IJANAL International Journal of Analysis 2314-4998 2314-498X Hindawi Publishing Corporation 10.1155/2016/9037692 9037692 Research Article Some Congruence Properties of a Restricted Bipartition Function cN(n) http://orcid.org/0000-0002-0288-8731 Saikia Nipen 1 Boruah Chayanika 1 Zayed Ahmed Department of Mathematics Rajiv Gandhi University Rono Hills Doimukh Arunachal Pradesh 791112 India rgu.ac.in 2016 882016 2016 12 04 2016 10 07 2016 2016 Copyright © 2016 Nipen Saikia and Chayanika 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 cN(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 cN(n) for N=7, 11, and 5l, for any integer l1, by employing Ramanujan’s theta-function identities.

Council of Scientific and Industrial Research of India 25(5498)/15
1. 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 cN(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 cN(n)  is given by (1)n=0cNnqn=1q;qqN;qN,where (2)a;q=n=01-aqn.The partition function cN(n) is first studied by Chan  for the particular case N=2 by considering the function c2(n) defined by (3)n=0c2nqn=1q;qq2;q2.Chan  proved that, for n0, (4)c23n+20mod3.Kim  gave a combinatorial interpretation (4). In a subsequent paper, Chan  showed that, for k1 and n0, (5)c23kn+sk0mod3k+δk,where sk is the reciprocal modulo 3k of 8 and δ(k)=1 if k is even and 0 otherwise. Inspired by the work of Ramanujan on the standard partition function p(n), Chan  asked whether there are any other congruence properties of the following form: c2(ln+k)0modl, where l is prime and 0kl. Sinick  answered Chan’s question in negative by considering restricted bipartition function cN(n) defined in (1). Liu and Wang  established several infinite families of congruence properties for c5(n) modulo 3. For example, they proved that (6)c532α+1n+7·32α+140mod3,α1,n0.

Baruah and Ojah  also proved some congruence properties for some particular cases of cN(n) by considering the generalised partition function p[cldm](n) defined by (7)n=0pcldmnqn=1qc;qclqd;qdmand using Ramanujan’s modular equations. Clearly, cN(n)=p[11N1](n). For example, Baruah and Ojah  proved that (8)p11314n+j0mod2,for  j=2,3,p11718n+70mod2.Ahmed et al.  investigated the function CN(n) for N= 3 and 4 and proved some congruence properties modulo 5. They also gave alternate proof of some congruence properties due to Chan .

In this paper, we investigate the restricted bipartition function cN(n) for n = 7, 11, and 5l, for any integer l1, and prove some congruence properties modulo 2, 3, and 5 by using Ramanujan’s theta-function identities. In Section 3, we prove congruence properties modulo 2 for c7(n). For example, we prove, for α0, (9)c722α+1n+5·22α+130mod2.In Section 4, we deal with the function c11(n) and establish the notion that if p is an odd prime, 1jp-1, and α0, then (10)c114p2α+1pn+j+p2α+2+120mod2.In Section 5, we show that, for any integer l1, c5l(5n+4)0(mod5). We also prove congruence properties modulo 3 for c15(n). Section 2 is devoted to listing some preliminary results.

2. Preliminary Results

Ramanujan’s general theta function f(a,b) is defined by (11)fa,b=n=0ann+1/2bnn-1/2,ab<1.Three important special cases of f(a,b) are (12)ϕqfq,q=n=-qn2=q2;q25q;q2q4;q42,(13)ψqfq,q3=n=0qnn+1/2=q2;q22q;q,(14)f-qf-q,-q2=n=--1nqn3n+1/2=q;q.Ramanujan also defined the function χ(q) as (15)χq=-q;q2.

Lemma 1.

For any prime p and positive integer m, one has (16)qpm;qpmqm;qmpmodp.

Proof.

It follows easily from the binomial theorem.

Lemma 2 (see [<xref ref-type="bibr" rid="B5">8</xref>, page 315]).

One has (17)ψqψq7=ϕq28ψq8+qψq14ψq2+q6ψq56ϕq4.

Lemma 3.

One has (18)ψqψq7q;q3q7;q73mod2.

