New congruences for k-tuples t-core partitions

Abstract

Let $A_{t,k}(n)$ denote the number of partition k-tuples of n where each partition is t-core. In this paper, we prove some Ramanujan-type congruences for the partition function $A_{t,k}(n)$ when $(t,k)=(3,4)$, (3,9), (4,8), (5, 6), (8, 4), (9, 3) and (9, 6) by employing q-series identities.

1 Introduction

A partition of a positive integer n is a non-increasing sequence of positive integers, called parts, whose sum equals n. Let p(n) denote the number of partition of n. For convenience, we shall set $p(0)=1$. The generating function for p(n) is given by

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

(1)

where

$$\begin{aligned} (a;q)_{\infty }=\prod _{n=0}^{\infty }(1-aq^n),~~|q|<1. \end{aligned}$$

(2)

Most popular congruences of p(n) that discovered by Ramanujan for $n\ge 0$,

$$\begin{aligned} p(5n+4)\equiv 0~\,(\text{mod}~\,5) \end{aligned}$$

(3)

$$\begin{aligned} p(7n+5)\equiv 0~\,(\text{mod}~\,7) \end{aligned}$$

(4)

$$\begin{aligned} p(11n+6)\equiv 0~\,(\text{mod}~\,11) \end{aligned}$$

(5)

Ramanujan’s work inspired scholars to study the arithmetic properties for the other types of partition functions such as t-core partition. A partition of n is called a t-core of n if none of its hook number is a multiple of t. Let $A_t(n)$ denote the number of t-core partitions of n, the generating function of $A_t(n)$ is given by

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

(6)

The arithmetic properties of t-core partition function have been studied by several authors, see [2, 4, 5, 8, 10, 15]. A partition k-tuple ($\lambda _1,\lambda _2,\ldots ,\lambda _k$) of a positive integer 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 partition k-tuple $(\lambda _1,\ldots ,\lambda _k)$ of n with t-cores means that each $\lambda _i$ is t-core. Let $A_{t,k}(n)$ denote the number of partition k-tuples of n with t-cores. The generating function of $A_{t,k}(n)$ can be obtained as

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

(7)

Wang [12] proved some arithmetic identities and congruences for partition triples with 3-cores. Recently, Chern [9] studied the function $A_{t,k}(n)$ and proved some identities by employing the method of modular form. In sequel, in this paper we study the arithmetic properties of $A_{t,k}(n)$ for $(t,k)=(3,4), (3,9), (4,8)$, (5, 6), (8, 4), (9, 3) and (9, 6) by using q-series identities and prove some Ramanujan-type congruences.

In Sect. 3, we prove some congruence and infinite family of congruences for $A_{3,4}$ for modulo 4. For example, we prove for $\alpha \ge 0$,

$$\begin{aligned} A_{3,4}\left( 2^{2(\alpha +1)+1}n+\frac{11\cdot2^{2\alpha +1}-4}{3}\right) \equiv 0~\,(\text{mod}~\,4). \end{aligned}$$

(8)

In Sect. 4, we prove arithmetic identities and congruences for $A_{3,9}$ modulo 3 and 9. For example, we prove, for $k\ge 0$,

$$\begin{aligned} A_{3,9}\left( 9\cdot2^{2k+2}n+30\cdot2^{2k}-3\right) \equiv 0~\,(\text{mod}~\,3). \end{aligned}$$

(9)

In Sect. 5, we prove congruences for $A_{4,8}$ modulo 4. In Sect. 6, we prove congruences for $A_{8,4}$ modulo 2. In Sects. 7 and 8, we prove some congruences for $A_{9,3}$ and $A_{9,6}$. Section 2 is devoted to record some preliminary results.

2 Preliminaries

Lemma 2.1

For any prime p, we have

$$\begin{aligned} (q^p;q^p)_\infty \equiv (q;q)_\infty ^p~(\text{mod}~\,p). \end{aligned}$$

Proof

Follows easily from binomial theorem.$\square$

Lemma 2.2

[1, Lemma 1.4] For any prime p, we have

$$\begin{aligned} (q;q)^{p^2}_\infty \equiv (q^p;q^p)_\infty ^p~(\text{mod}~\,p^2). \end{aligned}$$

Proof

Follows easily from binomial theorem.$\square$

Lemma 2.3

[13, Eq. (2.11)] We have

$$\begin{aligned} \frac{1}{(q;q)^4_{\infty }}=\frac{(q^4;q^4)^{14}_{\infty }}{(q^2;q^2)^{14}_{\infty }(q^8;q^8)^4_{\infty }}+4q\frac{(q^4;q^4)^2_{\infty }(q^8;q^8)^4_{\infty }}{(q^2;q^2)^{10}_{\infty }}. \end{aligned}$$

