Ramanujan-type Congruences for Overpartitions Modulo 5

TL;DR

This paper establishes new Ramanujan-type congruences for overpartition numbers modulo 5, using modular forms, theta function parametrizations, and Hecke operators, extending known results and discovering infinite families of congruences.

Contribution

It introduces novel congruences for overpartitions modulo 5, employing advanced modular form techniques and Hecke operators, expanding the understanding of overpartition congruences.

Findings

Proves overpartition congruences modulo 5 involving powers of 4 and 5.

Derives infinite families of congruences using Hecke eigenforms.

Establishes relations between overpartition counts modulo 8 and 5.

Abstract

Let $\overline{p}(n)$ denote the number of overpartitions of $n$. Hirschhorn and Sellers showed that $\overline{p}(4n+3)\equiv 0 \pmod{8}$ for $n\geq 0$. They also conjectured that $\overline{p}(40n+35)\equiv 0 \pmod{40}$ for $n\geq 0$. Chen and Xia proved this conjecture by using the $(p,k)$-parametrization of theta functions given by Alaca, Alaca and Williams. In this paper, we show that $\overline{p}(5n)\equiv (-1)^{n}\overline{p}(4\cdot 5n) \pmod{5}$ for $n \geq 0$ and $\overline{p}(n)\equiv (-1)^{n}\overline{p}(4n)\pmod{8}$ for $n \geq 0$ by using the relation of the generating function of $\overline{p}(5n)$ modulo $5$ found by Treneer and the $2$-adic expansion of the generating function of $\overline{p}(n)$ due to Mahlburg. As a consequence, we deduce that $\overline{p}(4^k(40n+35))\equiv 0 \pmod{40}$ for $n,k\geq 0$. Furthermore, applying the Hecke operator on $\phi(q)^3$ and the fact that $\phi(q)^3$ is a Hecke eigenform, we obtain an infinite family of congrences $\overline{p}(4^k \cdot5\ell^2n)\equiv 0 \pmod{5}$, where $k\ge 0$ and $\ell$ is a prime such that $\ell\equiv3 \pmod{5}$ and $\left(\frac{-n}{\ell}\right)=-1$. Moreover, we show that $\overline{p}(5^{2}n)\equiv \overline{p}(5^{4}n) \pmod{5}$ for $n \ge 0$. So we are led to the congruences $\overline{p}\big(4^k5^{2i+3}(5n\pm1)\big)\equiv 0 \pmod{5}$ for $n, k, i\ge 0$. In this way, we obtain various Ramanujan-type congruences for $\overline{p}(n)$ modulo $5$ such as $\overline{p}(45(3n+1))\equiv 0 \pmod{5}$ and $\overline{p}(125(5n\pm 1))\equiv 0 \pmod{5}$ for $n\geq 0$.