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On the parity of the number of (a,b,m)(a,b,m)-copartitions of nn

Hannah E. Burson and Dennis Eichhorn
Abstract

We continue the study of the (a,b,m)(a,b,m)-copartition function cpa,b,m(n)\mathrm{cp}_{a,b,m}(n), which arose as a combinatorial generalization of Andrews’ partitions with even parts below odd parts. The generating function of cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) has a nice representation as an infinite product. In this paper, we focus on the parity of cpa,b,m(n)\mathrm{cp}_{a,b,m}(n). As with the ordinary partition function, it is difficult to show positive density of either even or odd values of cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) for arbitrary a,ba,b, and mm. However, we find specific cases of a,b,ma,b,m such that cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) is even with density 1. Additionally, we show that the sequence {cpa,ma,m(n)}n=0\{\mathrm{cp}_{a,m-a,m}(n)\}_{n=0}^{\infty} takes both even and odd values infinitely often.

1 Introduction

In [4], the authors introduce and develop the theory of copartitions, which have connections to mock theta functions, Rogers-Ramanujan partitions, the capsids of Garvan and Schlosser [6], and many other classical partition-theoretic objects. Each (a,b,m)(a,b,m)-copartition is comprised of three partitions: a partition into parts a(modm)\equiv a\pmod{m}, a partition into parts b(modm)\equiv b\pmod{m}, and a rectangular partition that unites them. Additionally, the (a,b,m)(a,b,m)-copartition generating function can be written as an infinite product, further motivating interest in the copartition counting functions. As with any type of partitions, one of the first questions one may ask is about how frequently these counting functions take even and odd values.

For example, Kolberg showed that p(n)p(n), the ordinary partition function, takes both even and odd values infinitely often [7]. Parkin and Shanks studied the parity of p(n)p(n) computationally, and their evidence strongly suggests that p(n)p(n) is even about half of the time [8]. Sadly, the best known results on the parity of p(n)p(n) are spectacularly far from proving anything of the sort [2]. In fact, it is still an open problem to even show that p(n)p(n) is even or odd with positive density.

In this paper, we study the parity of the copartition counting functions. Although we conjecture that some copartition functions are equally often even and odd, we show that this is not the case for all such functions. Furthermore, in some special cases, we are able to demonstrate explicit sets with positive density on which cpa,b,m(n)\mathrm{cp}_{a,b,m}(n), the number of (a,b,m)(a,b,m)-copartitions of nn, is always even. These explicit sets allow us to give infinitely many arithmetic progressions on which certain copartition functions are always even. For example, we show that for r=3,17,24,31,38,45r=3,17,24,31,38,45, we have

cp3,1,4(49k+r)0(mod2),\mathrm{cp}_{3,1,4}(49k+r)\equiv 0\pmod{2},

and for s=9,14,19,24s=9,14,19,24, we have

cp5,1,6(25k+s)0(mod2)\mathrm{cp}_{5,1,6}(25k+s)\equiv 0\pmod{2}

for every nonnegative integer kk.

In Section 2, we recall the definition and generating function for copartitions. In Section 3, we recall conjugation, discuss self-conjugate copartitions, and give lower bounds on the number of even values of cpa,a,m(n)\mathrm{cp}_{a,a,m}(n). In Section 4, we focus on cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n); we show that cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) takes both odd and even values infinitely often, give explicit sets with density one on which cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) and cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) are even, and provide infinitely many congruences modulo two in arithmetic progressions for each of these functions. Sections 3 and 4 also include several open problems. In Appendix A, we provide computational data surrounding our open questions and conjectures.

2 Background on Copartitions

In this section, we review (a,b,m)(a,b,m)-copartitions and the function cpa,b,m(n),\mathrm{cp}_{a,b,m}(n), which were first introduced in [4].

Definition.

An (a,b,m)(a,b,m)-copartition is a triple of partitions (γ,ρ,σ)(\gamma,\rho,\sigma), where each of the parts of γ\gamma is at least aa and congruent to a(modm)a\pmod{m}, each of the parts of σ\sigma is at least bb and congruent to b(modm)b\pmod{m}, and ρ\rho has the same number of parts as σ\sigma, each of which have size equal to mm times the number of parts of γ\gamma.
When a,b,m1a,b,m\geq 1, we let cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) denote the number of (a,b,m)(a,b,m)-copartitions of size nn.

Although ρ\rho is completely determined by γ\gamma and σ\sigma, the graphical representation we use suggests that the natural way to write the triple is (γ,ρ,σ)(\gamma,\rho,\sigma). We refer to γ\gamma as the ground of a copartition and σ\sigma as the sky.

Example 2.1.

The (2,1,3)(2,1,3)-copartitions of size 9 are

({5,22},,),({5},{3},{1}),({2},{3},{4}),(,,{7+12}),\left(\{5,2^{2}\},\emptyset,\emptyset\right),\;\left(\{5\},\{3\},\{1\}\right),\;\left(\{2\},\{3\},\{4\}\right),\;\left(\emptyset,\emptyset,\{7+1^{2}\}\right),\;
(,,{42+1}),(,,{4+15}), and (,,{19}).\left(\emptyset,\emptyset,\{4^{2}+1\}\right),\left(\emptyset,\emptyset,\{4+1^{5}\}\right),\text{ and }\left(\emptyset,\emptyset,\{1^{9}\}\right).

Thus, cp2,1,3(9)=7\mathrm{cp}_{2,1,3}(9)=7.

To represent the (a,b,m)(a,b,m) copartition (σ,ρ,γ)(\sigma,\rho,\gamma) graphically, we append the mm-modular diagram for σ\sigma to the right of the mm-modular diagram for ρ\rho. Then, we append the conjugate of the mm-modular diagram for γ\gamma below ρ\rho.

Example 2.2.

The following diagram represents the (a,b,m)(a,b,m)-copartition ({3m+a,2m+a,2m+a,a},{4m,4m},{3m+b,2m+b})(\{3m+a,2m+a,2m+a,a\},\{4m,4m\},\{3m+b,2m+b\}).

