Congruences for partition functions related to mock theta functions

Abstract

Partitions associated with mock theta functions have received a great deal of attention in the literature. Recently, Choi and Kim derived several partition identities from the third- and sixth-order mock theta functions. In addition, three Ramanujan-type congruences were established by them. In this paper, we present some new congruences for these partition functions.

1 Introduction

A partition of a positive integer n is a finite nonincreasing sequence of positive integers whose sum equals n. Furthermore, a partition is called a t-core partition if there are no hook numbers being multiples of t. Let $a_t(n)$ be the number of t-core partitions of n. It is known [18] that

$$\begin{aligned} \sum _{n=0}^\infty a_t(n)q^n=\frac{(q^t;q^t)_\infty ^t}{(q;q)_\infty }. \end{aligned}$$

Here and in what follows, we make use of the standard q-series notation (cf. [19]).

$$\begin{aligned} (a)_n=(a;q)_n&:=\prod _{k=0}^{n-1}(1-aq^{k}),\\ (a)_\infty =(a;q)_{\infty }&:=\prod _{k= 0}^\infty (1-aq^{k}),\\ (a_1,a_2,\cdots ,a_m;q)_\infty&:=(a_1;q)_\infty (a_2;q)_\infty \cdots (a_m;q)_\infty . \end{aligned}$$

In addition, the cubic partition, which was introduced by Chan [11, 12] and named by Kim [21] in connection with Ramanujan’s cubic continued fractions, is a 2-color partition where the second color appears only in multiples of 2. Let a(n) denote the number of cubic partitions of n, then its generating function is

$$\begin{aligned} \sum _{n=0}^\infty a(n)q^n=\frac{1}{(q;q)_\infty (q^2;q^2)_\infty }. \end{aligned}$$

In his last letter to Hardy [9, pp. 220–223], Ramanujan defined 17 functions, which he called mock theta functions. Since then, there has been an intensive study of partition interpretations for mock theta functions; see [2,3,4,5,6].

Recently, Choi and Kim [15] obtained the following identity related to the third-order mock theta function,

$$\begin{aligned} \upsilon (q)+\upsilon _3(q,q;q)=2\frac{\left( q^4;q^4\right) _\infty ^3}{\left( q^2;q^2\right) _\infty ^2}, \end{aligned}$$

where $\upsilon (q)$ is the third mock theta function and $\upsilon _3(q,q;q)$ is defined by Choi [14],

$$\begin{aligned} \upsilon (q)=\sum _{n=0}^\infty \frac{q^{n(n+1)}}{(-q;q^2)_{n+1}},\quad \upsilon _3(q,q;q)=\sum _{n=0}^\infty q^n (-q;q^2)_n. \end{aligned}$$

We remark that $\upsilon _3(q,q;q)$ is, in fact, identical to $\upsilon (-q)$; see Fine’s book [17, Eq. (26.85)].

Choi and Kim also gave the following identities related to the sixth-order mock theta functions:

$$\begin{aligned} \varPsi (q)+2\varPsi _-(q)= & {} 3\frac{q(q^6;q^6)_\infty ^3}{(q;q)_\infty (q^2;q^2)_\infty },\\ 2\rho (q)+\lambda (q)= & {} 3\frac{(q^3;q^3)_\infty ^3}{(q;q)_\infty (q^2;q^2)_\infty }, \end{aligned}$$

where $\varPsi (q)$, $\varPsi _-(q)$, $\rho (q)$, and $\lambda (q)$ are the sixth-order mock theta functions,

$$\begin{aligned}&\varPsi (q)=\sum _{n=0}^\infty \frac{(-1)^{n}q^{(n+1)^2}(q;q^2)_{n}}{(-q;q)_{2n+1}}, \quad \varPsi _-(q)=\sum _{n=1}^\infty \frac{q^n(-q;q)_{2n-2}}{(q;q^2)_{n}},\\&\rho (q)=\sum _{n=0}^\infty \frac{q^{\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) }(-q;q)_n}{(q;q^2)_{n+1}}, \quad \lambda (q)=\sum _{n=0}^\infty \frac{(-1)^nq^n(q;q^2)_n}{(-q;q)_{n}}. \end{aligned}$$

Meanwhile, Choi and Kim studied three analogous partition functions defined by

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n&=\frac{(q^4;q^4)_\infty ^3}{(q^2;q^2)_\infty ^2}, \end{aligned}$$

