Arithmetic Properties of 5-Tuple Partitions with 3-Cores

Abstract

Let $A_{3,5}(n)$ denote the number of 5-tuple partitions of n with 3-cores. We establish some congruences modulo 2, 4, 5, 8 and 10 for $A_{3,5}(n)$ by employing q-series identities. For example, we prove for any prime $p\ge 5$, $\alpha \ge 1$, $\beta \ge 0$ and $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}p^{2\beta +2}n+\frac{(6j+p)\cdot 2^{2\alpha +1}p^{2\beta +1}-5}{3}\right) \equiv 0\pmod {2}, \end{aligned}$$

where $1\le j\le p-1$.

1 Introduction

A partition of a positive integer n is a non-increasing sequence of positive integers called parts whose sum is n. For convenience, we use the following notation

$$\begin{aligned} (a;q)_\infty =\prod _{n=0}^\infty (1-aq^n), \quad \text{ and }\quad f_k=(q^k;q^k)_\infty . \end{aligned}$$

A partition $\lambda$ of a positive integer n is said to be t-core if it has no hook numbers that are multiples of t. Let $a_t(n)$ denote the number of t-core partitions of n. In [7, Eq. (2.1)], the generating function of $a_t(n)$ is given by

$$\begin{aligned} \sum _{n=0}^\infty a_t(n)q^n=\frac{f_t^t}{f_1}. \end{aligned}$$

(1.1)

A k-tuple partition $(\lambda _1, \lambda _2,\ldots , \lambda _k)$ of n is a k-tuple of partitions $(\lambda _1, \lambda _2,\ldots , \lambda _k)$ such that the sum of all the parts equals n. A k-tuple partition $(\lambda _1, \lambda _2,\ldots , \lambda _k)$ of n with t-cores means that each $\lambda _i$ is t-core for $i=1,2,\cdots ,k$. Let $A_{t,k}(n)$ denote the number of k-tuple partitions of n with t-cores. The generating function of $A_{t,k}(n)$ is

$$\begin{aligned} \sum _{n=0}^\infty A_{t,k}(n)q^n=\frac{f_t^{kt}}{f_1^k}. \end{aligned}$$

(1.2)

Many authors have studied arithmetic properties of $A_{3,2}(n)$ and obtained some Ramanujan-type congruences. Lin [11] established some infinite families of congruences modulo 4, 5, 7 and 8 for $A_{3,2}(n)$. Based on Lin’s study, Xia [18] added proofs of several infinite families of congruences modulo 4, 8 and $\frac{4^k-1}{3}$ $(k\ge 2)$ for $A_{3,2}(n)$. For more results about $A_{3,2}(n)$, see [2, 5, 17, 18, 21].

Wang [16] proved some infinite families of identities and congruences for $A_{3,3}(n)$ by using some identities of q-series. In [17], Wang established some explicit formulas for $A_{3,2}(n)$ and $A_{3,3}(n)$. After that, Chern [5] extended the work of Wang [17] and studied some arithmetic identities of $A_{t,k}(n)$ for $(t,k)=(3,4),(3,6),(4,2)$,(5, 1),(5, 2) and (7, 1) by applying the theory of modular form.

Saikia and Boruah [13] proved some infinite families of congruences modulo 2, 3 for $A_{4,2}(n)$ and $A_{4,3}(n)$. Dasappa [6] discovered a nice congruence modulo $5^\alpha$ $(\alpha \ge 1)$ for $A_{5,2}(n)$. Saikia and Boruah [14] also studied arithmetic properties of $A_{5,2}(n)$ and proved some congruences modulo 2 and 5. In sequel, they [15] established some Ramanujan-type congruences for $A_{t,k}(n)$ when $(t,k)=(3,4),(3,9),(4,8),(5,6),(8,4),(9,3)$ and (9, 6) by employing q-series identities.

Zou [22] proved some congruences modulo 2 for $A_{t,2}(n)$; t is a prime such that $7\le t\le 23$. In 2020, Naika and Nayaka [12] established some Ramanujan-type congruences modulo 5, 7 and 8 for $A_{t,4}(n)$, $t=3, 5, 7, 25$.

In this paper, we mainly study arithmetic properties of $A_{3,5}(n)$. Its generating function is given by

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n=\frac{f_3^{15}}{f_1^5}. \end{aligned}$$

(1.3)

We establish some results about congruences modulo 2, 4, 5, 8 and 10 for $A_{3,5}(n)$.

To be specific, by using some dissection formulae, we obtain some infinite families of congruences modulo 2 for $A_{3,5}(n)$ as follows.

Theorem 1.1

For any integer $n\ge 0$, we have

$$\begin{aligned} A_{3,5}\left( 8n+5\right) \equiv 0\pmod {2}. \end{aligned}$$

(1.4)

and for $\alpha \ge 1$,

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +3}n+\frac{7\cdot 2^{2\alpha +1}-5}{3}\right)&\equiv 0\pmod {2}, \end{aligned}$$

(1.5)

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +4}n+\frac{13\cdot 2^{2\alpha +1}-5}{3}\right)&\equiv 0\pmod {2}. \end{aligned}$$

(1.6)

Theorem 1.2

For any prime $p\ge 5$, $\alpha \ge 1$, $\beta \ge 0$, and $n\ge 0$,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta }n+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n\equiv f(-q^4)\pmod {2}. \end{aligned}$$

(1.7)

We deduce the following infinite families of congruences with two parameters $\alpha , \beta$ modulo 2 for $A_{3,5}(n)$.

Corollary 1.1

For any prime $p\ge 5$, $\alpha \ge 1$, $\beta \ge 0$, if n cannot be represented as $2k(3k+1)$ for some integer k, then

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}p^{2\beta }n+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) \equiv 0 \pmod {2}. \end{aligned}$$

(1.8)

and for any integer $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}p^{2\beta +2}n+\frac{(6j+p)\cdot 2^{ 2\alpha +1}p^{2\beta +1}-5}{3}\right)&\equiv 0\pmod {2}, \quad j\in \{1,2\cdots p-1\}. \end{aligned}$$

(1.9)

$$\begin{aligned} A_{3,5}\left( 7^2\cdot 2^{2\alpha +2}p^{2\beta }n+\frac{(42k+7)\cdot 2^{2\alpha +1}p^{2\beta }-5}{3}\right)&\equiv 0\pmod {2}, \quad k\in \{0,2,3,4,5,6\}. \end{aligned}$$

(1.10)

and

$$\begin{aligned} A_{3,5}\left( 13^2\cdot 2^{2\alpha +2} p^{2\beta }n+\frac{(78k+13)\cdot 2^{2\alpha +1} p^{2\beta }-5}{3}\right)&\equiv 0\pmod {2}, \nonumber \\ \quad k\in \{0,1,3,4,5,6,7,8,9,10,11,12\}. \end{aligned}$$

(1.11)

We find the following congruences modulo 5 and 10 for $A_{3,5}(n)$ hold.