{ytableau} m&mmmbmmm mmmmbmm aaaa mmm mmm m

In [4], we also show that

𝐜𝐩a,b,m(q)\displaystyle{\mathbf{cp}}_{a,b,m}(q) :=n=0cpa,b,m(n)qn=(qa+b;qm)(qb;qm)(qa;qm).\displaystyle:=\sum_{n=0}^{\infty}\mathrm{cp}_{a,b,m}(n)q^{n}=\frac{(q^{a+b};q^{m})_{\infty}}{(q^{b};q^{m})_{\infty}(q^{a};q^{m})_{\infty}}. (1)

3 Conjugation, Self-Conjugate Copartitions, and the Parity of cpa,a,m(n)\mathrm{cp}_{a,a,m}(n)

In this section, we explore the parity of cpa,a,m(n)\mathrm{cp}_{a,a,m}(n). Many of our results follow from considering the fixed points of the conjugation involution on cpa,a,m(n)\mathrm{cp}_{a,a,m}(n). As in [4], we define the conjugate of a copartition (γ,ρ,σ)(\gamma,\rho,\sigma) as the copartition obtained by reflecting the graphical representation about the line y=xy=-x. Defining ν(λ)\nu(\lambda) to be the number of parts of a partition λ\lambda, the conjugate copartition is precisely (σ,ρ,γ),(\sigma,\rho^{\prime},\gamma), where ρ\rho^{\prime} consists of exactly ν(γ)\nu(\gamma) parts of size m×ν(σ)=m×ν(ρ)m\times\nu(\sigma)=m\times\nu(\rho). Equivalently, ρ\rho^{\prime} is the partition obtained by conjugating the mm-modular diagram representing ρ\rho.

{ytableau}

m&mmbmmm

mmmbm

aaa

mm

mm

m

\xrightarrow{\hskip 21.68121pt}\quad {ytableau} m&mammm mmamm mma bb mm m m

Figure 1: Conjugation of an (a,b,m)(a,b,m)-copartition.
Remark.

Conjugation is a size-preserving bijection from 𝒞𝒫a,b,m\mathcal{CP}_{a,b,m} to 𝒞𝒫b,a,m\mathcal{CP}_{b,a,m}.

Remark.

When aba\neq b, there are no self-conjugate (a,b,m)(a,b,m)-copartitions.

Theorem 3.1.

Let scpa,m(n)\mathrm{scp}_{a,m}(n) denote the number of self-conjugate (a,a,m)(a,a,m)-copartitions of size nn. Then, we have the following generating function:

n=0scpa,m(n)=(qm+2a;q2m).\sum_{n=0}^{\infty}\mathrm{scp}_{a,m}(n)=(-q^{m+2a};q^{2m})_{\infty}.
Proof.

We prove this theorem combinatorially by adapting Sylvester’s proof that self-conjugate partitions of nn are equinumerous with partitions of nn into distinct odd parts [9, p. 275]. Consider the graphical representation of a self-conjugate (a,a,m)(a,a,m)-copartition. For each cell on the line y=xy=-x, there is a corresponding hook that consists of that cell, all the cells directly below it, and all the cells directly to the right of it. Note that, because the diagram is self conjugate, each hook has an odd number of mm’s and exactly two aa’s. Thus, if we consider the hooks as the parts of a new partition, this new partition will have distinct parts that are congruent to m+2a(mod2m)m+2a\pmod{2m}. Similarly, given a partition into distinct parts congruent to m+2a(mod2m)m+2a\pmod{2m}, we can create a self-conjugate (a,a,m)(a,a,m) copartition. Since (qm+2a;q2m)(-q^{m+2a};q^{2m})_{\infty} generates partitions into distinct parts that are congruent to m+2a(mod2m)m+2a\pmod{2m}, our result follows. ∎

Since conjugation is an involution on (a,a,b)(a,a,b)-copartitions, cpa,a,m(n)\mathrm{cp}_{a,a,m}(n) is odd exactly when there are an odd number of self-conjugate (a,a,m)(a,a,m)-copartitions of size nn. Thus, by Theorem 3.1,

n=0cpa,a,m(n)qn(qm+2a;q2m)(mod2).\sum_{n=0}^{\infty}\mathrm{cp}_{a,a,m}(n)q^{n}\equiv(-q^{m+2a};q^{2m})_{\infty}\pmod{2}. (2)

This congruence immediately implies the following facts about the parity of cpa,a,m(n)\mathrm{cp}_{a,a,m}(n).

Corollary 3.2.

For even mm, cpa,a,m(2n+1)0(mod2)\mathrm{cp}_{a,a,m}(2n+1)\equiv 0\pmod{2}.

Corollary 3.3.

For even mm,

lim infn#{1kncpa,a,m(k) is even}n12.\liminf_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even}\}}{n}\geq\frac{1}{2}.
Corollary 3.4.

For m2(mod4)m\equiv 2\pmod{4} and odd aa,

lim infn#{1kncpa,a,m(k) is even}n34.\liminf_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even}\}}{n}\geq\frac{3}{4}.
Proof.

Note that, because m2(mod4)m\equiv 2\pmod{4} and aa is odd, m+2a0(mod4)m+2a\equiv 0\pmod{4} and 2m0(mod4)2m\equiv 0\pmod{4}. Thus, all self-conjugate copartitions must be of size 0(mod4)0\pmod{4}. ∎

Corollary 3.5.

For aa odd,

limn#{1kncpa,a,2a(k) is even}n=1.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,2a}(k)\text{ is even}\}}{n}=1.

Moreover, cpa,a,2a(k)\mathrm{cp}_{a,a,2a}(k) is odd if and only if kk is 4a4a times a pentagonal number; that is, if and only if k=2an(3n1)k=2an(3n-1) for some integer nn.

Proof.