Proof.

From (13), we have (19)ψqψq7=q2;q22q14;q142q;qq7;q7.Simplifying (19) using Lemma 1 with p=2, we arrive at the desired result.

Lemma 4 (see [<xref ref-type="bibr" rid="B2">9</xref>, page 286, Equation (<xref ref-type="disp-formula" rid="EEq3.19">60</xref>)]).

One has (20)ϕ-q=q;q2q2;q2,(21)ψ-q=q;qq4;q4q2;q2,(22)fq=q2;q23q;qq4;q4,(23)χq=q2;q22q;qq4;q4.

Lemma 5 (see [<xref ref-type="bibr" rid="B6">10</xref>, page 372]).

One has (24)ψqψq11=ϕq66ψq12+qfq44,q88fq2,q10+q22fq22,q110fq8,q4+q15ψq132ϕq6.

Lemma 6 (see [<xref ref-type="bibr" rid="B5">8</xref>, page 350, Equation (<xref ref-type="disp-formula" rid="EEq2.3">13</xref>)]).

One has (25)fq,q2=ϕ-q3χ-q,where (26)χ-q=q;qq2;q2.

Lemma 7.

One has (27)fq11;q22q11;q11mod2.

Proof.

Employing (20) in Lemma 6 and performing simplification using Lemma 1 with p=2, we obtain (28)fq;q2q;qmod2.Replacing q by q11 in (28), we arrive at the desired result.

Lemma 8 (see [<xref ref-type="bibr" rid="B5">8</xref>, page 51, Example (v)]).

One has (29)fq,q5=ψ-q3χq.

Lemma 9.

One has (30)fq,q5q3;q33q;qmod2.

Proof.

Employing (21) and (23) in Lemma 8, we obtain (31)fq,q5=q3;q3q12;q12q2;q22q6;q6q;qq4;q4.Simplifying (31) using Lemma 1 with p=2, we complete the proof.

Lemma 10 (see [<xref ref-type="bibr" rid="B10">11</xref>, page 5, Equation (<xref ref-type="disp-formula" rid="EEq2.5">15</xref>)]).

One has (32)q3;q33q;q=q4;q43q6;q62q2;q22q12;q12+qq12;q123q4;q4.

Lemma 11 (see [<xref ref-type="bibr" rid="B9">12</xref>, Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M91"><mml:mrow><mml:mn>2.1</mml:mn></mml:mrow></mml:math></inline-formula>]).

For any odd prime p, (33)ψq=k=0p-3/2qk2+k/2fqp2+2k+1p/2,qp2-2k+1p/2+qp2-1/8ψqp2,where, for 0k(p-3)/2, (34)k2+k2p2-18modp.

Lemma 12 (see [<xref ref-type="bibr" rid="B9">12</xref>, Theorem <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M96"><mml:mrow><mml:mn>2.2</mml:mn></mml:mrow></mml:math></inline-formula>]).

For any prime p5, one has (35)f-q=k=-p-1/2k±p-1/6p-1/2-1kq3k2+k/2f-q3p2+6k+1p/2,-q3p2-6k+1p/2+-1±p-1/6qp2-1/24f-qp2,where(36)±p-16p-16,ifp1mod6,-p-16,if  p-1mod6.

Lemma 13 (see [<xref ref-type="bibr" rid="B11">13</xref>]).

One has (37)1q;q=q25;q256q5;q56F4q5+qF3q5+2q2F2q5+3q3Fq5+5q4-3q5F-1q5+2q6F-2q5-q7F-3q5+q8F-4q5,where Fqq-1/5R(q) and R(q) is Rogers-Ramanujan continued fraction defined by (38)Rqq1/51+q1+q21+q31+,q<1.

Lemma 14 (see [<xref ref-type="bibr" rid="B5">8</xref>, page 345, Entry 1(iv)]).

One has (39)q;q3=q9;q934q3W2q3-3q+W-1q3,where W(q)=q-1/3G(q) and G(q) is Ramanujan’s cubic continued fraction defined by (40)Gqq1/31+q+q21+q2+q41+,q<1.

