Congruences modulo 4 for broken k-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, a number of parity results satisfied by $\Delta _k(n)$ for small values of k have been proved by Radu and Sellers and others. However, congruences modulo 4 for $\Delta _k(n)$ are unknown. In this paper, we will prove five congruences modulo 4 for $\Delta _5(n)$, four infinite families of congruences modulo 4 for $\Delta _7(n)$ and one congruence modulo 4 for $\Delta _{11}(n)$ by employing theta function identities. Furthermore, we will prove a new parity result for $\Delta _2(n)$.

1 Introduction

The aim of this paper is to establish congruences modulo 4 for broken 5-diamond, broken 7-diamond and broken 11-diamond partitions.

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

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

and often write

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

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

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

(1.1)

Jacobi’s triple product identity states that

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

(1.2)

Three special cases of (1.1) are defined by

$$\begin{aligned} \varphi (q)&=f(q,q) =\sum _{n=-\infty }^\infty q^{n^2},\end{aligned}$$

(1.3)

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

(1.4)

and

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

(1.5)

For any positive integer n, we use $f_n$ to denote $f(-q^n)$, that is,

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

By (1.2)–(1.5),

$$\begin{aligned} f(-q)=f_1, \qquad \varphi (q) =\frac{f_2^5}{f_1^2f_4^2},\qquad \psi (q)=\frac{f_2^2}{f_1}. \end{aligned}$$

(1.6)

MacMahon’s partition analysis guided Andrews and Paule [2] to introduce broken k-diamond partitions. For a fixed positive integer k, let $\Delta _k(n)$ denote the number of broken k-diamond partitions of n. Andrews and Paule [2] discovered the following generating function for $\Delta _k(n)$:

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

(1.7)

Various authors have obtained parity results for broken k-diamond partitions. See, Ahmed and Baruah [1], Chan [5], Cui and Gu [6], Hirschhorn and Sellers [8], Lin [10], Radu and Sellers [11, 12], Wang [14], Xia [15] and Yao [18].

However, Ramanujan-type congruences modulo 4 for $\Delta _k(n)$ are unknown. With this motivation, we will prove five congruences modulo 4 for $\Delta _5(n)$, four infinite families of congruences modulo 4 for $\Delta _7(n)$ and one congruence modulo 4 for $\Delta _{11}(n)$. The main results of this paper can be stated as follows.

Theorem 1.1

For $n\ge 0$,

$$\begin{aligned} \Delta _5(44n+j )\equiv 0 \ (\mathrm{mod}\ 4), \end{aligned}$$

(1.8)

where $j\in \{2,\ 14,\ 30,\ 34,\ 38\}$.

Theorem 1.2

For $n,\ \alpha \ge 0$,

$$\begin{aligned} \Delta _7\left( 16\times 5^{2\alpha +1} n+ \frac{4\times 5^{2\alpha }+2}{3} \right) \equiv&0\ (\mathrm{mod}\ 4), \end{aligned}$$

(1.9)

$$\begin{aligned} \Delta _7\left( 16\times 5^{2\alpha +1} n+ \frac{196\times 5^{2\alpha }+2}{3} \right) \equiv&0\ (\mathrm{mod}\ 4), \end{aligned}$$

(1.10)

$$\begin{aligned} \Delta _7\left( 16\times 5^{2\alpha +2} n+ \frac{460\times 5^{2\alpha }+2}{3} \right) \equiv&0\ (\mathrm{mod}\ 4), \end{aligned}$$

(1.11)

$$\begin{aligned} \Delta _7\left( 16\times 5^{2\alpha +2} n+ \frac{940\times 5^{2\alpha }+2}{3} \right) \equiv&0\ (\mathrm{mod}\ 4). \end{aligned}$$

(1.12)

Theorem 1.3

For $n,\alpha \ge 0$,

$$\begin{aligned} \Delta _{11}(4\times 23^\alpha n+1)- \Delta _{11}(2\times 23^\alpha n+1)\equiv \Delta _{11}(4n+1)-\Delta _{11}(2n+1)\ (\mathrm{mod} \ 4). \end{aligned}$$

(1.13)