From (2), we see that

n=0cpa,a,2a(n)qn\displaystyle\sum_{n=0}^{\infty}\mathrm{cp}_{a,a,2a}(n)q^{n} (q4a;q4a)(mod2)\displaystyle\equiv(-q^{4a};q^{4a})_{\infty}\pmod{2}
(q4a;q4a)\displaystyle\equiv(q^{4a};q^{4a})_{\infty}
=n=(1)nq2an(3n1).\displaystyle=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{2an(3n-1)}.

The final equality follows from Euler’s pentagonal number theorem:

(q;q)=n=(1)nqn(3n1)/2.(q;q)_{\infty}=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(3n-1)/2}.

Then, because the final series is lacunary, we have proved that cpa,a,2a(n)\mathrm{cp}_{a,a,2a}(n) is even with density 11. ∎

Corollaries 3.3 and 3.4 come from explicit sets upon which cpa,a,m(n)\mathrm{cp}_{a,a,m}(n) is always even. Outside of those sets, empirical evidence suggests that the parity is equally balanced, which leads us to the following conjecture.

Conjecture 3.6.

For even mm and odd aa,

limn#{1kncpa,a,m(k) is even}n=\displaystyle\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even}\}}{n}= 1\displaystyle 1 if m=2am=2a (3)
limn#{1kncpa,a,m(k) is even}n=\displaystyle\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even}\}}{n}= 34\displaystyle\frac{3}{4} if m0(mod4)m\equiv 0\pmod{4} (4)
limn#{1kncpa,a,m(k) is even}n=\displaystyle\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even}\}}{n}= 78\displaystyle\frac{7}{8} otherwise. (5)

Note that (3) is Corollary 3.5.

On the other hand, for mm odd, as is the case with p(n)p(n), we do not know explicit sets upon which cpa,a,m(n)\mathrm{cp}_{a,a,m}(n) is always even, and we make the following conjecture.

Conjecture 3.7.

For odd mm and gcd(a,m)=1\gcd(a,m)=1, cpa,a,m(n)\mathrm{cp}_{a,a,m}(n) is even (odd) with density 12\frac{1}{2}. That is,

limn#{1kncpa,a,m(k) is even (odd)}n=12.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,a,m}(k)\text{ is even (odd)}\}}{n}=\frac{1}{2}.

Table 3.1 provides some computational evidence for Conjecture 3.6.

nn #{1kncp3,3,4(k) is even}n\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{3,3,4}(k)\text{ is even}\}}{n} #{1kncp1,1,6(k) is even}n\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{1,1,6}(k)\text{ is even}\}}{n}
10001000 0.7650.765 0.8710.871
30003000 0.7520.752 0.8750.875
50005000 0.7530.753 0.8740.874
70007000 0.7490.749 0.8750.875
90009000 0.7480.748 0.8730.873
1100011000 0.7490.749 0.8740.874
1300013000 0.7500.750 0.8750.875
1500015000 0.7490.749 0.8750.875
Table 3.1: The proportion of even values of cpa,a,m(k)\mathrm{cp}_{a,a,m}(k) for 1kn1\leq k\leq n when (a,m)(a,m) is (3,4)(3,4) and (1,6)(1,6).

4 The Parity of cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n)

In this section, we explore the parity of the values of another special family of copartition functions: cpa,ma,m(n).\mathrm{cp}_{a,m-a,m}(n). Note that

𝐜𝐩a,ma,m(q)=(qm;qm)(qa;qm)(qma;qm)=(qm;qm)2f(qa,qma),\mathbf{cp}_{a,m-a,m}(q)=\frac{(q^{m};q^{m})_{\infty}}{(q^{a};q^{m})_{\infty}(q^{m-a};q^{m})_{\infty}}=\frac{(q^{m};q^{m})^{2}_{\infty}}{f(q^{a},q^{m-a})},

where f(x,y)=n=xn(n+1)/2yn(n1)/2f(x,y)=\sum_{n=-\infty}^{\infty}x^{n(n+1)/2}y^{n(n-1)/2} is Ramanujan’s theta function. This form for 𝐜𝐩a,ma,m(q)\mathbf{cp}_{a,m-a,m}(q) suggests that there is a wide range of analytic tools that we can use to study this family of functions.

A very basic question one may ask is simply, for which a,ma,m does the sequence {cpa,ma,m(n)}n=0\{\mathrm{cp}_{a,m-a,m}(n)\}_{n=0}^{\infty} take both even and odd values infinitely often?

Theorem 4.1.

For all a,ma,m, the sequence {cpa,ma,m(n)}n=0\{\mathrm{cp}_{a,m-a,m}(n)\}_{n=0}^{\infty} takes both even and odd values infinitely often.

Proof.

Since cpa,ma,m(n)=cpma,a,m(n)\mathrm{cp}_{a,m-a,m}(n)=\mathrm{cp}_{m-a,a,m}(n), without loss of generality, we assume 0<am/20<a\leq m/2.

If a=m/2a=m/2, by (1) we have

𝐜𝐩m/2,m/2,m(q)\displaystyle{\mathbf{cp}}_{m/2,m/2,m}(q) =(qm;qm)(qm/2;qm)(qm/2;qm)(qm;qm)(qm;q2m)(q2m;q2m)(mod2).\displaystyle=\frac{(q^{m};q^{m})_{\infty}}{(q^{m/2};q^{m})_{\infty}(q^{m/2};q^{m})_{\infty}}\equiv\frac{(q^{m};q^{m})_{\infty}}{(q^{m};q^{2m})_{\infty}}\equiv(q^{2m};q^{2m})_{\infty}\pmod{2}. (6)

By Euler’s Pentagonal Number Theorem, we see that the right-hand side of (6) takes both even and odd values infinitely often.

