Infinite families of congruences modulo 7 for broken 3-diamond partitions

Abstract

The notion of broken k-diamond partitions was introduced by Andrews and Paule. Let $\Delta _k(n)$ denote the number of broken k-diamond partitions of n for a fixed positive integer k. Recently, Paule and Radu conjectured that $\Delta _3(343n+82)\equiv \Delta _3(343n+278)\equiv \Delta _3(343n+327)\equiv 0\ (\mathrm{mod} \ 7)$. Jameson confirmed this conjecture and proved that $\Delta _3(343n+229)\equiv 0 \ (\mathrm{mod} \ 7)$ by using the theory of modular forms. In this paper, we prove several infinite families of Ramanujan-type congruences modulo 7 for $\Delta _3(n)$ by establishing a recurrence relation for a sequence related to $\Delta _3(7n+5)$. In the process, we also give new proofs of the four congruences due to Paule and Radu, and Jameson.

1 Introduction

The aim of this paper is to establish several infinite families of Ramanujan-type congruences modulo 7 for broken 3-diamond partitions. In the process, we also present new proofs of four congruences modulo 7 for broken 3-diamond partitions due to Paule and Radu [13], and Jameson [9].

Let us begin with some notation and terminology on q-series and partitions. We use the standard notation

$$\begin{aligned} (a;q)_\infty =\prod _{k=0}^{\infty }\big (1-aq^k\big ) \end{aligned}$$

(1.1)

and often write

$$\begin{aligned} (a_1,a_2,\ldots , a_n;q)_\infty&=(a_1;q)_\infty (a_2;q)_\infty \cdots (a_n;q)_\infty . \end{aligned}$$

(1.2)

Recall that the Ramanujan theta function f(a, b) is defined by

$$\begin{aligned} f(a,b)=\sum _{n=-\infty }^\infty a^{n(n+1)/2}b^{n(n-1)/2}, \end{aligned}$$

(1.3)

where $|ab|<1$. The Jacobi triple product identity can be restated as

$$\begin{aligned} f(a,b)=(-a,-b,ab;ab)_\infty . \end{aligned}$$

(1.4)

Two special cases of (1.3) are defined by

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

(1.5)

and

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

(1.6)

In this paper, for any positive integer n, we use $f_n$ to denote $f(-q^n)$, that is,

$$\begin{aligned} f_n=\big (q^n;q^n\big )_\infty =\prod _{k=1}^\infty \big (1-q^{nk}\big ). \end{aligned}$$

(1.7)

By (1.4), (1.5) and (1.6), we have

$$\begin{aligned} f(-q)=f_1, \qquad \psi (q)=\frac{f_2^2}{f_1}. \end{aligned}$$

(1.8)

A combinatorial study guided by MacMahon’s Partition Analysis led Andrews and Paule [2] to the construction of a new class of directed graphs called broken k-diamond partitions. Let $\Delta _k(n)$ denote the number of broken k-diamond partitions of n for a fixed positive integer k. Andrews and Paule [2] established the following generating function of $\Delta _k(n)$:

$$\begin{aligned} \sum _{n=0}^\infty \Delta _k(n) q^n= \frac{f_{2}f_{2k+1}}{f_1^3f_{4k+2}}. \end{aligned}$$

(1.9)

Employing generating function manipulations, Andrews and Paule [2] proved that for all integers $n \ge 0$,

$$\begin{aligned} \Delta _1 (2n+1) \equiv 0 \quad (\mathrm{mod}\ 3). \end{aligned}$$

(1.10)

They also gave three conjectures modulo 2, 5 and 25 for $\Delta _k(n)$. Since then, a number of congruences satisfied by $\Delta _k(n)$ for small values of k have been proved. Hirschhorn and Sellers [8] established an explicit representation of the generating function for $\Delta _1 (2n+1)$ which implied (1.10). Mortenson [12] reproved (1.10) by developing a statistic on the partitions enumerated by $\Delta _1 (2n+1)$ which naturally breaks these partitions into three subsets of equal size. In addition, Hirschhorn and Sellers [8] also provided elementary proofs of four congruences modulo 2 for $\Delta _1(n)$ and $\Delta _2(n)$ and one of which was a conjecture due to Andrews and Paule [2]. Radu and Sellers [15] established several infinite families of congruences modulo 3 for $\Delta _2(n)$. Lin and Wang [11] presented elementary proofs of some results of Radu and Sellers [15]. Chen, Fan and Yu [6] discovered two infinite families of congruences for $\Delta _2(n)$ modulo 3. Chan [5] found two infinite families of congruences modulo 5 for broken 2-diamond partitions. Radu and Sellers [14] have given numerous beautiful congruence properties for broken k-diamond partitions. Radu and Sellers [16] provided an extensive analysis of the parity of the function $\Delta _3(n)$, including a number of Ramanujan-like congruences modulo 2. Lin [10] gave elementary proofs of the results due to Radu and Sellers [16]. Cui and Gu [7] proved several infinite families of congruences modulo 2 for $\Delta _3(n)$. Xia [17] considered congruences modulo 4 for $\Delta _3(n)$ and proved a conjecture of Radu and Sellers [16]. Yao [19] proved several infinite families of congruences modulo 2 for $\Delta _{11}(n)$ and generalized some results due to Radu and Sellers [14]. Ahmed and Baruah [1] discovered some parity results for broken 5-diamond, 7-diamond and 11-diamond partitions. Paule and Radu [13] discovered two non-standard infinite families of congruences for broken 2-diamond partitions. They also presented four conjectures related to $\Delta _3(n)$ and $\Delta _5(n)$. Xiong [18] proved the following congruence which was a conjecture of Paule and Radu [13]:

