricted Bipartition

Transcription

1 International Analysis Volume 06, Article ID 90769, 7 pages Research Article Some Congruence Properties of a Restricted Bipartition Function c N (n) Nipen Saikia and Chayanika Boruah Department of Mathematics, Rajiv Gandhi University, Rono Hills, Doimukh, Arunachal Pradesh 79, India Correspondence should be addressed to Nipen Saikia; nipennak@yahoo.com Received April 06; Accepted 0 July 06 Academic Editor: Ahmed Zayed Copyright 06 N. Saikia and C. Boruah. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Let c N (n) denote the number of bipartitions (λ, μ) of a positive integer n subject to the restriction that each part of μ is divisible by N. In this paper, we prove some congruence properties of the function c N (n) for N7,, and 5l, for any integer l,by employing Ramanujan s theta-function identities.. Introduction A bipartition of a positive integer n is an ordered pair of partitions (λ, μ) such that the sum of all of the parts equals n. If c N (n) counts the number of bipartitions (λ, μ) of n subject to the restriction that each part of μ is divisible by N, then the generating function of c N (n) [] is given by where c N (n) q n (a; q) (q; q) (q N ;q N ), () ( aq n ). () The partition function c N (n) is first studied by Chan [] for the particular case Nby considering the function c (n) defined by c (n) q n Chan [] proved that, for n 0, (q; q) (q ;q ). () c (n + ) 0 (mod). (4) Kim [] gave a combinatorial interpretation (4). In a subsequent paper, Chan [4] showed that, for k and n 0, c ( k n+s k )0 (mod k+δ(k) ), (5) where s k is the reciprocal modulo k of 8 and δ(k) if k is even and 0 otherwise. Inspired by the work of Ramanujan on the standard partition function p(n), Chan [4] asked whether there are any other congruence properties of the following form: c (ln + k) 0 (modl), wherel is prime and 0 k l. Sinick [] answered Chan s question in negative by considering restricted bipartition function c N (n) defined in (). Liu and Wang [5] established several infinite families of congruence properties for c 5 (n) modulo. For example, they proved that c 5 ( α+ n+ 7 α + )0 (mod), 4 (6) α, n 0. Baruah and Ojah [6] also proved some congruence properties for some particular cases of c N (n) by considering the generalised partition function p [c l d m ](n) defined by p [c l d m ] (n) q n (q c ;q c ) l (qd ;q d ) m (7)

2 International Analysis and using Ramanujan s modular equations. Clearly, c N (n) p [ N ](n). For example, Baruah and Ojah [6] proved that p [ ] (4n + j) 0 (mod), for j,, (8) p [ 7 ] (8n + 7) 0 (mod). Ahmed et al. [7] investigated the function C N (n) for N and 4 and proved some congruence properties modulo 5. They also gave alternate proof of some congruence properties duetochan[]. In this paper, we investigate the restricted bipartition function c N (n) for n 7,, and 5l, for any integer l, and prove some congruence properties modulo,, and 5 by using Ramanujan s theta-function identities. In Section, we provecongruencepropertiesmoduloforc 7 (n). For example, we prove, for α 0, c 7 ( α+ n+ 5 α + )0 (mod). (9) In Section 4, we deal with the function c (n) and establish the notion that if p is an odd prime, j p,andα 0, then c (4p α+ (pn + j) + pα+ + )0 (mod). (0) In Section 5, we show that, for any integer l, c 5l (5n + 4) 0(mod5). We also prove congruence properties modulo for c 5 (n). Section is devoted to listing some preliminary results.. Preliminary Results Ramanujan s general theta function f(a, b) is defined by f (a, b) a n(n+)/ b n(n )/, ab <. () Three important special cases of f(a, b) are φ(q)fl f (q, q) ψ(q) fl f(q,q ) q n n f( q)fl f( q, q ) (q; q). (q ;q ) 5 (q; q) (q4 ;q 4 ), () q n(n+)/ (q ;q ), () (q; q) ( ) n q n(n+)/ n Ramanujan also defined the function χ(q) as (4) χ(q)( q;q ). (5) Lemma. For any prime p and positive integer m,onehas Proof. It follows easily from the binomial theorem. Lemma (see [8, page 5]). One has ψ(q) ψ(q 7 )φ(q 8 )ψ(q 8 )+qψ(q 4 )ψ(q ) Lemma. One has +q 6 ψ(q 56 )φ(q 4 ). ψ(q) ψ(q 7 )(q;q) (q7 ;q 7 ) Proof. From (), we have (7) (mod). (8) ψ(q) ψ(q 7 ) (q ;q ) (q4 ;q 4 ) (q; q) (q 7 ;q 7 ). (9) Simplifying (9) using Lemma with p,wearriveatthe desired result. Lemma 4 (see [9, page 86, Equation (60)]). One has φ ( q) (q; q) (q ;q ), (0) ψ( q) (q; q) (q4 ;q 4 ) (q ;q ), () f(q) χ(q) Lemma 5 (see [0, page 7]). One has ψ(q) ψ(q )φ(q 66 )ψ(q ) (q ;q ) (q; q) (q 4 ;q 4 ), () (q ;q ) (q; q) (q 4 ;q 4 ). () +qf(q 44,q 88 )f(q,q 0 ) +q f(q,q 0 )f(q 8,q 4 ) +q 5 ψ(q )φ(q 6 ). Lemma 6 (see [8, page 50, Equation ()]). One has where Lemma 7. One has (4) f (q, q ) φ( q ) χ( q), (5) χ( q) (q; q) (q ;q ). (6) (q pm ;q pm ) (q m ;q m ) p (modp). (6) f(q ;q )(q ;q ) (mod). (7)

