1. Introduction
Throughout this paper,
$f_m := \prod _{n\geq 1}(1-q^{mn})$
for a positive integer m and a complex number q with
$|q| < 1$
. A (unrestricted) partition of a positive integer n is a nonincreasing sequence of positive integers (called its parts) whose sum is n. The generating function due to Euler for the number
$p(n)$
of partitions of n with
$p(0):=1$
is
$$ \begin{align*} \sum_{n\geq 0} p(n)q^n = \frac{1}{f_1}. \end{align*} $$
Ramanujan [Reference Hardy, Seshu Aiyar and Wilson7, pages 210–213] showed that
$$ \begin{align} \sum_{n\geq 0} p(5n+4)q^n = 5\frac{f_5^5}{f_1^6}, \end{align} $$
which immediately yields the famous congruence
$$ \begin{align*} p(5n+4) \equiv 0\pmod{5} \end{align*} $$
for all
$n\geq 0$
.
Closely related to Ramanujan’s cubic continued fraction, Chan [Reference Chan3] and Kim [Reference Kim10] introduced a cubic partition of n as a partition of n whose even parts may appear in two colours. The generating function for the number
$a(n)$
of cubic partitions of n with
$a(0):=1$
is
$$ \begin{align*} \sum_{n\geq 0} a(n)q^n = \frac{1}{f_1f_2}. \end{align*} $$
Chan [Reference Chan3] established the identity
$$ \begin{align*} \sum_{n\geq 0} a(3n+2)q^n = 3\frac{f_3^3f_6^3}{f_1^4f_2^4}, \end{align*} $$
analogous to (1.1), which implies that
$$ \begin{align*} a(3n+2)\equiv 0\pmod{3} \end{align*} $$
for all
$n\geq 0$
. Chan and Toh [Reference Chan and Toh4] and Xiong [Reference Xiong11] independently obtained infinite families of congruences for
$a(n)$
modulo powers of
$5$
using modular forms. Subsequently, Hirschhorn [Reference Hirschhorn9] gave elementary proofs of these congruences. For more results on congruences for
$a(n)$
, we refer the reader to [Reference Chern and Dastidar5] and the references therein.
Recently, Amdeberhan et al. [Reference Amdeberhan, Sellers and Singh1] defined a generalised cubic partition of n as a partition of n whose even parts may appear in
$c\geq 1$
different colours. If we let
$a_c(n)$
be the number of such partitions of n with
$a_c(0):=1$
, then its generating function is
$$ \begin{align*} \sum_{n\geq 0} a_c(n)q^n = \frac{1}{f_1f_2^{c-1}}. \end{align*} $$
Using modular forms, they proved the congruences
$$ \begin{align}\ \,a_3(7n+4)&\equiv 0\pmod{7}, \end{align} $$
$$ \begin{align} a_5(11n+10)&\equiv 0\pmod{11}. \end{align} $$
Quite recently, the author [Reference Guadalupe6] provided another proof of (1.2) and (1.3) using classical q-series manipulations and established infinite families of congruences for
$a_c(n)$
modulo primes
$p\not \equiv 1\pmod {8}$
.
The goal of this note is to prove two new infinite families of congruences for
$a_c(n)$
modulo
$5$
.
Theorem 1.1. For all
$c\geq 0$
and
$n\geq 0$
,
$$ \begin{align} a_{25c+4}(25n+18)&\equiv 0\pmod{5},\ \, \end{align} $$
$$ \begin{align} a_{25c+9}(25n+8)&\equiv 0\pmod{5}. \end{align} $$
We prove Theorem 1.1 in Section 2 using elementary q-series techniques and
$5$
-dissections.
2. Proof of Theorem 1.1
To prove Theorem 1.1, we require the following
$5$
-dissections for
$f_1, 1/f_1$
and
$f_2^2/f_1$
. For complex numbers a and q with
$|q| < 1$
, we define
$$ \begin{align*} (a;q)_\infty &:= \prod_{n\geq 0}(1-aq^n), \\ (a_1,a_2,\ldots,a_n;q)_\infty &:= (a_1;q)_\infty(a_2;q)_\infty\cdots(a_n;q)_\infty \end{align*} $$
and
$$ \begin{align*} R_m := R(q^m)\quad\mbox{where } R(q) := \frac{(q,q^4;q^5)_\infty}{(q^2,q^3;q^5)_\infty}. \end{align*} $$
Lemma 2.1. We have the identities
$$ \begin{align*} f_1 &= f_{25}\bigg(\frac{1}{R_5}-q-q^2R_5\bigg),\\ \frac{1}{f_1} &= \frac{f_{25}^5}{f_5^6}\bigg(\frac{1}{R_5^4}+\frac{q}{R_5^3}+\frac{2q^2}{R_5^2}+\frac{3q^3}{R_5}+5q^4-3q^5R_5+2q^6R_5^2-q^7R_5^3+q^8R_5^4\bigg),\\ \frac{f_2^2}{f_1} &= (-q^{10},-q^{-15},q^{25};q^{25})_\infty+q(-q^5,-q^{-20},q^{25};q^{25})_\infty+q^3\frac{f_{50}^2}{f_{25}}. \end{align*} $$
Proof. See [Reference Hirschhorn8, (8.1.4), (8.4.4)] for the first two identities. The third identity follows from [Reference Berndt2, page 49, Corollary (ii)].