$$\begin{aligned} \sum _{n=0}^\infty \Delta _3(7n+5) q^n\equiv 6 f_1^4f_2^6\ (\mathrm{mod}\ 7). \end{aligned}$$

(1.11)

Employing the theory of modular forms, Jameson [9] proved the following theorem:

Theorem 1.1

For $n\ge 0$,

$$\begin{aligned} \Delta _3(343n+82)&\equiv 0 \ ( \mathrm{mod}\ 7), \end{aligned}$$

(1.12)

$$\begin{aligned} \Delta _3(343n+229)&\equiv 0 \ ( \mathrm{mod}\ 7), \end{aligned}$$

(1.13)

$$\begin{aligned} \Delta _3(343n+278)&\equiv 0 \ ( \mathrm{mod}\ 7),\end{aligned}$$

(1.14)

$$\begin{aligned} \Delta _3(343n+327)&\equiv 0 \ (\mathrm{mod}\ 7). \end{aligned}$$

(1.15)

Congruences (1.12), (1.14) and (1.15) were conjectured by Paule and Radu [13] and congruence (1.13) was discovered by Jameson [9].

In this paper, we prove several infinite families of Ramanujan-type congruences modulo 7 for $\Delta _3(n)$ by establishing a recurrence relation for the coefficients of $f_1^4f_2^6$. Furthermore, we give a new proof of Theorem 1.1. Our proof mainly relies on (1.11) and some identities involving theta functions due to Ramanujan. The main results of this paper can be stated as follows.

Theorem 1.2

For $n\ge 0$ and $k\ge 0$, we have

$$\begin{aligned} \Delta _3\left( 7\times 4^{7k+4}n+\frac{77\times 4^{7k+3}+1}{3}\right) \equiv 0 \ (\mathrm{mod}\ 7) \end{aligned}$$

(1.16)

and

$$\begin{aligned} \Delta _3\left( 14\times 4^{7k+7}n+\frac{14\times 4^{7k+6}+1}{3}\right) \equiv 0 \ (\mathrm{mod}\ 7). \end{aligned}$$

(1.17)

In order to state the following theorem, we introduce the Legendre symbol. Let $p\ge 3$ be a prime. The Legendre symbol $\left( \displaystyle \frac{a}{p}\right)$ is defined by

$$\begin{aligned} \left( \displaystyle \frac{a}{p}\right) := \left\{ \begin{array}{ll} 1 ,&{} \quad \mathrm{if \ } a \ \mathrm{is \ a \ quadratic \ residue \ modulo}\ p\ \mathrm{and} \ p\not \mid a, \\ -1 ,&{} \quad \mathrm{if \ } a \text { is a quadratic non-residue modulo}\ p,\\ 0, &{} \quad \mathrm{if}\ p|a. \end{array} \right. \end{aligned}$$

(1.18)

Theorem 1.3

Let $p\ge 5$ be a prime such that $\left( \displaystyle \frac{-7}{p}\right) =-1$. For $n\ge 0$, $j\ge 0$ and $k\ge 1$, we have

$$\begin{aligned} \Delta _3\left( 7\times 4^kp^{2j}(7n+s) + \frac{77\times 4^{k-1}p^{2j}+1}{3}\right) \equiv 0\ (\mathrm{mod}\ 7), \end{aligned}$$

(1.19)

where $s\in \{2,\ 3,\ 4,\ 6\}$.

Theorem 1.4

Let $p\ge 5$ be a prime such that $\left( \displaystyle \frac{-7}{p}\right) =-1$ and let $n\ge 0$, $j\ge 0$ and $k\ge 1$ be integers. If $p\not \mid n$, then

$$\begin{aligned} \Delta _3\left( 7\times 4^kp^{2j+1}n +\frac{77\times 4^{k-1} p^{2j+2} +1}{3} \right) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(1.20)

This paper is organized as follows: In Sect. 2, we present a new proof of Theorem 1.1 based on (1.11). In Sect. 3, we establish a recurrence relation for a(n) where the generating function of a(n) is $f_1^4f_2^6$. Moreover, we also derive some generating functions of $a(An+B)$ modulo 7 for some values of A and B. In Sect. 4, we prove Theorem 1.2 by using the recurrence relation given in Sect. 3. In Sect. 5, we prove Theorems 1.3 and 1.4 by employing the generating functions of $a(An+B)$ modulo 7.