3 International Analysis Proof. Employing (0) in Lemma 6 and performing simplificationusinglemmawithp,weobtain f(q;q )(q;q) (mod). (8) Replacing q by q in(8),wearriveatthedesiredresult. Lemma 8 (see [8, page 5, Example (v)]). One has Lemma 9. One has f(q,q 5 )ψ( q )χ(q). (9) f(q,q 5 ) (q ;q ) (mod). (0) (q; q) Proof. Employing () and () in Lemma 8, we obtain f(q,q 5 ) (q ;q ) (q ;q ) (q ;q ). () (q 6 ;q 6 ) (q; q) (q 4 ;q 4 ) Simplifying () using Lemma with p,wecompletethe proof. Lemma 0 (see [, page 5, Equation (5)]). One has (q ;q ) (q; q) (q 4 ;q 4 ) (q6 ;q 6 ) (q ;q ) +q (q (q ;q ) ;q ). () (q 4 ;q 4 ) Lemma (see [, Theorem.]). For any odd prime p, ψ(q) (p )/ k0 q (k +k)/ f(q (p +(k+)p)/,q (p (k+)p)/ ) +q (p )/8 ψ(q p ), where, for 0 k (p )/, k +k p 8 () (modp). (4) Lemma (see [, Theorem.]). For any prime p 5,one has where f( q) (p )/ k( p )/ k(±p )/6 ( ) k q (k +k)/ f( q (p +(6k+)p)/, q (p (6k+)p)/ ) + ( ) (±p )/6 q (p )/4 f( q p ), ±p 6 p {, if p (mod6), fl 6 { p, if p (mod6). { 6 (5) (6) Lemma (see []). One has (q 5 ;q 5 ) 6 (q; q) (q 5 ;q 5 ) 6 (F 4 (q 5 )+qf (q 5 ) +q F (q 5 )+q F(q 5 )+5q 4 q 5 F (q 5 ) +q 6 F (q 5 ) q 7 F (q 5 )+q 8 F 4 (q 5 )), (7) where F(q) fl q /5 R(q) and R(q) is Rogers-Ramanujan continued fraction defined by R(q)fl + + q /5 q q + q + Lemma 4 (see [8, page 45, Entry (iv)]). One has, q <. (8) (q; q) (q9 ;q 9 ) (4q W (q ) q+w (q )), (9) where W(q) q / G(q) and G(q) is Ramanujan s cubic continued fraction defined by G(q)fl + q / q+q + q +q 4 +. Congruence Identities for c 7 (n) Theorem 5. One has, q <. (40) c 7 (n + ) q n (q; q) (q 7 ;q 7 ) (mod). (4) Proof. For N7in (), we have c 7 (n) q n (q; q) (q 7 ;q 7 ). (4) Employing (9) in (4), we obtain c 7 (n) q n ψ(q) ψ(q 7 ) (q ;q ) (q4 ;q 4 ). (4) Employing Lemma in (4), we obtain c 7 (n) q n (q ;q ) (q4 ;q 4 ) [φ (q 8 )ψ(q 8 ) (44) +qψ(q 4 )ψ(q )+q 6 ψ(q 56 )φ(q 4 )]. Extracting the terms involving q n+, dividing by q, and replacing q by q in (44), we get c 7 (n + ) q n (q; q) (q7 ;q 7 ) [ψ(q 7 ) ψ(q)]. (45) EmployingLemmain(45),wecompletetheproof.

