Parity Indices and Two-Line Matrix Representation for Partitions

ABSTRACT

In this work we present a solution for Andrews's Problem 5¹ by establishing a bijection between the sets and defined in Fine's Theorem⁸ and the sets of partitions indexed by their lower parity index¹. We also solve Andrews's Problem 6, conjectured in¹.

Keywords:
partition; parity index; mock theta function

RESUMO

Neste trabalho apresentamos uma solução para o Problema 5 de Andrews¹ por meio de uma bijeção entre os conjuntos and definidos no Teorema de Fine⁽⁸ e os conjuntos das partições indexadas pelo índice de paridade inferior¹. Também resolvemos o Problema 6 de Andrews, conjecturado em¹.

Palavras-chave:
partição; índice de paridade; mock theta function

1 INTRODUCTION

In this section we present one of Fine's renowned theorems. Its combinatorial proof can be found in⁸. That proof does not use complicated analytic formulae and its popularization occurred because of its elegance and simplicity, being similar to the famous involutive proof by Franklin. Franklin's involution was used in many occasions to prove several refinements of Euler's Pentagonal Number Theorem⁷.

Theorem 1.1 (Fine). Letandbe the sets of partitions of n into distinct parts, such that the largest part a(λ) = λ₁is even and odd, respectively.Then

Problems that involve parity are related to classic partition identities such as Euler's, Rogers's, Ramanujan's and Gordon's identities. The theorem we just stated is an example of a problem that involves the concept of parity in the study of identities related to Partition Theory. Several authors studied questions that involve parity in partitions,^{4), (5), (6}. In¹, Andrews described several results on this subject. At the end of the same work he enumerated fifteen problems. Problems 1, 2 and 3 were solved in¹⁰. Problems 9 and 10 were solved in¹², with a purely analytic solution. In¹¹ other solutions for problems 9 and 10 are presented, and in the same paper a proof for problem 5 is given, which is based on some results related to generating functions for partitions and q-series. In⁹ there is also a solution to problem 5, based on the involutive proof by Franklin, 1881.

In section 2 we present a solution to Problem 5 establishing a bijection between the sets and defined on Fine's Theorem and the sets of partitions indexed by their lower odd parity index¹. In section 4 we present a solution for Problem 6, conjectured in¹

We adopt the standard q-series notation: for each integern > 0 define

(a)_n = (a;q)_n = (1 - aqj) and (a; q)0 = 1

2 LOWER INDEX AND FINE'S THEOREM

Definition 2.1. Let λ be a partition λ = λ1 + λ2 + ... + λj , where

λ1 ≥ λ2 ≥ ... ≥ λj > 0

The lower odd parity index of λ, denoted by Ilo, is defined as the maximum length of the subsequences of {λ1, λ2, ... ,λj} whose terms alternate in parity beginning with an odd λi.

Example. Consider λ = 8 + 7 + 7 + 6 + 5 + 4 + 4 + 2 + 2 + 1 a partition of 46. Thus, Ilo(λ) = 6.

We will denote by po(r, m, n) the number of partitions of n in m distinct parts with Ilo(λ) equals to r. Besides, Po(y, x; q) is the generating function for partitions λ of n into m distinct parts with Ilo(λ) = r, that is,

The next result, given in(¹), provides an explicit formula for Po(y, x; q).

Theorem 2.1. The generating function for partitions enumerated by po(r, m, n) is given by:

Evaluating (2.1) and (2.2) in y = −1 and x = 1 we have:

Problem 5 in(¹) asks for a combinatorial proof for:

The following lemma is very useful to solve Problem 5.

Lemma 2.1. Consider λ = λ1 + λ2 + ... + λm a partition of the positive integer N in m distinct parts, with Ilo(λ) = r and λ1 > λ2 > ... > λm Then r is even (odd) if, and only if, λ1 is even (odd).

Proof. Let λ = λ1 + λ2 + ... + λm be a partition of the positive integer N in m distinct parts, with Ilo(λ) = r, λ1 > λ2 > ... > λm and r even. Hence there exists an increasing subsequence of {λ1, λ2, ..., λm}, whose length is equal to r, (λi + r − 1, ...λi + 1, λi) with an odd λi +r − 1. Because λi + r − 1 is odd and r is even, then λi is even. In the same way, we can conclude that if λi is even, then r is even, because otherwise λi + r − 1 would be even, a contradiction.