For am/2a\neq m/2, by (1) we have

𝐜𝐩a,ma,m(q)\displaystyle{\mathbf{cp}}_{a,m-a,m}(q) =(qm;qm)2(qma;qm)(qa;qm)(qm;qm)(q2m;q2m)(qma;qm)(qa;qm)(qm;qm)(mod2).\displaystyle=\frac{(q^{m};q^{m})^{2}_{\infty}}{(q^{m-a};q^{m})_{\infty}(q^{a};q^{m})_{\infty}(q^{m};q^{m})_{\infty}}\equiv\frac{(q^{2m};q^{2m})_{\infty}}{(q^{m-a};q^{m})_{\infty}(q^{a};q^{m})_{\infty}(q^{m};q^{m})_{\infty}}\pmod{2}. (7)

By applying Jacobi’s triple product identity to the denominator of (7), multiplying both sides by that denominator, and applying Euler’s Pentagonal Number Theorem to the remaining right-hand side, we have

𝐜𝐩a,ma,m(q)n=(1)nqan+mn(n1)/2k=qmk(3k1)(mod2).\displaystyle{\mathbf{cp}}_{a,m-a,m}(q)\sum_{n=-\infty}^{\infty}(-1)^{n}q^{an+mn(n-1)/2}\equiv\sum_{k=-\infty}^{\infty}q^{mk(3k-1)}\pmod{2}. (8)

Now suppose cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) has finitely many odd values, and let dd be the largest integer such that cpa,ma,m(d)\mathrm{cp}_{a,m-a,m}(d) is odd. Notice that cpa,ma,m(0)=1\mathrm{cp}_{a,m-a,m}(0)=1 is also odd. If d=0d=0, then we have from (8) that {an+mn(n1)/2n}={mn(3n1)n}\{an+mn(n-1)/2\mid n\in\mathbb{Z}\}=\{mn(3n-1)\mid n\in\mathbb{Z}\}, which is not possible. For d>0d>0, we derive a contradiction by showing that if we go out far enough, the left-hand side of (8) has two close odd values, but the right-hand side does not. A short computation shows that since am/2a\neq m/2, an1+mn1(n11)/2=an2+mn2(n21)/2an_{1}+mn_{1}(n_{1}-1)/2=an_{2}+mn_{2}(n_{2}-1)/2 if and only if n1=n2n_{1}=n_{2}. Let N0N_{0} be so large that for |n|N0|n|\geq N_{0}, consecutive values of each set {an+mn(n1)/2n}\{an+mn(n-1)/2\mid n\in\mathbb{Z}\} and {mn(3n1)n}\{mn(3n-1)\mid n\in\mathbb{Z}\} are more than dd apart. Let N1>N0N_{1}>N_{0} be so large that aN1+mN1(N11)/2>mN0(3N01)aN_{1}+mN_{1}(N_{1}-1)/2>mN_{0}(3N_{0}-1). Then, since cpa,ma,m(0)\mathrm{cp}_{a,m-a,m}(0) and cpa,ma,m(d)\mathrm{cp}_{a,m-a,m}(d) are both odd and N1>N0N_{1}>N_{0}, the coefficients of qaN1+mN1(N11)/2q^{aN_{1}+mN_{1}(N_{1}-1)/2} and qaN1+mN1(N11)/2+dq^{aN_{1}+mN_{1}(N_{1}-1)/2+d} on the left-hand side of (8) are both odd. However, for exponents in that range, the terms of the right-hand side of (8) with odd coefficients are more than dd terms apart, a contradiction. Thus cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) has infinitely many odd values.

To show cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) is even infinitely often, define Ea,mE_{a,m} to be the set of nonnegative integers nn such that cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) is even, and define Ga,m(q)=nEa,mqnG_{a,m}(q)=\sum_{n\in E_{a,m}}q^{n}. Notice

11q𝐜𝐩a,ma,m(q)+Ga,m(q)(mod2).\frac{1}{1-q}\equiv{\mathbf{cp}}_{a,m-a,m}(q)+G_{a,m}(q)\pmod{2}.

Multiplying both sides by the denominator in (7) after applying Jacobi’s triple product identity, we have

n=(1)nqan+mn(n1)/21q\displaystyle\frac{\sum_{n=-\infty}^{\infty}(-1)^{n}q^{an+mn(n-1)/2}}{1-q} (qm;qm)2+Ga,m(q)n=(1)nqan+mn(n1)/2(mod2)\displaystyle\equiv(q^{m};q^{m})^{2}_{\infty}+G_{a,m}(q)\sum_{n=-\infty}^{\infty}(-1)^{n}q^{an+mn(n-1)/2}\pmod{2} (9)
(q2m;q2m)+Ga,m(q)n=qan+mn(n1)/2(mod2).\displaystyle\equiv(q^{2m};q^{2m})_{\infty}+G_{a,m}(q)\sum_{n=-\infty}^{\infty}q^{an+mn(n-1)/2}\pmod{2}. (10)

We can rewrite the left-hand side of (9) by pairing summands as

11q\displaystyle\frac{1}{1-q} n=0(1)n[qan+mn(n+1)/2qa(n+1)+mn(n+1)/2]=n=0(1)nqan+mn(n+1)/21qa(2n+1)1q\displaystyle\sum_{n=0}^{\infty}(-1)^{n}[q^{-an+mn(n+1)/2}-q^{a(n+1)+mn(n+1)/2}]=\sum_{n=0}^{\infty}(-1)^{n}q^{-an+mn(n+1)/2}\frac{1-q^{a(2n+1)}}{1-q}
n=0qan+mn(n+1)/2+qan+mn(n+1)/2+1++qa(n+1)+mn(n+1)/21(mod2),\displaystyle\equiv\sum_{n=0}^{\infty}q^{-an+mn(n+1)/2}+q^{-an+mn(n+1)/2+1}+\cdots+q^{a(n+1)+mn(n+1)/2-1}\pmod{2}, (11)

where each term of the sum in (11) has been expanded into a finite geometric series, the sum of a(2n+1)a(2n+1) consecutive powers of qq. These finite geometric series do not overlap, and so a short computation shows that (11) has aN2aN^{2} odd terms up through the qmN2/2q^{\lfloor mN^{2}/2\rfloor} term, so that the terms with odd coefficients in (11) and the left-hand side of (9) have density 2a/m2a/m.

