# Proof of a limited version of Mao’s partition rank inequality using a theta function identity

- Rupam Barman
^{1}Email author and - Archit Pal Singh Sachdeva
^{2}

**2**:22

**DOI: **10.1007/s40993-016-0055-5

© The Author(s) 2016

**Received: **20 May 2016

**Accepted: **25 August 2016

**Published: **10 October 2016

## Abstract

Ramanujan’s congruence \(p(5k+4) \equiv 0 \pmod 5\) led Dyson (Eureka 8:10–15, 1944) to define a measure “rank”, and then conjectured that \(p(5k+4)\) partitions of \(5k+4\) could be divided into subclasses with equal cardinality to give a direct proof of Ramanujan’s congruence. The notion of rank was extended to rank differences by Atkin and Swinnerton-Dyer (Some properties of partitions 4:84–106, 1954), who proved Dyson’s conjecture. More recently, Mao (Number Theory 133:3678–3702, 2013) proved several equalities and inequalities, leaving some as conjectures, for rank differences for partitions modulo 10 and for \(M_2\)-rank differences for partitions with no repeated odd parts modulo 6 and 10 (Mao in Ramanujan J 37:391–419, 2015). Alwaise et al. (Ramanujan J. doi:10.1007/s11139-016-9789-x, 2016) proved four of Mao’s conjectured inequalities, while leaving three open. Here, we prove a limited version of one of the inequalities conjectured by Mao.

### Keywords

Partitions Ranks Rank differences Theta functions### Mathematics Subject Classification

Primary 11P83## 1 Introduction and results

A *partition* of a positive integer *n* is a way of writing *n* as a sum of positive integers, usually written in non-increasing order of the summands or parts of the partition. The number of partitions of *n* is denoted by *p*(*n*). For a partition \(\lambda \), we denote the number of parts in the partition as \(n(\lambda )\) and the largest part as \(l(\lambda )\).

*s*in each residue class modulo 5 or 7, respectively

*N*(

*s*,

*m*,

*n*) denotes the number of partitions of

*n*with rank

*s*modulo

*m*. Atkin and Swinnerton-Dyer [2] proved Dyson’s conjecture by finding the generating functions for the rank differences \(N(s, m, mk+d) - N(s, m, mk+d)\) for \(k = 5, 7\). They obtained several other interesting identities apart from Ramanujan’s congruences.

Lovejoy and Osburn [5] expanded on the work by Atkin and Swinnerton-Dyer to find rank differences for overpartitions and \(M_2\)-rank differences for partitions without repeated odd parts, which is defined for such a partition \(\lambda \) by \( \left\lceil \frac{l(\lambda )}{2} \right\rceil - n(\lambda ).\) The corresponding count for number of partitions of *n* with no repeated odd parts having its \(M_2\)-rank congruent to *s* modulo *m* is given by \(N_2(s, m, n)\). They obtained all the rank difference formulas corresponding to \(m = 3, 5\).

### Conjecture 1.1

Alwaise et al. [1, Theorem 1.3] proved four of these seven inequalities conjectured by Mao, namely (1), (2), (3), and (4) by using elementary methods based on the number of solutions of Diophantine equations solving for the exponents in the generating functions in the corresponding rank differences. They also observed that in (2), the strict inequality holds. However, their methods weren’t strong enough to prove the remaining three conjectures, which are still open. Here, we prove a limited version of (7).

### Theorem 1.2

## 2 Preliminaries

*q*-series notation is employed which is defined as

*q*-series to prove equalities between expressions. For \(a, b \in \mathbb {Z}\), \(c \in \mathbb {C}\), and for \(k \in \mathbb {N}\), we have

### Theorem 2.1

*f*(

*a*,

*b*) is defined as

### Theorem 2.2

## 3 Proof of Theorem 1.2

We denote \(d(n) := N_2(0, 6, n) + N_2(1, 6, n) - N_2(2, 6, n) - N_2(3, 6, n)\) for simplicity. We will show that the generating function \(\sum _{n\ge 0}d(3n+2)q^n\) has strictly positive coefficients for all \(n \not \equiv 2 \pmod {3}\). We first compute the generating function \(\sum _{n\ge 0}d(3n+2)q^n\) using Theorem 2.1.

### Proposition 3.1

### Proof

*q*-powers which are 1 modulo 3, and the third and fourth only have

*q*-powers which are 2 modulo 3. Now, including only exponents congruent to 2 modulo 3 in the original generating function, and then letting \(q \mapsto q^{\frac{1}{3}}\), we deduce that

### Remark 3.2

Note that the while there is a *q* in the denominator of the common factor above, it is canceled because the constant term of the expression inside the parentheses in (17) is zero.

We will also need the following lemma which will tie together the proof.

### Lemma 3.3

### Proof

*q*-series and then use (11) to cancel common factors in both numerator and denominator. We find that

*q*-series by multiplying the missing factors in both numerator and denominator, and simplify the expression using (15 16) and Theorem 2.2 as follows:

We now prove our result Theorem 1.2.

### Proof of Theorem 1.2

*q*-series inside the parentheses in (17) are \(0 \pmod 3\). Hence,

*f*(

*x*). It now suffices to show that all coefficients of \(\frac{\varphi ^2(q) + \varphi ^2(q^3)}{J_{3,12}}\) are positive. This follows as

*q*power generated in the way described must have a positive coefficient, and so additional terms that arise with the same power would only add to the size of the coefficient. This completes our proof for Theorem 1.2 \(\square \)

## 4 Conclusion and remarks

The method employed by Alwaise et al. [1] doesn’t work for this inequality because the expression inside the parentheses in Proposition 3.1 does seem to have negative coefficients for an infinite number of coefficients.

This result is limited to \(3n+2\) when \(3 \not \mid n+1\), but computational evidence suggests that \( \frac{1}{1-q^{12}} \left( \frac{J_{2, 12}J_{6, 12}^2J_{12}^3}{2J_{1, 12}^2J_{5, 12}^2} - \sum \nolimits _{n=-\infty }^{\infty } \frac{(-1)^nq^{6n^2+3n}}{1+q^{6n}} \right) \) has non-negative coefficients, and given the simplification with help of Lemma 3.3, a stronger version of the method used in along with using properties of \(\varphi ^2(q)\), in which the coefficient of \(q^n\) counts number of Diophantine solutions to \(a^2 + b^2 = n\) might aid in proving the inequality when \(3 \mid n + 1\).

## Declarations

### Open Access

This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

### Acknowledgements

The authors would like to thank the anonymous referee for helpful suggestions and comments.

## Authors’ Affiliations

## References

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