If λ1 = λi then the proof is over. Suppose, then, that λ1 > λi.

For all i > j, we have that λj is even. If λj is odd we have one more term of the sequence that alternates parity with λi, which is even, and so the subsequence has length r + 1. However Ilo(λ) = r. Therefore λj is even. It follows from taking j = 1 that λ1 is even.

Conversely, suppose that λ1 is even. Suppose by absurd that Ilo(λ) = r with an odd r, then there exists a subsequence of λ, (λi + r − 1, ...λi + 1, λi), with an odd λi + r − 1, and r being the maximum with this property. Hence λi is odd. We have that λ1 ≠ λi, because λ1 is even. Since λi must alternate parity with λ1 we have that Ilo(λ) ≥ r + 1. Therefore r if even. It is similar in the case r is an odd number.

We will denote the set of partitions of a positive integer N, with distinct parts, whose lower odd parity index is Ilo(λ) = r by: . When r is even we use the notation , and when r is odd, the notation .

It follows from Lemma 2.1 that given a partition λ = λ1 + λ2 + ... + λm of the positive integer N in m distinct parts with Ilo(λ) = r and λ1 > λ2 > ... > λm, then:

λ ∈ if, and only if, λ ∈

In the same way, it follows from Lemma 2.1 that

λ ∈ if, and only if, λ ∈

This result establishes a 1-1 correspondence between the sets defined on Fine's Theorem and the set of partitions λ indexed by Ilo(λ) = r. So we can conclude that

and

It follows from (2.4) and (2.5) that

Therefore it follows from Theorem 1.1 that

Therefore we have another solution for Andrews's Problem 5,(¹).

Example. Let N = 11, 12, m denote the number of distinct parts of a partition λ of N and r be its lower odd parity index.

Table 1
Distinct partitions of 11 and 12.

11	12
10 + 1	11 + 1
9 + 2	10 + 2
8 + 3	9 + 3
7 + 4	8 + 4
6 + 5	7 + 5
8 + 2 + 1	9 + 2 + 1
7 + 3 + 1	8 + 3 + 1
6 + 4 + 1	7 + 4 + 1
6 + 3 + 2	7 + 3 + 2
5 + 4 + 2	6 + 5 + 1
5 + 3 + 2 + 1	6 + 4 + 2
	5 + 4 + 3
	6 + 3 + 2 + 1
	5 + 4 + 2 + 1

It follows from equation (2.7)

po(r, m, 11)(-1)r = 0.

and

po(r, m, 12)(-1)r = - 1.

In particular see cases n = 11 and n = 12, respectively in Table 2.

Table 2
The values of po(r, m, N), where N = 11 and N = 12, respectively.

m	r	po(r, m, 11)	po(r, m, 11)(-1)r	m	r	po(r, m, 12)	po(r, m, 12)(-1)r
1	0	0	0	1	0	1	1
1	1	1	-1	1	1	0	0
2	0	0	0	2	0	2	2
2	1	2	-2	2	1	3	-3
2	2	3	3	2	2	0	0
3	0	0	0	3	0	1	1
3	1	2	-2	3	1	1	-1
3	2	3	3	3	2	2	2
3	3	0	0	3	3	3	-3
4	0	0	0	4	0	0	0
4	1	0	0	4	1	0	0
4	2	0	0	4	2	0	0
4	3	1	-1	4	3	1	-1
4	4	0	0	4	4	1	1

3 PROBLEM 6 GIVEN IN(¹)

The main goal of this section is to present a solution to Andrews's Problem 6,(¹).

Definition 3.1. Given a partition λ of a positive integer n = λ1 + λ2 + ... + λm with λ1 ≥ λ2 ≥ ... ≥ λm, whose lower odd parity index equals to Ilo(λ) = r, the weight of λ, denoted by w(λ) is the real number given by (−1)r +m.

Lemma 3.1. Consider λ = λ1 + λ2 + ... + λm a partition of the positive integer N, with Ilo(λ) = r and λ1 ≥ λ2 ≥ ... ≥ λm. Then: r is even (odd) if, and only if, λ1 is even (odd).

Proof. The proof of this lemma is similar to the proof of Lemma 2.1.

