Congruences for the coefficients of the Gordon and McIntosh mock theta function $\xi (q)$

Abstract

Recently Gordon and McIntosh introduced the third order mock theta function $\xi (q)$ defined by

$$\begin{aligned} \xi (q)=1+2\sum _{n=1}^{\infty }\frac{q^{6n^2-6n+1}}{(q;q^6)_{n}(q^5;q^6)_{n}}. \end{aligned}$$

Our goal in this paper is to study arithmetic properties of the coefficients of this function. We present a number of such properties, including several infinite families of Ramanujan-like congruences.

1 Introduction

In his last letter to Hardy in 1920, Ramanujan introduced the notion of a mock theta function. He listed 17 such functions having orders 3, 5, and 7. Since then, other mock theta functions have been found. Gordon and McIntosh [8], for example, introduced many additional such functions, including the following of order 3:

$$\begin{aligned} \xi (q)=1+2\sum _{n=1}^{\infty }\frac{q^{6n^2-6n+1}}{(q;q^6)_{n}(q^5;q^6)_{n}}, \end{aligned}$$

(1)

where we use the standard q-series notation:

$$\begin{aligned} (a;q)_0&= 1, \\ (a;q)_n&= (1-a)(1-aq) \cdots (1-aq^{n-1}) \quad \forall n \ge 1, \\ (a;q)_{\infty }&= \lim _{n \rightarrow \infty } (a;q)_n, \quad |q|<1. \end{aligned}$$

Arithmetic properties of the coefficients of mock theta functions have received a great deal of attention. For instance, Zhang and Shi [15] recently proved seven congruences satisfied by the coefficients of the mock theta function $\beta (q)$ introduced by McIntosh. In a recent paper, Brietzke et al. [5] found a number of arithmetic properties satisfied by the coefficients of the mock theta function $V_0(q)$, introduced by Gordon and McIntosh [7]. Andrews et al. [2] prove a number of congruences for the partition functions $p_{\omega }(n)$ and $p_{\nu }(n)$, introduced in [1], associated with the third order mock theta functions $\omega (q)$ and $\nu (q)$, where $\omega (q)$ is defined below and

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

In a subsequent paper, Wang [14] presented some additional congruences for both $p_{\omega }(n)$ and $p_{\nu }(n)$.

This paper is devoted to exploring arithmetic properties of the coefficients $p_{\xi }(n)$ defined by

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(n)q^n = \xi (q). \end{aligned}$$

(2)

It is clear from (1) that $p_{\xi }(n)$ is even for all $n \ge 1$. In Sects. 4 and 5, we present other arithmetic properties of $p_{\xi }(n)$, including some infinite families of congruences.

2 Preliminaries

McIntosh [12, Theorem 3] proved a number of mock theta conjectures, including

$$\begin{aligned} \omega (q)&= g_3(q,q^2) \text {\ \ \ and} \end{aligned}$$

(3)

$$\begin{aligned} \xi (q)&= q^2g_3(q^3,q^6) + \frac{(q^2;q^2)_{\infty }^{4}}{(q;q)_{\infty }^2(q^6;q^6)_{\infty }}, \end{aligned}$$

(4)

where

$$\begin{aligned} g_3(a,q) = \sum _{n=0}^{\infty } \frac{(-q;q)_n q^{n(n+1)/2}}{(a;q)_{n+1}(a^{-1}q;q)_{n+1}} \end{aligned}$$

and $\omega (q)$ is the third order mock theta functions given by

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

It follows from (1), (3), and (4) that

$$\begin{aligned} \xi (q) = q^2\omega (q^3) + \frac{(q^2;q^2)_{\infty }^{4}}{(q;q)_{\infty }^2(q^6;q^6)_{\infty }}. \end{aligned}$$

(5)

Throughout the remainder of this paper, we define

$$\begin{aligned} f_k := (q^k;q^k)_{\infty } \end{aligned}$$

in order to shorten the notation. Combining (5) and (2), we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(n)q^n = q^2\omega (q^3) + \frac{f_2^{4}}{f_1^2f_6}. \end{aligned}$$

(6)

We recall Ramanujan’s theta functions

$$\begin{aligned} f(a,b)&:=\sum _{n=-\infty }^\infty a^\frac{n(n+1)}{2}b^\frac{n(n-1)}{2} \text{ for } |ab|<1, \nonumber \\ \phi (q)&:= f(q,q) = \sum _{n=-\infty }^{\infty } q^{n^2} = \frac{f_2^5}{f_1^2f_4^2}, \text {\ \ and} \end{aligned}$$

(7)

$$\begin{aligned} \psi (q)&:= f(q,q^3) = \sum _{n=0}^{\infty } q^{n(n+1)/2} = \frac{f_2^2}{f_1}. \end{aligned}$$

(8)

The function $\phi (q)$ satisfies many identities, including (see [3, (22.4)])

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

(9)

In some of the proofs, we employ the classical Jacobi’s identity (see [4, Theorem 1.3.9])

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

(10)

We note the following identities which will be used below.