2 A new proof of Theorem 1.1

By the binomial theory, it is easy to see that for any positive integer k,

$$\begin{aligned} f_k^7\equiv f_{7k}\ (\mathrm{mod}\ 7). \end{aligned}$$

(2.1)

Thanks to (1.11) and (2.1),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _3(7n+5) q^n \equiv 6 f_1^4f_2^6 \equiv 6 f_7 \psi ^3(q) \ (\mathrm{mod}\ 7), \end{aligned}$$

(2.2)

where $\psi (q)$ is defined by (1.8). From Entry 17 (iv) on page 303 in Berndt’s book [3], we have the 7-dissection

$$\begin{aligned} \psi (q)=A+qB +q^3C+q^6\psi \big (q^{49}\big ), \end{aligned}$$

(2.3)

where

$$\begin{aligned} A=f\big (q^{21},q^{28}\big ), \qquad B=f\big (q^{14},q^{35}\big ), \qquad C=f\big (q^7,q^{42}\big ). \end{aligned}$$

(2.4)

Therefore, combining (2.2) and (2.3), we get

$$\begin{aligned} \sum _{n=0}^\infty&\Delta _3(7n+5) q^n \nonumber \\ \equiv&\,6f_7 \left( A+q B+q^3C+q^6\psi \big (q^{49}\big )\right) ^3 \nonumber \\ \equiv&\, 6f_7\Big ( A^3+3qA^2B+3q^2AB^2+q^3B^3+3q^3A^2C+6q^4ABC\nonumber \\&+ 3q^5B^2C+3q^6AC^2+3q^6A^2 \psi \big (q^{49}\big ) +6q^7AB\psi \big (q^{49}\big ) + 3q^7BC^2 \nonumber \\&+ 3q^8B^2 \psi \big (q^{49}\big )+q^9C^3+6q^9AC\psi \big (q^{49}\big )+6q^{10}BC\psi \big (q^{49}\big ) \nonumber \\&+ 3q^{12}C^2\psi \big (q^{49}\big )+3 q^{12}A\psi ^2\big (q^{49}\big ) + 3q^{13}B\psi ^2\big (q^{49}\big )\nonumber \\&+ 3q^{15}C \psi ^2\big (q^{49}\big ) +q^{18}\psi ^3\big (q^{49}\big )\Big )\ (\mathrm{mod}\ 7). \end{aligned}$$

(2.5)

Extracting the terms in (2.5) that involves $q^{7n+4}$, dividing the resulting identity by $q^4$ and then replacing $q^7$ by q, we deduce that

$$\begin{aligned} \sum _{n=0}^\infty \Delta _3(49n+33) q^n \equiv f_1f\big (q^3,q^4\big )f\big (q^2,q^5\big )f\big (q,q^6\big ) + 6q^2 f_1 \psi ^3\big (q^7\big )\ (\mathrm{mod}\ 7). \end{aligned}$$

(2.6)

By (1.4), it is easy to check that

$$\begin{aligned} f\big (q^3,q^4\big )f\big (q^2,q^5\big )f\big (q,q^6\big )=\frac{f_2f_{7}^4}{f_1 f_{14}}. \end{aligned}$$

(2.7)

Thanks to (1.8), (2.6) and (2.7),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _3(49n+33) q^n \equiv \frac{f_2f_{7}^4}{ f_{14}} +6q^2 f_1\frac{f_{14}^6}{f_7^3}\ (\mathrm{mod}\ 7). \end{aligned}$$

(2.8)

From Entry 17 (v) on page 303 in [3], we have the 7-dissection

$$\begin{aligned} f_1 \!=\! f_{49}\frac{f\big (-q^{14},-q^{35}\big )}{f\big (-q^7,-q^{42}\big )} \!-\!qf_{49}\frac{f\big (-q^{21},-q^{28}\big )}{f\big (-q^{14},-q^{35}\big )}\!-\!q^2f_{49}\!+\!q^5f_{49}\frac{f\big (-q^7,-q^{42}\big )}{f\big (-q^{21},-q^{28}\big )}. \end{aligned}$$

(2.9)

Replacing q by $q^2$ in (2.9), we get

$$\begin{aligned} f_2 =&f_{98}\frac{f\big (-q^{28},-q^{70}\big )}{f\big (-q^{14},-q^{84}\big )} -q^2f_{98}\frac{f\big (-q^{42},-q^{56}\big )}{f\big (-q^{28},-q^{70}\big )}-q^4 f_{98} \nonumber \\&+ q^{10}f_{98}\frac{f\big (-q^{14},-q^{84}\big )}{f\big (-q^{42},-q^{56}\big )}. \end{aligned}$$

(2.10)

