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 l≥1, by employing Ramanujan’s theta-function identities.
Council of Scientific and Industrial Research of India25(5498)/151. 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) [1] is given by (1)∑n=0∞cNnqn=1q;q∞qN;qN∞,where (2)a;q∞=∏n=0∞1-aqn.The partition function cN(n) is first studied by Chan [2] for the particular case N=2 by considering the function c2(n) defined by (3)∑n=0∞c2nqn=1q;q∞q2;q2∞.Chan [2] proved that, for n≥0, (4)c23n+2≡0mod3.Kim [3] gave a combinatorial interpretation (4). In a subsequent paper, Chan [4] showed that, for k≥1 and n≥0, (5)c23kn+sk≡0mod3k+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 [4] asked whether there are any other congruence properties of the following form: c2(ln+k)≡0modl, where l is prime and 0≤k≤l. Sinick [1] answered Chan’s question in negative by considering restricted bipartition function cN(n) defined in (1). Liu and Wang [5] established several infinite families of congruence properties for c5(n) modulo 3. For example, they proved that (6)c532α+1n+7·32α+14≡0mod3,α≥1,n≥0.
Baruah and Ojah [6] 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=0∞pcldmnqn=1qc;qc∞lqd;qd∞mand using Ramanujan’s modular equations. Clearly, cN(n)=p[11N1](n). For example, Baruah and Ojah [6] proved that (8)p11314n+j≡0mod2,for j=2,3,p11718n+7≡0mod2.Ahmed et al. [7] 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 [2].
In this paper, we investigate the restricted bipartition function cN(n) for n = 7, 11, and 5l, for any integer l≥1, 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 α≥0, (9)c722α+1n+5·22α+13≡0mod2.In Section 4, we deal with the function c11(n) and establish the notion that if p is an odd prime, 1≤j≤p-1, and α≥0, then (10)c114p2α+1pn+j+p2α+2+12≡0mod2.In Section 5, we show that, for any integer l≥1, 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=0∞ann+1/2bnn-1/2,ab<1.Three important special cases of f(a,b) are (12)ϕq≔fq,q=∑n=-∞∞qn2=q2;q2∞5q;q∞2q4;q4∞2,(13)ψq≔fq,q3=∑n=0∞qnn+1/2=q2;q2∞2q;q∞,(14)f-q≔f-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;qpm∞≡qm;qm∞pmodp.
Proof.
It follows easily from the binomial theorem.
Lemma 2 (see [8, page 315]).
One has (17)ψqψq7=ϕq28ψq8+qψq14ψq2+q6ψq56ϕq4.
Lemma 3.
One has (18)ψqψq7≡q;q∞3q7;q7∞3mod2.
Proof.
From (13), we have (19)ψqψq7=q2;q2∞2q14;q14∞2q;q∞q7;q7∞.Simplifying (19) using Lemma 1 with p=2, we arrive at the desired result.
Lemma 4 (see [9, page 286, Equation (60)]).
One has (20)ϕ-q=q;q∞2q2;q2∞,(21)ψ-q=q;q∞q4;q4∞q2;q2∞,(22)fq=q2;q2∞3q;q∞q4;q4∞,(23)χq=q2;q2∞2q;q∞q4;q4∞.
Lemma 5 (see [10, page 372]).
One has (24)ψqψq11=ϕq66ψq12+qfq44,q88fq2,q10+q22fq22,q110fq8,q4+q15ψq132ϕq6.
Lemma 6 (see [8, page 350, Equation (13)]).
One has (25)fq,q2=ϕ-q3χ-q,where (26)χ-q=q;q∞q2;q2∞.
Lemma 7.
One has (27)fq11;q22≡q11;q11∞mod2.
Proof.
Employing (20) in Lemma 6 and performing simplification using Lemma 1 with p=2, we obtain (28)fq;q2≡q;q∞mod2.Replacing q by q11 in (28), we arrive at the desired result.
Lemma 8 (see [8, page 51, Example (v)]).
One has (29)fq,q5=ψ-q3χq.
Lemma 9.
One has (30)fq,q5≡q3;q3∞3q;q∞mod2.
Proof.