Lemma 1

The following 2-dissection identities hold:

$$\begin{aligned} \frac{1}{f_{1}^{2}}&= \frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}} + 2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}, \end{aligned}$$

(11)

$$\begin{aligned} {f_{1}^{2}}&= \frac{f_{2}f_8^5}{f_{4}^{2}f_{16}^{2}} -2q\frac{f_{2}f_{16}^{2}}{f_{8}}, \end{aligned}$$

(12)

$$\begin{aligned} \frac{1}{f_{1}^{4}}&= \frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}} + 4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}, \end{aligned}$$

(13)

$$\begin{aligned} \displaystyle \frac{f_{3}}{f_{1}}&= \displaystyle \frac{f_{4}f_{6}f_{16}f_{24}^{2}}{f_{2}^{2}f_{8}f_{12}f_{48}} + q\displaystyle \frac{f_{6}f_{8}^{2}f_{48}}{f_{2}^{2}f_{16}f_{24}}, \end{aligned}$$

(14)

$$\begin{aligned} \displaystyle \frac{f_{3}^2}{f_{1}^2}&= \displaystyle \frac{f_{4}^4f_{6}f_{12}^{2}}{f_{2}^{5}f_{8}f_{24}} + 2q\displaystyle \frac{f_{4}f_{6}^{2}f_{8}f_{24}}{f_{2}^{4}f_{12}}, \end{aligned}$$

(15)

$$\begin{aligned} \displaystyle \frac{f_{1}^3}{f_{3}}&= \displaystyle \frac{f_{4}^3}{f_{12}} - 3q\displaystyle \frac{f_{2}^{2}f_{12}^3}{f_{4}f_{6}^2} \end{aligned}$$

(16)

$$\begin{aligned} \displaystyle \frac{f_{3}}{f_{1}^3}&= \displaystyle \frac{f_{4}^6f_{6}^3}{f_{2}^{9}f_{12}^2} + 3q\displaystyle \frac{f_{4}^2f_{6}f_{12}^2}{f_{2}^{7}}, \end{aligned}$$

(17)

$$\begin{aligned} \displaystyle \frac{1}{f_{1}f_{3}}&= \displaystyle \frac{f_{8}^{2}f_{12}^{5}}{f_{2}^{2}f_{4}f_{6}^{4}f_{24}^{2}} + q\displaystyle \frac{f_{4}^{5}f_{24}^{2}}{f_{2}^{4}f_{6}^{2}f_{8}^{2}f_{12}} \end{aligned}$$

(18)

Proof

By Entry 25 (i), (ii), (v), and (vi) in [3, p. 40], we have

$$\begin{aligned} \phi (q)&= \phi (q^4) + 2q\psi (q^8), \end{aligned}$$

(19)

$$\begin{aligned} \phi (q)^2&= \phi (q^2)^2 + 4q\psi (q^4)^2. \end{aligned}$$

(20)

Using (7) and (8) we can rewrite (19) in the form

$$\begin{aligned} \frac{f_{2}^{5}}{f_{1}^{2}f_{4}^{2}} = \frac{f_{8}^{5}}{f_{4}^{2}f_{16}^{2}} + 2q\frac{f_{16}^{2}}{f_{8}}, \end{aligned}$$

from which we obtain (11) after multiplying both sides by $\frac{f_4^2}{f_2^5}$. Identity (12) can be easily deduced from (11) using the procedure described in Section 30.10 of [9].

By (7) and (8) we can rewrite (20) in the form

$$\begin{aligned} \frac{f_{2}^{10}}{f_{1}^{4}f_{4}^{4}} = \frac{f_{4}^{10}}{f_{2}^{4}f_{8}^{4}} + 4q\frac{f_{8}^{4}}{f_{4}^{2}}, \end{aligned}$$

from which we obtain (13).

Identities (14), (15), and (18) are equations (30.10.3), (30.9.9), and (30.12.3) of [9], respectively. Finally, for proofs of (16) and (17) see [13, Lemma 4]. $\square$

The next lemma exhibits the 3-dissections of $\psi (q)$ and $1/\phi (-q)$.

Lemma 2

We have

$$\begin{aligned} \psi (q)= & {} \displaystyle \frac{f_{6}f_{9}^{2}}{f_{3}f_{18}} + q\displaystyle \frac{f_{18}^{2}}{f_{9}}, \end{aligned}$$

(21)

$$\begin{aligned} \displaystyle \frac{1}{\phi (-q)}= & {} \displaystyle \frac{f_{6}^4f_{9}^6}{f_{3}^8f_{18}^3} + 2q\displaystyle \frac{f_{6}^3f_{9}^3}{f_{3}^7} + 4q^2\displaystyle \frac{f_{6}^2f_{18}^3}{f_{3}^6}. \end{aligned}$$

(22)

Proof

Identity (21) is Eq. (14.3.3) of [9]. A proof of (22) can be seen in [10]. $\square$

3 Dissections for $p_{\xi }(n)$

This section is devoted to proving the 2-, 3-, and 4-dissections of (2). We begin with the 2-dissection.

Theorem 1

We have

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(2n+1)q^{n+1}&= \frac{f_{6}^6f_{12}}{f_3^4f_{24}^2} - f(q^{12}) +4q\frac{f_2^2f_{8}^2}{f_1f_3f_4}, \text {\ \ and} \end{aligned}$$

(23)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(2n)q^{n}&= q\frac{f_{6}^8f_{24}^2}{f_3^4f_{12}^5} - q^4\omega (-q^{6}) + \frac{f_4^5}{f_1f_3f_{8}^2}. \end{aligned}$$

(24)

Proof

We start with equation (4) of [2]:

$$\begin{aligned} f(q^8) +2q\omega (q) + 2q^3\omega (-q^4) = F(q), \end{aligned}$$

where f(q) is the mock theta function

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

and

$$\begin{aligned} F(q) = \frac{\phi (q)\phi (q^2)^2}{f_4^2} = \frac{f_2f_{4}^6}{f_1^2f_{8}^4}. \end{aligned}$$

Thus,

$$\begin{aligned} f(q^{24}) +2q^3\omega (q^3) + 2q^9\omega (-q^{12}) = F(q^3). \end{aligned}$$

Using (5), it follows that

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(n)q^{n+1} = F(q^3) - f(q^{24}) - 2q^9\omega (-q^{12}) + 2q\frac{f_2^4}{f_1^2f_6}. \end{aligned}$$

(25)