If we substitute (2.9) and (2.10) into (2.8) and compare relevant powers of q, we obtain (1.12), (1.14), (1.15) and

$$\begin{aligned} \sum _{n=0}^\infty \Delta _3(343n+229) q^n \equiv 6\frac{f_1^4f_{14}}{f_2}+\frac{f_2^6f_7}{f_1^3} \ (\mathrm{mod}\ 7). \end{aligned}$$

(2.11)

Congruence (1.13) follows from (2.1) and (2.11). This completes the proof. $\square$

3 A recurrence relation of a sequence

In this section, we establish a recurrence relation of a(n), where a(n) is related to $\Delta _3(7n+5)$ and the generating function of a(n) is $f_1^4f_2^6$. The recurrence relation plays an important role in this paper. In the process, we also find generating functions for $a(mn+k)$ for some small values of m and k by using an iterative method, which are helpful in proving Theorems 1.2, 1.3 and 1.4.

Theorem 3.1

Let a(n) be defined by

$$\begin{aligned} \sum _{n=0}^\infty a(n) q^n=f_1^4f_2^6. \end{aligned}$$

(3.1)

For $n\ge 0$, we have

$$\begin{aligned} a(128n+ 42 )=-240a(8n+2)+1024a(2n). \end{aligned}$$

(3.2)

Proof

The following relations are the consequences of dissection formulas of Ramanujan collected in Entry 25 in Berndt’s book [3, p.40]:

$$\begin{aligned} f_1^4&=\frac{f_4^{10}}{f_2^2f_8^4}-4q\frac{f_2^2f_8^4}{f_4^2} \end{aligned}$$

(3.3)

and

$$\begin{aligned} \frac{1}{f_1^4}&= \frac{f_4^{14}}{f_2^{14}f_8^4} +4q\frac{f_4^2f_8^4}{f_2^{10}}. \end{aligned}$$

(3.4)

Substituting (3.3) into (3.1), we have

$$\begin{aligned} \sum _{n=0}^\infty a(n) q^n = \left( \frac{f_4^{10}}{f_2^2f_8^4}-4q\frac{f_2^2f_8^4}{f_4^2} \right) f_2^6 =\frac{f_2^4f_4^{10}}{ f_8^4}-4q\frac{f_2^8f_8^4}{f_4^2}, \end{aligned}$$

(3.5)

which yields

$$\begin{aligned} \sum _{n=0}^\infty a(2n)q^n&= \frac{f_1^4f_2^{10}}{f_4^4} ,\end{aligned}$$

(3.6)

$$\begin{aligned} \sum _{n=0}^\infty a(2n+1)q^n&=-4\frac{f_1^8f_4^{4}}{f_2^2}. \end{aligned}$$

(3.7)

If we substitute (3.3) into (3.6) and (3.7), and then extract the even parts and the odd parts in the resulting identity, we get

$$\begin{aligned} \sum _{n=0}^\infty a(4n)q^n&= \frac{f_1^8f_2^6}{f_4^4},\end{aligned}$$

(3.8)

$$\begin{aligned} \sum _{n=0}^\infty a(4n+2)q^n&= -4\frac{f_1^{12}f_4^4}{f_2^6}, \end{aligned}$$

(3.9)

$$\begin{aligned} \sum _{n=0}^\infty a(4n+3)q^n&= 32\frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(3.10)

Similarly, substituting (3.3) into (3.8) and (3.9), and then extracting the even parts and the odd parts in the resulting identity, we have

$$\begin{aligned} \sum _{n=0}^\infty a(8n)q^n&= \frac{f_1^2f_2^{16}}{f_4^8}+ 16q\frac{f_1^{10}f_4^8}{f_2^8}, \end{aligned}$$

(3.11)

$$\begin{aligned} \sum _{n=0}^\infty a(8n+2)q^n&= -4\frac{f_2^{34}}{f_1^{12}f_4^{12}} -192q\frac{f_2^{10}f_4^4}{f_1^4} , \end{aligned}$$

(3.12)

$$\begin{aligned} \sum _{n=0}^\infty a(8n+6)q^n&= 48\frac{f_2^{22}}{f_1^8f_4^4}+ 256q\frac{f_4^{12}}{f_2^2}. \end{aligned}$$

(3.13)

Substituting (3.4) into (3.12) and (3.13), and then extracting the even parts and the odd parts in the resulting identity, we see that

$$\begin{aligned} \sum _{n=0}^\infty a(16n+2)q^n&= -4\frac{f_2^{30}}{f_1^8f_4^{12}} -960qf_2^6f_4^4, \end{aligned}$$

(3.14)

$$\begin{aligned} \sum _{n=0}^\infty a(16n+10)q^n&= -240\frac{f_2^{18}}{f_1^4f_4^4}-256q\frac{f_1^4f_4^{12}}{f_2^6} ,\end{aligned}$$