4 4 International Analysis Theorem 6. One has (i) c 7 (4n + ) q n (q ;q ) (q 4 ;q 4 ) (mod), (ii) c 7 (8n + 7) 0 (mod). Proof. From Theorem 5, we obtain c 7 (n + ) q n (q7 ;q 7 ) (q; q) (q 7 ;q 7 ) (q; q) Employing Lemma in (47), we obtain (46) (mod). (47) c 7 (n + ) q n ψ(q) ψ(q 7 ) (q ;q ) (q 4 ;q 4 (mod). (48) ) Employing Lemma in (48), extracting the terms involving q n+, dividing by q, and replacing q by q,weobtain c 7 (4n + ) q n (q; q) (q 7 ;q 7 ) ψ(q) ψ(q 7 ) (mod). (49) Employing Lemma in (49) and performing simplification using Lemma with p, we arrive at (i). All the terms on the right hand side of (i) are of the form q n. Extracting the terms involving q n+ on both sides of (i), we complete the proof of (ii). Theorem 7. For all n 0,onehas (i) c 7 (4n + 7) 0 (mod), (ii) c 7 (4n + 9) 0 (mod), (iii) c 7 (4n + ) 0 (mod). Proof. Employing (4) in Theorem 5, we obtain c 7 (n + ) q n (q 7 ;q 7 ) ( ) n q n(n+)/ (mod). (50) Extracting those terms on each side of (50) whose power of q is of the forms 7n +, 7n + 4, and7n + 6 and employing the fact that there exists no integer n such that n(n+)/is congruentto,4,and6modulo7,weobtain c 7 (4n + 7) q 7n+ c 7 (4n + 9) q 7n+4 c 7 (4n + ) q 7n+6 0 (mod). Now, (i), (ii), and (iii) are obvious from (5). (5) Theorem 8. For α,onehas c 7 ( α+ n+ α+ + (q; q) (q 7 ;q 7 ) (mod). (5) Proof. We proceed by induction on α. Extracting the terms involving q n and replacing q by q in Theorem 6(i), we obtain c 7 (8n + ) q n (q; q) (q 7 ;q 7 ) (mod), (5) which corresponds to the case α. Assume that the result is true for αk,sothat c 7 ( k+ n+ k+ + (q; q) (q 7 ;q 7 ) (mod). Employing Lemma in (54), we obtain c 7 ( k+ n+ k+ + ψ(q) ψ(q 7 ) (q; q) (q7 ;q 7 ) (mod). (54) (55) Employing Lemma in (55) and extracting the terms involving q n+, dividing by q, and replacing q by q,weobtain c 7 ( k+ (n + ) + k+ + ψ(q) ψ(q 7 ) (q; q) (q 7 ;q 7 ) (mod). (56) Simplifying (56) using Lemmas and with p,weobtain c 7 ( (k+) n+ (k+)+ + (q ;q ) (q 4 ;q 4 ) (mod). (57) Extracting the terms involving q n and replacing q by q in (57), we obtain c 7 ( (k+)+ n+ (k+)+ + (q; q) (q 7 ;q 7 ) (mod), which is the αk+case.hence,theproofiscomplete. Theorem 9. For α 0,onehas (58) c 7 ( α+ n+ 5 α + )0 (mod). (59)