By (11), we have

$$\begin{aligned} F(q^3) = \frac{f_{12}^6f_{24}}{f_6^4f_{48}^2} + 2q^3\frac{f_{12}^8f_{48}^2}{f_6^4f_{24}^5}, \end{aligned}$$

which along with (11) allows us to rewrite (25) as

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(n)q^{n+1}&= \frac{f_{12}^6f_{24}}{f_6^4f_{48}^2} + 2q^3\frac{f_{12}^8f_{48}^2}{f_6^4f_{24}^5} - f(q^{24}) - 2q^9\omega (-q^{12}) \\&\quad + 2q\frac{f_8^5}{f_2f_6f_{16}^2} +4q^2\frac{f_4^2f_{16}^2}{f_2f_6f_8}. \end{aligned}$$

Thus,

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(2n+1)q^{2n+2}&= \frac{f_{12}^6f_{24}}{f_6^4f_{48}^2} - f(q^{24}) +4q^2\frac{f_4^2f_{16}^2}{f_2f_6f_8}, \text {\ \ and} \end{aligned}$$

(26)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(2n)q^{2n+1}&= q^3\frac{f_{12}^8f_{48}^2}{f_6^4f_{24}^5} - q^9\omega (-q^{12}) + q\frac{f_8^5}{f_2f_6f_{16}^2}. \end{aligned}$$

(27)

Dividing (27) by q and replacing $q^2$ by q in the resulting identity and in (26), we obtain (23) and (24). $\square$

The next theorem exhibits the 3-dissection of (2).

Theorem 2

We have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&= \frac{f_{2}f_{3}^4}{f_{1}^2f_{6}^2}, \end{aligned}$$

(28)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n+1)q^n&= 2\frac{f_{3}f_{6}}{f_{1}}, \text {\ \ and} \end{aligned}$$

(29)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n+2)q^n&= \omega (q) + \frac{f_{6}^4}{f_2f_{3}^2}. \end{aligned}$$

(30)

Proof

In view of (8), we rewrite (6) as

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(n)q^n = q^2\omega (q^3) + \frac{\psi (q)^2}{f_6}. \end{aligned}$$

Using (21), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(n)q^n = q^2\omega (q^3) + \frac{f_{6}f_{9}^4}{f_{3}^2f_{18}^2} +2q\frac{f_{9}f_{18}}{f_{3}} + q^2\frac{f_{18}^4}{f_6f_{9}^2}. \end{aligned}$$

(31)

Extracting the terms of the form $q^{3n+r}$ on both sides of (31), for $r \in \{0,1,2\}$, dividing both sides of the resulting identity by $q^r$ and then replacing $q^3$ by q, we obtain the desired results. $\square$

We close this section with the 4-dissection of (2).

Theorem 3

We have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n)q^n&= 4q^2\frac{f_{12}^6}{f_{3}^2f_{6}^3} - q^2 \omega (-q^3) + \frac{f_2^4f_6^5}{f_1^2f_3^4f_{12}^2}, \end{aligned}$$

(32)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+1)q^n&= 2q\frac{f_{6}^3f_{12}^2}{f_{3}^4} + 2\frac{f_{4}^4f_{6}^5}{f_{2}^2f_{3}^4f_{12}^2}, \end{aligned}$$

(33)

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+2)q^n&= \frac{f_{6}^9}{f_{3}^6f_{12}^2} +\frac{f_{2}^{10}f_{12}^2}{f_{1}^4f_{3}^2f_{4}^4f_{6}}, \text { and} \end{aligned}$$

(34)

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(4n+3)q^{n+1}&= \frac{f_{6}^{15}}{f_3^8f_{12}^6} - f(q^6) + 4q\frac{f_{2}^4f_{12}^2}{f_{1}^2f_{3}^2f_{6}}. \end{aligned}$$

(35)

Proof

In order to prove (32), we use (13) and (18) to obtain the even part of (24), which is given by

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n)q^{2n}&= 4q^4\frac{f_{24}^6}{f_6^2f_{12}^3} -q^4\omega (-q^6) + \frac{f_4^{4}f_{12}^5}{f_2^2f_6^4f_{24}^2}. \end{aligned}$$

Replacing $q^2$ by q we obtain (32).

Using (13) and (18) we can extract the odd part of (23):

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(4n+1)q^{2n+1}&= 4q^3\frac{f_{12}^3f_{24}^2}{f_6^4} + 4q\frac{f_8^4f_{12}^5}{f_4^2f_6^4f_{24}^2}. \end{aligned}$$

After simplifications we arrive at (33).

Next, extracting the odd part of (24) with the help of (13) and (18) yields

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+2)q^{2n+1}&= q\frac{f_{12}^9}{f_6^6f_{24}^2} + q\frac{f_4^{10}f_{24}^2}{f_2^4f_6^2f_{8}^4f_{12}}, \end{aligned}$$

which, after simplifications, gives us (34).

In order to obtain (35), we use (13) and (18) in (23) to extract its even part:

$$\begin{aligned} 2\sum _{n=0}^{\infty } p_{\xi }(4n+3)q^{2n+2}&= \frac{f_{12}^{15}}{f_6^8f_{24}^6} -f(q^{12}) + 4q^2\frac{f_4^4f_{24}^2}{f_2^2f_6^2f_{12}}. \end{aligned}$$

Replacing $q^2$ by q in this identity, we obtain (35). $\square$

4 Arithmetic properties of $p_{\xi }(n)$

Our first observation provides a characterization of $p_{\xi }(3n) \pmod {4}.$

Theorem 4

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(3n) \equiv {\left\{ \begin{array}{ll} 1 \pmod {4} &{} \hbox { if}\ n=0,\\ 2 \pmod {4} &{} \text{ if } n \hbox { is a square}, \\ 0 \pmod {4} &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Proof

By (28), using (9) and the fact that $f_k^4 \equiv f_{2k}^2 \pmod {4}$ for all $k\ge 1,$ it follows that

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&= \frac{f_{2}f_{3}^4}{f_{1}^2f_{6}^2} \equiv \frac{f_2}{f_1^2} = \frac{f_1^2f_2}{f_1^4} \equiv \frac{f_1^2}{f_ 2} = \phi (-q) \pmod {4}. \end{aligned}$$

By (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&\equiv \sum _{n=-\infty }^{\infty } (-1)^n q^{n^2} \equiv 1 + 2\sum _{n=1}^{\infty } q^{n^2} \pmod {4}, \end{aligned}$$

which completes the proof. $\square$

Theorem 4 yields an infinite family of Ramanujan-like congruences modulo 4.