(1)

$$\begin{aligned} \sum _{n=0}^\infty c(n)q^n&=\frac{q(q^6;q^6)_\infty ^3}{(q;q)_\infty (q^2;q^2)_\infty },\end{aligned}$$

(2)

$$\begin{aligned} \sum _{n=0}^\infty d(n)q^n&=\frac{(q^3;q^3)_\infty ^3}{(q;q)_\infty (q^2;q^2)_\infty }, \end{aligned}$$

(3)

where b(n) denotes the number of partition pairs $(\lambda ,\sigma )$; $\sigma$ is a partition into distinct even parts; and $\lambda$ is a partition into even parts of which 2-modular diagram is 2-core, and both c(n) and d(n) can be regarded as 3-core cubic partitions.

In this paper, we mainly study Ramanujan-type congruences for these partition functions. This paper is organized as follows: In Sect. 2, we introduce some preliminary results. In the next two sections, we will prove some Ramanujan-type congruences for b(n) and c(n), respectively. In Sect. 5, by employing p-dissection formulas of Ramanujan’s theta functions $\psi (q)$ and $f(-q)$ established by Cui and Gu [16] as well as (p, k)-parameter representations due to Alaca and Williams [1], we show some congruences for d(n). Finally, we end this paper with several open problems.

2 Preliminaries

Let f(a, b) be Ramanujan’s general theta function given by

$$\begin{aligned} f(a,b)=\sum _{n=-\infty }^{\infty }a^{\frac{n(n+1)}{2}}b^{\frac{n(n-1)}{2}},\quad |ab|<1. \end{aligned}$$

We now introduce the following Ramanujan’s classical theta functions:

$$\begin{aligned} \varphi (q)&:=f(q,q)=\sum _{n=-\infty }^{\infty }q^{n^2} =\frac{f^5_2}{f^2_1f^2_4}, \end{aligned}$$

(4)

$$\begin{aligned} \psi (q)&:=f(q,q^3)=\sum _{n=0}^{\infty }q^{\frac{n(n+1)}{2}} =\frac{f^2_2}{f_1},\end{aligned}$$

(5)

$$\begin{aligned} f(-q)&:=f(-q,-q^2)=\sum _{n=-\infty }^{\infty }(-1)^nq^{\frac{n(3n+1)}{2}}=f_1. \end{aligned}$$

(6)

One readily verifies

$$\begin{aligned} \varphi (-q)=\frac{f_1^2}{f_2}. \end{aligned}$$

(7)

Here and in the sequel, we write $f_k:=(q^k;q^k)_\infty$ for positive integers k for convenience.

We first require the following 2-dissections.

Lemma 1

It holds that

$$\begin{aligned} \frac{1}{f_1^2}&=\frac{f_8^5}{f_2^5f_{16}^2} +2q\frac{f_4^2f_{16}^2}{f_2^5f_8}, \end{aligned}$$

(8)

$$\begin{aligned} \frac{f_3}{f_1^3}&=\frac{f_4^6f_6^3}{f_2^9f_{12}^2} +3q\frac{f_4^2f_6f_{12}^2}{f_2^7},\end{aligned}$$

(9)

$$\begin{aligned} \frac{f_3^3}{f_1}&=\frac{f_4^3f_6^2}{f_2^2f_{12}} +q\frac{f_{12}^3}{f_4}. \end{aligned}$$

(10)

Proof

Here (8) comes from the 2-dissection of $\varphi (q)$ (cf. [8, p. 40, Entry 25]). For (9) and (10), see [26]. $\square$

The following 3-dissections are also necessary.

Lemma 2

It holds that

$$\begin{aligned} \frac{1}{\varphi (-q)}&=\frac{\varphi ^3(-q^9)}{\varphi ^4(-q^3)}\left( 1+2qw(q^3)+4q^2 w^2(q^3)\right) , \end{aligned}$$

(11)

$$\begin{aligned} \frac{1}{\psi (q)}&=\frac{\psi ^3(q^9)}{\psi ^4(q^3)} \left( \frac{1}{w^2(q^3)}-\frac{q}{w(q^3)}+q^2\right) , \end{aligned}$$

(12)

where

$$\begin{aligned} w(q)=\frac{f_1f_6^3}{f_2f_3^3}. \end{aligned}$$

(13)

Furthermore,

$$\begin{aligned} \frac{1}{f_1^3}=\frac{f_9^3}{f_3^{12}}\left( P^2(q^3)+3qP(q^3)f_9^3+9q^2f_9^6\right) , \end{aligned}$$