Employing (21) and (23) in Lemma 8, we obtain (31)fq,q5=q3;q3∞q12;q12∞q2;q2∞2q6;q6∞q;q∞q4;q4∞.Simplifying (31) using Lemma 1 with p=2, we complete the proof.
Lemma 10 (see [11, page 5, Equation (15)]).
One has (32)q3;q3∞3q;q∞=q4;q4∞3q6;q6∞2q2;q2∞2q12;q12∞+qq12;q12∞3q4;q4∞.
Lemma 11 (see [12, Theorem 2.1]).
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 0≤k≤(p-3)/2, (34)k2+k2≢p2-18modp.
Lemma 12 (see [12, Theorem 2.2]).
For any prime p≥5, 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-16≔p-16,ifp≡1mod6,-p-16,if p≡-1mod6.
Lemma 13 (see [13]).
One has (37)1q;q∞=q25;q25∞6q5;q5∞6F4q5+qF3q5+2q2F2q5+3q3Fq5+5q4-3q5F-1q5+2q6F-2q5-q7F-3q5+q8F-4q5,where Fq≔q-1/5R(q) and R(q) is Rogers-Ramanujan continued fraction defined by (38)Rq≔q1/51+q1+q21+q31+⋯,q<1.
Lemma 14 (see [8, page 345, Entry 1(iv)]).
One has (39)q;q∞3=q9;q9∞34q3W2q3-3q+W-1q3,where W(q)=q-1/3G(q) and G(q) is Ramanujan’s cubic continued fraction defined by (40)Gq≔q1/31+q+q21+q2+q41+⋯,q<1.
3. Congruence Identities for c7(n)Theorem 15.
One has (41)∑n=0∞c72n+1qn≡q;q∞q7;q7∞mod2.
Proof.
For N=7 in (1), we have (42)∑n=0∞c7nqn=1q;q∞q7;q7∞.Employing (19) in (42), we obtain (43)∑n=0∞c7nqn=ψqψq7q2;q2∞2q14;q14∞2.Employing Lemma 2 in (43), we obtain (44)∑n=0∞c7nqn=1q2;q2∞2q14;q14∞2ϕq28ψq8+qψq14ψq2+q6ψq56ϕq4.Extracting the terms involving q2n+1, dividing by q, and replacing q2 by q in (44), we get (45)∑n=0∞c72n+1qn=1q;q∞2q7;q7∞2ψq7ψq.Employing Lemma 3 in (45), we complete the proof.
Theorem 16.
One has (46)(i)∑n=0∞c74n+3qn≡q2;q2∞q14;q14∞mod2,(ii)c78n+7≡0mod2.
Proof.
From Theorem 15, we obtain (47)∑n=0∞c72n+1qn≡q7;q7∞3q;q∞3q7;q7∞2q;q∞2mod2.Employing Lemma 3 in (47), we obtain (48)∑n=0∞c72n+1qn≡ψqψq7q2;q2∞q14;q14∞mod2.Employing Lemma 2 in (48), extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (49)∑n=0∞c74n+3qn≡1q;q∞q7;q7∞ψqψq7mod2.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 n≥0, 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=0∞c72n+1qn≡q7;q7∞∑n=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=0∞c714n+7q7n+3≡∑n=0∞c714n+9q7n+4≡∑n=0∞c714n+13q7n+6≡0mod2.Now, (i), (ii), and (iii) are obvious from (51).
Theorem 18.
For α≥1, one has (52)∑n=0∞c722α+1n+22α+1+13qn≡q;q∞q7;q7∞mod2.
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=0∞c78n+3qn≡q;q∞q7;q7∞mod2,which corresponds to the case α=1. Assume that the result is true for α=k≥1, so that (54)∑n=0∞c722k+1n+22k+1+13qn≡q;q∞q7;q7∞mod2.Employing Lemma 3 in (54), we obtain (55)∑n=0∞c722k+1n+22k+1+13qn≡ψqψq7q;q∞2q7;q7∞2mod2.Employing Lemma 2 in (55) and extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (56)∑n=0∞c722k+12n+1+22k+1+13qn≡ψqψq7q;q∞q7;q7∞mod2.Simplifying (56) using Lemmas 3 and 1 with p=2, we obtain (57)∑n=0∞c722k+1n+22k+1+1+13qn≡q2;q2∞q14;q14∞mod2.Extracting the terms involving q2n and replacing q2 by q in (57), we obtain (58)∑n=0∞c722k+1+1n+22k+1+1+13qn≡q;q∞q7;q7∞mod2,which is the α=k+1 case. Hence, the proof is complete.