(3.15)

$$\begin{aligned} \sum _{n=0}^\infty a(16n+14)q^n&= 640\frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(3.16)

Substituting (3.3) and (3.4) into (3.14) and (3.15), and then extracting the even parts and the odd parts in the resulting identity, we deduce that

$$\begin{aligned} \sum _{n=0}^\infty a(32n+2)q^n&= -4\frac{f_1^2f_2^{16}}{f_4^8} -64q\frac{f_1^{10}f_4^8}{f_2^8},\end{aligned}$$

(3.17)

$$\begin{aligned} \sum _{n=0}^\infty a(32n+10)q^n&= -240\frac{f_1^4f_2^{10}}{f_4^4} +1024q\frac{f_2^{10}f_4^4}{f_1^4},\end{aligned}$$

(3.18)

$$\begin{aligned} \sum _{n=0}^\infty a(32n+26)q^n&= -256\frac{f_2^{22}}{f_1^8f_4^4}-960\frac{f_1^8f_4^4}{f_2^2}. \end{aligned}$$

(3.19)

If we substitute (3.3) and (3.4) into (3.18) and (3.19), and then extract the even parts and the odd parts in the resulting identity, we find that

$$\begin{aligned} \sum _{n=0}^\infty a(64n+10 )q^n&= -240\frac{f_1^8f_2^6}{f_4^4} + 4096qf_2^6f_4^4 ,\end{aligned}$$

(3.20)

$$\begin{aligned} \sum _{n=0}^\infty a(64n+42 )q^n&= 1024\frac{f_2^{18}}{f_1^4f_4^4}+960\frac{f_1^{12}f_4^4}{f_2^6} ,\end{aligned}$$

(3.21)

$$\begin{aligned} \sum _{n=0}^\infty a(64n+58 )q^n&= 5632\frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(3.22)

Substituting (3.3) and (3.4) into (3.20) and (3.21), and then extracting the even parts and the odd parts in the resulting identity, we have

$$\begin{aligned} \sum _{n=0}^\infty a(128n+10 )q^n&= -240\frac{f_1^2f_2^{16}}{f_4^8} -3840q\frac{f_1^{10}f_4^8}{f_2^8}, \end{aligned}$$

(3.23)

$$\begin{aligned} \sum _{n=0}^\infty a(128n+ 42 )q^n&= 1024\frac{f_1^4f_2^{10}}{f_4^4}+960\frac{f_2^{34}}{f_1^{12}f_4^{12}} +46080q\frac{f_2^{10}f_4^4}{f_1^4} ,\end{aligned}$$

(3.24)

$$\begin{aligned} \sum _{n=0}^\infty a(128n+ 106 )q^n&= \,4096\frac{f_1^8f_4^4}{f_2^2} -11520\frac{f_2^{22}}{f_1^8f_4^4} -61440q\frac{f_4^{12}}{f_2^2}. \end{aligned}$$

(3.25)

Theorem 3.1 follows from (3.6), (3.12) and (3.24). The proof is complete. $\square$

4 Proof of Theorem 1.2

In this section, we present a proof of Theorem 1.2. We first prove the following two lemmas.

Lemma 4.1

For $k\ge 1$,

$$\begin{aligned} \sum _{n=0}^\infty a\left( 4^kn+\frac{11\times 4^{k-1}-2}{3}\right) q^n=f(k)\frac{f_2^{12}}{f_1^2}, \end{aligned}$$

(4.1)

where $f(1)=32$, $f(2)=640$, $f(3)=5632$, $f(4)=-186368$ and for $k\ge 5$,

$$\begin{aligned} f(k)= -240f(k-2)+1024f(k-3). \end{aligned}$$

(4.2)

Proof

We are ready to prove Lemma 4.1 by induction on k. Substituting (3.3) and (3.4) into (3.25), we find that

$$\begin{aligned} \sum _{n=0}^\infty a(128n+106)q^n =&\,4096\frac{ f_4^4}{f_2^2} \left( \frac{f_4^{10}}{f_2^2f_8^4}-4q\frac{f_2^2f_8^4}{f_4^2}\right) ^2 \nonumber \\&-11520\frac{f_2^{22}}{ f_4^4} \left( \frac{f_4^{14}}{f_2^{14}f_8^4} +4q\frac{f_4^2f_8^4}{f_2^{10}} \right) ^2 -61440q\frac{f_4^{12}}{f_2^2} \nonumber \\ =&-7424\frac{f_4^{24}}{f_2^6f_8^8}-186368q\frac{f_4^{12}}{f_2^2} -118784q^2f_2^2f_8^8, \end{aligned}$$

(4.3)

which yields

$$\begin{aligned} \sum _{n=0}^\infty a(256n+234)q^n=-186368 \frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(4.4)