(14)

where

$$\begin{aligned} P(q)=f_1\left( \frac{\varphi ^3(-q^3)}{\varphi (-q)}+4q\frac{\psi ^3(q^3)}{\psi (q)}\right) . \end{aligned}$$

(15)

Proof

For (11) and (12), see Baruah and Ojah [7]. For (14), see Wang [24]. Note that Wang [24] showed

$$\begin{aligned} P(q)=f_1\left( 1+6\sum _{n\ge 0} \left( \frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right) \right) . \end{aligned}$$

We know from [23, Eqs. (3.2) and (3.5)] that

$$\begin{aligned} 4q\frac{\psi ^3(q^3)}{\psi (q)}= & {} 4\sum _{n\ge 0} \left( \frac{q^{3n+1}}{1-q^{6n+2}}-\frac{q^{3n+2}}{1-q^{6n+4}}\right) ,\\ \frac{\varphi ^3(-q^3)}{\varphi (-q)}= & {} 1+2\sum _{n\ge 0} \left( \frac{q^{6n+1}}{1-q^{6n+1}} +\frac{q^{6n+2}}{1-q^{6n+2}}-\frac{q^{6n+4}}{1-q^{6n+4}} -\frac{q^{6n+5}}{1-q^{6n+5}}\right) . \end{aligned}$$

Hence, (15) follows immediately by the following trivial identity:

$$\begin{aligned} \frac{x}{1-x^2}=\frac{x}{1-x}-\frac{x^2}{1-x^2}. \end{aligned}$$

$\square$

Furthermore, we need

Lemma 3

([16, Theorem 2.1]) For any odd prime p,

$$\begin{aligned} \psi (q)=q^{\frac{p^{2}-1}{8}}\psi \big (q^{p^{2}}\big ) +\sum _{k=0}^{\frac{p-3}{2}}q^{\frac{k^{2}+k}{2}} f\left( q^{\frac{p^{2}+(2k+1)p}{2}},q^{\frac{p^{2}-(2k+1)p}{2}}\right) . \end{aligned}$$

We further claim that for $0\le k\le (p-3)/2$,

$$\begin{aligned} \frac{k^{2}+k}{2} \not \equiv \frac{p^{2}-1}{8} \pmod p. \end{aligned}$$

Lemma 4

([16, Theorem 2.2]) For any prime $p\ge 5$,

$$\begin{aligned} f(-q)&=(-1)^{\frac{\pm p-1}{6}}q^{\frac{p^{2}-1}{24}}f\big (-q^{p^{2}}\big )\\&\quad \quad +\sum _{\begin{array}{c} k=-\frac{p-1}{2}\\ k\ne \frac{\pm p-1}{6} \end{array}}^{\frac{p-1}{2}}(-1)^{k}q^{\frac{3k^{2}+k}{2}} f\left( -q^{\frac{3p^{2}+(6k+1)p}{2}},-q^{\frac{3p^{2}-(6k+1)p}{2}}\right) . \end{aligned}$$

We further claim that for $-(p-1)/2\le k\le (p-1)/2$ and $k\ne (\pm p-1)/6$,

$$\begin{aligned} \frac{3k^{2}+k}{2} \not \equiv \frac{p^{2}-1}{24} \pmod {p}. \end{aligned}$$

Here for any prime $p\ge 5$,

$$\begin{aligned} \frac{\pm p-1}{6}:=\left\{ \begin{array}{ll} \frac{p-1}{6},&{}\quad \ p \equiv 1 \pmod {6},\\ \frac{-p-1}{6},&{}\quad \ p \equiv -1 \pmod {6}.\end{array}\right. \end{aligned}$$

At last, we require the following relations due to Alaca and Williams [1].

Lemma 5

Let

$$\begin{aligned} p=p(q):=\frac{\varphi ^2(q)-\varphi ^2(q^3)}{2\varphi ^2(q^3)}, \end{aligned}$$