Theorem 1.3

For $n\ge 0$, we have

$$\begin{aligned} A_{3,5}(4n+3)&\equiv 0 \pmod {10}, \end{aligned}$$

(1.12)

$$\begin{aligned} A_{3,5}(5n+k)&\equiv 0 \pmod {5}, \quad k\in \{1, 2, 3 ,4\}. \end{aligned}$$

(1.13)

$$\begin{aligned} A_{3,5}(20n+15)&\equiv 0 \pmod {5}, \end{aligned}$$

(1.14)

and for $\alpha \ge 0$, we get

$$\begin{aligned} A_{3,5}\left( 5\cdot 2^{2\alpha }n+\frac{5\cdot 2^{2\alpha }-5}{3}\right)&\equiv A_{3,5}(5n) \pmod {5}, \end{aligned}$$

(1.15)

$$\begin{aligned} A_{3,5}\left( 5\cdot 2^{2\alpha +2}n+\frac{5^2\cdot 2^{2\alpha +1}-5}{3}\right)&\equiv 0 \pmod {5}. \end{aligned}$$

(1.16)

We also establish the following congruences modulo 4, 8 for $A_{3,5}(n)$.

Theorem 1.4

For $\alpha \ge 2$ and $n\ge 0$, we have

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +1}n+\frac{5\cdot 2^{2\alpha }-5}{3}\right) \equiv 0\pmod {4}. \end{aligned}$$

(1.17)

Let $p\ge 3$ be a prime and a be an integer. The Legendre symbol is defined by

$$\begin{aligned} \left( \frac{a}{p}\right) ={\left\{ \begin{array}{ll} 1,&{}\ \text{ if } a ~\text{ is } \text{ a } \text{ quadratic } \text{ residue } \text{ modulo } p~ \text{ and }~ a\not \equiv 0\pmod {p},\\ 0,&{}\ \text{ if }~ a\equiv 0\pmod {p},\\ -1,&{}\ \text{ if }~ a ~\text{ is } \text{ a } \text{ quadratic } \text{ non-residue } \text{ modulo } p. \end{array}\right. } \end{aligned}$$

Theorem 1.5

For any prime $p\ge 5$ such that $\left( \frac{-6}{p}\right) =-1$, $\alpha \ge 0$, and $n\ge 0$,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{14\cdot p^{2\alpha }-5}{3}\right) q^n \equiv 2f(-q)\psi (q^2)\pmod {4}. \end{aligned}$$

(1.18)

Corollary 1.2

For any prime $p\ge 5$ such that $\left( \frac{-6}{p}\right) =-1$, $\alpha \ge 0$, if n cannot be represented as the sum of a pentagonal number and twice a triangular number, then

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{14\cdot p^{2\alpha }-5}{3}\right) \equiv 0 \pmod {4}. \end{aligned}$$

(1.19)

and for any integer $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha +2}n+\frac{(48j+14p)\cdot p^{2\alpha +1}-5}{3}\right) \equiv 0\pmod {4}, \end{aligned}$$

(1.20)

where $1\le j\le p-1.$

Theorem 1.6

For any prime $p\ge 5$ such that $\left( \frac{-18}{p}\right) =-1$, $\alpha \ge 0$, and $n\ge 0$,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{38\cdot p^{2\alpha }-5}{3}\right) q^n\equiv 2f(-q)\psi (q^6)\pmod {4}. \end{aligned}$$

(1.21)

Corollary 1.3

For any prime $p\ge 5$ such that $\left( \frac{-18}{p}\right) =-1$, $\alpha \ge 0$, if n cannot be represented as the sum of a pentagonal number and six times a triangular number, then

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{38\cdot p^{2\alpha }-5}{3}\right) \equiv 0 \pmod {4}. \end{aligned}$$

(1.22)

and for any integer $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha +2}n+\frac{(48j+38p)\cdot p^{2\alpha +1}-5}{3}\right) \equiv 0\pmod {4}, \end{aligned}$$

(1.23)

where $1\le j\le p-1$.

Theorem 1.7

For $n\ge 0$, we have

$$\begin{aligned} A_{3,5}(4n+2)&\equiv 0\pmod {4}, \end{aligned}$$

(1.24)

$$\begin{aligned} A_{3,5}(16n+13)&\equiv 0 \pmod {8}. \end{aligned}$$

(1.25)

Theorem 1.8

For any prime $p\ge 5$ such that $\left( \frac{-9}{p}\right) =-1$, $\alpha \ge 0$, and $n\ge 0$,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2\cdot p^{2\alpha }n+\frac{5\cdot p^{2\alpha }-5}{3}\right) q^n \equiv (-1)^{\alpha (\frac{\pm p-1}{6})}f(-q^2)\psi (q^6)\pmod {4}. \end{aligned}$$

(1.26)

Corollary 1.4

For any prime $p\ge 5$ such that $\left( \frac{-9}{p}\right) =-1$, $\alpha \ge 0$, if n cannot be represented as the sum of a pentagonal number and three times a triangular number, then

$$\begin{aligned} A_{3,5}\left( 4\cdot p^{2\alpha }n+\frac{5\cdot p^{2\alpha }-5}{3}\right) \equiv 0 \pmod {4}. \end{aligned}$$

(1.27)

and for any integer $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 2\cdot p^{2\alpha +2}n+\frac{(6j+5p)\cdot p^{2\alpha +1}-5}{3}\right) \equiv 0\pmod {4}, \end{aligned}$$

(1.28)

where $1\le j\le p-1$.

Theorem 1.9

For any prime $p\ge 5$ such that $\left( \frac{-9}{p}\right) =-1$, $\alpha \ge 0$ and $n\ge 0$,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 8\cdot p^{2\alpha }n+\frac{20\cdot p^{2\alpha }-5}{3}\right) q^n \equiv 6(-1)^{\alpha (\frac{\pm p-1}{6})}f(-q^2)\psi (q^6)\pmod {8}. \end{aligned}$$

(1.29)

Corollary 1.5

For any prime $p\ge 5$ such that $\left( \frac{-9}{p}\right) =-1$, $\alpha \ge 0$, if n cannot be represented as the sum of a pentagonal number and three times a triangular number, then

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{20\cdot p^{2\alpha }-5}{3}\right) \equiv 0 \pmod {8}. \end{aligned}$$

(1.30)

and for any integer $n\ge 0$,

$$\begin{aligned} A_{3,5}\left( 8\cdot p^{2\alpha +2}n+\frac{(24j+20p)\cdot p^{2\alpha +1}-5}{3}\right) \equiv 0\pmod {8}, \end{aligned}$$

(1.31)

where $1\le j\le p-1$.