5 International Analysis 5 Proof. All the terms in the right hand side of (57) are of the form q n, so, extracting the coefficients of q n+ on both sides of (57) and replacing k by α,weobtain c 7 ( (α+)+ n+ 5 (α+) + ) 0 (mod). (60) Replacing α+by α in (60) completes the proof. Theorem 0. If any prime p 5, ( 7/p), andα 0, then c 7 ( α+ p n+ α+ p(j+p)+ )0 (mod), (6) where j p. Proof. Employing Lemma in (5), we obtain c 7 ( α+ n+ α+ + ( (p )/ k( p )/ k(±p )/6 ( ) k q (k +k)/ f( q (p +(6k+)p)/, q (p (6k+)p)/ ) + ( ) (±p )/6 q (p )/4 f( q p )) q pn+(p )/ from both sides of (6) and replacing q p by q,we obtain c 7 ( α+ pn + α+ p + (q p ;q p ) (q 7p ;q 7p ) (mod). (65) Extracting the coefficients of q pn+j,for j p,onboth sides of (65) and performing simplification, we arrive at the desired result. 4. Congruence Identities for c (n) Theorem. One has c (4n + ) q n (q ;q ) ψ(q) (mod). (66) (q; q) Proof. Setting Nin (), we obtain c (n) q n Employing () in (67), we obtain (q; q) (q ;q ). (67) c (n) q n ψ(q) ψ(q ) (q ;q ) (q ;q ). (68) ( (p )/ k( p )/ k(±p )/6 ( ) m q 7 (m +m)/ f( q 7 (p +(6m+)p)/, q 7 (p (6m+)p)/ )+( ) (±p )/6 q 7 (p )/4 f( q 7p )) (mod). We consider the congruence (6) Employing Lemma 5 in (68), extracting the terms involving q n+, dividing by q, and replacing q by q,weobtain c (n + ) q n (q; q) (q ;q ) [f (q,q 44 )f(q,q 5 ) +q 7 ψ(q 66 )φ(q )]. EmployingLemmas9and0in(69),wefindthat c (n + ) q n (69) k +k +7 m +m 8p 8 4 (modp), (6) where (p )/ k,m (p )/.Thecongruence(6)is equivalent to (6k + ) +7(6m + ) 0 (modp) (64) and, for ( 7/p), the congruence (64) has unique solution k m (±p )/6. Extracting terms containing [ f(q,q 44 ) (q ;q ) (q ;q ) [ ( (q4 ;q 4 ) (q6 ;q 6 ) (q ;q ) +q (q ;q ) ) (q ;q ) (q 4 ;q 4 ) +q 7 ψ(q 66 )φ(q )] (mod). ] (70)

6 6 International Analysis Extracting the terms involving q n and replacing q by q on both sides of (70) and performing simplification using Lemma with p,weobtain c (4n + ) q n (q; q) (q ;q ) f(q ;q ) (q ;q ) (mod). (7) Employing Lemma 7 in (7) and using (), we complete the proof. Theorem. For any odd prime p and any integer α 0,one has c (4p α n+ pα + ψ(q) (mod). (7) Proof. We proceed by induction on α. Thecaseα0corresponds to the congruence theorem (Theorem ). Suppose that the theorem holds for αk 0,sothat c (4p k n+ pk + ) q n ψ(q) (mod). (7) Employing Lemma in (7), extracting the terms involving q pn+(p )/8 on both sides of (7), dividing by q (p )/8,and replacing q p by q,weobtain c (4p k+ n+ p(k+) + ψ(q p ) (mod). (74) Extracting the terms containing q pn from both sides of (74) and replacing q p by q, we arrive at c (4p (k+) n+ p(k+) + ψ(q) (mod), (75) which shows that the theorem is true for α k+.hence, the proof is complete. Theorem. For any odd prime p and integers α 0 and j p,onehas Proof. Setting N5lin (), we obtain c 5l (n) q n (q; q) (q 5l ;q 5l ). (78) Using Lemma in (78) and extracting the terms involving q 5n+4, dividing by q 4, and replacing q 5 by q,weobtain (q 5 ;q 5 ) 6 c 5l (5n + 4) q n 5 (q l,q l ) (q; q) 6. (79) The desired result follows easily from (79). Theorem 5. For all n 0,onehas (i) c 5 (5n + 4) 0 (mod5), (ii) c 5 (5n + 9) 0 (mod), (iii) c 5 (5n + 4) 0 (mod). Proof. Setting N5in (), we obtain c 5 (n) q n (q; q) (q 5 ;q 5 ). (80) Employing Lemma in (80), extracting terms involving q 5n+4, dividing by q 4, and replacing q 5 by q,weobtain (q 5 ;q 5 ) 6 c 5 (5n + 4) q n 5 (q ;q ) (q; q) 6. (8) Now, (i) follows from (8). Simplifying (8) by using Lemma with p,weobtain c 5 (5n + 4) q n (q5 ;q 5 ) (q; q) (q; q) (q; q)6 (q; q) (q5 ;q 5 ) (q; q) (q ;q ) 4 (mod). (8) Employing Lemma 4 in (8) and performing simplification, we obtain c 5 (5n + 4) q n c (4p α+ (pn + j) + pα+ + ) 0 (mod). (76) (q5 ;q 5 ) (q9 ;q 9 ) (q ;q ) 4 [q W (q ) (8) Proof. Extracting the coefficients of q pn+j for j p on both sides of (74) and replacing k by α, we arrive at the desired result. 5. Congruence Identities for c 5l (n) Theorem 4. For any positive integer l,one has c 5l (5n + 4) 0 (mod5). (77) +W (q )] (mod). Extracting terms involving q n+ and q n+ on both sides of (8), we arrive at (ii) and (iii), respectively. Competing Interests The authors declare that there are no competing interests regarding the publication of this paper.