Corollary 1

For all primes $p \ge 3$ and all $n\ge 0,$ we have

$$\begin{aligned} p_{\xi }(3(pn+r)) \equiv 0 \pmod {4}, \end{aligned}$$

if r is a quadratic nonresidue modulo p.

Proof

If $pn+r = k^2$, then $r \equiv k^2 \pmod {p}$, which contradicts the fact that r is a quadratic nonresidue modulo p. $\square$

Since $\gcd (3, p) = 1$, among the $p-1$ residues modulo p, we have $\frac{p-1}{2}$ residues r for which r is a quadratic nonresidue modulo p. Thus, for instance, the above corollary yields the following congruences:

$$\begin{aligned} p_{\xi }(9n+6)&\equiv 0 \pmod {4}, \\ p_{\xi }(15n+k)&\equiv 0 \pmod {4}, \text{ for } k \in \{ 6, 9 \}, \\ p_{\xi }(21n+k)&\equiv 0 \pmod {4}, \text{ for } k \in \{ 9, 15, 18 \}, \\ p_{\xi }(33n+k)&\equiv 0 \pmod {4}, \text{ for } k \in \{ 6, 18, 21, 24, 30 \}. \end{aligned}$$

Theorem 5

For all $n \ge 0$, we have

$$\begin{aligned}p_{\xi }(3n+1) \equiv {\left\{ \begin{array}{ll} 2 \pmod {4} &{} \text{ if } 3n+1 \hbox { is a square}, \\ 0 \pmod {4} &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Proof

From Theorem 2,

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n+1)q^n&= 2\frac{f_{3}f_{6}}{f_{1}}. \end{aligned}$$

(36)

So we only need to consider the parity of

$$\begin{aligned} \frac{f_{3}f_{6}}{f_{1}}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{f_{3}f_{6}}{f_{1}} \equiv \frac{f_{3}^3}{f_{1}} = \sum _{n=0}^{\infty } a_3(n)q^n \pmod {2}, \end{aligned}$$

where $a_3(n)$ is the number of 3-core partitions of n (see [11, Theorem 1]). Thanks to [6, Theorem 7], we know that

$$\begin{aligned} a_{3}(n) \equiv {\left\{ \begin{array}{ll} 1 \pmod {2} &{} \text {if }3n+1 \hbox { is a square,}\\ 0 \pmod {2} &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

This completes the proof. $\square$

Theorem 5 yields an infinite family of congruences modulo 4.

Corollary 2

For all primes $p > 3$ and all $n\ge 0,$ we have

$$\begin{aligned} p_{\xi }(3(pn+r)+1) \equiv 0 \pmod {4}, \end{aligned}$$

if $3r+1$ is a quadratic nonresidue modulo p.

Proof

If $3(pn+r)+1 = k^2$, then $3r+1 \equiv k^2 \pmod {p}$, which would be a contradiction with $3r+1$ being a quadratic nonresidue modulo p. $\square$

For example, the following congruences hold for all $n\ge 0$:

$$\begin{aligned} p_{\xi }(15n+k)&\equiv 0 \pmod {4} \text{ for } k \in \{ 7, 13 \}, \\ p_{\xi }(21n+k)&\equiv 0 \pmod {4} \text{ for } k \in \{ 10, 13, 19 \}, \\ p_{\xi }(33n+k)&\equiv 0 \pmod {4} \text{ for } k \in \{ 7, 10, 13, 19, 28 \}. \end{aligned}$$

We next turn our attention to the arithmetic progression $4n+2$ to yield an additional infinite family of congruences.

Theorem 6

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(4n+2) \equiv {\left\{ \begin{array}{ll} 2 \pmod {4} &{} \text{ if } n=6k(3k \pm 1), \\ 0 \pmod {4} &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Proof

From (34), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+2)q^n&\equiv \frac{f_{6}^7}{f_{3}^2f_{12}^2} +\frac{f_{12}^2}{f_{3}^2f_{6}} \equiv 2\frac{f_6^3}{f_3^2} \equiv 2f_6^2 \equiv 2f_{12} \pmod {4}. \end{aligned}$$

(37)

Using Euler’s identity (see [9, Eq. (1.6.1)])

$$\begin{aligned} f_1 = \sum _{n=-\infty }^{\infty } (-1)^n q^{n(3n-1)/2}, \end{aligned}$$

(38)

we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+2)q^n \equiv 2\sum _{n=-\infty }^{\infty } (-1)^n q^{6n(3n-1)} \pmod {4}, \end{aligned}$$

which concludes the proof. $\square$

Theorem 6 yields an infinite family of congruences modulo 4.