By Theorem 1.3 and the facts that $\Delta _{11}(5)\equiv \Delta _{11}(3) \ (\mathrm{mod\ 4})$, $\Delta _{11}(9)-\Delta _{11}(5) \equiv -1\ (\mathrm{mod\ 4})$, $\Delta _{11}(13)-\Delta _{11}(7) \equiv 2\ (\mathrm{mod\ 4})$ and $\Delta _{11}(65)-\Delta _{11}(33) \equiv 1\ (\mathrm{mod\ 4})$, we obtain the following corollary:

Corollary 1.4

For $\alpha \ge 0$,

$$\begin{aligned} \Delta _{11}(4\times 23^\alpha +1)-\Delta _{11}(2\times 23^\alpha +1)&\equiv 0\ (\mathrm{mod}\ 4),\\ \Delta _{11}(8\times 23^\alpha +1)-\Delta _{11}(4\times 23^\alpha +1)&\equiv -1 \ (\mathrm{mod}\ 4),\\ \Delta _{11}(12 \times 23^\alpha +1)-\Delta _{11}(6 \times 23^\alpha +1)&\equiv 2\ (\mathrm{mod}\ 4),\\ \Delta _{11}(64\times 23^\alpha +1)-\Delta _{11}(32\times 23^\alpha +1)&\equiv 1\ (\mathrm{mod}\ 4). \end{aligned}$$

Moreover, we will prove the following congruence modulo 2 for $\Delta _2(n)$ in Sect. 5.

Theorem 1.5

For $n\ge 0$,

$$\begin{aligned} \Delta _2(2n+1)\equiv \Delta _2(10n+4)\ (\mathrm{mod}\ 2). \end{aligned}$$

(1.14)

2 Proof of Theorem 1.1

In order to prove Theorem 1.1, we first prove three lemmas.

Lemma 2.1

We have

$$\begin{aligned} \frac{f_{2}^2f_{22}^2}{f_{1}f_{11}}&= \frac{f_{24}^2f_{132}^5}{f_{12}f_{66}^2f_{264}^2} \nonumber \\&\quad \quad +q\frac{f_4^2f_6f_{24}f_{88}f_{132}^2}{f_2f_8f_{12}f_{44}f_{264}} \nonumber \\&\quad \quad +q^6\frac{f_8f_{12}^2f_{44}^2f_{66}f_{264}}{f_4f_{22}f_{24}f_{88}f_{132}} +q^{15}\frac{f_{12}^5f_{264}^2}{f_{6}^2f_{24}^2f_{132}}. \end{aligned}$$

(2.1)

Proof

From (36.8) in Berndt’s book [3, p. 69], we see that if $\mu$ is even, then

$$\begin{aligned} \psi \big (q^{\mu +\nu }\big )\psi \big (q^{\mu -\nu }\big )=&\varphi \Big (q^{\mu (\mu ^2-\nu ^2)}\Big ) \psi \big (q^{2\mu }\big ) +\sum _{m=1}^{\mu /2-1}q^{\mu m^2-\nu m}\nonumber \\&\times f\Big (q^{(\mu +2m)(\mu ^2-\nu ^2)}, q^{(\mu -2m)(\mu ^2-\nu ^2)}\Big )f\Big (q^{2\nu m},q^{2\mu -2\nu m}\Big )\nonumber \\&+q^{\mu ^3/4-\mu \nu /2}\psi \Big (q^{2\mu (\mu ^2-\nu ^2)}\Big ) f\Big (q^{\mu \nu },q^{2\mu -\mu \nu }\Big ). \end{aligned}$$

(2.2)

Setting $\mu =6$ and $\nu =5$ in (2.2), we get

$$\begin{aligned} \psi (q)\psi \Big (q^{11}\Big )=&\varphi \Big (q^{66}\Big )\psi \Big (q^{12}\Big )+qf\Big (q^{88},q^{44}\Big )f\Big (q^{10},q^2\Big )\nonumber \\&+q^{14}f\Big (q^{110},q^{22}\Big )f\Big (q^{20},q^{-8}\Big )+q^{39}\psi \Big (q^{132}\Big )f\Big (q^{30},q^{-18}\Big ). \end{aligned}$$

(2.3)

By (1.2),

$$\begin{aligned} f\Big (q^{88},q^{44}\Big )&=\frac{f_{88}f_{132}^2}{f_{44}f_{264}}, \end{aligned}$$