and

$$\begin{aligned} k=k(q):=\frac{\varphi ^3(q^3)}{\varphi (q)}. \end{aligned}$$

Then

$$\begin{aligned} f_1&=2^{-\frac{1}{6}}q^{-\frac{1}{24}}p^{\frac{1}{24}}(1-p)^{\frac{1}{2}} (1+p)^{\frac{1}{6}}(1+2p)^{\frac{1}{8}}(2+p)^{\frac{1}{8}}k^{\frac{1}{2}},\\ f_2&=2^{-\frac{1}{3}}q^{-\frac{1}{12}}p^{\frac{1}{12}}(1-p)^{\frac{1}{4}} (1+p)^{\frac{1}{12}}(1+2p)^{\frac{1}{4}}(2+p)^{\frac{1}{4}}k^{\frac{1}{2}},\\ f_3&=2^{-\frac{1}{6}}q^{-\frac{1}{8}}p^{\frac{1}{8}}(1-p)^{\frac{1}{6}} (1+p)^{\frac{1}{2}}(1+2p)^{\frac{1}{24}}(2+p)^{\frac{1}{24}}k^{\frac{1}{2}},\\ f_4&=2^{-\frac{2}{3}}q^{-\frac{1}{6}}p^{\frac{1}{6}}(1-p)^{\frac{1}{8}} (1+p)^{\frac{1}{24}}(1+2p)^{\frac{1}{8}}(2+p)^{\frac{1}{2}}k^{\frac{1}{2}},\\ f_6&=2^{-\frac{1}{3}}q^{-\frac{1}{4}}p^{\frac{1}{4}}(1-p)^{\frac{1}{12}} (1+p)^{\frac{1}{4}}(1+2p)^{\frac{1}{12}}(2+p)^{\frac{1}{12}}k^{\frac{1}{2}},\\ f_{12}&=2^{-\frac{2}{3}}q^{-\frac{1}{2}}p^{\frac{1}{2}}(1-p)^{\frac{1}{24}} (1+p)^{\frac{1}{8}}(1+2p)^{\frac{1}{24}}(2+p)^{\frac{1}{6}}k^{\frac{1}{2}}. \end{aligned}$$

3 Congruences for b(n)

Theorem 1

For $n\ge 0$, $\alpha \ge 1$, and prime $p\ge 5$, we have

$$\begin{aligned} b\left( p^{2\alpha }n+\frac{(3j+p)p^{2\alpha -1}-1}{3}\right) \equiv 0\pmod 2, \end{aligned}$$

(16)

where $j=1$, 2, $\ldots$, $p-1$.

Proof

In light of (1), we derive that

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n=\frac{f_4^3}{f_2^2}\equiv f_8\pmod 2. \end{aligned}$$

Applying Lemma 4, we deduce that, for any prime $p\ge 5$,

$$\begin{aligned} \sum _{n=0}^\infty b\left( pn+\frac{p^2-1}{3}\right) q^n \equiv (-1)^{\frac{\pm p-1}{6}} f(-q^{8p})\pmod 2, \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^\infty b\left( p^2n+\frac{p^2-1}{3}\right) q^n\equiv (-1)^{\frac{\pm p-1}{6}} f(-q^{8})\pmod 2. \end{aligned}$$

Moreover,

$$\begin{aligned} \sum _{n=0}^\infty b\left( p^{3}n+\frac{p^{4}-1}{3}\right) q^n&\equiv f(-q^{8p})\pmod 2. \end{aligned}$$

Hence, by induction on $\alpha$, we derive that, for $\alpha \ge 1$,

$$\begin{aligned} \sum _{n=0}^\infty b\left( p^{2\alpha -1}n+\frac{p^{2\alpha }-1}{3}\right) q^n&\equiv (-1)^{\alpha \left( \frac{\pm p-1}{6}\right) } f(-q^{8p})\pmod 2. \end{aligned}$$

This immediately leads to

$$\begin{aligned} b\left( p^{2\alpha -1}(pn+j)+\frac{p^{2\alpha }-1}{3}\right) \equiv 0\pmod 2, \end{aligned}$$

for $j=1$, 2, $\ldots$, $p-1$. $\square$

Remark 1

When studying the 1-shell totally symmetric plane partition function f(n) (which is different to Ramanujan’s theta function $f(-q)$ given in Sect. 2) introduced by Blecher [10], Hirschhorn and Sellers [20] proved that, for $n\ge 1$,

$$\begin{aligned} f(3n-2)=h(n), \end{aligned}$$

with

$$\begin{aligned} \sum _{n=0}^\infty h(2n+1)q^n =\frac{f_2^3}{f_1^2}. \end{aligned}$$

A couple of congruences modulo powers of 2 and 5 for h(n) have been obtained subsequently; see [13, 25, 27]. We see from (1) that

$$\begin{aligned} b(2n)=h(2n+1). \end{aligned}$$

One therefore may obtain some congruences for b(n) as well. For example,

$$\begin{aligned} b(8n+6)\equiv 0\pmod 4. \end{aligned}$$

4 Congruences for c(n)

Theorem 2

For $n\ge 0$, we have

$$\begin{aligned} c(27n+24)\equiv 0\pmod 9. \end{aligned}$$

(17)