Corollary 3

Let $p > 3$ be a prime and r an integer such that $2r+1$ is a quadratic nonresidue modulo p. Then, for all $n \ge 0$,

$$\begin{aligned} p_{\xi }(4(pn +r)+2) \equiv 0 \pmod {4}. \end{aligned}$$

Proof

If $pn+r = 6k(3k\pm 1)$, then $r \equiv 18k^2 \pm 6k \pmod {p}$. Thus, $2r+1 \equiv (6k \pm 1)^2 \pmod {p}$, which contradicts the fact that $2r+1$ is a quadratic nonresidue modulo p. $\square$

Thanks to Corollary 3, the following example congruences hold for all $n\ge 0$:

$$\begin{aligned} p_{\xi }(20n + j)&\equiv 0 \pmod {4} \text {\ \ for\ \ } j \in \{ 6, 14 \}, \\ p_{\xi }(28n + j)&\equiv 0 \pmod {4} \text {\ \ for\ \ } j \in \{ 6, 10, 26 \}, \\ p_{\xi }(44n + j)&\equiv 0 \pmod {4} \text {\ \ for\ \ } j \in \{ 14, 26, 34, 38, 42 \}, \\ p_{\xi }(52n + j)&\equiv 0 \pmod {4} \text {\ \ for\ \ } j \in \{ 10, 14, 22, 30, 38, 42 \}. \end{aligned}$$

We now provide a mod 8 characterization for $p_{\xi }(3n).$

Theorem 7

For all $n \ge 0$, we have

$$\begin{aligned}p_{\xi }(3n) \equiv {\left\{ \begin{array}{ll} 1 \pmod {8} &{} \text{ if } n=0 ,\\ 6(-1)^k \pmod {8} &{} \text{ if } n=k^2, \\ 4 \pmod {8} &{} \text{ if } n = 2k^2, n = 3k^2, \hbox { or }n = 6k^2, \\ 0 \pmod {8} &{} \text{ otherwise }. \end{array}\right. }\end{aligned}$$

Proof

By (28), using (7) and (9), we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&= \frac{f_1^6f_{2}f_{3}^4}{f_{1}^8f_{6}^2} \equiv \left( \frac{f_1^2}{f_{2}} \right) ^3 \left( \frac{f_3^2}{f_{6}} \right) ^2 \equiv \phi (-q)^3 \phi (-q^3)^2 \\&\equiv \left( \sum _{n=-\infty }^{\infty } (-1)^n q^{n^2} \right) ^3 \left( \sum _{n=-\infty }^{\infty } (-1)^n q^{3n^2} \right) ^2 \\&\equiv \left( 1+ 2\sum _{n=1}^{\infty } (-1)^n q^{n^2} \right) ^3 \left( 1+ 2\sum _{n=1}^{\infty } (-1)^n q^{3n^2} \right) ^2 \pmod {8}, \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&\equiv 1 + 6\sum _{n=1}^{\infty } (-1)^n q^{n^2} + 4 \left( \sum _{n=1}^{\infty } (-1)^n q^{n^2} \right) ^2 \\&\quad + 4\sum _{n=1}^{\infty } (-1)^n q^{3n^2} + 4 \left( \sum _{n=1}^{\infty } (-1)^n q^{3n^2} \right) ^2 \pmod {8}. \end{aligned}$$

Since

$$\begin{aligned} \left( \sum _{n=1}^{\infty } (-1)^n q^{n^2} \right) ^2 \equiv \sum _{n=1}^{\infty } q^{2n^2} \pmod {2}, \end{aligned}$$

we have

$$\begin{aligned} \left( \sum _{n=1}^{\infty } (-1)^n q^{3n^2} \right) ^2 \equiv \sum _{n=1}^{\infty } q^{6n^2} \pmod {2}. \end{aligned}$$

Therefore

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n)q^n&\equiv 1 + 6\sum _{n=1}^{\infty } (-1)^n q^{n^2} + 4\sum _{n=1}^{\infty } q^{2n^2} \\&\quad + 4\sum _{n=1}^{\infty } (-1)^n q^{3n^2} + 4 \sum _{n=1}^{\infty } q^{6n^2} \pmod {8}, \end{aligned}$$

which completes the proof. $\square$

As with the prior results, Theorem 7 provides an effective way to yield an infinite family of congruences modulo 8.

Corollary 4

Let p be a prime such that $p \equiv \pm 1 \pmod {24}$. Then

$$\begin{aligned} p_{\xi }(3(pn+r)) \equiv 0 \pmod {8}, \end{aligned}$$

if r is a quadratic nonresidue modulo p.

Proof

Since $p \equiv \pm 1 \pmod {8}$ and $p \equiv \pm 1 \pmod {12}$, it follows that 2 and 3 are quadratic residues modulo p. Thus, r, 2r, 3r, and 6r are quadratic nonresidues modulo p. Indeed, according to the properties of Legendre’s symbol, for $j\in \{1,2,3,6\}$, we have

$$\begin{aligned} \left( \frac{jr}{p}\right) = \left( \frac{j}{p}\right) \left( \frac{r}{p}\right) = \left( \frac{r}{p}\right) = -1. \end{aligned}$$

It follows that we cannot have $3(pn+r) = jk^2,$ for some $k \in {\mathbb {N}}$ and $j \in \{ 1, 2, 3, 6 \}$. In fact, $3(pn+r) = jk^2$ would imply $3(pn+r) \equiv 3r \equiv jk^2 \pmod {p}$. However, for $j = 1, 2, 3, 6$, this would imply that 3r, 6r, r, or 2r, respectively, is a quadratic residue modulo p, which would be a contradiction since 2, 3, and 6 are quadratic residues modulo p. The result follows from Theorem 7. $\square$

As an example, we note that, for $p=23$ and all $n\ge 0,$ we have

$$\begin{aligned} p_{\xi }(69n+k)&\equiv 0 \pmod {8} \text{ for } k \in \{ 15, 21, 30, 33, 42, 45, 51, 57, 60, 63, 66 \}. \end{aligned}$$

Theorem 8

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(12n+4) \equiv p_{\xi }(3n+1) \pmod {8}. \end{aligned}$$

Proof

Initially we use (14) to extract the odd part on both sides of (29). The resulting identity is

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(6n+4)q^n = 2\frac{f_{3}^2f_4^2f_{24}}{f_{1}^2f_8f_{12}}. \end{aligned}$$

(39)

Using (15) in (39), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+4)q^n = 2\frac{f_{2}^6f_3f_{6}}{f_{1}^5f_4^2} = 2\frac{f_1^3f_{2}^6f_3f_{6}}{f_{1}^8f_4^2} \equiv 2\frac{f_3f_{6}}{f_{1}} \pmod {8}. \end{aligned}$$

The result follows using (29). $\square$

Now we present complete characterizations of $p_{\xi }(48n+4)$ and $p_{\xi }(12n+1)$ modulo 8.