(10)

Lemma 2.4

[3, p. 648, Eq. (2.9)] For any integer $k\ge 1$, we have

$$\begin{aligned} p_8\left( 2^{2k}n+\frac{2^{2k}-1}{3}\right) =(-8)^k p_8(n), \end{aligned}$$

(11)

where

$$\begin{aligned} \sum _{n=0}^{\infty }p_8(n)q^n=(q;q)^8_{\infty }. \end{aligned}$$

(12)

Lemma 2.5

[13, Eq. (3.75)] We have

$$\begin{aligned} \frac{(q^3;q^3)^3_{\infty }}{(q;q)_{\infty }}=\frac{(q^4;q^4)^3_{\infty }(q^6;q^6)_{\infty }^2}{(q^2;q^2)^2_{\infty }(q^{12};q^{12})_{\infty }}+q\frac{(q^{12};q^{12})^3_{\infty }}{(q^4;q^4)_{\infty }}. \end{aligned}$$

(13)

Lemma 2.6

[14, Lemma 2.1, Eq. (2.3)] We have

$$\begin{aligned} (q;q)^8_{\infty }=\frac{(q^4;q^4)_{\infty }^{20}}{(q^2;q^2)^4_{\infty }(q^8;q^8)^8_{\infty }}+16q^2\frac{(q^2;q^2)^4_{\infty }(q^8;q^8)^8_{\infty }}{(q^4;q^4)^4_{\infty }}-8q(q^4;q^4)^8_{\infty }. \end{aligned}$$

(14)

Lemma 2.7

[11] We have

$$\begin{aligned} \frac{1}{(q;q)_{\infty }} &= {} \frac{(q^{25};q^{25})^5_{\infty }}{(q^5;q^5)^6_{\infty }}\big(F^{-4}(q^5)+qF^{-3}(q^5)+2q^2F^{-2}(q^5)\nonumber \\&\quad+\,3q^3F^{-1}(q^5)+5q^4-3q^5F(q^5)\nonumber \\&+\,2q^6F^2(q^5)-q^7F^3(q^5)+q^8F^4(q^5)\big), \end{aligned}$$

(15)

where $F(q):=q^{-1/5}R(q)$ and R(q) is Roger’s Ramanujan continued fraction defined by

$$\begin{aligned} R(q):=\frac{q^{1/5}}{1}_{+}\frac{q}{1}_{+}\frac{q^2}{1}_{+}\frac{q^3}{1}_{+ \cdots }, \quad \vert q\vert <1. \end{aligned}$$

Lemma 2.8

[7, p. 345, Entry 1(iv)] We have

$$\begin{aligned} (q;q)^3_{\infty }=(q^9;q^9)^3_{\infty }\left( 4q^3W^2(q^3)-3q+W^{-1}(q^3)\right) , \end{aligned}$$

(16)

where $W(q)=q^{-1/3}G(q)$ and G(q) is the Ramanujan’s cubic continued fraction defined by

$$\begin{aligned} G(q):=\frac{q^{1/3}}{1}_{+}\frac{q+q^2}{1}_{+}\frac{q^2+q^4}{1}_{+ \cdots }, \quad \vert q\vert <1. \end{aligned}$$

Lemma 2.9

[6, Eq. (3.9)] We have

$$\begin{aligned} \frac{1}{(q;q)^3_{\infty }} &= {} \frac{(q^9;q^9)^9_{\infty }}{(q^3;q^3)^{12}_{\infty }}\left( \frac{1}{w^2(q^3)}+\frac{3q}{w(q^3)}+9q^2\right. \nonumber \\& \left. \quad +8q^3w(q^3)+12q^4w^2(q^3)+16q^6w^4(q^3)\right) , \end{aligned}$$

(17)

where $w(q)=\frac{(q;q)_{\infty }(q^6;q^6)^3_{\infty }}{(q^2;q^2)_{\infty }(q^3;q^3)^3_{\infty }}.$

3 Congruences for $A_{3,4}(n)$ modulo 4

Theorem 3.1

For $n\ge 0$, we have

$$\begin{aligned} (i)A_{3,4}(2n+1)\equiv 0~\,(\text{mod}~\,4),\\ (ii)A_{3,4}(8n+6)\equiv 0~\,(\text{mod}~\,4),\\ (iii)A_{3,4}(16n+8)\equiv 0~\,(\text{mod}~\,4). \end{aligned}$$

Proof

Setting $t=3$ and $k=4$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(n)q^n=\frac{(q^3;q^3)_{\infty }^{12}}{(q;q)^4_{\infty }}. \end{aligned}$$

(18)

Using Lemma 2.2 in (18), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(n)q^n\equiv \frac{(q^6;q^6)_{\infty }^6}{(q^2;q^2)^2_{\infty }}~\,(\text{mod}~\,\,4). \end{aligned}$$

(19)

Since there are no terms containing $q^{2n+1}$ in (19), we complete the proof (i).