By Euler’s Pentagonal Number Theorem, the odd values of (q2m;q2m)(q^{2m};q^{2m})_{\infty} have density zero, thus the nonzero values of the product Ga,m(q)n=(1)nqan+mn(n1)/2G_{a,m}(q)\sum_{n=-\infty}^{\infty}(-1)^{n}q^{an+mn(n-1)/2} must have density 2a/m2a/m. Since the nonzero values of n=(1)nqan+mn(n1)/2\sum_{n=-\infty}^{\infty}(-1)^{n}q^{an+mn(n-1)/2} have density zero, there must be infinitely many nonzero terms of Ga,m(q)G_{a,m}(q). ∎

Remark.

Some of the analysis above is very similar to that of Berndt, Yee, and Zaharescu [3] in treating the parity of p(r,s;n)p(r,s;n), the number of partitions of nn into parts congruent to rr, ss, or r+sr+s modulo r+sr+s. In the second half of the proof above, tracking the number of odd terms as they did, one can arrive at the quantitative result that #{n<Ncpa,ma,m(n) is even}>(a/2mo(1))N\#\{n<N\mid\mathrm{cp}_{a,m-a,m}(n)\text{ is even}\}>(a/\sqrt{2m}-o(1))\sqrt{N} for all am/2a\leq m/2. In the special case a=m/2a=m/2, we have #{n<Ncpa,ma,m(n) is even}>(1o(1))N\#\{n<N\mid\mathrm{cp}_{a,m-a,m}(n)\text{ is even}\}>(1-o(1))N.

Empirical evidence suggests that the parity of cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) may be balanced for many but not all a,ma,m with gcd(a,m)=1\gcd(a,m)=1. This leads us to the following conjecture and question.

Conjecture 4.2.

For odd mm and gcd(a,m)=1\gcd(a,m)=1, the sequence {cpa,ma,m(n)}n=0\{\mathrm{cp}_{a,m-a,m}(n)\}_{n=0}^{\infty} takes both even and odd values with density 12\frac{1}{2}. That is,

limn#{1kncpa,ma,m(k) is even (odd)}n=12.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,m-a,m}(k)\text{ is even (odd)}\}}{n}=\frac{1}{2}.
Question 4.3.

For which a,ma,m with gcd(a,m)=1\gcd(a,m)=1 is

limn#{1kncpa,ma,m(k) is even}n=12?\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,m-a,m}(k)\text{ is even}\}}{n}=\frac{1}{2}?

Furthermore, when the limit is not 1/21/2, what is

limn#{1kncpa,ma,m(k) is even}n?\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,m-a,m}(k)\text{ is even}\}}{n}?

When the limit is not 1/21/2, is it always 11?

In Appendix A, we provide computational data surrounding Conjecture 4.2 and Question 4.3.

4.1 The parity of cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) and cp5,1,6(n)\mathrm{cp}_{5,1,6}(n)

In two special cases, we can answer Question 4.3 and give the exact asymptotic density of nn for which cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) is even. The generating functions for cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) and cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) have nice properties that allow us to apply the classical theory of binary quadratic forms to show that cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) and cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) are even on sets with arithmetic density one.

We now provide an explicit set with arithmetic density one on which cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) is even.

Theorem 4.4.

When a=3,b=1,a=3,b=1, and m=4m=4,

limn#{1kncp3,1,4(k) is even}n=1.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{3,1,4}(k)\text{ is even}\}}{n}=1.

In particular, cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) is even if the prime factorization of 24n+524n+5 has a prime 3(mod4)\equiv 3\pmod{4} occurring with an odd exponent.

Remark.

Note that 24n+524n+5 having a prime 3(mod4)\equiv 3\pmod{4} occurring with an odd exponent is a sufficient condition, but is not necessary.

Proof.

We can rewrite the generating function for cp3,1,4\mathrm{cp}_{3,1,4} in a very useful form. Consider

n=0cp3,1,4(n)qn\displaystyle\sum_{n=0}^{\infty}\mathrm{cp}_{3,1,4}(n)q^{n} =(q4;q4)(q3;q4)(q;q4)=(q4;q4)(q;q2)\displaystyle=\frac{(q^{4};q^{4})_{\infty}}{(q^{3};q^{4})_{\infty}(q;q^{4})_{\infty}}=\frac{(q^{4};q^{4})_{\infty}}{(q;q^{2})_{\infty}}
=(q4;q4)(q2;q2)(q;q)(q;q)(q4;q4)(mod2).\displaystyle=\frac{(q^{4};q^{4})_{\infty}(q^{2};q^{2})_{\infty}}{(q;q)_{\infty}}\equiv(q;q)_{\infty}(q^{4};q^{4})_{\infty}\pmod{2}. (12)

Applying Euler’s Pentagonal Number Theorem, we then have that

n=0cp3,1,4(n)qn(j=qj(3j+1)/2)(k=q2k(3k+1))(mod2).\sum_{n=0}^{\infty}\mathrm{cp}_{3,1,4}(n)q^{n}\equiv\left(\sum_{j=-\infty}^{\infty}q^{j(3j+1)/2}\right)\left(\sum_{k=-\infty}^{\infty}q^{2k(3k+1)}\right)\pmod{2}.

Thus, cp3,1,4(n)\mathrm{cp}_{3,1,4}(n) must be even unless n=j(3j+1)/2+2k(3k+1)n=j(3j+1)/2+2k(3k+1) for some integers jj and kk, or equivalently, unless 24n+5=(6j+1)2+4(6k+1)224n+5=(6j+1)^{2}+4(6k+1)^{2}. Analyzing this modulo 2424, this occurs if and only if 24n+524n+5 is represented by the form A2+B2A^{2}+B^{2}. From the classical theory of binary quadratic forms, we know that the integers 24n+524n+5 representable by the form A2+B2A^{2}+B^{2} are precisely those with prime factorizations having all powers of primes 3(mod4)\equiv 3\pmod{4} occurring with an even exponent. Since that set of representable (24n+5)(24n+5)s has density zero, we have

limn#{1kncp3,1,4(k) is even}n=1\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{3,1,4}(k)\text{ is even}\}}{n}=1

as desired. ∎

Considering the set of even values of cp3,1,4\mathrm{cp}_{3,1,4} guaranteed by Theorem 4.4, it is straightforward to write down arithmetic progressions on which cp3,1,4\mathrm{cp}_{3,1,4} is always even.