This paper is organized as follows. In Sect. 2, we shall prove some theorems about congruences for $A_{3,5}(n)$ modulo 2, 5 and 10. In Sect. 3, we give the proofs of remaining theorems about congruences modulo 4, 8 for $A_{3,5}(n)$.

2 Proofs of Theorems 1.1–1.3 and Corollary 1.1

Proof of Theorem 1.1

In order to prove Theorem 1.1, we first prove the following lemma.

Lemma 2.1

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

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha }n+\frac{2^{2\alpha +1}-5}{3}\right) q^n\equiv \frac{f_3^6}{f_1^2}\pmod {2}. \end{aligned}$$

(2.1)

Proof

Hirschhorn, Garvan and Borwein [9] proved that

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

(2.2)

By the binomial theorem, for any positive integer k and any prime p,

$$\begin{aligned} f_1^{p^k}\equiv f_p^{p^{k-1}} \pmod {p^k}. \end{aligned}$$

(2.3)

Substituting (2.2) into (1.3) and employing (2.3), we obtain that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n&=\frac{f_3^{12}}{f_1^4}\frac{f_3^3}{f_1} \equiv \frac{f_6^{6}}{f_2^2}\left( \frac{f_4^3f_6^2}{f_2^2f_{12}}+q\frac{f_{12}^3}{f_4}\right) \nonumber \\&\equiv \frac{f_4^3f_6^8}{f_2^4f_{12}}+q\frac{f_6^6f_{12}^3}{f_2^2f_4}\pmod {2}. \end{aligned}$$

(2.4)

Extracting the terms involving $q^{2n+1}$ from both sides of (2.4) , dividing by q and replacing $q^2$ by q, then employing (2.3), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n+1)q^n\equiv \frac{f_3^6f_6^3}{f_1^2f_2}\equiv \frac{f_6^6}{f_2^2}\pmod {2}. \end{aligned}$$

(2.5)

Hence,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(4n+1)q^n \equiv \frac{f_3^6}{f_1^2} \pmod {2}, \end{aligned}$$

(2.6)

which is the case $\alpha =1$ of (2.1).

Suppose that (2.1) holds for $\alpha \ge 1$. Utilizing (2.2), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha }n+\frac{2^{2\alpha +1}-5}{3}\right) q^n&\equiv \frac{f_3^6}{f_1^2}\equiv \left( \frac{f_4^3f_6^2}{f_2^2f_{12}}+q\frac{f_{12}^3}{f_4}\right) ^2\nonumber \\&\equiv \frac{f_4^6f_6^4}{f_2^4f_{12}^2}+q^2\frac{f_{12}^6}{f_4^2}\pmod {2}. \end{aligned}$$

(2.7)

Hence,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +1}n+\frac{2^{2\alpha +1}-5}{3}\right) q^n \equiv \frac{f_2^6f_3^4}{f_1^4f_{6}^2}+q\frac{f_{6}^6}{f_2^2}\equiv f_8+q\frac{f_6^6}{f_2^2}\pmod {2}. \end{aligned}$$

(2.8)

Extracting the terms involving $q^{2n+1}$ from both sides of (2.8), dividing by q and replacing $q^2$ by q, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}n+\frac{2^{2\alpha +3}-5}{3}\right) q^n \equiv \frac{f_3^6}{f_1^2}\pmod {2}, \end{aligned}$$

(2.9)

which is the case $\alpha +1$ of (2.1). The proof of Lemma 2.1 is completed by induction on $\alpha$.

Now, we turn to prove Theorem 1.1. It follows from (2.3) that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(4n+1)q^n \equiv \frac{f_3^6}{f_1^2}\equiv \frac{f_6^3}{f_2}\pmod {2}, \end{aligned}$$

(2.10)

which implies (1.4).

Extracting the terms involving $q^{2n}$ from both sides of (2.8) and replacing $q^2$ by q, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}n+\frac{2^{2\alpha +1}-5}{3}\right) q^n\equiv f_4 \pmod {2}. \end{aligned}$$

(2.11)

which implies that

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}(2n+1)+\frac{2^{2\alpha +1}-5}{3}\right)&\equiv 0\pmod {2}, \end{aligned}$$

(2.12)

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}(4n+2)+\frac{2^{2\alpha +1}-5}{3}\right)&\equiv 0 \pmod {2}. \end{aligned}$$

(2.13)

We thus obtain (1.5) and (1.6). This completes the proof of Theorem 1.1. $\square$

Proof of Theorem 1.2 and Corollary 1.1.

We first recall that Ramanujan’s general theta function f(a, b) is defined by

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

(2.14)

Two important cases of f(a, b) are the theta functions $\psi (q)$ and $f(-q)$ [1, p.36. Entry 22], which are given by

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

(2.15)

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

(2.16)

Cui and Gu [4] proved the following p-dissection identities for $\psi (q)$ and $f(-q)$.

Lemma 2.2

[4, Theorem 2.1] For any odd prime p,

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

(2.17)

Furthermore, for $0\le k\le \frac{p-3}{2},$

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

Lemma 2.3

[4, Theorem 2.2] For any prime $p\ge 5$,

$$\begin{aligned} f(-q)= & {} \sum _{k=-\frac{p-1}{2},k\ne \frac{\pm p-1}{6}}^{\frac{p-1}{2}}(-1)^kq^{\frac{3k^2+k}{2}}f(-q^{\frac{3p^2+(6k+1)p}{2}},-q^{\frac{3p^2-(6k+1)p}{2}})\nonumber \\&\quad +(-1)^{\frac{\pm p-1}{6}}q^{\frac{p^2-1}{24}}f(-q^{p^2}), \end{aligned}$$

(2.18)

where

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

Furthermore, for $-\frac{p-1}{2}\le k\le \frac{p-1}{2}$ and $k\ne \frac{\pm p-1}{6}$,

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

It follows from (2.11) and (2.16) that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}n+\frac{2^{2\alpha +1}-5}{3}\right) q^n\equiv f(-q^4) \pmod {2}, \end{aligned}$$

(2.19)

which is the case $\beta =0$ of (1.7).

Now, suppose that (1.7) holds for $\beta \ge 0$. Invoking Lemma 2.3, we get that for any prime $p\ge 5$,

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta }(pn+\frac{p^2-1}{6})+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n\nonumber \\&\quad = \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta +1}n+\frac{2^{2\alpha +1}p^{2\beta +2}-5}{3}\right) q^n \equiv f(-q^{4p})\pmod {2}, \end{aligned}$$

(2.20)

which implies that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta +2}n+\frac{2^{2\alpha +1}p^{2\beta +2}-5}{3}\right) q^n\equiv f(-q^4)\pmod {2}, \end{aligned}$$

(2.21)

which is the case $\beta +1$ of (1.7). The proof of Theorem 1.2 is completed by induction on $\beta$.