It follows from Lemma 3.1 that given any partition λ of a positive integer N,

w(λ) = (-1)m + r =

where m is the number of parts and λ1 is its greatest part.

In the following table we describe partitions λ of 7 with its respectives Ilo(λ) = r, number of parts m and weights w(λ).

It follows from(¹) that

is the generating function for the partitions λ of the positive integer n in exactly m parts, with Ilo(λ) = r. Evaluating (3.1) in x = y = −1 we have:

Observe that the coefficient of q7 in (3.2) is given by Σr, m, ≥ 0uo(r, m, 7)(-1)m + r. This is equivalent to sum the elements in the fourth column of Table 3, which in this case equals to 7.

Table 3
Partitions of 7 with its respectives Ilo(λ) = r.

partitions λ of 7	Ilo(λ) = r	m	w(λ) = (-1)m + r
7	1	1	(-1)1 + 1 = 1
6 + 1	2	2	(-1)2 + 2 = 1
5 + 2	1	2	(-1)1 + 2 = -1
5 + 1 + 1	1	3	(-1)1 + 3 = 1
4 + 3	2	2	(-1)2 + 2 = 1
4 + 2 + 1	2	3	(-1)3 + 2 = -1
4 + 1 + 1 + 1	2	4	(-1)4 + 2 = 1
3 + 3 + 1	1	3	(-1)3 + 1 = 1
3 + 2 + 2	1	3	(-1)3 + 1 = 1
3 + 2 + 1 + 1	3	4	(-1)4 + 3 = -1
3 + 1 + 1 + 1 + 1	1	5	(-1)5 + 1 = 1
2 + 2 + 2 + 1	2	4	(-1)4 + 2 = 1
2 + 2 + 1 + 1 + 1	2	5	(-1)5 + 2 = -1
2 + 1 + 1 + 1 + 1 + 1	2	6	(-1)6 + 2 = 1
1 + 1 + 1 + 1 + 1 + 1 + 1	1	7	(-1)7 + 1 = 1

In(²) it is given a bijection between the set of unrestricted partitions and two-line matrices A defined as:

whose entries satisfy

with ct, dt ≥ 0 and n = (ci + di). Matrices defined in this way are presented as:

l ≥ 1, s, d1 , ..., dl , j1, j2, ..., jl ∈ N.

Given a partition λ of a integer n, n = λ1 + λ2 + ... + λm, the procedure to obtain the matrix A from λ(²) is given below:

• the side l of the Durfee square is the number of columns of the matrix;

• the parameters d1, d2, ..., dl indicate the amount of parts 1's, 2's, ...l's, respectively, that appear below the Durfee square of side l.

• 

• where j1, j2, ..., jl are obtained in the solution of the system:

• 

In(³), the weight of the matrix A, given in (3.3) is defined by w(A) = . We will check that the weight of the matrix A has the same weight of the partition λ, associated to A by means of the bijection given in(³). Indeed, c1 + d1 + 1 = (2l − 1) + j1 + ...jl + d2 + ... + dl + d1 + 1 = 2l + j1 + ...jl + d1 + d2 + ... + dl = (l + j1 + ...jl) + (l + d1 + d2 + ... + dl). Observe that the greatest part of the partition is λ1 = l + j1 + ...jl and the number of parts of the partition is: m = l + d1 + d2 + ... + dl. Therefore, w(A) = = = (−1)^r+m =w(λ), where r =I_lo(λ). It follows from³ that

Theorem 3.1. The third order mock theta function f(q) , is the generating n = 0 (−q;q)²function for the weighted number of matrices A of the form (3.3). Where matrix is to be counted with the weight w(A) = .

With these observations and together with Theorem 3.1 we have:

Theorem 3.2. The third order mock theta function f(q) = , is the generating n = 0 (−q;q)²function for the weighted number of unrestricted partitions λ = λ₁ + λ₂ + ... + λ_m with I_lo(λ) =r. Where each partition is to be counted with the weight w(λ) = (−1)^m+r and m is the number of parts. From this from Theorem we have then, that

REFERENCES

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  » http://imartinjak.files.wordpress.com/2013/06/refinementslecturenotes1.pdf
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Publication Dates

• Publication in this collection
  Dec 2015

History

• Received
  06 Nov 2014
• Accepted
  28 Oct 2015