Proof of Theorem 1.1.
We first prove (1.4). Note that, by the binomial theorem,
$f_k^5\equiv f_{5k}\pmod {5}$
for any
$k\geq 1$
. Thus, by Lemma 2.1, modulo
$5$
,
$$ \begin{align*} \sum_{n\geq 0} a_{25c+4}(n)q^n &= \frac{1}{f_1f_2^{25c+3}}\equiv \frac{1}{f_{10}f_{50}^c}\cdot\frac{f_2^2}{f_1}\\ &\equiv \frac{1}{f_{10}f_{50}^c}\bigg((-q^{10},-q^{-15},q^{25};q^{25})_\infty+q(-q^5,-q^{-20},q^{25};q^{25})_\infty+q^3\frac{f_{50}^2}{f_{25}}\bigg). \end{align*} $$
Extracting the terms involving
$q^{5n+3}$
, dividing both sides by
$q^3$
and then replacing
$q^5$
with q yields
$$ \begin{align*} \sum_{n\geq 0} a_{25c+4}(5n+3)q^n &\equiv \frac{1}{f_2f_{10}^c}\cdot\frac{f_{10}^2}{f_5}\\ &\equiv\frac{1}{f_5f_{10}^{c-2}}\cdot \frac{f_{50}^5}{f_{10}^6}\bigg(\frac{1}{R_{10}^4}+\frac{q^2}{R_{10}^3}+ \frac{2q^4}{R_{10}^2}+\frac{3q^6}{R_{10}}+5q^8-3q^{10}R_{10} \\ &\qquad\qquad\qquad\qquad +2q^{12}R_{10}^2-q^{14}R_{10}^3+q^{16}R_{10}^4\bigg)\pmod{5}. \end{align*} $$
Extracting again the terms involving
$q^{5n+3}$
, dividing both sides by
$q^3$
and then replacing
$q^5$
with q, we arrive at
$$ \begin{align*} \sum_{n\geq 0} a_{25c+4}(25n+18)q^n &\equiv 5q\frac{f_{10}^5}{f_1f_2^{c+4}}\equiv 0\pmod{5}, \end{align*} $$
which yields (1.4).
We next prove (1.5). Modulo
$5$
,
$$ \begin{align*} \sum_{n\geq 0} a_{25c+9}(n)q^n &= \frac{1}{f_1f_2^{25c+8}}\equiv \frac{1}{f_{10}^2f_{50}^c}\cdot\frac{f_2^2}{f_1}\\ &\equiv \frac{1}{f_{10}^2f_{50}^c}\bigg((-q^{10},-q^{-15},q^{25};q^{25})_\infty+q(-q^5,-q^{-20},q^{25};q^{25})_\infty+q^3\frac{f_{50}^2}{f_{25}}\bigg). \end{align*} $$
We consider the terms involving
$q^{5n+3}$
, divide both sides by
$q^3$
and then replace
$q^5$
with q giving
$$ \begin{align*} \sum_{n\geq 0} a_{25c+9}(5n+3)q^n &\equiv \frac{1}{f_2^2f_{10}^c}\cdot\frac{f_{10}^2}{f_5}\equiv\frac{f_2^3}{f_5f_{10}^{c-1}} \equiv \frac{f_{50}^3}{f_5f_{10}^{c-1}}\bigg(\frac{1}{R_{10}}-q^2-q^4R_{10}\bigg)^3\\ &\equiv \frac{f_{50}^3}{f_5f_{10}^{c-1}}\bigg(\frac{1}{R_{10}^3}-\frac{3q^2}{R_{10}^2}+5q^6-3q^{10}R_{10}^2-q^{12}R_{10}^3\bigg)\pmod{5}. \end{align*} $$
We finally consider the terms involving
$q^{5n+1}$
, divide both sides by q and then replace
$q^5$
with q, arriving at
$$ \begin{align*} \sum_{n\geq 0} a_{25c+9}(25n+8)q^n &\equiv 5q\frac{f_{10}^3}{f_1f_2^{c-1}}\equiv 0\pmod{5}, \end{align*} $$
which gives (1.5).
Acknowledgement
The author would like to thank the anonymous referee for helpful comments that improved this paper.