Combining (1.7) and (2.16), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta }n+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n \equiv \sum _{k=-\infty }^{\infty }(-1)^kq^{2k(3k+1)} \pmod {2}, \end{aligned}$$

(2.22)

which implies (1.8).

Applying (2.20) yields that for $j=1,2,\ldots ,p-1$,

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha +2}p^{2\beta +1}(pn+j)+\frac{2^{2\alpha +1}p^{2\beta +2}-5}{3}\right) \equiv 0\pmod {2}, \end{aligned}$$

which implies (1.9).

Following Hirschhorn [10], for a positive integer k and a power series $\sum _{n=0}^\infty a(n)q^n$, we introduce an operator $H_k$, which acts on series of (positive and negative) powers of a single variable and picks out those terms in which the power is congruent to 0 modulo k. That is

$$\begin{aligned} H_k\left( \sum _{n=0}^\infty a(n)q^n\right) :=\sum _{n=0}^\infty a(kn)q^{kn}. \end{aligned}$$

(2.23)

Garvan [8] proved the following results. Let

$$\begin{aligned} \xi =\frac{f(-q)}{q^2f(-q^{49})} \quad \mathrm{and} \quad T=\frac{f(-q^7)^4}{q^7f(-q^{49})^4}. \end{aligned}$$

(2.24)

We have

$$\begin{aligned}&H_7(\xi )=-1, \quad \quad \quad \quad H_7(\xi ^2)=1, \quad \quad \quad \quad H_7(\xi ^3)=-7,\nonumber \\&H_7(\xi ^4)=-4T-7, \quad H_7(\xi ^5)=10T+49, \quad H_7(\xi ^6)=49. \end{aligned}$$

(2.25)

Applying (2.3) and (2.24) in Theorem 1.2, for any prime $p\ge 5$, $\alpha \ge 1$ and $\beta \ge 0$, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2} p^{2\beta }n+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n\equiv f(-q)^4 \equiv q^8f(-q^{49})^4\xi ^4\pmod {2}. \end{aligned}$$

(2.26)

Then, we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2} p^{2\beta }(7n+1)+\frac{2^{2\alpha +1} p^{2\beta }-5}{3}\right) q^{7n+1}&\equiv q^8f(-q^{49})^4H_7(\xi ^4)\nonumber \\&\equiv q^8f(-q^{49})^4\pmod {2}. \end{aligned}$$

(2.27)

      Furthermore,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 7\cdot 2^{2\alpha +2} p^{2\beta }n+\frac{7\cdot 2^{2\alpha +1} p^{2\beta }-5}{3}\right) q^n\equiv qf(-q^{7})^4\pmod {2}. \end{aligned}$$

(2.28)

Therefore, for $k\in \{0,2,3,4,5,6\}$, we get

$$\begin{aligned} A_{3,5}\left( 7\cdot 2^{2\alpha +2} p^{2\beta }(7n+k)+\frac{7\cdot 2^{2\alpha +1} p^{2\beta }-5}{3}\right) \equiv 0\pmod {2}, \end{aligned}$$

(2.29)

which implies (1.10).

Hirschhorn [11] stated the following results. Denote

$$\begin{aligned} \zeta =\frac{f(-q)}{q^7f(-q^{169})} \quad and \quad S=\frac{f(-q^{13})^2}{q^{13}f(-q^{169})^2}. \end{aligned}$$

(2.30)

We have

$$\begin{aligned}&H_{13}(\zeta )=1,\quad \quad \quad H_{13}(\zeta ^2)=-2S-1, \nonumber \\&H_{13}(\zeta ^3)=13, \quad \quad H_{13}(\zeta ^4)=2S^2-13. \end{aligned}$$

(2.31)

Applying (2.3) and (2.30) in Theorem 1.2, for any prime $p\ge 5$, $\alpha \ge 1$ and $\beta \ge 0$, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2} p^{2\beta }n+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n\equiv f(-q)^4 \equiv q^{28}f(-q^{169})^4\zeta ^4\pmod {2}. \end{aligned}$$

(2.32)

Then,

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}p^{2\beta }(13n+2)+\frac{2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^{13n+2}&\equiv q^{28}f(-q^{169})^4H_{13}(\zeta ^4)\nonumber \\&\equiv q^{28}f(-q^{169})^4\pmod {2}. \end{aligned}$$

(2.33)

      Furthermore, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 13\cdot 2^{2\alpha +2}p^{2\beta }n+\frac{13\cdot 2^{2\alpha +1}p^{2\beta }-5}{3}\right) q^n\equiv q^2f(-q^{13})^4\pmod {2}. \end{aligned}$$

(2.34)

Therefore, for $k\in \{0,1,3,4,5,6,7,8,9,10,11,12\}$,

$$\begin{aligned} A_{3,5}\left( 13\cdot 2^{2\alpha +2}p^{2\beta }(13n+k)+\frac{13\cdot 2^{2\alpha +1}p^{2\beta }-5}{3}\right) \equiv 0\pmod {2}, \end{aligned}$$

(2.35)

which implies (1.11). This completes the proof of Theorem 1.2 and Corollary 1.1.

Proof of Theorem 1.3

Using (2.2) in (1.3), we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n&=\frac{f_3^{15}}{f_1^5}=\left( \frac{f_3^3}{f_1}\right) ^5 =\left( \frac{f_4^3f_6^2}{f_2^2f_{12}}+q\frac{f_{12}^3}{f_4}\right) ^5\nonumber \\&\equiv \frac{f_4^{15}f_6^{10}}{f_2^{10}f_{12}^5}+5q\frac{f_4^{11}f_6^8}{f_2^8f_{12}} +5q^4\frac{f_6^2f_{12}^{11}}{f_2^2f_4}+q^5\frac{f_{12}^{15}}{f_4^5} \pmod {10}. \end{aligned}$$

(2.36)

Extracting the terms involving $q^{2n+1}$ from both sides of (2.36), dividing by q and replacing $q^2$ by q, then employing (2.3), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n+1)q^n\equiv 5\frac{f_2^{11}f_3^8}{f_1^8f_{6}}+q^2\frac{f_{6}^{15}}{f_2^5} \equiv 5f_2^7f_6^3+q^2\frac{f_{6}^{15}}{f_2^5}\pmod {10}, \end{aligned}$$

(2.37)

which implies (1.12).

Utilizing (2.3) in (1.3), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n=\frac{f_3^{15}}{f_1^5}\equiv \frac{f_{15}^3}{f_5} \pmod {5}. \end{aligned}$$

(2.38)

This yields that for $k\in \{1,2,3,4\}$, $A_{3,5}(5n+k)\equiv 0 \pmod {5}$, which is (1.13).

