A Parity Result for Some $p$-Regular Partitions

1. Introduction

Let $n\in {\mathbb{Z}}_{\ge 0}$. A partition of $n$ is a sequence $({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{\ell })$ satisfying ${\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_{\ell }\ge 1$ and $\sum _{i=1}^{\ell }{\lambda }_i=n$. The number of partitions of $n$ is usually denoted by $p(n)$ and $p(0)$ is defined to be 1. For instance, there are 7 partitions of 5, namely; $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1)$ and $(1,1,1,1,1)$. Thus $p(5)=7$. The function $p(n)$ is called the (unrestricted) partition function.

At times, further restrictions on parts of partitions are imposed. If that happens, the partition enumerating function in question is called restricted partition function. One of such examples is the number of $p$-regular partitions. A partition is $p$-regular if none of its parts is divisible by $p$.

Parity results for the number of $p$-regular partitions have been recorded. The interested reader is referred to [1-3] for related work. We shall use the following $q$-series notation $$(a;q{)}_{\mathrm{\infty }}=\prod _{n=1}^{\mathrm{\infty }}(1–a{q}^n).$$ Using the notation above, the generating function for $p(n)$ is given by $$\begin{array}{}\text{(1)}& \sum _{n=0}^{\mathrm{\infty }}p(n)q^n=\frac{1}{(q;q{)}_{\mathrm{\infty }}}.\end{array}$$ See [4]. Furthermore, we recall the following $q$-identities: $$\begin{array}{}\text{(2)}& \sum _{j=0}^{\mathrm{\infty }}(-1{)}^j(2j+1)q^{\frac{j(j+1)}{2}}=(q;q{)}_{\mathrm{\infty }}^3,\end{array}$$ $$\begin{array}{}\text{(3)}& \sum _{j=-\mathrm{\infty }}^{\mathrm{\infty }}(-1{)}^j{q}^{\frac{j(3j–1)}{2}}=(q;q{)}_{\mathrm{\infty }}.\end{array}$$ See [4,5]. For a prime $p$ greater than 3, Sellers [6] gave parity results for $p$-regular partitions into distinct parts. We relax this condition of parts being distinct, and allow repetitions of parts up to three times. Let $c_p(n)$ denote the number of $p$-regular partitions of $n$ in which each part cannot appear more than three times. In this note, we prove the following result.

2. Proof of Theorem 1

The generating function of $c_p(n)$ is $$\sum _{n=0}^{\mathrm{\infty }}{c}_p(n)q^n=\prod _{n=1}^{\mathrm{\infty }}\frac{1+q^n+q^{2n}+q^{3n}}{1+q^{pn}+q^{2pn}+q^{3pn}}.$$ Thus, by (1), (2) and (3), $$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{c}_p(n)q^n& =\frac{(q^4;q^4{)}_{\mathrm{\infty }}}{(q;q{)}_{\mathrm{\infty }}}\frac{(q^p;q^p{)}_{\mathrm{\infty }}}{(q^{4p};q^{4p}{)}_{\mathrm{\infty }}}\\ & \equiv \frac{(q;q{)}_{\mathrm{\infty }}^4}{(q;q{)}_{\mathrm{\infty }}}\frac{(q^p;q^p{)}_{\mathrm{\infty }}}{(q^{4p};q^{4p}{)}_{\mathrm{\infty }}}(\mathrm{mod}2)\\ & =\frac{(q;q{)}_{\mathrm{\infty }}^3(q^p;q^p{)}_{\mathrm{\infty }}}{(q^{4p};q^{4p}{)}_{\mathrm{\infty }}}\\ & \equiv \sum _{j=0}^{\mathrm{\infty }}{q}^{\frac{j(j+1)}{2}}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{q}^{\frac{pk(3k–1)}{2}}\sum _{l=0}^{\mathrm{\infty }}p(l)q^{4pl}(\mathrm{mod}2).\end{array}$$ Suppose $pn+r=\frac{j(j+1)}{2}+\frac{pk(3k–1)}{2}+4pl$ for some integers $j,l\ge 0$ and $k$. Then $$r\equiv \frac{j(j+1)}{2}(\mathrm{mod}p)$$ which implies $$8r+1\equiv (2j+1{)}^2(\mathrm{mod}p).$$ Hence, if $8r+1$ is a quadratic non-residue modulo $p$, $pn+r$ cannot be represented as a sum $\frac{j(j+1)}{2}+\frac{pk(3k–1)}{2}+4pl$ for some integers $j,l\ge 0$ and $k$. Therefore, we must have $$c_p(pn+r)\equiv 0(\mathrm{mod}2).$$

Conflict of Interest

The authors declare no conflict of interest