(2.4)

$$\begin{aligned} f\Big (q^{10},q^{2}\Big )&=\frac{f_4^2f_{6}f_{24}}{f_2f_8f_{12}}, \end{aligned}$$

(2.5)

$$\begin{aligned} f\Big (q^{20},q^{-8}\Big )&=q^{-8}\frac{f_{8}f_{12}^2}{f_4f_{24}}, \end{aligned}$$

(2.6)

and

$$\begin{aligned} f\Big (q^{30},q^{-18}\Big )&=q^{-24}\frac{f_{12}^5}{f_6^2f_{24}^2}. \end{aligned}$$

(2.7)

Substituting (1.6), (2.4)–(2.7) into (2.3), we arrive at (2.1). This completes the proof. $\square$

Lemma 2.2

Define

$$\begin{aligned} \sum _{n=0}^\infty a(n)q^n:= \frac{f_1^2f_3^2}{f_2f_6}. \end{aligned}$$

(2.8)

Then for $n\ge 0$,

$$\begin{aligned} a(11n+i)\equiv 0 \ (\mathrm{mod}\ 4), \end{aligned}$$

(2.9)

where $i\in \{2,\ 6,\ 7,\ 8,\ 10\}$.

Proof

It is well known that

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

Thus, combining (2.8) and the above identity yields

$$\begin{aligned} \sum _{n=0}^\infty a(n)q^n&=\left( 1+2\sum _{n=1}^\infty (-1)^n q^{n^2}\right) \left( 1+2\sum _{n=1}^\infty (-1)^n q^{3n^2}\right) \nonumber \\&\equiv 1+2\sum _{n=1}^\infty (-1)^n q^{n^2} +2\sum _{n=1}^\infty (-1)^n q^{3n^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(2.10)

It is easy to check that for any integer n,

$$\begin{aligned} n^2\equiv 0,\ 1,\ 3,\ 4,\ 5,\ 9\ (\mathrm{mod}\ 11) \end{aligned}$$

(2.11)

and

$$\begin{aligned} 3n^2\equiv 0,\ 1,\ 3,\ 4,\ 5,\ 9\ (\mathrm{mod}\ 11). \end{aligned}$$

(2.12)

Congruence (2.9) follows from (2.10), (2.11) and (2.12). The proof is complete. $\square$

Lemma 2.3

Define

$$\begin{aligned} \sum _{n=0}^\infty b(n)q^n:= \frac{f_3^3}{f_1}. \end{aligned}$$

(2.13)

Then for $n\ge 0$,

$$\begin{aligned} b(11n+k)\equiv 0\ (\mathrm{mod}\ 2), \end{aligned}$$

(2.14)

where $k\in \{2,\ 3,\ 4,\ 6,\ 9\}$.

Proof

Hirschhorn and Sellers [9] proved that if

$$\begin{aligned} 3n+1=\prod _{p_i\equiv 1\ (\mathrm{mod}\ 3)}p_i^{\alpha _i} \prod _{q_i\equiv 2\ (\mathrm{mod}\ 3)}q_i^{\beta _i} \end{aligned}$$

(2.15)

is the prime factorization of $3n + 1$, then

(2.16)

where b(n) is defined by (2.13). By (2.15) and (2.16), we find that for $n\ge 0$, b(n) is odd if and only if $3n + 1$ is a square of an integer. From (2.11), we know that $33n + 7$, $33n + 10$, $33n + 13$, $33n + 19$ and $33n + 28$ are not squares, which implies that for $n\ge 0$,

$$\begin{aligned} b(11n+k)\equiv 0\ (\mathrm{mod}\ 2), \end{aligned}$$

where $k\in \{2,\ 3,\ 4,\ 6,\ 9\}$. This completes the proof. $\square$

Now, we turn to prove Theorem 1.1.