Combining (2.2) and (2.38), we have that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(5n)q^n\equiv \frac{f_{3}^3}{f_1}=\frac{f_4^3f_6^2}{f_2^2f_{12}}+q\frac{f_{12}^3}{f_4} \pmod {5}. \end{aligned}$$

(2.39)

Extracting the terms involving $q^{2n+1}$ from both sides of (2.39), dividing by q and replacing $q^2$ by q, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(10n+5)q^n\equiv \frac{f_{6}^3}{f_2} \pmod {5}. \end{aligned}$$

(2.40)

Then, we have

$$\begin{aligned} A_{3,5}(20n+15)\equiv 0 \pmod {5}, \end{aligned}$$

(2.41)

which is (1.14). And we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(20n+5)q^n\equiv \frac{f_{3}^3}{f_1} \pmod {5}. \end{aligned}$$

(2.42)

In view of (2.39) and (2.42), we have

$$\begin{aligned} A_{3,5}(20n+5)\equiv A_{3,5}(5n) \pmod {5}. \end{aligned}$$

(2.43)

      Utilizing (2.43) and by mathematical induction on $\alpha$, we get (1.15). Combining (2.41) and (1.15), we thus arrive at (1.16). This completes the proof of Theorem 1.3. $\square$

3 Proofs of Theorems 1.4–1.9 and Corollaries 1.2–1.5

In this section, we prove the remaining theorems and corollaries about congruences modulo 4, 8 for $A_{3,5}(n)$.

Proof of Theorem 1.4

In order to prove Theorem 1.4, we first prove the following lemma.

Lemma 3.1

For $\alpha \ge 2$ and $n\ge 0$, we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha }n+\frac{2^{2\alpha +1}-5}{3}\right) q^n\equiv -\frac{f_6^3}{f_2}\pmod {4}. \end{aligned}$$

(3.1)

Proof

Baruah and Ojah [3] established that

$$\begin{aligned} \frac{1}{f_1f_3}&=\frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2}+q\frac{f_4^5f_{24}^2}{f_2^4f_6^2f_8^2f_{12}}. \end{aligned}$$

(3.2)

      Using (3.2) in (1.3), we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n&=\frac{f_3^{15}}{f_1^5}=\frac{f_3^{16}}{f_1^4}\frac{1}{f_1f_3}\nonumber \\&\equiv \frac{f_6^8}{f_2^2}\left( \frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2}+q\frac{f_4^5f_{24}^2}{f_2^4f_6^2f_8^2f_{12}}\right) \nonumber \\&\equiv \frac{f_6^4f_8^2f_{12}^5}{f_2^4f_4f_{24}^2}+q\frac{f_4^5f_6^6f_{24}^2}{f_2^6f_8^2f_{12}}\pmod {4}. \end{aligned}$$

(3.3)

Xia and Yao [19] proved the following identity

$$\begin{aligned} \frac{f_3^2}{f_1^2}=\frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}. \end{aligned}$$

(3.4)

Extracting the terms involving $q^{2n+1}$ from both sides of (3.3), dividing by q and replacing $q^2$ by q, then employing (3.4), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n+1)q^n&\equiv \frac{f_2^5f_3^6f_{12}^2}{f_1^6f_4^2f_6} \equiv \frac{f_2^5f_{12}^2}{f_4^2f_6}\left( \frac{f_{3}^2}{f_1^2}\right) ^3\nonumber \\&\equiv \frac{f_2^5f_{12}^2}{f_4^2f_6}\left( \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}\right) ^3\nonumber \\&\equiv \frac{f_4^{10}f_6^2f_{12}^8}{f_2^{10}f_8^3f_{24}^3} +2q\frac{f_4^7f_6^3f_{12}^5}{f_2^9f_8f_{24}}\pmod {4}. \end{aligned}$$

(3.5)

Extracting the terms involving $q^{2n}$ from both sides of (3.5), replacing $q^2$ by q and using (3.4), we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(4n+1)q^n&\equiv \frac{f_2^{10}f_3^2f_{6}^8}{f_1^{10}f_4^3f_{12}^3} \equiv \frac{f_2^6f_6^8}{f_4^3f_{12}^3}\frac{f_3^2}{f_1^2}\nonumber \\&\equiv \frac{f_2^6f_6^8}{f_4^3f_{12}^3}\left( \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}}+2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}\right) \nonumber \\&\equiv \frac{f_2f_4f_6^9}{f_8f_{12}f_{24}}+2q\frac{f_2^2f_6^{10}f_8f_{24}}{f_4^2f_{12}^4}\pmod {4}. \end{aligned}$$

(3.6)

Xia and Yao [20] proved the following 2-dissection formula.

$$\begin{aligned} f_1f_3=\frac{f_2f_8^2f_{12}^4}{f_4^2f_6f_{24}^2}-q\frac{f_4^4f_6f_{24}^2}{f_2f_8^2f_{12}^2}. \end{aligned}$$

(3.7)

Extracting the terms involving $q^{2n}$ from both sides of (3.6), replacing $q^2$ by q and using (3.7), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(8n+1)q^n&\equiv \frac{f_1f_2f_3^9}{f_4f_{6}f_{12}}\equiv \frac{f_2f_6^3}{f_4f_{12}}\cdot f_1f_3\nonumber \\&\equiv \frac{f_2f_6^3}{f_4f_{12}}\cdot \left( \frac{f_2f_8^2f_{12}^4}{f_4^2f_6f_{24}^2}-q\frac{f_4^4f_6f_{24}^2}{f_2f_8^2f_{12}^2}\right) \nonumber \\&\equiv \frac{f_2^2f_6^2f_8^2f_{12}^3}{f_4^3f_{24}^2}-q\frac{f_4^3f_6^4f_{24}^2}{f_8^2f_{12}^3}\pmod {4}. \end{aligned}$$

(3.8)

Extracting the terms involving $q^{2n+1}$ from both sides of (3.8), dividing by q and replacing $q^2$ by q, then using (2.3), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(16n+9)q^n\equiv - \frac{f_2^3f_3^4f_{12}^2}{f_4^2f_{6}^3} \equiv -\frac{f_6^3}{f_2}\pmod {4}, \end{aligned}$$

(3.9)

which is the case $\alpha =2$ of (3.1).

Suppose that (3.1) is true for $\alpha \ge 2$. According to (3.1) and employing (2.2), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +1}n+\frac{2^{2\alpha +1}-5}{3}\right) q^n \equiv -\frac{f_3^3}{f_1}\equiv -\frac{f_4^3f_6^2}{f_2^2f_{12}}-q\frac{f_{12}^3}{f_4}\pmod {4}. \end{aligned}$$

(3.10)

Extracting the terms involving $q^{2n+1}$ from both sides of (3.10), dividing by q and replacing $q^2$ by q, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 2^{2\alpha +2}n+\frac{2^{2\alpha +3}-5}{3}\right) q^n \equiv -\frac{f_6^3}{f_2}\pmod {4}, \end{aligned}$$

(3.11)

which is the case $\alpha +1$ of (3.1). The proof of Lemma 3.1 is completed by induction on $\alpha$.

