# Comment on “Some Congruence Properties of a Restricted Bipartition FunctioncN(n)”

Authors: Qing Zou
Journal: International Journal of Analysis
 (2017)

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## Body

## 1. Introduction

This commentary is intended to correct the errors that happened in [1]. Recall that a partition of a positive integer n is a nonincreasing sequence of positive integers whose sum is n. A bipartition (λ,μ) of a positive integer n is a pair of partitions (λ,μ) such that the sum of all the parts is n. Let p-2(n) be the number of bipartitions of n. This function has some beautiful arithmetic properties (see, e.g., [2]). Recently, there is more and more research on the bipartitions with certain restrictions on each partition, for example, [3–7]. In this short review, we consider the number of bipartitions (λ,μ) with the restriction that each part of μ is divisible by N. Let cN(n) denote the number of bipartitions (λ,μ) of n such that part of μ is divisible by N. The generating function for cN(n) is given by [8](1)∑n=0∞cNnqn=1q;q∞qN;qN∞, where(2)a;q∞=∏n=0∞1-aqn.We will mainly focus on the case whenN=5l, where l is any positive integer. In [1], the discussion on congruence identities for c5l(n) was based on an incorrect lemma. We will point out all the incorrect points in that paper in the next section.


## 2. Comments on the Results of Saikia and Boruah

In [1], Saikia and Boruah considered the congruence identities for c5l(n) by employing the following lemma.Lemma
13 of [1]
(3)
1
q
;
q
∞
=
q
25
;
q
25
∞
6
q
5
;
q
5
∞
6
F
4
q
5
+
q
F
3
q
5
+
2
q
2
F
2
q
5
+
3
q
3
F
q
5
+
5
q
4
-
3
q
5
F
-
1
q
5
+
2
q
6
F
-
2
q
5
-
q
7
F
-
3
q
5
+
q
8
F
-
4
q
5
,
where
(4)
R
q
≔
q
1
/
5
1
+
q
1
+
q
2
1
+
q
3
1
+
⋯
,
q
<
1
is the Rogers-Ramanujan continued fraction and
(5)
F
q
≔
q
-
1
/
5
R
q
.However, the lemma of this version is wrong, The correct version should be as follows.Lemma 1.
LetF(q) be defined as above: (6)1q;q∞=q25;q25∞5q5;q5∞61F4q5+qF3q5+2q2F2q5+3q3Fq5+5q4-3q5Fq5+2q6F2q5-q7F3q5+q8F4q5.There are several proofs of this famous identity (see, e.g., [9, 10]).In [1], all the discussion in Section 5 was based on Lemma 13 of [1] which has been shown above.Since Lemma13 of [1] is not correct, the discussion in Section 5 is not correct.First, we can say that the conclusion shown in Theorem24 of [1], which says(7)c5l5n+4≡0mod5, is correct. However, in the course of the proof, there is one mistake. The equation(8)∑n=0∞c5l5n+4qn=5q5;q5∞6ql;ql∞q;q∞6, which is equation (79) of [1], is not correct. It should be(9)∑n=0∞c5l5n+4qn=5q5;q5∞5ql;ql∞q;q∞6.The first result of Theorem25 of [1] is also correct, but the proof has a problem. In the course of the proof, equation (81) of [1] should be(10)∑n=0∞c155n+4qn=5q5;q5∞5q3;q3∞q;q∞6, rather than(11)∑n=0∞c155n+4qn=5q5;q5∞6q3;q3∞q;q∞6.However, the second conclusion and the third conclusion of Theorem25 of [1] are not correct. The proof of Theorem 5 (ii, iii) of [1] relies on the claim that(12)q5;q5∞6≡q15;q15∞2mod3. However, as we have pointed out above, we do not have the factor q5;q5∞6 in ∑n=0∞c15(5n+4)qn. This is the reason why the second conclusion and the third conclusion of Theorem 25 of [1] cannot hold.Actually, we do have some 3-dissection formulas to find how∑n=0∞c15(15n+9)qn and ∑n=0∞c15(15n+14)qn modulo-3 look like. Let us show them now.Thanks to [11], we have(13)q5;q5∞2q;q∞2≡q3;q3∞4+qq3;q3∞2q15;q15∞2-q2q15;q15∞4mod3. If we let a=-q,b=q2 and n=3 in Entry 31 of [12, p. 48], we can obtain(14)q;q∞=J12,27-qJ6,27-q2J3,27, where(15)Ja,b=qa;qb∞qb-a;qb∞qb;qb∞. According to these two formulas, we get that (16)∑n=0∞c155n+4qn≡5q5;q5∞5q3;q3∞q;q∞6≡2·q5;q5∞3q5;q5∞2q3;q3∞q;q∞6≡2q15;q15∞q5;q5∞2q;q∞2q3;q3∞4·q;q∞≡2·q15;q15∞+qq15;q15∞3q3;q3∞2-q2q15;q15∞5q3;q3∞4×J12,27-qJ6,27-q2J3,27≡2·J12,27q15;q15∞+qJ12,27q15;q15∞3q3;q3∞2-q2J12,27q15;q15∞5q3;q3∞4-qJ6,27q15;q15∞-q2J6,27q15;q15∞3q3;q3∞2+q3J6,27q15;q15∞5q3;q3∞4-q2J3,27q15;q15∞-q3J3,27q15;q15∞3q3;q3∞2+q4J3,27q15;q15∞5q3;q3∞4mod3.When extracting terms involving q3n+1, dividing by q and replacing q3 by q, we obtain (17)∑n=0∞c1515n+9qn≡2·J4,9q5;q5∞3q;q∞2-J2,9q5;q5∞+qJ1,9q5;q5∞5q;q∞4≡J2,9q5;q5∞+2J4,9q5;q5∞3q;q∞2+2qJ1,9q5;q5∞5q;q∞4mod3. When extracting terms involving q3n+2, dividing by q2 and replacing q3 by q, we obtain (18)∑n=0∞c1515n+14qn≡-2·J4,9q5;q5∞5q;q∞4+J2,9q5;q5∞3q;q∞3+J1,9q5;q5∞≡J4,9q5;q5∞5q;q∞4+J2,9q5;q5∞3q;q∞3+J1,9q5;q5∞mod3. Up to now, we have obtained the forms of ∑n=0∞c15(15n+9)qn and ∑n=0∞c15(15n+14)qn modulo-3. However, we cannot get any congruence properties of c15(15n+9) or c15(15n+14) by the forms of ∑n=0∞c15(15n+9)qn and ∑n=0∞c15(15n+14)qn modulo-3 above.



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*Source: 9721671-2017-07-13.xml*