Corollary 4.5.

For any prime p>3,p3(mod4)p>3,p\equiv 3\pmod{4}, let 24δ1(modp2)24\delta\equiv 1\pmod{p^{2}}. Then

cp3,1,4(p2k+pt5δ)0(mod2)\mathrm{cp}_{3,1,4}(p^{2}k+pt-5\delta)\equiv 0\pmod{2}

for t=1,2,,p1t=1,2,\dots,p-1 and every nonnegative integer kk.

Below we give two specific examples.

Corollary 4.6.

For r=3,17,24,31,38,45r=3,17,24,31,38,45, we have

cp3,1,4(49k+r)0(mod2)\mathrm{cp}_{3,1,4}(49k+r)\equiv 0\pmod{2}

for every nonnegative integer kk.

Corollary 4.7.

For r=3,14,36,47,58,69,80,91,102,113r=3,14,36,47,58,69,80,91,102,113, we have

cp3,1,4(121k+r)0(mod2)\mathrm{cp}_{3,1,4}(121k+r)\equiv 0\pmod{2}

for every nonnegative integer kk.

In order to treat cp5,1,6(n)\mathrm{cp}_{5,1,6}(n), we use the same techniques, but the argument is slightly more complicated. We first require a lemma showing the relationship between two binary quadratic forms.

Lemma 4.8.

An integer N1(mod6)N\equiv 1\pmod{6} is representable by the binary quadratic form A2+3B2A^{2}+3B^{2} if and only if 4N4N is representable by the binary quadratic form (6J+1)2+3(6K+1)2(6J+1)^{2}+3(6K+1)^{2}.

Proof.

Suppose N1(mod6)N\equiv 1\pmod{6} is representable by the form A2+3B2A^{2}+3B^{2}. Then there exist a,ba,b\in\mathbb{Z} such that

N=a2+3b2.N=a^{2}+3b^{2}. (13)

Specifically, because a2=(a)2a^{2}=(-a)^{2} and a0(mod3)a\not\equiv 0\pmod{3}, if ab(mod3)a\equiv b\pmod{3}, let us instead choose aa to be a-a in (13) to obtain a representation where ab(mod3)a\not\equiv b\pmod{3}. Now notice

(a+3b)2+3(ab)2=4a2+12b2\displaystyle(a+3b)^{2}+3(a-b)^{2}=4a^{2}+12b^{2} =4N\displaystyle=4N
(a+3b)2+3(ba)2=4a2+12b2\displaystyle(a+3b)^{2}+3(b-a)^{2}=4a^{2}+12b^{2} =4N\displaystyle=4N
(a3b)2+3(ab)2=4a2+12b2\displaystyle(-a-3b)^{2}+3(a-b)^{2}=4a^{2}+12b^{2} =4N, and\displaystyle=4N,\text{\ and}
(a3b)2+3(ba)2=4a2+12b2\displaystyle(-a-3b)^{2}+3(b-a)^{2}=4a^{2}+12b^{2} =4N.\displaystyle=4N. (14)

Since aa and bb must be of opposite parity by (13), we have one representation of 4N4N in (4.1) of the form (6J+1)2+3(6K+1)2(6J+1)^{2}+3(6K+1)^{2}.

Now instead suppose NN is any positive integer such that 4N=(6j+1)2+3(6k+1)24N=(6j+1)^{2}+3(6k+1)^{2}. If jk(mod2)j\equiv k\pmod{2}, let a=(3j+9k)/2+1a=(3j+9k)/2+1 and b=(3j3k)/2b=(3j-3k)/2. Then

a2+3b2=36j2+12j+14+336k2+12k+14=(6j+1)24+3(6k+1)24=N.a^{2}+3b^{2}=\frac{36j^{2}+12j+1}{4}+3\frac{36k^{2}+12k+1}{4}=\frac{(6j+1)^{2}}{4}+3\frac{(6k+1)^{2}}{4}=N. (15)

Otherwise, if jk(mod2)j\not\equiv k\pmod{2}, let a=(3j9k1)/2a=(3j-9k-1)/2 and b=(3j+3k+1)/2b=(3j+3k+1)/2. Then

a2+3b2=36j2+12j+14+336k2+12k+14=(6j+1)24+3(6k+1)24=N.a^{2}+3b^{2}=\frac{36j^{2}+12j+1}{4}+3\frac{36k^{2}+12k+1}{4}=\frac{(6j+1)^{2}}{4}+3\frac{(6k+1)^{2}}{4}=N. (16)

In either case, we see that NN is of the form A2+3B2A^{2}+3B^{2}, and the lemma follows.

We now provide an explicit set with arithmetic density one on which cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) is even.

Theorem 4.9.

When a=5,b=1,a=5,b=1, and m=6m=6,

limn#{1kncp5,1,6(k) is even}n=1.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{5,1,6}(k)\text{ is even}\}}{n}=1.

In particular, cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) is even if the prime factorization of 6n+16n+1 has a prime 2(mod3)\equiv 2\pmod{3} occurring with an odd exponent.

Remark.

Note that 6n+16n+1 having a prime 2(mod3)\equiv 2\pmod{3} occurring with an odd exponent is a sufficient condition, but is not necessary. For example, cp5,1,6(5)=2\mathrm{cp}_{5,1,6}(5)=2 is even, even though 6(5)+1=316(5)+1=31.

Proof.