3. Congruence Identities for <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M108"><mml:msub><mml:mrow><mml:mi>c</mml:mi></mml:mrow><mml:mrow><mml:mn>7</mml:mn></mml:mrow></mml:msub><mml:mo mathvariant="bold">(</mml:mo><mml:mi>n</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> Theorem 15.

One has (41)n=0c72n+1qnq;qq7;q7mod2.

Proof.

For N=7 in (1), we have (42)n=0c7nqn=1q;qq7;q7.Employing (19) in (42), we obtain (43)n=0c7nqn=ψqψq7q2;q22q14;q142.Employing Lemma 2 in (43), we obtain (44)n=0c7nqn=1q2;q22q14;q142ϕq28ψq8+qψq14ψq2+q6ψq56ϕq4.Extracting the terms involving q2n+1, dividing by q, and replacing q2 by q in (44), we get (45)n=0c72n+1qn=1q;q2q7;q72ψq7ψq.Employing Lemma 3 in (45), we complete the proof.

Theorem 16.

One has (46)(i)n=0c74n+3qnq2;q2q14;q14mod2,(ii)c78n+70mod2.

Proof.

From Theorem 15, we obtain (47)n=0c72n+1qnq7;q73q;q3q7;q72q;q2mod2.Employing Lemma 3 in (47), we obtain (48)n=0c72n+1qnψqψq7q2;q2q14;q14mod2.Employing Lemma 2 in (48), extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (49)n=0c74n+3qn1q;qq7;q7ψqψq7mod2.Employing Lemma 3 in (49) and performing simplification using Lemma 1 with p=2, we arrive at (i).

All the terms on the right hand side of (i) are of the form q2n. Extracting the terms involving q2n+1 on both sides of (i), we complete the proof of (ii).

Theorem 17.

For all n0, one has

c7(14n+7)0(mod2),

c7(14n+9)0(mod2),

c7(14n+13)0(mod2).

Proof.

Employing (14) in Theorem 15, we obtain (50)n=0c72n+1qnq7;q7n=0-1nqn3n+1/2mod2.Extracting those terms on each side of (50) whose power of q is of the forms 7n+3, 7n+4, and 7n+6 and employing the fact that there exists no integer n such that n(3n+1)/2 is congruent to 3, 4, and 6 modulo 7, we obtain (51)n=0c714n+7q7n+3n=0c714n+9q7n+4n=0c714n+13q7n+60mod2.Now, (i), (ii), and (iii) are obvious from (51).

Theorem 18.

For α1, one has (52)n=0c722α+1n+22α+1+13qnq;qq7;q7mod2.

Proof.

We proceed by induction on α. Extracting the terms involving q2n and replacing q2 by q in Theorem 16(i), we obtain (53)n=0c78n+3qnq;qq7;q7mod2,which corresponds to the case α=1. Assume that the result is true for α=k1, so that (54)n=0c722k+1n+22k+1+13qnq;qq7;q7mod2.Employing Lemma 3 in (54), we obtain (55)n=0c722k+1n+22k+1+13qnψqψq7q;q2q7;q72mod2.Employing Lemma 2 in (55) and extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (56)n=0c722k+12n+1+22k+1+13qnψqψq7q;qq7;q7mod2.Simplifying (56) using Lemmas 3 and 1 with p=2, we obtain (57)n=0c722k+1n+22k+1+1+13qnq2;q2q14;q14mod2.Extracting the terms involving q2n and replacing q2 by q in (57), we obtain (58)n=0c722k+1+1n+22k+1+1+13qnq;qq7;q7mod2,which is the α=k+1 case. Hence, the proof is complete.

Theorem 19.

For α0, one has (59)c722α+1n+5·22α+130mod2.

Proof.

All the terms in the right hand side of (57) are of the form q2n, so, extracting the coefficients of q2n+1 on both sides of (57) and replacing k by α, we obtain (60)c722α+1+1n+5·22α+1+130mod2.Replacing α+1 by α in (60) completes the proof.

Theorem 20.

If any prime p5, -7/p=-1, and α0, then (61)c722α+1p2n+22α+1p3j+p+130mod2,where 1jp-1.