Now, we turn to prove Theorem 1.4. According to (3.1), we have

$$\begin{aligned} A_{3,5}\left( 2^{2\alpha }(2n+1)+\frac{2^{2\alpha +1}-5}{3}\right) \equiv 0 \pmod {4}, \end{aligned}$$

(3.12)

which is (1.17). This completes the proof of Theorem 1.4. $\square$

Proof of Theorem 1.5 and Corollary 1.2

Hirschhorn, Garvan and Borwein [9] proved that

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

(3.13)

Extracting the terms involving $q^{2n+1}$ from both sides of (3.5), dividing by q and replacing $q^2$ by q, then using (3.13), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(4n+3)q^n&\equiv 2\frac{f_2^7f_6^5}{f_4f_{12}}\frac{f_3^3}{f_1^9} \equiv 2\frac{f_2^7f_6^5}{f_4f_{12}}\left( \frac{f_4^6f_6^3}{f_2^9f_{12}^2}+3q\frac{f_4^2f_6f_{12}^2}{f_2^7}\right) ^3 \nonumber \\&\equiv 2\frac{f_4^{17}f_6^{14}}{f_2^{20}f_{12}^7}+2q\frac{f_4^{13}f_6^{12}}{f_2^{18}f_{12}^3} +2q^2\frac{f_4^9f_6^{10}f_{12}}{f_2^{16}}+2q^3\frac{f_4^5f_6^8f_{12}^5}{f_2^{14}} \pmod {4}. \end{aligned}$$

(3.14)

Extracting the terms involving $q^{2n}$ from both sides of (3.14), replacing $q^2$ by q, we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(8n+3)q^n\equiv 2\frac{f_2^{17}f_3^{14}}{f_1^{20}f_{6}^7}+2q\frac{f_2^9f_3^{10}f_6}{f_1^{16}}\pmod {4}. \end{aligned}$$

(3.15)

By (2.3), we see that

$$\begin{aligned} \frac{f_2^{17}f_3^{14}}{f_1^{20}f_{6}^7}\equiv f_2f_4^3 \pmod {2}, \end{aligned}$$

(3.16)

$$\begin{aligned} \frac{f_2^9f_3^{10}f_6}{f_1^{16}}\equiv f_2f_{12}^3 \pmod {2}. \end{aligned}$$

(3.17)

Combining (3.15), (3.16) and (3.17), we deduce that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(8n+3)q^n\equiv 2f_2f_4^3+2qf_2f_{12}^3 \pmod {4}, \end{aligned}$$

(3.18)

which yields

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(16n+3)q^n&\equiv 2f_1f_2^3\pmod {4}, \end{aligned}$$

(3.19)

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(16n+11)q^n&\equiv 2 f_1f_{6}^3\pmod {4}. \end{aligned}$$

(3.20)

Using (2.3) in (2.15), we obtain

$$\begin{aligned} \psi (q)\equiv f_1^3 \pmod {4}. \end{aligned}$$

(3.21)

Furthermore, in view of (2.16) and (3.21), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(16n+3)q^n&\equiv 2 f(-q)\psi (q^2)\pmod {4}, \end{aligned}$$

(3.22)

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(16n+11)q^n&\equiv 2f(-q)\psi (q^6)\pmod {4}. \end{aligned}$$

(3.23)

By (3.22), we find (1.18) holds for $\alpha =0$. Assume that (1.18) holds for $\alpha \ge 0$. Employing Lemmas 2.2 and 2.3 in (1.18), we consider the congruence

$$\begin{aligned} \frac{3k^2+k}{2}+2\cdot \frac{m^2+m}{2}\equiv \frac{7p^2-7}{24} \pmod {p}, \end{aligned}$$

(3.24)

where $-\frac{p-1}{2}\le k\le \frac{p-1}{2}$ and $0\le m\le \frac{p-3}{2}$.

The congruence (3.24) is equivalent to

$$\begin{aligned} (6k+1)^2+6(2m+1)^2\equiv 0 \pmod {p}. \end{aligned}$$

(3.25)

For $\left( \frac{-6}{p}\right) =-1$, the congruence (3.25) holds if and only if $k=\frac{\pm p-1}{6}$ and $m=\frac{p-1}{2}.$

Extracting the terms containing $q^{pn+\frac{7p^2-7}{24}}$ from both sides of (1.18), dividing by $q^{\frac{7p^2-7}{24}}$ and replacing $q^p$ by q, we arrive at

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha +1}n+\frac{14\cdot p^{2\alpha +2}-5}{3}\right) q^n\equiv 2f(-q^p)\psi (q^{2p})\pmod {4}. \end{aligned}$$

(3.26)

This implies that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha +2}n+\frac{14\cdot p^{2\alpha +2}-5}{3}\right) q^n\equiv 2f(-q)\psi (q^{2})\pmod {4}, \end{aligned}$$

(3.27)

which is the case $\alpha +1$ of (1.18). The proof of Theorem 1.5 is completed by induction on $\alpha$.

Combining (1.18), (2.15) and (2.16), we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{14\cdot p^{2\alpha }-5}{3}\right) q^n \equiv 2\sum _{n=-\infty }^\infty \sum _{k=0}^\infty q^{\frac{n(3n+1)}{2}+k(k+1)}\pmod {4}, \end{aligned}$$

(3.28)

which implies (1.19).

Applying (3.26) yields that for $j=1, 2, \ldots ,p-1$,

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha +1}(pn+j)+\frac{14\cdot p^{2\alpha +2}-5}{3}\right) q^n\equiv 0\pmod {4}, \end{aligned}$$

(3.29)

which implies (1.20). This completes the proof of Theorem 1.5 and Corollary 1.2. $\square$

Proof of Theorem 1.6 and Corollary 1.3

It follows from (3.23) that (1.21) holds for $\alpha =0$. Suppose that (1.21) holds for $\alpha \ge 0$. Employing Lemma 2.2 and Lemma 2.3 in (1.21), we consider the congruence

$$\begin{aligned} \frac{3k^2+k}{2}+6\cdot \frac{m^2+m}{2}\equiv \frac{19p^2-19}{24} \pmod {p}, \end{aligned}$$

(3.30)

where $-\frac{p-1}{2}\le k\le \frac{p-1}{2}$ and $0\le m\le \frac{p-3}{2}$.