Theorem 9

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(48n+4) \equiv p_{\xi }(12n+1) \equiv {\left\{ \begin{array}{ll} 2(-1)^k \pmod {8} &{} \text{ if } n=k(3k\pm 1), \\ 0 \pmod {8} &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Proof

The first congruence follows directly from Theorem 8. Replacing (14) in (29), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(3n+1)q^n = 2\displaystyle \frac{f_{4}f_{6}^2f_{16}f_{24}^{2}}{f_{2}^{2}f_{8}f_{12}f_{48}} + 2q\displaystyle \frac{f_{6}^2f_{8}^{2}f_{48}}{f_{2}^{2}f_{16}f_{24}}. \end{aligned}$$

Extracting the terms of the form $q^{2n}$, we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(6n+1)q^{2n} = 2\displaystyle \frac{f_{4}f_{6}^2f_{16}f_{24}^{2}}{f_{2}^{2}f_{8}f_{12}f_{48}}, \end{aligned}$$

which, after replacing $q^2$ by q, yields

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(6n+1)q^{n} = 2\displaystyle \frac{f_{2}f_{3}^2f_{8}f_{12}^{2}}{f_{1}^{2}f_{4}f_{6}f_{24}}. \end{aligned}$$

(40)

Now we use (15) to obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+1)q^{n}&= 2\displaystyle \frac{f_{2}^3f_{6}^4}{f_{1}^4f_{12}^2}&\\&\equiv 2f_2 \equiv 2 \sum _{n=-\infty }^{\infty } (-1)^n q^{n(3n-1)} \pmod {8}&(\text{ by } (38)), \end{aligned}$$

which completes the proof. $\square$

Theorem 9 also provides an effective way to yield an infinite family of congruences modulo 8.

Corollary 5

For all primes $p > 3$ and all $n\ge 0$, we have

$$\begin{aligned} p_{\xi }(48(pn+r)+4) \equiv p_{\xi }(12(pn+r)+1) \equiv 0 \pmod {8} \end{aligned}$$

if $12r+1$ is a quadratic nonresidue modulo p.

Proof

Let $p>3$ be a prime and $12r+1$ a quadratic nonresidue modulo p. If $pn+r = k(3k\pm 1)$, then $r \equiv 3k^2\pm k \pmod {p}$, which implies that $12r+1 \equiv (6k\pm 1)^2 \pmod {p}$, a contradiction. The result follows from Theorem 9. $\square$

5 Additional congruences

In this section, we prove several additional Ramanujan-like congruences that are not included in the results of the previous section.

Theorem 10

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(24n+19)&\equiv 0 \pmod {3}, \end{aligned}$$

(41)

$$\begin{aligned} p_{\xi }(27n+18)&\equiv 0 \pmod {3}, \text { and} \end{aligned}$$

(42)

$$\begin{aligned} p_{\xi }(72n+51)&\equiv 0 \pmod {3}. \end{aligned}$$

(43)

Proof

Using (15) we can now 2-dissect (40) to obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(6n+1)q^{n} = 2\displaystyle \frac{f_{4}^3f_{12}^4}{f_{2}^4f_{24}^2} + 4q\displaystyle \frac{f_{6}f_{8}^2f_{12}}{f_{2}^3}, \end{aligned}$$

from which we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+7)q^{2n+1} = 4q\displaystyle \frac{f_{6}f_{8}^2f_{12}}{f_{2}^3}. \end{aligned}$$

Now, dividing both sides of the above expression by q and replacing $q^2$ by q, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+7)q^{n} = 4\displaystyle \frac{f_{3}f_{4}^2f_{6}}{f_{1}^3}. \end{aligned}$$

(44)

Using (17) we rewrite (44) as

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+7)q^{n} = 4\displaystyle \frac{f_{4}^8f_{6}^4}{f_{2}^{9}f_{12}^2} + 12q\displaystyle \frac{f_{4}^4f_{6}^2f_{12}^2}{f_{2}^{7}}. \end{aligned}$$

Taking the odd parts on both sides of the last equation, we are left with

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(24n+19)q^{n} = 12\displaystyle \frac{f_{2}^4f_{3}^2f_{6}^2}{f_{1}^{7}}, \end{aligned}$$

which proves (41).

In order to prove (42), we use (22) to extract the terms of the form $q^{3n}$ of (28). The resulting identity is

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n)q^{3n}&= \frac{f_{6}^2f_{9}^6}{f_{3}^4f_{18}^3}, \end{aligned}$$

which, after replacing $q^3$ by q and using (8), yields

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n)q^{n}&= \frac{f_{2}^2f_{3}^6}{f_{1}^4f_{6}^3} \equiv \frac{f_{2}^2f_{3}^5}{f_{1}f_{6}^3} = \psi (q)\frac{f_{3}^5}{f_{6}^3} \pmod {3}. \end{aligned}$$

By (8), we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n)q^{n}&\equiv \frac{f_{3}^5}{f_{6}^3} \sum _{n=0}^{\infty } q^{n(n+1)/2} \pmod {3}. \end{aligned}$$

Since $n(n+1)/2 \not \equiv 2 \pmod {3}$ for all $n\ge 0,$ all terms of the form $q^{3n+2}$ in the last expression have coefficients congruent to $0 \pmod {3}$, which proves (42).