Proof.

Employing Lemma 12 in (52), we obtain (62)n=0c722α+1n+22α+1+13qnk=-p-1/2k±p-1/6p-1/2-1kq3k2+k/2f-q3p2+6k+1p/2,-q3p2-6k+1p/2+-1±p-1/6qp2-1/24f-qp2k=-p-1/2k±p-1/6p-1/2-1mq7·3m2+m/2f-q7·3p2+6m+1p/2,-q7·3p2-6m+1p/2+-1±p-1/6q7·p2-1/24f-q7p2mod2.We consider the congruence (63)3k2+k2+7·3m2+m28p2-824modp,where -(p-1)/2k,m(p-1)/2. The congruence (63) is equivalent to (64)6k+12+76m+120modpand, for (-7/p)=-1, the congruence (64) has unique solution k=m=(±p-1)/6. Extracting terms containing qpn+(p2-1)/3 from both sides of (62) and replacing qp by q, we obtain (65)n=0c722α+1pn+22α+1p2+13qnqp;qpq7p;q7pmod2.Extracting the coefficients of qpn+j, for 1jp-1, on both sides of (65) and performing simplification, we arrive at the desired result.

4. Congruence Identities for <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M191"><mml:msub><mml:mrow><mml:mi>c</mml:mi></mml:mrow><mml:mrow><mml:mn>11</mml:mn></mml:mrow></mml:msub><mml:mo mathvariant="bold">(</mml:mo><mml:mi>n</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> Theorem 21.

One has (66)n=0c114n+1qnq2;q22q;q=ψqmod2.

Proof.

Setting N=11 in (1), we obtain (67)n=0c11nqn=1q;qq11;q11.Employing (13) in (67), we obtain (68)n=0c11nqn=ψqψq11q2;q22q22;q222.Employing Lemma 5 in (68), extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (69)n=0c112n+1qn=1q;q2q11;q112fq22,q44fq,q5+q7ψq66ϕq3.Employing Lemmas 9 and 10 in (69), we find that (70)n=0c112n+1qn1q2;q2q22;q22fq22,q44q4;q43q6;q62q2;q22q12;q12+qq12;q123q4;q4+q7ψq66ϕq3mod2.Extracting the terms involving q2n and replacing q2 by q on both sides of (70) and performing simplification using Lemma 1 with p=2, we obtain (71)n=0c114n+1qn1q;qq11;q11fq11;q22q2;q22mod2.Employing Lemma 7 in (71) and using (13), we complete the proof.

Theorem 22.

For any odd prime p and any integer α0, one has (72)n=0c114p2αn+p2α+12qnψqmod2.

Proof.

We proceed by induction on α. The case α=0 corresponds to the congruence theorem (Theorem 21). Suppose that the theorem holds for α=k0, so that (73)n=0c114p2kn+p2k+12qnψqmod2.Employing Lemma 11 in (73), extracting the terms involving qpn+(p2-1)/8 on both sides of (73), dividing by q(p2-1)/8, and replacing qp by q, we obtain (74)n=0c114p2k+1n+p2k+1+12qnψqpmod2.Extracting the terms containing qpn from both sides of (74) and replacing qp by q, we arrive at (75)n=0c114p2k+1n+p2k+1+12qnψqmod2,which shows that the theorem is true for α=k+1. Hence, the proof is complete.

Theorem 23.

For any odd prime p and integers α0 and 1jp-1, one has (76)c114p2α+1pn+j+p2α+2+120mod2.

Proof.

Extracting the coefficients of qpn+j for 1jp-1 on both sides of (74) and replacing k by α, we arrive at the desired result.

5. Congruence Identities for <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M232"><mml:msub><mml:mrow><mml:mi>c</mml:mi></mml:mrow><mml:mrow><mml:mn>5</mml:mn><mml:mi>l</mml:mi></mml:mrow></mml:msub><mml:mo mathvariant="bold">(</mml:mo><mml:mi>n</mml:mi><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula> Theorem 24.

For any positive integer l, one has (77)c5l5n+40mod5.

Proof.