The congruence (3.30) is equivalent to

$$\begin{aligned} (6k+1)^2+18(2m+1)^2\equiv 0 \pmod {p}. \end{aligned}$$

(3.31)

For $\left( \frac{-18}{p}\right) =-1$, the congruence (3.31) holds if and only if $k=\frac{\pm p-1}{6}$ and $m=\frac{p-1}{2}.$

Extracting the terms containing $q^{pn+\frac{19p^2-19}{24}}$ from both sides of (1.21), dividing by $q^{\frac{19p^2-19}{24}}$ and replacing $q^p$ by q, we arrive at

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha +1}n+\frac{38\cdot p^{2\alpha +2}-5}{3}\right) q^n\equiv 2f(-q^p)\psi (q^{6p})\pmod {4}. \end{aligned}$$

(3.32)

This implies that

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha +2}n+\frac{38\cdot p^{2\alpha +2}-5}{3}\right) q^n\equiv 2f(-q)\psi (q^{6})\pmod {4}, \end{aligned}$$

(3.33)

which is the case $\alpha +1$ of (1.21). The proof of Theorem 1.6 is completed by induction on $\alpha$.

Combining (1.21), (2.15) and (2.16), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( 16\cdot p^{2\alpha }n+\frac{38\cdot p^{2\alpha }-5}{3}\right) q^n \equiv 2\sum _{n=-\infty }^\infty \sum _{k=0}^\infty q^{\frac{n(3n+1)}{2}+3k(k+1)}\pmod {4}, \end{aligned}$$

(3.34)

which implies (1.22).

Applying (3.32) yields that for $1\le j\le p-1$,

$$\begin{aligned} A_{3,5}\left( 16\cdot p^{2\alpha +1}(pn+j)+\frac{38\cdot p^{2\alpha +2}-5}{3}\right) \equiv 0\pmod {4}, \end{aligned}$$

(3.35)

which implies (1.23). This completes the proof of Theorem 1.6 and Corollary 1.3. $\square$

Proof of Theorem 1.7

Xia and Yao [19] proved the following 2-dissection formulae.

$$\begin{aligned} \frac{f_1^3}{f_3}&=\frac{f_4^3}{f_{12}}-3q\frac{f_2^2f_{12}^3}{f_4f_6^2}, \end{aligned}$$

(3.36)

$$\begin{aligned} \frac{1}{f_1^2f_3^2}&=\frac{f_8^5f_{24}^5}{f_2^5f_6^5f_{16}^2f_{48}^2}+2q\frac{f_4^4f_{12}^4}{f_2^6f_6^6} +4q^4\frac{f_4^2f_{12}^2f_{16}^4f_{48}^2}{f_2^5f_6^5f_8f_{24}}. \end{aligned}$$

(3.37)

In view of (3.36), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(n)q^n&=\frac{f_3^{16}}{f_1^8}\frac{f_1^3}{f_3}\equiv \frac{f_6^8}{f_2^4}\left( \frac{f_4^3}{f_{12}}-3q\frac{f_2^2f_{12}^3}{f_4f_6^2}\right) \nonumber \\&\equiv \frac{f_6^8f_4^3}{f_2^4f_{12}}-3q\frac{f_6^6f_{12}^3}{f_2^2f_4}\pmod {8}. \end{aligned}$$

(3.38)

      Utilizing (2.3), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n)q^n\equiv \frac{f_3^8f_2^3}{f_1^4f_{6}} \equiv f_2f_6^3\pmod {4}, \end{aligned}$$

(3.39)

which implies (1.24).

Extracting the terms involving $q^{2n+1}$ from both sides of (3.38), dividing by q and replacing $q^2$ by q, then using (3.37), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n+1)q^n&\equiv -3\frac{f_3^6f_{6}^3}{f_1^2f_2}\equiv 5\frac{f_3^6f_{6}^3}{f_1^2f_2}\equiv 5\frac{f_3^8f_{6}^3}{f_2f_1^2f_3^2} \equiv 5\frac{f_6^7}{f_2}\frac{1}{f_1^2f_3^2}\nonumber \\&\equiv 5\frac{f_6^7}{f_2}\left( \frac{f_8^5f_{24}^5}{f_2^5f_6^5f_{16}^2f_{48}^2}+2q\frac{f_4^4f_{12}^4}{f_2^6f_6^6} +4q^4\frac{f_4^2f_{12}^2f_{16}^4f_{48}^2}{f_2^5f_6^5f_8f_{24}}\right) \nonumber \\&\equiv 5\frac{f_6^2f_8^5f_{24}^5}{f_2^6f_{16}^2f_{48}^2}+2q\frac{f_4^4f_6f_{12}^4}{f_2^7} +4q^4\frac{f_4^2f_6^2f_{12}^2f_{16}^4f_{48}^2}{f_2^6f_8f_{24}}\pmod {8}. \end{aligned}$$

(3.40)

In view of (2.3), we obtain

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(4n+1)q^n&\equiv 5\frac{f_3^2f_4^5f_{12}^5}{f_1^6f_{8}^2f_{24}^2}+4q^2\frac{f_2^2f_3^2f_{6}^2f_{8}^4f_{24}^2}{f_1^6f_4f_{12}} \nonumber \\&\equiv 5\frac{f_3^2f_4^5f_{12}^5}{f_1^6f_{8}^2f_{24}^2}+4q^2\frac{f_6f_{8}^4f_{24}^2}{f_2f_4}\pmod {8}. \end{aligned}$$

(3.41)

Using (3.13), we see that

$$\begin{aligned} 5\frac{f_3^2f_4^5f_{12}^5}{f_1^6f_{8}^2f_{24}^2}&\equiv 5\frac{f_4^5f_{12}^5}{f_8^2f_{24}^2}\left( \frac{f_3}{f_1^3}\right) ^2 \equiv 5\frac{f_4^5f_{12}^5}{f_8^2f_{24}^2}\left( \frac{f_4^6f_6^3}{f_2^9f_{12}^2}+3q\frac{f_4^2f_6f_{12}^2}{f_2^7}\right) ^2\nonumber \\&\equiv 5\frac{f_4^5f_{12}^5}{f_8^2f_{24}^2} \left( \frac{f_4^{12}f_6^6}{f_2^{18}f_{12}^4}+6q\frac{f_4^8f_6^4}{f_2^{16}}+9q^2\frac{f_4^4f_6^2f_{12}^4}{f_2^{14}}\right) \nonumber \\&\equiv 5\frac{f_4^{17}f_6^6f_{12}}{f_2^{18}f_8^2f_{24}^2} +6q\frac{f_4^{13}f_6^4f_{12}^5}{f_2^{16}f_8^2f_{24}^2}+5q^2\frac{f_4^9f_6^2f_{12}^9}{f_2^{14}f_8^2f_{24}^2}\pmod {8}. \end{aligned}$$

(3.42)

Combining (3.41) and (3.42), extracting the terms involving $q^{2n+1}$ from both sides of (3.41), dividing by q and replacing $q^2$ by q, we find

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(8n+5)q^n&\equiv 6\frac{f_2^{13}f_3^4f_{6}^5}{f_1^{16}f_4^2f_{12}^2}\equiv 6f_2f_6^3\pmod {8}, \end{aligned}$$

(3.43)

which implies (1.25). This completes the proof of Theorem 1.7. $\square$

Proof of Theorem 1.8 and Corollary 1.4

By (3.39), using (2.16) and (3.21), we get

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(2n)q^n\equiv f(-q^2)\psi (q^6) \pmod {4}, \end{aligned}$$

(3.44)

which is the case $\alpha =0$ of (1.26).