Setting $k=5$ in (1.7), we have

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(n)q^n= \frac{f_2f_{11}}{f_1^3f_{22}}. \end{aligned}$$

(2.17)

By the binomial theorem, for positive integers u and v,

$$\begin{aligned} \big (q^u;q^v\big )_\infty ^2\equiv \big (q^{2u};q^{2v}\big )_\infty \ (\mathrm{mod}\ 2) \end{aligned}$$

(2.18)

and

$$\begin{aligned} \big (q^u;q^v\big )_\infty ^4\equiv \big (q^{2u};q^{2v}\big )_\infty ^2 \ (\mathrm{mod}\ 4). \end{aligned}$$

(2.19)

In particular,

$$\begin{aligned} f_1^2\equiv f_2\ (\mathrm{mod}\ 2) \end{aligned}$$

(2.20)

and

$$\begin{aligned} f_1^4\equiv f_2^2\ (\mathrm{mod}\ 4). \end{aligned}$$

(2.21)

Thanks to (2.17) and (2.21),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(n)q^n \equiv \frac{f_1f_{11}}{f_2f_{22}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(2.22)

Replacing q by $-q$ in (2.1) and using the fact that

$$\begin{aligned} (-q;-q)_\infty =\frac{f_2^3}{f_1f_4}, \end{aligned}$$

(2.23)

then multiplying $\frac{1}{f_4f_{44}}$ on both sides, we obtain

$$\begin{aligned} \frac{f_{1}f_{11}}{f_{2}f_{22}}=&\frac{f_{24}^2f_{132}^5}{f_4f_{12}f_{44}f_{66}^2 f_{264}^2} - q\frac{f_4f_6f_{24}f_{88}f_{132}^2}{f_2f_8f_{12}f_{44}^2f_{264}} \nonumber \\&\quad +q^6\frac{f_8f_{12}^2f_{44}f_{66}f_{264}}{f_4^2f_{22}f_{24}f_{88}f_{132}} -q^{15}\frac{f_{12}^5f_{264}^2}{f_4f_{6}^2f_{24}^2f_{44}f_{132}}. \end{aligned}$$

(2.24)

Combining (2.22) and (2.24) yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(2n)q^n\equiv \frac{f_{12}^2f_{66}^5 }{f_2f_{6}f_{22}f_{33}^2 f_{132}^2} +q^3\frac{f_4f_6^2f_{22}f_{33}f_{132} }{f_2^2f_{11}f_{12}f_{44}f_{66}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(2.25)

The following relation is a consequence of dissection formulas of Ramanujan collected in Entry 25 in Berndt’s book [3, p. 40]:

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

(2.26)

Xia and Yao [16] proved that

$$\begin{aligned} \frac{f_3}{f_1}=\frac{f_4f_6f_{16}f_{24}^2}{f_2^2f_8f_{12}f_{48}} +q\frac{f_6f_8^2f_{48}}{f_2^2f_{16}f_{24}}. \end{aligned}$$

(2.27)

By substituting (2.26) and (2.27) into (2.25),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(2n)q^n\equiv&\frac{f_{12}^2f_{66}^3f_{264}^5}{ f_2f_6f_{22}f_{132}^2f_{528}^5}+ q^3\frac{f_4f_6^2f_{176}f_{264}^2}{ f_2^2f_{12}f_{22}f_{88}f_{528}} \\&+q^{14}\frac{f_4f_6^2f_{88}^2f_{132} f_{528}}{f_2^2 f_{12}f_{22}f_{44}f_{176}f_{264}} +2q^{33}\frac{f_{12}^2f_{528}^2}{ f_2f_6f_{22}f_{264}} \ (\mathrm{mod}\ 4). \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(4n+2)q^n\equiv q\frac{f_2f_3^2f_{88}f_{132}^2 }{f_1^2f_6f_{11}f_{44}f_{264}}+2q^{16} \frac{f_6^2f_{264}^2}{f_1f_3f_{11}f_{132}} \ (\mathrm{mod}\ 4). \end{aligned}$$

(2.28)

By (2.20), (2.21) and (2.28),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(4n+2)q^n\equiv q\frac{f_1^2 f_3^2f_{88}f_{132}^2 }{f_2f_6f_{11}f_{44}f_{264}}+2q^{16} \frac{f_3^3f_{264}^2}{f_1 f_{11}f_{132}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(2.29)

Therefore, we can rewrite (2.29) as

$$\begin{aligned} \sum _{n=0}^\infty \Delta _5(4n+2)q^n\equiv q\frac{ f_{88}f_{132}^2 }{ f_{11}f_{44}f_{264}}\sum _{n=0}^\infty a(n)q^n +2q^{16} \frac{ f_{264}^2}{ f_{11}f_{132}}\sum _{n=0}^\infty b(n)q^n\ (\mathrm{mod}\ 4), \end{aligned}$$

(2.30)

where a(n) and b(n) are defined by (2.8) and (2.13), respectively. Theorem 1.1 follows from (2.9), (2.14) and (2.30). This completes the proof. $\square$

3 Proof of Theorem 1.2

In this section, we present a proof of Theorem 1.2.