Proof

We see from (2) and Lemma 2 that

$$\begin{aligned} \sum _{n=0}^\infty c(n)q^n&=\frac{qf_6^3}{\varphi (-q)\psi (q)}\nonumber \\&=qf_6^3\frac{\varphi ^3(-q^9)\psi ^3(q^9)}{\varphi ^4(-q^3)\psi ^4(q^3)} \big (1+2qw(q^3)+4q^2w^2(q^3)\big )\nonumber \\&\quad \times \left( \frac{1}{w^2(q^3)} -\frac{q}{w(q^3)}+q^2\right) . \end{aligned}$$

Employing Lemma 2, we deduce that

$$\begin{aligned} \sum _{n=0}^\infty c(3n)q^n&=\frac{3q\varphi ^3(-q^3)\psi ^3(q^3)}{f_1^3\varphi (-q)\psi (q)}\nonumber \\&=\frac{3q\varphi ^3(-q^9)\psi ^3(q^9)f_9^3}{\varphi (-q^3)\psi (q^3)f_3^{12}} \left( P^2(q^3)+3qP(q^3)f_9^3+9q^2f_9^6\right) \nonumber \\&\quad \times \left( 1+2qw(q^3)+4q^2 w^2(q^3)\right) \left( \frac{1}{w^2(q^3)}-\frac{q}{w(q^3)}+q^2\right) . \end{aligned}$$

(18)

Extracting terms involving $q^{3n+2}$ and replacing $q^3$ by q in (18), it follows that

$$\begin{aligned} \sum _{n=0}^\infty c(9n+6)q^n=12\frac{f_2^2f_3^{21}}{f_1^{16}f_6^6}+135q\frac{f_3^{12}f_6^3}{f_1^{13}f_2} +72q^2\frac{f_3^3f_6^{12}}{f_1^{10}f_2^4}+192q^3\frac{f_6^{21}}{f_1^7f_2^7f_3^6}. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{n=0}^\infty c(9n+6)q^n&\equiv 3\frac{f_2^2f_3^{21}}{f_1^{16}f_6^6}+3q^3\frac{f_6^{21}}{f_1^7f_2^7f_3^6}\\&\equiv 3 \frac{f_2^2}{f_1}\left( \frac{f_3^{16}}{f_6^6}+ q^3\frac{f_6^{18}}{f_3^8}\right) \pmod {9}. \end{aligned}$$

Noting that $f_2^2/f_1$ contains no terms of the form $q^{3n+2}$, we have

$$\begin{aligned} \sum _{n=0}^\infty c(27n+24)q^n\equiv 0\pmod 9. \end{aligned}$$

$\square$

Theorem 3

For $n\ge 0$, we have

$$\begin{aligned} c(45n+t)\equiv 0\pmod 5, \end{aligned}$$

(19)

where $t=9$ and 18.

Proof

Referring to (18), we have

$$\begin{aligned} \sum _{n=0}^\infty c(9n)q^n=45q\frac{f_2f_3^{18}}{f_1^{15}f_6^3}+90q^2 \frac{f_3^9f_6^6}{f_1^{12}f_2^2}+288q^3\frac{f_6^{15}}{f_1^9f_2^5}. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{n=0}^\infty c(9n)q^n\equiv 3q^3f_1\frac{f_{30}^{3}}{f_5^2f_{10}}\pmod 5. \end{aligned}$$

Since $f_1$ contains no terms of the form $q^{5n+3}$ and $q^{5n+4}$, we have

$$\begin{aligned} c(9(5n+1))=c(45n+9)\equiv 0\pmod 5, \end{aligned}$$

and

$$\begin{aligned} c(9(5n+2))=c(45n+18)\equiv 0\pmod 5. \end{aligned}$$

This yields that (19). $\square$

Corollary 1

For $n\ge 0$, we have

$$\begin{aligned} c(45n+t)\equiv 0\pmod {15}, \end{aligned}$$

(20)

where $t=9$ and 18.