We can rewrite the generating function for cp5,1,6\mathrm{cp}_{5,1,6} in a very useful form. Consider

(q2;q2)(q3;q3)(q;q)(q6;q6)\displaystyle\frac{(q^{2};q^{2})_{\infty}(q^{3};q^{3})_{\infty}}{(q;q)_{\infty}(q^{6};q^{6})_{\infty}} =(q2;q2)(q3;q6)(q6;q6)(q;q2)(q2;q2)(q6;q6)\displaystyle=\frac{(q^{2};q^{2})_{\infty}(q^{3};q^{6})_{\infty}(q^{6};q^{6})_{\infty}}{(q;q^{2})_{\infty}(q^{2};q^{2})_{\infty}(q^{6};q^{6})_{\infty}}
=(q3;q6)(q;q2)\displaystyle=\frac{(q^{3};q^{6})_{\infty}}{(q;q^{2})_{\infty}}
=1(q;q6)(q5;q6).\displaystyle=\frac{1}{(q;q^{6})_{\infty}(q^{5};q^{6})_{\infty}}.

Thus,

n=0cp5,1,6(n)qn\displaystyle\sum_{n=0}^{\infty}\mathrm{cp}_{5,1,6}(n)q^{n} =(q6;q6)(q5;q6)(q;q6)=(q2;q2)(q3;q3)(q;q)\displaystyle=\frac{(q^{6};q^{6})_{\infty}}{(q^{5};q^{6})_{\infty}(q;q^{6})_{\infty}}=\frac{(q^{2};q^{2})_{\infty}(q^{3};q^{3})_{\infty}}{(q;q)_{\infty}}
=(q;q)(q3;q3)(q;q)(q3;q3)(mod2).\displaystyle=(-q;q)_{\infty}(q^{3};q^{3})_{\infty}\equiv(q;q)_{\infty}(q^{3};q^{3})_{\infty}\pmod{2}. (18)

Applying Euler’s Pentagonal Number Theorem, we then have that

n=0cp5,1,6(n)qn(j=qj(3j+1)/2)(k=q3k(3k+1)/2)(mod2).\sum_{n=0}^{\infty}\mathrm{cp}_{5,1,6}(n)q^{n}\equiv\left(\sum_{j=-\infty}^{\infty}q^{j(3j+1)/2}\right)\left(\sum_{k=-\infty}^{\infty}q^{3k(3k+1)/2}\right)\pmod{2}.

Thus, cp5,1,6(n)\mathrm{cp}_{5,1,6}(n) must be even unless n=j(3j+1)/2+3k(3k+1)/2n=j(3j+1)/2+3k(3k+1)/2 for some integers jj and kk, or equivalently, unless (6j+1)2+3(6k+1)2=24n+4(6j+1)^{2}+3(6k+1)^{2}=24n+4. By Lemma 4.8, this occurs if and only if 6n+16n+1 is represented by the form A2+3B2A^{2}+3B^{2}. From the classical theory of binary quadratic forms [5, Chapter 1], we know that the integers 6n+16n+1 representable by the form A2+3B2A^{2}+3B^{2} are precisely those with prime factorizations having all powers of primes 2(mod3)\equiv 2\pmod{3} occurring with an even exponent. Since that set of representable (6n+1)(6n+1)s has density zero, we have

limn#{1kncp5,1,6(k) is even}n=1\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{5,1,6}(k)\text{ is even}\}}{n}=1

as desired. ∎

Considering the set of even values of cp5,1,6\mathrm{cp}_{5,1,6} guaranteed by Theorem 4.9, it is straightforward to write down arithmetic progressions on which cp5,1,6\mathrm{cp}_{5,1,6} is always even.

Corollary 4.10.

For any prime p>2,p2(mod3)p>2,p\equiv 2\pmod{3}, let 6δ1(modp2)6\delta\equiv 1\pmod{p^{2}}. Then

cp5,1,6(p2k+ptδ)0(mod2)\mathrm{cp}_{5,1,6}(p^{2}k+pt-\delta)\equiv 0\pmod{2}

for t=1,2,,p1t=1,2,\dots,p-1 and every nonnegative integer kk.

Below we give two specific examples.

Corollary 4.11.

For r=9,14,19,24r=9,14,19,24, we have

cp5,1,6(25k+r)0(mod2)\mathrm{cp}_{5,1,6}(25k+r)\equiv 0\pmod{2}

for every nonnegative integer kk.

Corollary 4.12.

For r=9,31,42,53,64,75,86,97,108,119r=9,31,42,53,64,75,86,97,108,119, we have

cp5,1,6(121k+r)0(mod2)\mathrm{cp}_{5,1,6}(121k+r)\equiv 0\pmod{2}

for every nonnegative integer kk.

5 Conclusion

Conjectures 3.6, 3.7, and 4.2 and Question 4.3 remain open. Although we treat the parity of cpa,a,m(n)\mathrm{cp}_{a,a,m}(n) and cpa,ma,m(n)\mathrm{cp}_{a,m-a,m}(n) in some detail above, the parity of cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) when aba\neq b, and a+bma+b\neq m is largely unexplored. With only a small amount of evidence, we make the following somewhat bold conjecture.

Conjecture 5.1.

When gcd(a,b,m)=1\gcd(a,b,m)=1, aba\neq b, and a+bma+b\neq m, cpa,b,m(n)\mathrm{cp}_{a,b,m}(n) is even (odd) with density 12\frac{1}{2}. That is,

limn#{1kncpa,b,m(k) is even}n=12.\lim_{n\to\infty}\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{a,b,m}(k)\text{ is even}\}}{n}=\frac{1}{2}.

In another direction, we give several congruences modulo 2 in Corollaries 4.5-4.7 and 4.10-4.12, In [1], Andrews gave a congruence modulo 5, namely 𝒪(10n+8)0(mod5)\mathcal{EO}^{*}(10n+8)\equiv 0\pmod{5}, which, when written in the language of copartitions, becomes cp1,1,2(5n+4)0(mod5)\mathrm{cp}_{1,1,2}(5n+4)\equiv 0\pmod{5}. Additionally, he defined an even-odd partition crank that witnesses this congruence. The equivalent copartition crank is the number of ground parts minus the number of sky parts, and the copartition crank witnesses the congruence cp1,1,2(5n+4)0(mod5)\mathrm{cp}_{1,1,2}(5n+4)\equiv 0\pmod{5}. A combinatorial proof of these congruences would be a welcome addition to the literature.