Taking $k=7$ in (1.7), we get

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(n)q^n= \frac{f_2f_{15}}{f_1^3f_{30}}. \end{aligned}$$

(3.1)

In view of (2.21) and (3.1),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(n)q^n\equiv \frac{f_1f_{15}}{f_2f_{30}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.2)

Replacing q by $-q$ in (3.2) and utilizing the relation (2.23), we get

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(n)(-1)^nq^n\equiv \frac{\psi (q)\psi \big (q^{15}\big )}{f_4f_{60}}\ (\mathrm{mod}\ 4), \end{aligned}$$

(3.3)

where $\psi (q)$ is defined by (1.6). From Entry 9 in Berndt’s book [3, p. 377],

$$\begin{aligned} \psi (q)\psi \big (q^{15}\big )+ \psi (-q)\psi \big (-q^{15}\big ) =2\psi \big (q^6\big )\psi \big (q^{10}\big ) \end{aligned}$$

(3.4)

Based on (3.3) and (3.4),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(n)\big (1+(-1)^n\big ) q^n&\equiv \frac{\psi (q)\psi \big (q^{15}\big )+ \psi (-q)\psi \big (-q^{15}\big )}{f_4f_{60}}\\&\equiv 2\frac{\psi \big (q^6\big )\psi \big (q^{10}\big ) }{f_4f_{60}}\ (\mathrm{mod}\ 4), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(2n) q^n\equiv \frac{\psi \big (q^3\big )\psi \big (q^{5}\big )}{f_2f_{30}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.5)

From Entry 9 in Berndt’s book [3, p. 377],

$$\begin{aligned} \psi \big (q^3\big )\psi \big (q^{5}\big )- \psi \big (-q^3\big )\psi \big (-q^{5}\big ) =2q^3\psi \big (q^2\big )\psi \big (q^{30}\big ). \end{aligned}$$

(3.6)

Combining (3.5) and (3.6), we deduce that

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(2n)\big (1-(-1)^n\big ) q^n&\equiv \frac{\psi \big (q^3\big )\psi \big (q^{5}\big )-\psi \big (-q^3\big )\psi \big (-q^{5}\big ) }{f_2f_{30}}\\&\equiv 2q^3\frac{ \psi \big (q^2\big )\psi \big (q^{30}\big )}{f_2f_{30}}\ (\mathrm{mod}\ 4), \end{aligned}$$

which implies

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(4n+2) q^n\equiv q\frac{\psi (q)\psi \big (q^{15}\big )}{f_1f_{15}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.7)

In view of (1.6), (2.21) and (3.7),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(4n+2) q^n\equiv q\frac{f_2^2f_{30}^2 }{f_1^2f_{15}^2}\equiv qf_1^2f_{15}^2\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.8)

Ramanujan [13] stated the following identity without proof:

$$\begin{aligned} f_1=f_{25}\left( R\big (q^5\big )-q-\frac{q^2}{R\big (q^5\big )}\right) , \end{aligned}$$

(3.9)

where

$$\begin{aligned} R(q)=\frac{\big (q^2,q^3;q^5\big )_\infty }{\big (q,q^4;q^5\big )_\infty }. \end{aligned}$$

(3.10)

Hirschhorn [7] gave a simple proof of (3.9) by using Jacobi’s triple product identity. Substituting (3.9) into (3.8), we have

$$\begin{aligned}&\sum _{n=0}^\infty \Delta _7(4n+2) q^n \\&\equiv f_{15}^2f_{25}^2\left( qR^2\big (q^5\big )-2q^2R\big (q^5\big ) -q^3+\frac{2q^4}{R\big (q^5\big )}+\frac{q^5}{R^2\big (q^5\big )}\right) \ (\mathrm{mod}\ 4), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(20n+6) q^n\equiv&f_3^2f_5^2R^2(q)\ (\mathrm{mod}\ 4), \end{aligned}$$