By (3.10), (3.16), (3.22) and (4.4), we see that Lemma 4.1 holds when $k=1,\ 2,\ 3,\ 4$. Now suppose $k\ge 5$ and that the lemma is true for $l< k$. Then

$$\begin{aligned} \sum _{n=0}^\infty a\left( 4^{k-3}n+\frac{11\times 4^{k-4}-2}{3}\right) q^n=f(k-3)\frac{f_2^{12}}{f_1^2} \end{aligned}$$

(4.5)

and

$$\begin{aligned} \sum _{n=0}^\infty a\left( 4^{k-2}n+\frac{11\times 4^{k-3}-2}{3}\right) q^n=f(k-2)\frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(4.6)

If in Theorem 3.1 we replace 2n by $4^{k-3}n+\frac{11\times 4^{k-4}-2}{3}$, we obtain

$$\begin{aligned}&\sum _{n=0}^\infty a\left( 4^{k}n+\frac{11\times 4^{k-1}-2}{3}\right) q^n \nonumber \\ =&-240\sum _{n=0}^\infty a\left( 4^{k-2}n+\frac{11\times 4^{k-3}-2}{3}\right) q^n\nonumber \\&\quad +1024 \sum _{n=0}^\infty a\left( 4^{k-3}n+\frac{11\times 4^{k-4}-2}{3}\right) q^n \nonumber \\ =&\, \Big (-240f(k-2)+1024f(k-3)\Big )\frac{f_2^{12}}{f_1^2} \nonumber \\ =&f(k)\frac{f_2^{12}}{f_1^2}. \end{aligned}$$

(4.7)

This lemma is proved by induction. $\square$

Employing (3.11), (3.17), (3.23) and Theorem 3.1, we can prove the following lemma:

Lemma 4.2

For $k\ge 1$,

$$\begin{aligned} \sum _{n=0}^\infty a\left( 2\times 4^{k }n+\frac{ 2\times 4^{k-1}-2}{3}\right) q^n= h(k) \left( \frac{f_1^2f_2^{16}}{f_4^8}+16q\frac{f_1^{10}f_4^8}{f_2^8}\right) , \end{aligned}$$

(4.8)

where $h(1)=1$, $h(2)=-4$, $h(3)=-240$, $h(4)=1984$ and for $k\ge 5$,

$$\begin{aligned} h(k)=-240h(k-2)+1024h(k-3). \end{aligned}$$

(4.9)

The proof of Lemma 4.2 is analogous to the proof of Lemma 4.1, and hence is omitted.

Now, we turn to prove Theorem 1.2.

It is easy to check that $f(2)\equiv 3 \ (\mathrm{mod}\ 7)$, $f(3)\equiv 4 \ (\mathrm{mod}\ 7)$, $f(4)\equiv 0 \ (\mathrm{mod}\ 7)$, and for $k\ge 5$, $f(k)\equiv 5f(k-2)+2f(k-3)\ (\mathrm{mod}\ 7)$. It follows that the sequence $\{f(k)\ (\mathrm{mod}\ 7)\}$ is, for $k \ge 2$,

$$\begin{aligned}&3, \ 4,\ 0, \ 5,\ 1,\ 4, \ 1, \ 1, \ 6,\ 0,\ 4,\ 5,\ 6,\ 5, \ 5,\ 2,\ 0,\ 6,\ 4,\ 2,\\&4, \ 4,\ 3,\ 0,\ 2,\ 6,\ 3,\ 6,\ 6,\ 1,\ 0, \ 3,\ 2,\ 1,\ 2, \ 2, \ 5,\ 0, \ 3,\ 5,\\&3, \ 4,\ 0, \ 5,\ 1, \cdots \end{aligned}$$

which is periodic with period 42, and for any integer $k\ge 0$,

$$\begin{aligned} f(7 k+4) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(4.10)

By Lemma 4.1 and (4.10), we see that for $k\ge 0$ and $n\ge 0$,

$$\begin{aligned} a\left( 4^{7k+4}n+\frac{11\times 4^{7k+3}-2}{3}\right) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(4.11)

In view of (1.11) and (3.1), we see that for $n\ge 0$,

$$\begin{aligned} \Delta _3(7n+5)\equiv 6a(n) \ (\mathrm{mod}\ 7). \end{aligned}$$

(4.12)

Replacing n by $4^{7k+4}n+\frac{11\times 4^{7k+3}-2}{3}$ in (4.12) and utilizing (4.11), we arrive at the congruence (1.16).