Setting N=5l in (1), we obtain (78)n=0c5lnqn=1q;qq5l;q5l.Using Lemma 13 in (78) and extracting the terms involving q5n+4, dividing by q4, and replacing q5 by q, we obtain (79)n=0c5l5n+4qn=5q5;q56ql,qlq;q6.The desired result follows easily from (79).

Theorem 25.

For all n0, one has

c15(5n+4)0(mod5),

c15(15n+9)0(mod3),

c15(15n+14)0(mod3).

Proof.

Setting N=15 in (1), we obtain (80)n=0c15nqn=1q;qq15;q15.Employing Lemma 13 in (80), extracting terms involving q5n+4, dividing by q4, and replacing q5 by q, we obtain (81)n=0c155n+4qn=5q5;q56q3;q3q;q6.Now, (i) follows from (81).

Simplifying (81) by using Lemma 1 with p=3, we obtain (82)n=0c155n+4qn2q15;q152q;q3q;q6q;q3q;q3=2q15;q152q;q3q3;q34mod3.Employing Lemma 14 in (82) and performing simplification, we obtain (83)n=0c155n+4qn2q15;q152q9;q93q3;q34q3W2q3+W-1q3mod3.Extracting terms involving q3n+1 and q3n+2 on both sides of (83), we arrive at (ii) and (iii), respectively.

Competing Interests

The authors declare that there are no competing interests regarding the publication of this paper.

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. 25(0241)/15/EMR-II (F. no. 25(5498)/15).

Sinick J. Ramanujan congruences for a class of eta quotients International Journal of Number Theory 2010 6 4 835 847 10.1142/s1793042110003253 MR2661284 2-s2.0-77954203659 Chan H.-C. Ramanujan's cubic continued fraction and an analog of his ‘most beautiful identity’ International Journal of Number Theory 2010 6 3 673 680 10.1142/s1793042110003150 MR2652901 2-s2.0-77952836092 Kim B. A crank analog on a certain kind of partition function arising from the cubic continued fraction Acta Arithmetica 2011 148 1 19 Chan H.-C. Ramanujan's cubic continued fraction and Ramanujan type congruences for a certain partition function International Journal of Number Theory 2010 6 4 819 834 10.1142/S1793042110003241 MR2661283 ZBL1205.11111 2-s2.0-77954201782 Liu J. Wang A. Y. Z. Arithmetic properties of a restricted bipartition function The Electronic Journal of Combinatorics 2015 22 3 1 11 MR3367857 Baruah N. D. Ojah K. K. Analogues of Ramanujan's partition identities and congruences arising from his theta functions and modular equations Ramanujan Journal 2012 28 3 385 407 10.1007/s11139-011-9296-z MR2950513 2-s2.0-84864414658 Ahmed Z. Baruah N. D. Dastidar M. G. New congruences modulo 5 for the number of 2-color partitions Journal of Number Theory 2015 157 184 198 10.1016/j.jnt.2015.05.002 MR3373236 2-s2.0-84936879723 Berndt B. C. Ramanujan's Notebooks, Part III 1991 New York, NY, USA Springer 10.1007/978-1-4612-0965-2 MR1117903 Baruah N. D. Bora J. Saikia N. Some new proofs of modular relations for the Göllnitz-Gordon functions Ramanujan Journal 2008 15 2 281 301 10.1007/s11139-007-9079-8 MR2377581 2-s2.0-43249089302 Berndt B. C. Ramanujan's Notebooks 1998 Part V New York, NY, USA Springer 10.1007/978-1-4612-1624-7 MR1486573 Hirschhorn M. Garvan F. Borwein J. Cubic analogues of the Jacobian theta function θ(z;q) Canadian Journal of Mathematics 1993 45 4 673 694 10.4153/cjm-1993-038-2 Cui S.-P. Gu N. S. S. Arithmetic properties of l-regular partitions Advances in Applied Mathematics 2013 51 4 507 523 10.1016/j.aam.2013.06.002 MR3097009 2-s2.0-84883866261 Hirschhorn M. D. An identity of Ramanujan, and application in q-series from a contemporary perspective Contemporary Mathematics 2000 254 229 234