Proof

We know from [15, Theorem 4.2] that

$$\begin{aligned} c(3n)\equiv 0 \pmod {3}, \end{aligned}$$

which is indeed a direct consequence of (18). Hence, Corollary 1 follows by Theorem 3. $\square$

5 Congruences for d(n)

Theorem 4

For $n\ge 0$, $\alpha \ge 1$, and prime $p\ge 3$, we have

$$\begin{aligned} d\left( 2p^{2\alpha }+\frac{(8j+p)p^{2\alpha -1}-1}{4}\right) \equiv 0\pmod 2, \end{aligned}$$

(21)

where $j=1$, 2, $\ldots$, $p-1$.

Proof

From (3), one can see

$$\begin{aligned} \sum _{n=0}^\infty d(n)q^n=\frac{f_3^3}{f_1 f_2}\equiv f_6\frac{f_3}{f_1^3}\pmod 2. \end{aligned}$$

With the help of (9), we have

$$\begin{aligned} \sum _{n=0}^\infty d(n)q^n\equiv \frac{f_4^6f_6^4}{f_2^9f_{12}^2}+3q\frac{f_4^2f_6^2f_{12}^2}{f_2^7}\pmod 2. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{n=0}^\infty d(2n)q^n\equiv \frac{f_2^6f_3^4}{f_1^9f_{6}^2}\equiv \psi (q)\pmod 2. \end{aligned}$$

Invoking Lemma 3, for any odd prime p, we derive that

$$\begin{aligned} \sum _{n=0}^\infty d \left( 2\left( pn+\frac{p^2-1}{8}\right) \right) q^n\equiv \psi (q^p)\pmod 2, \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^\infty d \left( 2\left( p^2n+\frac{p^2-1}{8}\right) \right) q^n\equiv \psi (q)\pmod 2. \end{aligned}$$

Furthermore,

$$\begin{aligned} \sum _{n=0}^\infty d \left( 2p^{3}n+\frac{p^4-1}{4}\right) q^n\equiv \psi (q^p)\pmod 2. \end{aligned}$$

It therefore follows by induction on $\alpha$ that for $\alpha \ge 1$,

$$\begin{aligned} \sum _{n=0}^\infty d \left( 2p^{2\alpha -1}n+\frac{p^{2\alpha }-1}{4}\right) q^n\equiv \psi (q^p)\pmod 2. \end{aligned}$$

Thus, for $j=1$, 2, $\ldots$, $p-1$,

$$\begin{aligned} d \left( 2p^{2\alpha -1}(pn+j)+\frac{p^{2\alpha }-1}{4}\right) \equiv 0\pmod 2, \end{aligned}$$

which is the desired result. $\square$

Theorem 5

For $n\ge 0$, $\alpha \ge 1$, and prime $p\ge 5$, we have

$$\begin{aligned} d\left( 6p^{2\alpha }n+\frac{(24j+p)p^{2\alpha -1}-1}{4}\right) \equiv 0\pmod 3, \end{aligned}$$

(22)

where $j=1$, 2, $\ldots$, $p-1$.

Proof

It follows by (11) and (12) that

$$\begin{aligned} \sum _{n=0}^\infty d(n)&=\frac{f_3^3}{\varphi (-q)\psi (q)}\nonumber \\&=f_3^3 \frac{\varphi ^3(-q^9)\psi ^3(q^9)}{\varphi ^4(-q^3)\psi ^4(q^3)}\left( 1+2qw(q^3)+4q^2 w^2(q^3)\right) \nonumber \\&\quad \times \left( \frac{1}{w^2(q^3)} -\frac{q}{w(q^3)}+q^2\right) . \end{aligned}$$

(23)

So we get

$$\begin{aligned} \sum _{n=0}^\infty d(3n)q^n&=f_1^3\frac{\varphi ^3(-q^3)\psi ^3(q^3)}{\varphi ^4(-q)\psi ^4(q)} \left( \frac{1}{w^2(q)}-2qw(q)\right) \\&=\frac{1}{f_2^2f_6^3}\left( \frac{f_3^3}{f_1}\right) ^3-2q\frac{f_6^6}{f_2^5}. \end{aligned}$$

Based on (10), we derive that

$$\begin{aligned} \sum _{n=0}^\infty d(6n)q^n=\frac{f_2^9f_3^3}{f_1^8f_6^3}+3q\frac{f_2f_6^5}{f_1^2f_3}\equiv \frac{f_2^9f_3^3}{f_1^8f_6^3} \equiv f_1\pmod 3. \end{aligned}$$