References

  • [1] George E. Andrews, Integer partitions with even parts below odd parts and the mock theta functions, Ann. Comb. 22 (2018), no. 3, 433–445.
  • [2] Joël Bellaïche, Ben Green, and Kannan Soundararajan, Nonzero coefficients of half-integral weight modular forms mod{\rm mod}\,\ell, Res. Math. Sci. 5 (2018), no. 1, Paper No. 6, 10. MR 3757164
  • [3] Bruce C. Berndt, Ae Ja Yee, and Alexandru Zaharescu, On the parity of partition functions, Internat. J. Math. 14 (2003), no. 4, 437–459. MR 1984662
  • [4] Hannah E. Burson and Dennis Eichhorn, Copartitions, arXiv:2111.04171 (2021).
  • [5] David A. Cox, Primes of the form x2+ny2x^{2}+ny^{2}, A Wiley-Interscience Publication, John Wiley & Sons, Inc., New York, 1989, Fermat, class field theory and complex multiplication.
  • [6] Frank Garvan and Michael J. Schlosser, Combinatorial interpretations of Ramanujan’s tau function, Discrete Math. 341 (2018), no. 10, 2831 – 2840.
  • [7] O. Kolberg, Note on the parity of the partition function, Math. Scand. 7 (1959), 377–378. MR 117213
  • [8] Thomas R. Parkin and Daniel Shanks, On the distribution of parity in the partition function, Math. Comp. 21 (1967), 466–480. MR 227126
  • [9] J. J. Sylvester and F. Franklin, A constructive theory of partitions, arranged in three acts, an interact and an exodion, American Journal of Mathematics 5, 251.

Appendix A Data tables

For readers interested in thinking about Conjecture 4.2 and Question 4.3, we tabulate the proportion of even values of cpa,ma,m(k)\mathrm{cp}_{a,m-a,m}(k) for kk up to 3200032000 for several values of aa and mm. In general, we find that the growth/behavior of the proportion seems roughly consistent when fixing mm and varying aa. For example, in Table A.1, we see that the behavior in each column is roughly the same.

n\quad n #{1kncp1,11,14(k) is even}n\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{1,11,14}(k)\text{ is even}\}}{n} #{1kncp3,11,14(k) is even}n\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{3,11,14}(k)\text{ is even}\}}{n} #{1kncp5,9,14(k) is even}n\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{5,9,14}(k)\text{ is even}\}}{n}
1000\quad 1000 0.5350.535 0.5430.543 0.5300.530
2000\quad 2000 0.5360.536 0.5450.545 0.5360.536
4000\quad 4000 0.5490.549 0.5520.552 0.5430.543
8000\quad 8000 0.5570.557 0.5530.553 0.5540.554
16000\quad 16000 0.5680.568 0.5640.564 0.5650.565
32000\quad 32000 0.5760.576 0.5720.572 0.5730.573
Table A.1: The proportion of even values of cpa,14a,14(k)\mathrm{cp}_{a,14-a,14}(k) for 1kn1\leq k\leq n when aa is 11, 33, and 55.

Since this is the case, in Table A.2 below, we just give #{1kncp1,m1,m(k) is even}n\displaystyle\frac{\#\{1\leq k\leq n\mid\mathrm{cp}_{1,m-1,m}(k)\text{ is even}\}}{n} for several mm between 33 and 3232.

n\mn\backslash m 33 44 55 66 77 88 99 1010 1212
10001000 0.5040.504 0.6020.602 0.5030.503 0.5810.581 0.5000.500 0.6320.632 0.5190.519 0.5530.553 0.4980.498
20002000 0.4950.495 0.6300.630 0.5110.511 0.5990.599 0.5050.505 0.6570.657 0.5000.500 0.5770.577 0.4950.495
40004000 0.4950.495 0.6560.656 0.5090.509 0.6230.623 0.5000.500 0.6810.681 0.5050.505 0.5930.593 0.4990.499
80008000 0.5060.506 0.6810.681 0.5090.509 0.6410.641 0.4930.493 0.7000.700 0.4970.497 0.6080.608 0.4940.494
1600016000 0.5030.503 0.7010.701 0.5080.508 0.6530.653 0.4960.496 0.7190.719 0.4970.497 0.6250.625 0.4990.499
3200032000 0.5070.507 0.7200.720 0.5010.501 0.6710.671 0.4960.496 0.7360.736 0.5020.502 0.6380.638 0.4980.498
n\mn\backslash m 1414 1616 1818 2020 2222 2424 2626 2828 3030 3232
10001000 0.5350.535 0.4800.480 0.5430.543 0.5230.523 0.5400.540 0.4890.489 0.4650.465 0.4900.490 0.4840.484 0.4880.488
20002000 0.5360.536 0.4880.488 0.5500.550 0.5020.502 0.5440.544 0.5080.508 0.5070.507 0.4980.498 0.4950.495 0.5010.501
40004000 0.5490.549 0.4970.497 0.5470.547 0.4990.499 0.5500.550 0.5030.503 0.5170.517 0.5020.502 0.5030.503 0.5030.503
80008000 0.5570.557 0.5010.501 0.5510.551 0.4940.494 0.5660.566 0.4960.496 0.5130.513 0.5020.502 0.5070.507 0.5040.504
1600016000 0.5680.568 0.5040.504 0.5540.554 0.4960.496 0.5660.566 0.4990.499 0.5090.509 0.5010.501 0.5090.509 0.5020.502
3200032000 0.5760.576 0.5040.504 0.5560.556 0.4980.498 0.5750.575 0.5010.501 0.5050.505 0.5000.500 0.5060.506 0.5000.500
Table A.2: The proportion of even values of cp1,m1,m(k)\mathrm{cp}_{1,m-1,m}(k) for 1kn1\leq k\leq n.