Similarly, we can prove that for $k\ge 1$,

$$\begin{aligned} h(7 k ) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(4.13)

The proof of (4.13) is analogous to the proof of (4.10), and hence is omitted. It follows from (4.8) and (4.13) that for $n\ge 0$ and $k\ge 1$,

$$\begin{aligned} a\left( 2\times 4^{7k }n+\frac{ 2\times 4^{7k-1}-2}{3}\right) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(4.14)

Congruence (1.17) follows from (4.12) and (4.14). This completes the proof. $\square$

5 Proofs of Theorems 1.3 and 1.4

In this section, we present proofs of Theorem 1.3 and 1.4. We first prove the following lemma:

Lemma 5.1

Define

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n=f_{14}\frac{f_2^5}{f_1^2}. \end{aligned}$$

(5.1)

Let $p\ge 5$ be a prime such that $\big (\frac{-7}{p}\big )=-1$. For $n\ge 0$,

$$\begin{aligned} b\left( pn+\frac{11(p^2-1)}{12}\right) =\left( \displaystyle \frac{-1}{p}\right) pb(n/p). \end{aligned}$$

(5.2)

Proof

The following identity is Euler’s pentagonal number theorem

$$\begin{aligned} f_1=\sum _{\alpha \equiv 1 \ (\mathrm{mod}\ 6)}(-1)^{\frac{\alpha -1}{6}} q^{\frac{\alpha ^2-1}{24}}, \end{aligned}$$

(5.3)

which is a direct consequence of Jacobi’s triple product identity; see Corollary 1.3.5 on page 12 of Berndt’s book [4]. The following identity is a consequence of the quintuple product identity:

$$\begin{aligned} \frac{f_2^5}{f_1^2}=\sum _{\beta \equiv 1 \ (\mathrm{mod}\ 3)} (-1)^{\beta -1}\beta q^{\frac{\beta ^2-1}{3}}, \end{aligned}$$

(5.4)

which is Corollary 1.3.22 on page 21 of Berndt’s book [4]. In view of (5.1), (5.3) and (5.4),

$$\begin{aligned} b(n)&=\sum _{\begin{array}{c} \alpha \equiv 1 \ (\mathrm{mod}\ 6),\ \beta \equiv 1 \ (\mathrm{mod}\ 3), \\ \frac{14(\alpha ^2-1)}{24}+\frac{\beta ^2-1}{3}=n \end{array}} (-1)^{\frac{\alpha -1}{6}}(-1)^{\beta -1}\beta \nonumber \\&= \sum _{\begin{array}{c} \alpha \equiv 1 \ (\mathrm{mod}\ 6),\ \beta \equiv 1 \ (\mathrm{mod}\ 3),\\ 7\alpha ^2+4\beta ^2=12n+11 \end{array}} (-1)^{\frac{\alpha -1}{6}}(-1)^{\beta -1} \beta . \end{aligned}$$

(5.5)

Hence,

$$\begin{aligned} b\left( pn+\frac{11(p^2-1)}{12}\right) =\sum _{\begin{array}{c} \alpha \equiv 1 \ (\mathrm{mod}\ 6),\ \beta \equiv 1 \ (\mathrm{mod}\ 3), \\ 7\alpha ^2+4\beta ^2=12pn+11p^2 \end{array}} (-1)^{\frac{\alpha -1}{6}}(-1)^{\beta -1}\beta . \end{aligned}$$

(5.6)

Note that $7\alpha ^2+(2\beta )^2\equiv 0\ (\mathrm{mod}\ p)$. Since $p\ge 5$ is a prime and $\big (\frac{-7}{p}\big ) = -1$, then $\alpha \equiv \beta \equiv 0 \ (\mathrm{mod}\ p)$. Set $\alpha = \big (\frac{-3}{p}\big )p\alpha ^{'}$ and $\beta = \big (\frac{-3}{p}\big )p\beta ^{'}$. The facts $\alpha \equiv 1 \ (\mathrm{mod}\ 6)$ and $\beta \equiv 1 \ (\mathrm{mod}\ 3)$ imply that $\alpha ^{'} \equiv 1 \ (\mathrm{mod}\ 6)$ and $\beta ^{'} \equiv 1 \ (\mathrm{mod}\ 3)$. Therefore,

$$\begin{aligned} (-1)^{\frac{\alpha -1}{6}}=\left( \displaystyle \frac{3}{p}\right) (-1)^{\frac{\alpha ^{'}-1}{6}} \end{aligned}$$

(5.7)

and

$$\begin{aligned} (-1)^{\beta -1}= (-1)^{\beta ^{'}-1}. \end{aligned}$$

(5.8)

Combining (5.5), (5.6), (5.7) and (5.8), we get

$$\begin{aligned} b\left( pn\!+\!\frac{11(p^2-1)}{12}\right) \!=&\!\!\sum _{\begin{array}{c} \alpha ^{'} \equiv 1 \ (\mathrm{mod}\ 6),\ \beta ^{'} \equiv 1 \ (\mathrm{mod}\ 3),\\ 7p^2{\alpha ^{'}}^2+4p^2{\beta ^{'}}^2=12pn+11p^2 \end{array}} \left( \displaystyle \frac{3}{p}\right) (-1)^{\frac{\alpha ^{'}-1}{6}} \left( \displaystyle \frac{-3}{p}\right) (-1)^{\beta ^{'}-1} p\beta ^{'}\nonumber \\ =&\left( \displaystyle \frac{-1}{p}\right) p\sum _{\begin{array}{c} \alpha ^{'} \equiv 1 \ (\mathrm{mod}\ 6),\ \beta ^{'} \equiv 1 \ (\mathrm{mod}\ 3),\\ 7 {\alpha ^{'}}^2+4 {\beta ^{'}}^2=12n/p+11 \end{array}} (-1)^{\frac{\alpha ^{'}-1}{6}} (-1)^{\beta ^{'}-1} \beta ^{'}\nonumber \\ =&\left( \displaystyle \frac{-1}{p}\right) pb(n/p). \end{aligned}$$