We now prove (43). Replacing (22) in (28) and extracting the terms of the form $q^{3n+2}$, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n+6)q^{3n+2} = 4q^2\displaystyle \frac{f_{18}^3}{f_{3}^2}. \end{aligned}$$

(45)

Dividing both sides of (45) by $q^2$ and replacing $q^3$ by q, we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n+6)q^{n} = 4\displaystyle \frac{f_{6}^3}{f_{1}^2}. \end{aligned}$$

(46)

Now we use (11) to extract the odd part of (46) and obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(18n+15)q^{n} = 8\displaystyle \frac{f_{2}^2f_3^3f_8^2}{f_{1}^5f_4}. \end{aligned}$$

Since $f_1^3 \equiv f_3 \pmod {3}$, we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(18n+15)q^{n} \equiv 2\displaystyle \frac{f_{2}^2f_3^2f_8^2}{f_{1}^2f_4} \pmod {3}. \end{aligned}$$

Using (15) we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(36n+15)q^{n} \equiv 2\displaystyle \frac{f_{2}^3f_3f_4f_6^2}{f_{1}^3f_{12}} \pmod {3}. \end{aligned}$$

Since the odd part of (17) is divisible by 3, then the coefficients of the terms of the form $q^{2n+1}$ in $\sum _{n=0}^{\infty } p_{\xi }(36n+15)q^{n}$ are congruent to 0 modulo 3. This completes the proof of (43). $\square$

We now prove a pair of unexpected congruences modulo 5 satisfied by $p_{\xi }(n).$

Theorem 11

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(45n+33)&\equiv 0 \pmod {5}, \end{aligned}$$

(47)

$$\begin{aligned} p_{\xi }(45n+42)&\equiv 0 \pmod {5}. \end{aligned}$$

(48)

Proof

By (46), we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(9n+6)q^{n}&= 4\frac{f_{6}^3}{f_{1}^2} = 4\frac{f_1^3f_{6}^3}{f_{1}^5} \equiv 4\frac{f_1^3f_{6}^3}{f_{5}} \pmod {5}. \end{aligned}$$

Thanks to Jacobi’s identity (10) we know

$$\begin{aligned} f_1^3f_6^3 = \sum _{j,k=0}^{\infty } (-1)^{j+k} (2j+1)(2k+1) q^{3j(j+1)+k(k+1)/2}. \end{aligned}$$

Note that, for all integers j and k, $3j(j+1)$ and $k(k+1)/2$ are congruent to either 0, 1 or 3 modulo 5. The only way to obtain $3j(j+1)+k(k+1)/2 = 5n+3$ is the following:

• $3j(j+1) \equiv 0 \pmod {5}$ and $k(k+1)/2 \equiv 3 \pmod {5}$, or

• $3j(j+1) \equiv 3 \pmod {5}$ and $k(k+1)/2 \equiv 0 \pmod {5}$.

Thus, $j \equiv 2 \pmod {5}$ or $k \equiv 2 \pmod {5}$ in all possible cases, and this means

$$\begin{aligned} (2j+1)(2k+1) \equiv 0 \pmod {5}. \end{aligned}$$

Therefore, for all $n \ge 0$, $p_{\xi }(45n+33) = p_{\xi }(9(5n+3)+6) \equiv 0 \pmod {5}$, which is (47).

In order to complete the proof of (48), we want to see when

$$\begin{aligned} 3j(j+1)+k(k+1)/2 = 5n+4. \end{aligned}$$

Four possible cases arise:

• $k \equiv 1 \pmod {5}$ and $j \equiv 2 \pmod {5}$,

• $k \equiv 3 \pmod {5}$ and $j \equiv 2 \pmod {5}$,

• $j \equiv 1 \pmod {5}$ and $k \equiv 2 \pmod {5}$, or

• $j \equiv 3 \pmod {5}$ and $k \equiv 2 \pmod {5}$.

In all four cases above, either $j \equiv 2 \pmod {5}$ or $k \equiv 2 \pmod {5}$. So

$$\begin{aligned} (2j+1)(2k+1) \equiv 0 \pmod {5} \end{aligned}$$

in all these cases. Therefore,

$$\begin{aligned} p_{\xi }(45n+42) = p_{\xi }(9(5n+4)+6) \equiv 0 \pmod {5}, \end{aligned}$$

which completes the proof of (48). $\square$

Next, we prove three congruences modulo 8 which are not covered by the above results.

Theorem 12

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(16n+14)&\equiv 0 \pmod {8}, \end{aligned}$$

(49)

$$\begin{aligned} p_{\xi }(24n+13)&\equiv 0 \pmod {8}, \end{aligned}$$

(50)

$$\begin{aligned} p_{\xi }(24n+22)&\equiv 0 \pmod {8}. \end{aligned}$$

(51)

Proof

Initially we prove (49). From (34) and (7) we have

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(4n+2)q^n&\equiv \frac{f_{3}^2f_6^5}{f_{12}^2} +\frac{f_{12}^2}{f_{3}^2f_{6}} \phi (q)^2 \pmod {8}. \end{aligned}$$

Now we can use (11), (12), and (20) to extract the terms involving $q^{2n+1}$ from both sides of the previous congruence:

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(8n+6)q^{2n+1}&\equiv -2q^3\frac{f_{6}^6f_{48}^2}{f_{12}^2f_{24}} +2q^3\frac{f_4^{10}f_{12}^4f_{48}^2}{f_{2}^4f_{6}^6f_8^4f_{24}} +4q\frac{f_{8}^4f_{12}^2f_{24}^5}{f_{4}^2f_6^6f_{48}^2} \pmod {8}. \end{aligned}$$

After dividing both sides by q and then replacing $q^2$ by q, we are left with

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(8n+6)q^{n}&\equiv -2q\frac{f_{3}^6f_{24}^2}{f_{6}^2f_{12}} +2q\frac{f_2^{10}f_{6}^4f_{24}^2}{f_{1}^4f_{3}^6f_4^4f_{12}} +4\frac{f_{4}^4f_{6}^2f_{12}^5}{f_{2}^2f_3^6f_{24}^2} \\&\equiv -2q\frac{f_{3}^6f_{24}^2}{f_{6}^2f_{12}} +2q\frac{f_3^{6}f_{6}^4f_{24}^2}{f_{3}^{12}f_{12}} +4\frac{f_{4}^4f_{12}^5}{f_{2}^2f_6f_{24}^2} \\&\equiv 4\frac{f_{4}^3f_{12}}{f_6} \pmod {8}, \end{aligned}$$

whose odd part is congruent to 0 modulo 8, which implies (49).