Invoking Lemma 4, we arrive at that, for any prime $p\ge 5$,

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( pn+\frac{p^2-1}{24}\right) \right) q^n\equiv (-1)^{\frac{\pm p-1}{6}} f(-q^p)\pmod 3, \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( p^2n+\frac{p^2-1}{24}\right) \right) q^n\equiv (-1)^{\frac{\pm p-1}{6}} f(-q)\pmod 3. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( p^2\left( pn+\frac{p^2-1}{24}\right) +\frac{p^2-1}{24}\right) \right) q^n\equiv f(-q^p)\pmod 3. \end{aligned}$$

Namely,

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6p^3n+\frac{p^4-1}{4}\right) q^n\equiv f(-q^p)\pmod 3. \end{aligned}$$

Thus, by induction on $\alpha$, we derive that, for $\alpha \ge 1$,

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6p^{2\alpha -1}n+\frac{p^{2\alpha }-1}{4}\right) q^n\equiv (-1)^{\alpha \left( \frac{\pm p-1}{6}\right) } f(-q^p)\pmod 3. \end{aligned}$$

This yields that, for $j=1$, 2, $\ldots$, $p-1$,

$$\begin{aligned} d\left( 6p^{2\alpha -1}(pn+j)+\frac{p^{2\alpha }-1}{4}\right) \equiv 0\pmod 3, \end{aligned}$$

which implies (22). $\square$

Theorem 6

For $n\ge 0$, $\alpha \ge 1$, and prime $p\ge 5$, we have

$$\begin{aligned} d\left( 6p^{2\alpha }n+\frac{(24j+9p)p^{2\alpha -1}-1}{4}\right) \equiv 0\pmod 9, \end{aligned}$$

(24)

where $j=1$, 2, $\ldots$, $p-1$.

Proof

Extracting terms involving $q^{3n+2}$ and replace $q^3$ by q in (23), then we derive that

$$\begin{aligned} \sum _{n=0}^\infty d(3n+2)q^n=\frac{3f_3^3 f_6^3}{\varphi (-q)\psi (q)f_2^3}=\frac{3f_3^3 f_6^3}{f_1 f_2^4}. \end{aligned}$$

(25)

It follows by (10) that

$$\begin{aligned} \sum _{n=0}^\infty d(3n+2)q^n=3\frac{f_3^3f_6^3}{f_1f_2^4}=3\frac{f_4^3f_6^5}{f_2^6f_{12}} +3q\frac{f_6^3f_{12}^3}{f_2^4f_4}. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{n=0}^\infty d(6n+2)q^n=3\frac{f_2^3f_3^5}{f_1^6f_6}\equiv 3f_9\pmod {9}. \end{aligned}$$

In view of Lemma 4, for any prime $p\ge 5$, we deduce that

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( pn+\frac{3(p^2-1)}{8}\right) +2\right) q^n\equiv 3(-1)^{\frac{\pm p-1}{6}} f(-q^{9p})\pmod 9, \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( p^2n+\frac{3(p^2-1)}{8}\right) +2\right) q^n\equiv 3(-1)^{\frac{\pm p-1}{6}} f(-q^9)\pmod 9. \end{aligned}$$

Moreover,

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( p^3n+\frac{3(p^4-1)}{8}\right) +2\right) q^n&\equiv 3f(-q^{9p})\pmod 9. \end{aligned}$$

Hence, by induction on $\alpha \ge 1$, we arrive at

$$\begin{aligned} \sum _{n=0}^\infty d\left( 6\left( p^{2\alpha -1}n+\frac{3(p^{2\alpha }-1)}{8}\right) +2\right) q^n\equiv 3(-1)^{\alpha \left( \frac{\pm p-1}{6}\right) } f(-q^{9p})\pmod 9, \end{aligned}$$

which implies that for $j=1$, 2, $\ldots$, $p-1$,

$$\begin{aligned} d\left( 6\left( p^{2\alpha -1}(pn+j)+\frac{3(p^{2\alpha }-1)}{8}\right) +2\right) \equiv 0\pmod 9. \end{aligned}$$

This leads to (24). $\square$

Theorem 7

For $n\ge 0$, we have

$$\begin{aligned} d(45n+t)\equiv 0 \pmod {5}, \end{aligned}$$

(26)

where $t=17$ and 35.