(5.9)

This completes the proof of this lemma. $\square$

Now, we are ready to prove Theorems 1.3 and 1.4. Replacing n by np in (5.2), we have

$$\begin{aligned} b\left( p^2n+\frac{11(p^2-1)}{12}\right) =\left( \displaystyle \frac{-1}{p}\right) pb(n). \end{aligned}$$

(5.10)

By (5.10) and mathematical induction, we find that for $j\ge 0$,

$$\begin{aligned} b\left( p^{2j}n+\frac{11(p^{2j}-1)}{12}\right) =\left( \displaystyle \left( \frac{-1}{p}\right) p\right) ^jb(n). \end{aligned}$$

(5.11)

It follows from (5.2) that if $p\not \mid n$, then

$$\begin{aligned} b\left( pn+\frac{11(p^2-1)}{12}\right) =0. \end{aligned}$$

(5.12)

Replacing n by $pn+\frac{11(p^2-1)}{12}$ in (5.11) and employing (5.12), we see that if $p\not \mid n$, then

$$\begin{aligned} b\left( p^{2j+1}n+\frac{11\big (p^{2j+2}-1\big )}{12}\right) =0. \end{aligned}$$

(5.13)

In view of (5.1) and (5.4), we find that

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n=f_{14} \sum _{\beta \equiv 1 \ (\mathrm{mod}\ 3)} (-1)^{\beta -1}\beta q^{\frac{\beta ^2-1}{3}}. \end{aligned}$$

(5.14)

It is easy to check that if $\beta \equiv 1 \ (\mathrm{mod}\ 3),$ then

$$\begin{aligned} \frac{\beta ^2-1}{3}\equiv 0, \ 1,\ 2,\ 5\ (\mathrm{mod \ 7}). \end{aligned}$$

(5.15)

Thus,

$$\begin{aligned} b(7n+3)=b(7n+4)=b(7n+6)=0. \end{aligned}$$

(5.16)

Furthermore, $\frac{\beta ^2-1}{3}\equiv 2 \ (\mathrm{mod}\ 7)$ holds if and only if $\beta \equiv 0\ (\mathrm{mod}\ 7)$. Therefore,

$$\begin{aligned} b(7n+2)\equiv 0 \ (\mathrm{mod \ 7}). \end{aligned}$$

(5.17)

Replacing n by $7n+s$ $(s\in \{2,\ 3,\ 4,\ 6\})$ in (5.11) and using (5.16) and (5.17), we deduce that

$$\begin{aligned} b\left( p^{2j}(7n+s)+\frac{11\big (p^{2j}-1\big )}{12}\right) \equiv 0 \ (\mathrm{mod \ 7}). \end{aligned}$$

(5.18)

By (2.1) and (5.1), we see that

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n \equiv \frac{f_2^{12}}{f_1^2}\ (\mathrm{mod}\ 7). \end{aligned}$$

(5.19)

In view of (4.1) and (5.19), we see that for $n\ge 0$,

$$\begin{aligned} a\left( 4^kn+\frac{11\times 4^{k-1}-2}{3}\right) \equiv f(k)b(n)\ (\mathrm{mod}\ 7), \end{aligned}$$

(5.20)

where f(k) is defined by (4.2). Replacing n by $p^{2j}(7n+s)+\frac{11(p^{2j}-1)}{12}$ $(s\in \{2,\ 3,\ 4,\ 6\})$ in (5.20) and utilizing (5.18), we see that for $n\ge 0$, $k\ge 1$ and $j\ge 0$,

$$\begin{aligned} a\left( 4^k\left( p^{2j}(7n+s)+\frac{11\big (p^{2j}-1\big )}{12} \right) +\frac{11\times 4^{k-1}-2}{3}\right) \equiv 0\ (\mathrm{mod}\ 7). \end{aligned}$$

(5.21)

Similarly, it follows from (5.13) and (5.20) that for $n\ge 0$, $k\ge 1$ and $j\ge 0$,

$$\begin{aligned} a\left( 4^k\left( p^{2j+1}n+\frac{11\big (p^{2j+2}-1\big )}{12}\right) +\frac{11\times 4^{k-1}-2}{3}\right) \equiv 0 \ (\mathrm{mod}\ 7), \qquad p\not \mid n. \end{aligned}$$

(5.22)

Congruences (1.19) and (1.20) follow from (4.12), (5.21) and (5.22). This completes the proof. $\square$.