(3.11)

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(20n+14) q^n\equiv&-f_3^2f_5^2\ (\mathrm{mod}\ 4) \end{aligned}$$

(3.12)

and

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(20n+2) q^n\equiv&q \frac{f_3^2f_5^2}{R^2(q)}\ (\mathrm{mod}\ 4), \end{aligned}$$

(3.13)

where R(q) is defined by (3.10). We can rewrite (3.13) as

$$\begin{aligned}&\sum _{n=0}^\infty \Delta _7(20n+2) q^n \nonumber \\&\quad \equiv qf_3^2f_5^2 \frac{\big (q,q^4;q^5\big )_\infty ^2}{\big (q^2,q^3;q^5\big )_\infty ^2}\nonumber \\&\quad \equiv q\frac{f_3^2f_5^4\big (q,q^4;q^5\big )_\infty ^4}{f_1^2} \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.14)

By (2.19), (2.21) and (3.14),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(20n+2) q^n \equiv q\frac{f_3^2f_{10}^2\big (q^2,q^8;q^{10}\big )_\infty ^2}{f_1^2} \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.15)

Xia and Yao [17] proved that

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

(3.16)

Thanks to (2.20), (2.21) and (3.16),

$$\begin{aligned} \frac{f_3^2}{f_1^2}\equiv \frac{ f_6f_8f_{12}^2}{f_2^5f_{24}} +2q f_4 f_{24} \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.17)

Substituting (3.17) into (3.15), we get

$$\begin{aligned}&\sum _{n=0}^\infty \Delta _7(20n+2) q^n \\&\quad \equiv q\frac{f_6 f_8 f_{10}^2f_{12}^2 \big (q^2,q^8;q^{10}\big )_\infty ^2}{ f_2^5f_{24}}\\&\quad \quad + 2 q^2f_4 f_{10}^2f_{24}\big (q^2,q^8;q^{10}\big )_\infty ^2 \ (\mathrm{mod}\ 4), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(40n+2) q^n\equiv 2qf_2f_5^2f_{12} \big (q,q^4;q^{5}\big )_\infty ^2 \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.18)

By (2.18), (2.20) and (3.18), we obtain

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(40n+2) q^n\equiv 2qf_2f_{10}f_{12}\big (q^2,q^8;q^{10}\big )_\infty \ (\mathrm{mod}\ 4), \end{aligned}$$

which implies that for $n\ge 0$,

$$\begin{aligned} \Delta _7(80n+2)\equiv 0\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.19)

Similarly, by (2.19), (2.21), (3.10) and (3.11),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(20n+6) q^n\equiv&f_3^2f_5^2\frac{\big (q^2,q^3;q^5\big )_\infty ^2 }{\big (q,q^4;q^5\big )_\infty ^2} \equiv \frac{f_3^2f_5^4\big (q^2,q^3;q^5\big )_\infty ^4 }{f_1^2}\nonumber \\ \equiv&\frac{f_3^2}{f_1^2}f_{10}^2\big (q^4,q^6;q^{10} \big )_\infty ^2\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.20)

By substituting (3.17) into (3.20) and extracting the terms containing odd powers of q,

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(40n+26) q^n\equiv 2f_2f_5^2f_{12}\big (q^2,q^3 ;q^5\big )_\infty ^2 \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.21)

In view of (2.18), (2.20) and (3.21),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(40n+26) q^n \equiv 2f_2f_{10}f_{12}\big (q^4,q^6 ;q^{10}\big )_\infty \ (\mathrm{mod}\ 4), \end{aligned}$$

which implies that for $n\ge 0$,

$$\begin{aligned} \Delta _7(80n+66)\equiv 0\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.22)

Substituting (3.9) into (3.12), we obtain

$$\begin{aligned}&\sum _{n=0}^\infty \Delta _7(20n+14) q^n \\&\quad \equiv f_5^2f_{75}^2\left( -R^2\big (q^{15}\big )+2q^3R\big (q^{15}\big )+q^6 -\frac{2q^9}{R\big (q^{15}\big )}-\frac{q^{12}}{R^2\big (q^{15}\big )}\right) \ (\mathrm{mod}\ 4), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(100n+14) q^n\equiv&-f_1^2f_{15}^2 R^2\big (q^3\big )\ (\mathrm{mod}\ 4), \end{aligned}$$