In order to prove (50), we use (15) to obtain the even part of identity (40), which is

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+1)q^{n} = 2\frac{f_{2}^3f_6^4}{f_1^4f_{12}^2}. \end{aligned}$$

Now, employing (13), we obtain the odd part of the last identity, which is

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(24n+13)q^{n} = 8\frac{f_{2}^2f_3^4f_4^4}{f_1^7f_{6}^2}, \end{aligned}$$

which implies (50).

Now we prove (51). We employ (15) in (39) to obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+10)q^{n} = 4\frac{f_{2}^3f_3^2f_{12}^2}{f_{1}^4f_6^2}. \end{aligned}$$

(52)

By (12) and (13), we rewrite (52) in the form

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+10)q^{n} = 4\frac{f_{2}^3f_{12}^2}{f_{6}^2} \left( \frac{f_4^{14}}{f_2^{14}f_8^4} +4q \frac{f_4^2f_8^4}{f_2^{10}} \right) \left( \frac{f_6f_{24}^5}{f_{12}^2f_{48}^2} -2q^3 \frac{f_6f_{48}^2}{f_{24}} \right) , \end{aligned}$$

from which we obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(24n+22)q^{2n+1} = 4\frac{f_{2}^3f_{12}^2}{f_{6}^2} \left( -2q^3\frac{f_4^{14}f_6f_{48}^2}{f_2^{14}f_8^4f_{24}} +4q \frac{f_4^2f_6f_8^4f_{24}^5}{f_2^{10}f_{12}^2f_{48}^2} \right) . \end{aligned}$$

Dividing both sides by q and replacing $q^2$ by q, we are left with

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(24n+22)q^{n} = -8q\frac{f_2^{14}f_6^2f_{24}^2}{f_1^{11}f_3f_4^4f_{12}} +16 \frac{f_2^2f_4^4f_{12}^5}{f_1^{7}f_3f_{24}^2}, \end{aligned}$$

which implies (51). $\square$

We close this section by proving a congruence modulo 9.

Theorem 13

For all $n \ge 0$, we have

$$\begin{aligned} p_{\xi }(96n+76)&\equiv 0 \pmod {9}. \end{aligned}$$

(53)

Proof

We use (21) to extract the terms of the form $q^{3n+1}$ from (32). The resulting identity is

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+4)q^{3n+1} = 2q\frac{f_{6}^6f_9f_{18}}{f_3^5f_{12}^2}, \end{aligned}$$

which, after dividing by q and replacing $q^3$ by q, yields

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(12n+4)q^{n} = 2\frac{f_{2}^6f_3f_{6}}{f_1^5f_{4}^2} = 2\frac{f_{2}^6f_{6}}{f_{4}^2} \frac{f_3}{f_1} \frac{1}{f_1^4}. \end{aligned}$$

Using (13) and (14), we extract the even part on both sides of the above identity to obtain

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(24n+4)q^{n}&= 2\frac{f_{2}^{13}f_3^2f_{8}f_{12}^2}{f_1^{10}f_{4}^5f_6f_{24}} + 8q\frac{f_{3}^2f_{4}^6f_{24}}{f_1^6f_8f_{12}} \\&\equiv 2\frac{f_{2}^{13}f_{8}f_{12}^2}{f_{4}^5f_6f_{24}}\frac{1}{f_1f_3} + 8q\frac{f_{4}^6f_{24}}{f_8f_{12}} \frac{f_1^3}{f_3} \pmod {9}. \end{aligned}$$

Now we employ (18) and (16) to extract the odd part on both sides of the last congruence:

$$\begin{aligned} \sum _{n=0}^{\infty } p_{\xi }(48n+28)q^{n}&\equiv 2\frac{f_{1}^{9}f_6f_{12}}{f_{3}^3f_4} + 8\frac{f_{2}^9f_{12}}{f_4f_{6}^2} \equiv \frac{f_6f_{12}}{f_4} \pmod {9}, \end{aligned}$$

which implies (53). $\square$

6 Concluding remarks

Computational evidence indicates that $p_{\xi }(n)$ satisfies many other congruences. The interested reader may wish to consider the following two conjectures.

Conjecture 1

$$\begin{aligned} \sum _{n=0}^\infty p_{\xi }(8n+3)q^n \equiv 2\sum _{n=0}^\infty q^{3n(n+1)/2} \pmod {3} \end{aligned}$$

Conjecture 2

$$\begin{aligned} \sum _{n=0}^\infty p_{\xi }(32n+12)q^n \equiv 6\sum _{n=0}^\infty q^{3n(n+1)/2} \pmod {9} \end{aligned}$$

Clearly, once proven, Conjectures 1 and 2 would immediately lead to infinite families of Ramanujan-like congruences. Morever, Conjecture 2 would immediately imply Theorem 13 since $96n+76 = 32(3n+2)+12$ while the right-hand side of Conjecture 2 is clearly a function of $q^3.$ The same argument would imply that, for all $n\ge 0,$

$$\begin{aligned} p_{\xi }(96n+44) \equiv 0 \pmod {9}. \end{aligned}$$

since $96n+44 = 32(3n+1)+12.$