Proof

From (25), we have

$$\begin{aligned} \sum _{n=0}^\infty d(3n+2)q^n=\frac{3f_3^3 f_6^3}{\varphi (-q)\psi (q)f_2^3}. \end{aligned}$$

Again by (11), (12), and (14), we have

$$\begin{aligned} \sum _{n=0}^\infty d(9n+8)q^n=f_2\cdot H, \end{aligned}$$

where

$$\begin{aligned} H&=\left( \frac{9f_3^9 f_4 f_6^9}{f_1^3 f_2^{13} f_{12}^3}+\frac{9 f_3^3 f_4^2 f_6^{18}}{f_1 f_2^{16} f_{12}^6}\right) +q\left( \frac{27 f_3^6 f_6^9}{f_1^2 f_2^{13}}-\frac{18 f_4 f_6^{18}}{f_2^{16} f_{12}^3}\right) \\&\quad +q^2\left( \frac{36 f_3^9 f_{12}^6}{f_1^{3} f_2^{10} f_4^2}+\frac{72 f_3^3 f_6^9 f_{12}^3}{f_1 f_2^{13} f_4}+\frac{108 f_1 f_6^{18}}{f_{2}^{16} f_3^3}\right) \\&\quad -q^3 \frac{72 f_6^9 f_{12}^6}{f_{2}^{13} f_4^2}+q^4\frac{144 f_3^3 f_{12}^{12}}{f_1 f_2^{10} f_4^4}. \end{aligned}$$

$\square$

We next show a surprising congruence.

Lemma 6

It holds that

$$\begin{aligned} H\equiv 3 \frac{f_{15}^3}{f_5 f_{10}^2} \pmod {5}. \end{aligned}$$

(27)

Proof (Proof of Lemma 6)

To prove (27), it suffices to show

$$\begin{aligned} H-3 \frac{f_3^{15}}{f_1^5 f_2^{10}}\equiv 0 \pmod {5}, \end{aligned}$$

or equivalently,

$$\begin{aligned} \left( H-3 \frac{f_3^{15}}{f_1^5 f_2^{10}}\right) \frac{f_1^5 f_3 f_4^{10} f_6^{10}}{f_2^4 f_{12}^6}\equiv 0 \pmod {5}, \end{aligned}$$

since $\frac{f_1^5 f_3 f_4^{10} f_6^{10}}{f_2^4 f_{12}^6}$ is invertible in the ring $\mathbb {Z}/5\mathbb {Z}[[q]]$. According to Lemma 5, it becomes

$$\begin{aligned} \frac{15 p^2 (1-p) (1+p)^5 (2+p)^2 (2+5p +12 p^2+ 5 p^3+ 2 p^4) k^8}{32 q^2 (1+2p)}\equiv 0 \pmod {5}. \end{aligned}$$

Lemma 6 follows obviously. $\square$

We know from Lemma 6 that

$$\begin{aligned} \sum _{n=0}^\infty d(9n+8)q^n\equiv 3 f_2 \frac{f_{15}^3}{f_5 f_{10}^2} \pmod {5}. \end{aligned}$$

Since $f_2=(q^2;q^2)_\infty$ contains no terms of the form $q^{5n+1}$ and $q^{5n+3}$, we have

$$\begin{aligned} d(9(5n+1)+8)=d(45n+17)\equiv 0 \pmod {5}, \end{aligned}$$

and

$$\begin{aligned} d(9(5n+3)+8)=d(45n+35)\equiv 0 \pmod {5}, \end{aligned}$$

which leads to Theorem 7. $\square$

Corollary 2

For $n\ge 0$, we have

$$\begin{aligned} d(45n+t)\equiv 0 \pmod {15}, \end{aligned}$$

(28)

where $t=17$ and 35.

Proof

Again, we know from [15, Theorem 4.2] that

$$\begin{aligned} d(3n+2)\equiv 0 \pmod {3}. \end{aligned}$$

It indeed follows directly from (25). We thus prove Corollary 2 by Theorem 7. $\square$

6 Final remarks

We end this paper by raising the following congruences.

Question 1

We have

$$\begin{aligned} c(45n+21)&\equiv 0\pmod 5, \end{aligned}$$

(29)

$$\begin{aligned} c(63n+t)&\equiv 0\pmod 7, \end{aligned}$$

(30)

where $t=30$, 48, and 57.

Question 2

We have

$$\begin{aligned} d(45n+41)&\equiv 0\pmod 5, \end{aligned}$$

(31)

$$\begin{aligned} d(63n+t)&\equiv 0\pmod 7, \end{aligned}$$

(32)

where $t=32$, 50, and 59.

All these congruences have been verified by the authors using an algorithm due to Radu and Sellers [22]. However, since the modular form proofs are very routine and tedious, we here want to ask if there exist elementary proofs of these congruences.