(3.23)

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(100n+34) q^n\equiv&qf_1^2f_{15}^2\ (\mathrm{mod}\ 4) \end{aligned}$$

(3.24)

and

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(100n+54) q^n\equiv -q^2 \frac{f_1^2f_{15}^2}{ R^2\big (q^3\big )}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.25)

By (2.19), (2.21), (3.10) and (3.23),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(100n+14) q^n&\equiv -f_1^2f_{15}^2 \frac{\big (q^6,q^9;q^{15}\big )_\infty ^2}{ \big (q^3,q^{12};q^{15}\big )_\infty ^2}\nonumber \\&\equiv -\frac{f_1^2f_{15}^4\big (q^6,q^9;q^{15}\big )_\infty ^4}{f_3^2} \nonumber \\&\equiv -\frac{f_1^2f_{30}^2\big (q^{12},q^{18};q^{30}\big )_\infty ^2 }{f_3^2}\nonumber \\&\equiv - \frac{f_2^2f_3^2f_{30}^2\big (q^{12},q^{18};q^{30}\big )_\infty ^2}{ f_1^2f_6^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.26)

By substituting (3.17) into (3.26) and extracting the terms containing odd powers of q,

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(200n+114) q^n\equiv 2\frac{f_1^2f_2f_{12}f_{15}^2\big (q^6,q^9 ;q^{15}\big )_\infty ^2 }{f_3^2} \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.27)

Thanks to (2.18), (2.20) and (3.27),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(200n+114) q^n\equiv 2f_4f_6f_{30}\big (q^{12},q^{18};q^{30}\big )_\infty \ (\mathrm{mod}\ 4), \end{aligned}$$

which implies that for $n\ge 0$,

$$\begin{aligned} \Delta _7(400n+314)\equiv 0\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.28)

It follows from (3.8) and (3.24) that for $n\ge 0$,

$$\begin{aligned} \Delta _7(100n+34)\equiv \Delta _7(4n+2) \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.29)

By (3.29) and mathematical induction, we deduce that for $n\ge 0$ and $\alpha \ge 0$,

$$\begin{aligned} \Delta _7\left( 4\left( 5^{2\alpha }n+\frac{5^{2\alpha }-1}{3} \right) +2\right) \equiv \Delta _7(4n+2) \ (\mathrm{mod}\ 4). \end{aligned}$$

(3.30)

By (2.19), (2.21), (3.10) and (3.25),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(100n+54) q^n&\equiv -q^2 \frac{f_1^2f_{15}^2\big (q^3,q^{12};q^{15}\big )_\infty ^2 }{ \big (q^6,q^9;q^{15}\big )_\infty ^2}\nonumber \\&\equiv -q^2 \frac{f_1^2f_{15}^4\big (q^3,q^{12};q^{15}\big )_\infty ^4 }{ f_3^2} \nonumber \\&\equiv -q^2\frac{f_1^2f_{30}^2\big (q^6,q^{24};q^{30}\big )_\infty ^2}{f_3^2} \nonumber \\&\equiv -q^2\frac{f_2^2f_3^2 f_{30}^2\big (q^6,q^{24};q^{30}\big )_\infty ^2}{f_1^2f_6^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.31)

By substituting (3.17) into (3.31) and extracting the terms containing odd powers of q,

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(200n+154) q^n \equiv 2q\frac{f_1^2f_2f_{12}f_{15}^2\big (q^3,q^{12};q^{15}\big )_\infty ^2 }{f_3^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.32)

Based on (2.18), (2.20) and (3.32),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _7(200n+154) q^n \equiv 2qf_4f_6f_{30}\big (q^6,q^{24};q^{30}\big )_\infty \ (\mathrm{mod}\ 4), \end{aligned}$$

which implies that for $n\ge 0$,

$$\begin{aligned} \Delta _7(400n+154)\equiv 0\ (\mathrm{mod}\ 4). \end{aligned}$$

(3.33)