Extracting terms involving $q^{2n}$ and replacing $q^2$ by q from (19), we have

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(2n)q^n\equiv \frac{(q^3;q^3)^6_{\infty }}{(q;q)^2_{\infty }}~\,(\text{mod}~\,\,4). \end{aligned}$$

(20)

Using Lemma 2.5 in (20) and squaring, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(2n)q^n & \equiv {} \left( \frac{(q^4;q^4)^6_{\infty }(q^6;q^6)^4_{\infty }}{(q^2;q^2)^4_{\infty }(q^{12};q^{12})_{\infty }}+2q\frac{(q^4;q^4)^3_{\infty }(q^6;q^6)^2_{\infty }(q^{12};q^{12})^2_{\infty }}{(q^2;q^2)^2_{\infty }(q^{12};q^{12})_{\infty }(q^2;q^2)^2_{\infty }}\right. \nonumber \\&\left. \quad +\,q^2\frac{(q^{12};q^{12})^6_{\infty }}{(q^4;q^4)^2_{\infty }}\right) ~(\text{mod}~\,\,4). \end{aligned}$$

(21)

Extracting terms involving $q^{2n+1}$ in (21), dividing by q and replacing $q^2$ by q, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(4n+2)q^n\equiv 2(q;q)^2_{\infty }(q^3;q^3)^6_{\infty }~(\text{mod}~\,\,4). \end{aligned}$$

(22)

Equation (22) can be written as

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(4n+2)q^n\equiv 2(q;q)^4_{\infty }\frac{(q^3;q^3)^6_{\infty }}{(q;q)^2_{\infty }}~(\text{mod}~\,4). \end{aligned}$$

(23)

Again using Lemma 2.5 in (23), we have

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(4n+2)q^n\equiv 2\frac{(q^4;q^4)^6_{\infty }}{(q^2;q^2)^2_{\infty }}+2q^2\frac{(q^2;q^2)^2_{\infty }(q^{12};q^{12})^6_{\infty }}{(q^4;q^4)^2_{\infty }}~(\text{mod}~\,\,4). \end{aligned}$$

(24)

Extracting terms containing $q^{2n+1}$ in (24), we arrive at (ii).

Extracting terms involving $q^{2n}$ in (21) and replacing $q^2$ by q, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(4n)q^n\equiv (q^2;q^2)^4_{\infty }+q\frac{(q^6;q^6)^6_{\infty }}{(q^2;q^2)^2_{\infty }}~(\text{mod}~\,\,4). \end{aligned}$$

(25)

Extracting terms involving $q^{2n}$ in (25) and replacing $q^2$ by q , we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(8n)q^n\equiv (q;q)^4_{\infty }\equiv (q^2;q^2)_{\infty }^2~(\text{mod}~\,\,4). \end{aligned}$$

(26)

Again, extracting terms containing $q^{2n+1}$ in (26), we arrive at (iii).$\square$

Theorem 3.2

For any positive integer n,  and $\alpha \ge 0,$ we have

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

(27)

Proof

Extracting terms involving $q^{2n+1}$ in (25), dividing by q,  and replacing $q^2$ by q, we have

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,4}(8n+4)q^n\equiv \frac{(q^3;q^3)^6_{\infty }}{(q;q)^2_{\infty }}~(\text{mod}~\,\,4). \end{aligned}$$

(28)

From (20) and (28), we can deduce that

$$\begin{aligned} A_{3,4}(2n)\equiv A_{3,4}(8n+4)~(\text{mod}~\,\,4). \end{aligned}$$

(29)

Replacing n by $4n+2$ in (29) and iterating, we arrive at the desired result.$\square$

Theorem 3.3

For any positive integer n,  and $\alpha \ge 0,$ we have

$$\begin{aligned} A_{3,4}\left( 2^{2(\alpha +1)+1}n+\frac{11\cdot2^{2\alpha +1}-4}{3}\right) \equiv 0~(\text{mod}~\,\,4). \end{aligned}$$

(30)

Proof

Replacing $n$ by $4n+3$ in (27) and employing Theorem 3.1(ii), we complete the proof.$\square$