Theorem 19.
For α≥0, one has (59)c722α+1n+5·22α+13≡0mod2.
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+13≡0mod2.Replacing α+1 by α in (60) completes the proof.
Theorem 20.
If any prime p≥5, -7/p=-1, and α≥0, then (61)c722α+1p2n+22α+1p3j+p+13≡0mod2,where 1≤j≤p-1.
Proof.
Employing Lemma 12 in (52), we obtain (62)∑n=0∞c722α+1n+22α+1+13qn≡∑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∑k=-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+m2≡8p2-824modp,where -(p-1)/2≤k,m≤(p-1)/2. The congruence (63) is equivalent to (64)6k+12+76m+12≡0modpand, 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=0∞c722α+1pn+22α+1p2+13qn≡qp;qp∞q7p;q7pmod2.Extracting the coefficients of qpn+j, for 1≤j≤p-1, on both sides of (65) and performing simplification, we arrive at the desired result.
4. Congruence Identities for c11(n)Theorem 21.
One has (66)∑n=0∞c114n+1qn≡q2;q2∞2q;q∞=ψqmod2.
Proof.
Setting N=11 in (1), we obtain (67)∑n=0∞c11nqn=1q;q∞q11;q11∞.Employing (13) in (67), we obtain (68)∑n=0∞c11nqn=ψqψq11q2;q2∞2q22;q22∞2.Employing Lemma 5 in (68), extracting the terms involving q2n+1, dividing by q, and replacing q2 by q, we obtain (69)∑n=0∞c112n+1qn=1q;q∞2q11;q11∞2fq22,q44fq,q5+q7ψq66ϕq3.Employing Lemmas 9 and 10 in (69), we find that (70)∑n=0∞c112n+1qn≡1q2;q2∞q22;q22∞fq22,q44q4;q4∞3q6;q6∞2q2;q2∞2q12;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=0∞c114n+1qn≡1q;q∞q11;q11∞fq11;q22q2;q2∞2mod2.Employing Lemma 7 in (71) and using (13), we complete the proof.
Theorem 22.
For any odd prime p and any integer α≥0, one has (72)∑n=0∞c114p2αn+p2α+12qn≡ψqmod2.
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=k≥0, so that (73)∑n=0∞c114p2kn+p2k+12qn≡ψqmod2.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=0∞c114p2k+1n+p2k+1+12qn≡ψqpmod2.Extracting the terms containing qpn from both sides of (74) and replacing qp by q, we arrive at (75)∑n=0∞c114p2k+1n+p2k+1+12qn≡ψqmod2,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 α≥0 and 1≤j≤p-1, one has (76)c114p2α+1pn+j+p2α+2+12≡0mod2.
Proof.
Extracting the coefficients of qpn+j for 1≤j≤p-1 on both sides of (74) and replacing k by α, we arrive at the desired result.
5. Congruence Identities for c5l(n)Theorem 24.
For any positive integer l, one has (77)c5l5n+4≡0mod5.
Proof.
Setting N=5l in (1), we obtain (78)∑n=0∞c5lnqn=1q;q∞q5l;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=0∞c5l5n+4qn=5q5;q5∞6ql,ql∞q;q∞6.The desired result follows easily from (79).
Theorem 25.
For all n≥0, 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=0∞c15nqn=1q;q∞q15;q15∞.Employing Lemma 13 in (80), extracting terms involving q5n+4, dividing by q4, and replacing q5 by q, we obtain (81)∑n=0∞c155n+4qn=5q5;q5∞6q3;q3∞q;q∞6.Now, (i) follows from (81).
Simplifying (81) by using Lemma 1 with p=3, we obtain (82)∑n=0∞c155n+4qn≡2q15;q15∞2q;q∞3q;q∞6q;q∞3q;q∞3=2q15;q15∞2q;q∞3q3;q3∞4mod3.Employing Lemma 14 in (82) and performing simplification, we obtain (83)∑n=0∞c155n+4qn≡2q15;q15∞2q9;q9∞3q3;q3∞4q3W2q3+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).
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