7 International Analysis 7 Acknowledgments The first author (Nipen Saikia) is thankful to the Council of Scientific and Industrial Research of India for partially supporting the research work under Research Scheme no. 5(04)/5/EMR-II (F. no. 5(5498)/5). References [] J. Sinick, Ramanujan congruences for a class of eta quotients, International Number Theory, vol.6,no.4,pp , 00. [] H.-C. Chan, Ramanujan s cubic continued fraction and an analog of his most beautiful identity, International Number Theory,vol.6,no.,pp ,00. [] B. Kim, A crank analog on a certain kind of partition function arising from the cubic continued fraction, Acta Arithmetica, vol. 48, pp. 9, 0. [4] H.-C. Chan, Ramanujan s cubic continued fraction and Ramanujan type congruences for a certain partition function, International Number Theory, vol.6,no.4,pp.89 84, 00. [5] J.LiuandA.Y.Z.Wang, Arithmeticpropertiesofarestricted bipartition function, The Electronic Combinatorics, vol., no., pp., 05. [6] N. D. Baruah and K. K. Ojah, Analogues of Ramanujan s partition identities and congruences arising from his theta functions and modular equations, Ramanujan Journal, vol. 8, no., pp , 0. [7]Z.Ahmed,N.D.Baruah,andM.G.Dastidar, Newcongruences modulo 5 for the number of -color partitions, Number Theory, vol. 57, pp , 05. [8] B. C. Berndt, Ramanujan s Notebooks, Part III, Springer, New York, NY, USA, 99. [9] N.D.Baruah,J.Bora,andN.Saikia, Somenewproofsofmodular relations for the Göllnitz-Gordon functions, Ramanujan Journal,vol.5,no.,pp.8 0,008. [0] B. C. Berndt, Ramanujan s Notebooks, PartV,Springer,New York, NY, USA, 998. [] M. Hirschhorn, F. Garvan, and J. Borwein, Cubic analogues of the Jacobian theta function θ(z;q), Canadian Mathematics,vol.45,no.4,pp ,99. [] S.-P. Cui and N. S. S. Gu, Arithmetic properties of l-regular partitions, Advances in Applied Mathematics,vol.5,no.4,pp , 0. [] M. D. Hirschhorn, An identity of Ramanujan, and application in q-series from a contemporary perspective, Contemporary Mathematics,vol.54,pp.9 4,000.

8 Advances in Operations Research Advances in Decision Sciences Applied Mathematics Algebra Probability and Statistics The Scientific World Journal International Differential Equations Submit your manuscripts at International Advances in Combinatorics Mathematical Physics Complex Analysis International Mathematics and Mathematical Sciences Mathematical Problems in Engineering Mathematics Discrete Mathematics Discrete Dynamics in Nature and Society Function Spaces Abstract and Applied Analysis International Stochastic Analysis Optimization