4 Congruences for $A_{3,9}(n)$ modulo 3 and 9

Theorem 4.1

For any positive integer n and $k\ge 0$, we have

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}\left( 9\cdot 2^{2k}n+3(2^{2k}-1)\right) q^n\equiv (q;q)^8_{\infty }~(\text{mod}~\,\,3).\end{aligned}$$

(31)

Proof

Setting $t=3$ and $k=9$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(n)q^n=\frac{(q^3;q^3)_{\infty }^{27}}{(q;q)^9_{\infty }}. \end{aligned}$$

(32)

Using Lemma 2.1 in (32), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(n)q^n\equiv \frac{(q^9;q^9)_{\infty }^9}{(q^9;q^9)_{\infty }}~(\text{mod}~\,\,3). \end{aligned}$$

(33)

Extracting terms involving $q^{9n}$ in (33) and replacing $q^9$ by q, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(9n)q^n\equiv (q;q)^8_{\infty }~(\text{mod}~\,\,3). \end{aligned}$$

(34)

Using (12) in (34), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(9n)q^n\equiv p_8(n)~(\text{mod}~\,\,3). \end{aligned}$$

(35)

Employing Lemma 2.4 in (35), we arrive at the desired result. $\square$

Theorem 4.2

For $n\ge 0,$ we have

$$\begin{aligned} A_{3,9}(9n+j)\equiv 0~\,(\text{mod}~\,\,3), \end{aligned}$$

where $j=1, 2, 3, 4 ,5 ,6, 7, 8$

Proof

Extracting terms containing $q^{9n+j}$ for $1\le j\le 8,$ from both sides of (33), we arrive at the desired result.$\square$

Theorem 4.3

For any positive integer n and $k\ge 0$, we have

$$\begin{aligned} A_{3,9}(9 \cdot 2^{2k+2}n+30\cdot2^{2k}-3)\equiv 0~\,(\text{mod}~\,\,3). \end{aligned}$$

(36)

Proof

Using Lemma 2.6 in (31) and then extracting terms involving $q^{4n+3}$, dividing by $q^3$ and replacing $q^4$ by q ,we complete the proof.$\square$

Theorem 4.4

For $n\ge 0$, we have

$$\begin{aligned} A_{3,9}(3n)\equiv \tau (n+1)~\,(\text{mod}~\,\,9) \end{aligned}$$

where $\tau$ is the Ramanujan’s tau function defined by

$$\begin{aligned} q(q;q)_{\infty }^{24}=\sum _{n=1}^{\infty }\tau (n)q^n \end{aligned}$$

(37)

Proof

Setting $t=3$ and $k=9$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(n)q^n=\frac{(q^3;q^3)_{\infty }^{27}}{(q;q)^9_{\infty }}. \end{aligned}$$

(38)

Using Lemma 2.2 in (38), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(n)q^n\equiv \frac{(q^3;q^3)_{\infty }^{27}}{(q^3;q^3)^3_{\infty }}~\,(\text{mod}~\,\,9). \end{aligned}$$

(39)

Extracting terms involving $q^{3n}$ in (39) and replacing $q^3$ by q, we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(3n)q^n\equiv (q;q)^{24}_{\infty }~\,(\text{mod}~\,\,9). \end{aligned}$$

(40)

From (40) and (37), we deduce that

$$\begin{aligned} \sum _{n=0}^{\infty }A_{3,9}(3n)q^{n+1}\equiv \sum _{n=0}^{\infty }\tau (n+1)~\,(\text{mod}~\,\,9). \end{aligned}$$

(41)

From (41), we easily arrive at the desired result.$\square$

5 Congruences for $A_{4,8}(n)$ modulo 4

Theorem 5.1

For $n\ge 0$, we have

$$\begin{aligned} A_{4,8}(4n+j)\equiv 0\;(\text{mod}~\,4); \quad j=1, 2, 3. \end{aligned}$$

Proof

Setting $t=4$ and $k=8$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{4,8}(n)q^n=\frac{(q^4;q^4)_{\infty }^{32}}{(q;q)^8_{\infty }}. \end{aligned}$$

(42)

Applying Lemma 2.2 in (42), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{4,8}(n)q^n\equiv (q^4;q^4)^{30}_{\infty }\;(\text{mod}~\,4). \end{aligned}$$

(43)

Extracting terms involving $q^{4n+j}$ for $j=1, 2,$ and 3 in (43), we complete the proof. $\square$

6 Congruences for $A_{5,6}(n)$ modulo 3 and 5

Theorem 6.1

For $n\ge 0$, we have

$$\begin{aligned} (i)\;A_{5,6}(3n+1)\equiv 0\;(\text{mod}~\,3)\\ (ii)\;A_{5,6}(3n+2)\equiv 0\;(\text{mod}~\,3)\\ (iii)\;A_{5,6}(5n+4)\equiv 0\;(\text{mod}~\,\,5) \end{aligned}$$