Suppose that (1.26) holds for $\alpha \ge 0$. Employing Lemma 2.2 and Lemma 2.3 in (1.26), we consider the congruence

$$\begin{aligned} 2\cdot \frac{3k^2+k}{2}+6\cdot \frac{m^2+m}{2}\equiv \frac{20p^2-20}{24} \pmod {p}, \end{aligned}$$

(3.45)

where $-\frac{p-1}{2}\le k\le \frac{p-1}{2}$ and $0\le m\le \frac{p-3}{2}$.

The congruence (3.45) is equivalent to

$$\begin{aligned} (6k+1)^2+9(2m+1)^2\equiv 0 \pmod {p}. \end{aligned}$$

(3.46)

For $\left( \frac{-9}{p}\right) =-1$, the congruence (3.46) holds if and only if $k=\frac{\pm p-1}{6}$ and $m=\frac{p-1}{2}.$

Extracting the terms containing $q^{pn+\frac{5p^2-5}{6}}$ from both sides of (1.26), dividing by $q^{\frac{5p^2-5}{6}}$ and replacing $q^p$ by q, we arrive at

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 2\cdot p^{2\alpha +1}n+\frac{5\cdot p^{2\alpha +2}-5}{3}\right) q^n\nonumber \\&\quad \equiv (-1)^{(\alpha +1)(\frac{\pm p-1}{6})}f(-q^{2p})\psi (q^{6p})\pmod {4}. \end{aligned}$$

(3.47)

This implies that

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 2\cdot p^{2\alpha +2}n+\frac{5\cdot p^{2\alpha +2}-5}{3}\right) q^n\nonumber \\&\quad \equiv (-1)^{(\alpha +1)(\frac{\pm p-1}{6})}f(-q^2)\psi (q^6)\pmod {4}, \end{aligned}$$

(3.48)

which is the case $\alpha +1$ of (1.26). The proof of Theorem 1.8 is completed by induction on $\alpha$.

Combining (1.26), (2.15) and (2.16), we obtain

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 4\cdot p^{2\alpha }n+\frac{5\cdot p^{2\alpha }-5}{3}\right) q^n\equiv \sum _{n=-\infty }^\infty \sum _{k=0}^\infty (-1)^{\alpha (\frac{\pm p-1}{6})+n}q^{\frac{n(3n+1)}{2}+3\frac{k(k+1)}{2}}\pmod {4}, \end{aligned}$$

which implies (1.27).

Applying (3.47) yields that for $j=1,2, \cdots , p-1$,

$$\begin{aligned} A_{3,5}\left( 2\cdot p^{2\alpha +1}(pn+j)+\frac{5\cdot p^{2\alpha +2}-5}{3}\right) \equiv 0\pmod {4}. \end{aligned}$$

which implies (1.28). This finishes the proof of Theorem 1.8 and Corollary 1.4. $\square$

Proof of Theorem 1.9 and Corollary 1.5

By (3.43), according to (2.16) and (3.21), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}(8n+5)q^n\equiv 6f(-q^2)\psi (q^6)\pmod {8}, \end{aligned}$$

(3.49)

which is the case $\alpha =0$ of (1.29).

Suppose that (1.29) holds for $\alpha \ge 0$. Employing Lemma 2.2 and Lemma 2.3 in (1.29), we consider the congruence

$$\begin{aligned} 2\cdot \frac{3k^2+k}{2}+6\cdot \frac{m^2+m}{2}\equiv \frac{20p^2-20}{24} \pmod {p}, \end{aligned}$$

(3.50)

where $-\frac{p-1}{2}\le k\le \frac{p-1}{2}$ and $0\le m\le \frac{p-3}{2}$.

The congruence (3.50) is equivalent to

$$\begin{aligned} (6k+1)^2+9(2m+1)^2\equiv 0 \pmod {p}. \end{aligned}$$

(3.51)

For $\left( \frac{-9}{p}\right) =-1$, the congruence (3.51) holds if and only if $k=\frac{\pm p-1}{6}$ and $m=\frac{p-1}{2}.$

Extracting the terms containing $q^{pn+\frac{5p^2-5}{6}}$ from both sides of (1.29), dividing by $q^{\frac{5p^2-5}{6}}$ and replacing $q^p$ by q, we arrive at

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 8\cdot p^{2\alpha +1}n+\frac{20\cdot p^{2\alpha +2}-5}{3}\right) q^n\nonumber \\&\quad \equiv 6(-1)^{(\alpha +1) (\frac{\pm p-1}{6})}f(-q^{2p})\psi (q^{6p})\pmod {8}. \end{aligned}$$

(3.52)

This implies that

$$\begin{aligned}&\sum _{n=0}^\infty A_{3,5}\left( 8\cdot p^{2\alpha +2}n+\frac{20\cdot p^{2\alpha +2}-5}{3}\right) q^n\nonumber \\&\quad \equiv 6(-1)^{(\alpha +1) (\frac{\pm p-1}{6})}f(-q^{2})\psi (q^{6})\pmod {8}, \end{aligned}$$

(3.53)

which is the case $\alpha +1$ of (1.29). The proof of Theorem 1.9 is completed by induction on $\alpha$.

Combining (1.29), (2.16) and (3.21), we have

$$\begin{aligned} \sum _{n=0}^\infty A_{3,5}\left( \!16\cdot p^{2\alpha }n+\frac{20\cdot p^{2\alpha }-5}{3}\!\right) q^n \equiv 6 \sum _{n=-\infty }^\infty \sum _{k=0}^\infty (-1)^{\alpha (\frac{\pm p-1}{6})+n}q^{\frac{n(3n+1)}{2}+3\frac{k(k+1)}{2}}\pmod {8}, \end{aligned}$$

which implies (1.30).

Applying (3.52) yields that for $1\le j\le p-1$,

$$\begin{aligned} A_{3,5}\left( 8\cdot p^{2\alpha +1}(pn+j)+\frac{20\cdot p^{2\alpha +2}-5}{3}\right) \equiv 0\pmod {8}, \end{aligned}$$

(3.54)

which implies (1.31). This completes the proof of Theorem 1.9 and Corollary 1.5. $\square$