# Some Congruence Properties of a Restricted Bipartition FunctioncN(n)

Authors: Nipen Saikia; Chayanika Boruah
Journal: International Journal of Analysis
 (2016)

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## Abstract

LetcN(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.


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## Body

## 1. Introduction

A bipartition of a positive integern 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≡0mod⁡3.Kim [3] gave a combinatorial interpretation (4). In a subsequent paper, Chan [4] showed that, for k≥1 and n≥0, (5)c23kn+sk≡0mod⁡3k+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)≡0mod⁡l, 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≡0mod⁡3,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≡0mod⁡2,forj=2,3,p11718n+7≡0mod⁡2.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 functioncN(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≡0mod⁡2.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≡0mod⁡2.In Section 5, we show that, for any integer l≥1, c5l(5n+4)≡0(mod⁡5). 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 functionf(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 primep and positive integer m, one has (16)qpm;qpm∞≡qm;qm∞pmod⁡p.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∞3mod⁡2.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∞mod⁡2.Proof.
Employing (20) in Lemma 6 and performing simplification using Lemma 1 with p=2, we obtain (28)fq;q2≡q;q∞mod⁡2.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∞mod⁡2.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 primep, (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≢p2-18mod⁡p.Lemma 12 (see [12, Theorem 2.2]).
For any primep≥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≡1mod⁡6,-p-16,ifp≡-1mod⁡6.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 forc7(n)

Theorem 15.
One has(41)∑n=0∞c72n+1qn≡q;q∞q7;q7∞mod⁡2.Proof.
ForN=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∞mod⁡2,(ii)c78n+7≡0mod⁡2.Proof.
From Theorem15, we obtain (47)∑n=0∞c72n+1qn≡q7;q7∞3q;q∞3q7;q7∞2q;q∞2mod⁡2.Employing Lemma 3 in (47), we obtain (48)∑n=0∞c72n+1qn≡ψqψq7q2;q2∞q14;q14∞mod⁡2.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∞ψqψq7mod⁡2.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 formq2n. Extracting the terms involving q2n+1 on both sides of (i), we complete the proof of (ii).Theorem 17.
For alln≥0, one has (i)
c7(14n+7)≡0(mod⁡2),(ii)
c7(14n+9)≡0(mod⁡2),(iii)
c7(14n+13)≡0(mod⁡2).Proof.
Employing (14) in Theorem 15, we obtain (50)∑n=0∞c72n+1qn≡q7;q7∞∑n=0∞-1nqn3n+1/2mod⁡2.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≡0mod⁡2.Now, (i), (ii), and (iii) are obvious from (51).Theorem 18.
Forα≥1, one has (52)∑n=0∞c722α+1n+22α+1+13qn≡q;q∞q7;q7∞mod⁡2.Proof.
We proceed by induction onα. 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∞mod⁡2,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∞mod⁡2.Employing Lemma 3 in (54), we obtain (55)∑n=0∞c722k+1n+22k+1+13qn≡ψqψq7q;q∞2q7;q7∞2mod⁡2.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∞mod⁡2.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∞mod⁡2.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∞mod⁡2,which is the α=k+1 case. Hence, the proof is complete.Theorem 19.
Forα≥0, one has (59)c722α+1n+5·22α+13≡0mod⁡2.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≡0mod⁡2.Replacing α+1 by α in (60) completes the proof.Theorem 20.
If any primep≥5, -7/p=-1, and α≥0, then (61)c722α+1p2n+22α+1p3j+p+13≡0mod⁡2,where 1≤j≤p-1.Proof.
Employing Lemma12 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-q7p2mod⁡2.We consider the congruence (63)3k2+k2+7·3m2+m2≡8p2-824mod⁡p,where -(p-1)/2≤k,m≤(p-1)/2. The congruence (63) is equivalent to (64)6k+12+76m+12≡0mod⁡pand, 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;q7pmod⁡2.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 forc11(n)

Theorem 21.
One has(66)∑n=0∞c114n+1qn≡q2;q2∞2q;q∞=ψqmod⁡2.Proof.
SettingN=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∞+q7ψq66ϕq3mod⁡2.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∞2mod⁡2.Employing Lemma 7 in (71) and using (13), we complete the proof.Theorem 22.
For any odd primep and any integer α≥0, one has (72)∑n=0∞c114p2αn+p2α+12qn≡ψqmod⁡2.Proof.
We proceed by induction onα. 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≡ψqmod⁡2.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≡ψqpmod⁡2.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≡ψqmod⁡2,which shows that the theorem is true for α=k+1. Hence, the proof is complete.Theorem 23.
For any odd primep and integers α≥0 and 1≤j≤p-1, one has (76)c114p2α+1pn+j+p2α+2+12≡0mod⁡2.Proof.
Extracting the coefficients ofqpn+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 forc5l(n)

Theorem 24.
For any positive integerl, one has (77)c5l5n+4≡0mod⁡5.Proof.
SettingN=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 alln≥0, one has (i)
c15(5n+4)≡0(mod⁡5),(ii)
c15(15n+9)≡0(mod⁡3),(iii)
c15(15n+14)≡0(mod⁡3).Proof.
SettingN=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∞4mod⁡3.Employing Lemma 14 in (82) and performing simplification, we obtain (83)∑n=0∞c155n+4qn≡2q15;q15∞2q9;q9∞3q3;q3∞4q3W2q3+W-1q3mod⁡3.Extracting terms involving q3n+1 and q3n+2 on both sides of (83), we arrive at (ii) and (iii), respectively.



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*Source: 9037692-2016-08-08.xml*