Proof

Setting $t=5$ and $k=6$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{5,6}(n)q^n=\frac{(q^5;q^5)_{\infty }^{30}}{(q;q)^6_{\infty }}. \end{aligned}$$

(44)

Applying Lemma 2.1 in (44) we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{5,6}(n)q^n\equiv \frac{(q^{15};q^{15})^{10}_{\infty }}{(q^3;q^3)^3_{\infty }}(q;q)^3_{\infty }~(\text{mod}~\,\,3). \end{aligned}$$

(45)

Using Lemma 2.8 in (45), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{5,6}(n)q^n\equiv \frac{(q^{15};q^{15})^{10}_{\infty }}{(q^3;q^3)^3_{\infty }}\left( 4q^3w^2(q^3)-3q+w^{-1}(q^3)\right) ~(\text{mod}~\,3). \end{aligned}$$

(46)

Extracting terms containing $q^{3n+1}$ and $q^{3n+2}$ in (46), we complete the proof of (i), (ii) and (iii), respectively.

Applying Lemma 2.1 in (44), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{5,6}(n)q^n\equiv \frac{(q^5;q^5)^{30}_{\infty }}{(q^5;q^5)_{\infty }(q;q)_{\infty }}~(\text{mod}~\,5). \end{aligned}$$

(47)

Using Lemma 2.7 in (47) and extracting terms involving $q^{5n+4}$, dividing by $q^4$ and replacing $q^5$ by q, we can easily obtain (iii). $\square$

7 Congruences for $A_{8,4}(n)$ modulo 2

Theorem 7.1

For $n\ge 0$, we have

$$\begin{aligned} & (i)\,A_{8,4}(4n+1)\equiv 0~(\text{mod}~\,2)\\ &(ii)\,A_{8,4}(4n+2)\equiv 0~(\text{mod}~\,2)\\&(iii)\,A_{8,4}(4n+3)\equiv0~(\text{mod}~\,2)\end{aligned}$$

Proof

Setting $t=8$ and $k=4$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{8,4}(n)q^n=\frac{(q^8;q^8)_{\infty }^{32}}{(q;q)^4_{\infty }}. \end{aligned}$$

(48)

Applying Lemma 2.3 in (48) and extracting the terms involving $q^{4n+1}$, $q^{4n+2}$ and $q^{4n+3}$, we complete the proof of (i), (ii), and (iii), respectively.$\square$

8 Congruences for $A_{9,3}(n)$ modulo 3

Theorem 8.1

For $n\ge 0$, we have

$$\begin{aligned} (i)\;A_{9,3}(3n+1)\equiv 0\;(\text{mod}~\,3)\\ (ii)\;A_{9,3}(3n+2)\equiv 0\;(\text{mod}~\,3) \end{aligned}$$

Proof

Setting $t=9$ and $k=3$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{9,3}(n)q^n=\frac{(q^9;q^9)_{\infty }^{27}}{(q;q)^3_{\infty }}. \end{aligned}$$

(49)

Using Lemma 2.9 in (49) and extracting the terms involving $q^{3n+1}$ and $q^{3n+2}$, we complete the proof of (i) and (ii), respectively.$\square$

9 Congruences for $A_{9,6}(n)$ modulo 3

Theorem 9.1

For $n\ge 0$, we have

$$\begin{aligned} (i)\;A_{9,6}(3n+1)\equiv 0\;(\text{mod}~\,3)\\ (ii)\;A_{9,6}(3n+2)\equiv 0\;(\text{mod}~\,3) \end{aligned}$$

Proof

Setting $t=9$ and $k=6$ in (7), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{9,6}(n)q^n=\frac{(q^9;q^9)_{\infty }^{54}}{(q;q)^6_{\infty }} =\frac{(q^9;q^9)_{\infty }^{54}(q;q)^3_{\infty }}{(q;q)^9_{\infty }}. \end{aligned}$$

(50)

Using Lemma 2.8 in (50), we obtain

$$\begin{aligned} \sum _{n=0}^{\infty }A_{9,6}(n)q^n\equiv \frac{(q^9;q^9)^{54}_{\infty }}{(q^3;q^3)^3_{\infty }}\left( 4q^3w^2(q^3)-3q+w^{-1}(q^3)\right) ~(\text{mod}~\,3). \end{aligned}$$

(51)

Extracting terms involving $q^{3n+1}$ and $q^{3n+2}$ in (51), we complete the proof of (i) and (ii), respectively.$\square$