Replacing n by 20n in (3.30) and using (3.19), we get (1.9). Replacing n by $20n+16$ in (3.30) and using (3.22), we obtain (1.10). Replacing n by $100n+38$ in (3.30) and using (3.33), we arrive at (1.11). Replacing n by $100n+78$ in (3.30) and using (3.28), we deduce (1.12). The proof of Theorem 1.2 is complete. $\square$

4 Proof of Theorem 1.3

Setting $k=11$ in (1.7) and employing (2.21), we have

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(n) q^n=\frac{f_2f_{23}}{f_1^3f_{46}}\equiv \frac{f_1f_{23}}{f_2f_{46}} \ (\mathrm{mod}\ 4). \end{aligned}$$

(4.1)

Replacing q by $-q$ in (4.1) and using (2.23) yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(n) (-1)^n q^n \equiv \frac{f_2^2f_{46}^2 }{f_1f_4f_{23}f_{92}} \ (\mathrm{mod}\ 4). \end{aligned}$$

(4.2)

Chan and Toh [4] proved that

$$\begin{aligned} \frac{1}{f_1f_{23}}-\frac{f_1f_4f_{23}f_{92}}{ f_2^3f_{46}^3} =2q\frac{f_4f_{92}}{f_2^2f_{46}^2}+2q^3\frac{f_4^2f_{92}^2 }{f_2^3f_{46}^3}. \end{aligned}$$

(4.3)

Multiplying $-\frac{f_2^2f_{46}^2}{f_4f_{92}}$ on both sides of (4.3) yields

$$\begin{aligned} \frac{f_1f_{23}}{f_2f_{46}}-\frac{f_2^2f_{46}^2}{ f_1f_4f_{23}f_{92}} =-2q-2q^3\frac{f_4f_{92}}{f_2f_{46}}. \end{aligned}$$

(4.4)

Combining (4.1), (4.2) and (4.4) yields

$$\begin{aligned}&\sum _{n=0}^\infty \Delta _{11}(n)\big (1-(-1)^n\big )q^n\\&\quad \equiv \frac{f_1f_{23}}{f_2f_{46}}-\frac{f_2^2f_{46}^2}{ f_1f_4f_{23}f_{92}} \\&\quad \equiv -2q-2q^3\frac{f_4f_{92}}{f_2f_{46}} \ (\mathrm{mod}\ 4). \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(2n+1)q^n\equiv -1-q\frac{f_2f_{46}}{f_1f_{23}}\ (\mathrm{mod}\ 4). \end{aligned}$$

(4.5)

Replacing q by $-q$ in (4.5) and using (2.23) yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(2n+1)(-1)^n q^n\equiv -1+q\frac{f_1 f_4f_{23} f_{92}}{f_2^2f_{46}^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(4.6)

In view of (4.4), (4.5) and (4.6),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(2n+1)\big (1+(-1)^n\big )q^n&\equiv -2 + q\frac{f_4f_{92}}{f_2f_{46}} \left( \frac{f_1f_{23}}{f_2f_{46}}-\frac{f_2^2f_{46}^2}{ f_1f_4f_{23}f_{92}}\right) \nonumber \\&\equiv -2-2q^2\frac{f_4f_{92}}{f_2f_{46}} -2q^4\frac{f_4^2f_{92}^2}{f_2^2f_{46}^2}\ (\mathrm{mod}\ 4), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(4n+1) q^n \equiv -1-q\frac{f_2f_{46}}{f_1f_{23}} -q^2\frac{f_2^2f_{46}^2}{f_1^2f_{23}^2}\ (\mathrm{mod}\ 4). \end{aligned}$$

(4.7)

By (2.21), (4.5) and (4.7),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(4n+1) q^n \equiv \sum _{n=0}^\infty \Delta _{11}(2n+1) q^n -q^2f_1^2f_{23}^2\ (\mathrm{mod}\ 4). \end{aligned}$$

(4.8)

From Berndt’s book [3, Entry 31, p. 48],

$$\begin{aligned} f(U_1,V_1)=\sum _{r=0}^{n-1}U_r f\left( \frac{U_{n+r}}{U_r}, \frac{V_{n-r}}{ U_r}\right) , \end{aligned}$$

(4.9)

where $U_n=a^{\frac{(n+1)n}{2}}b^{\frac{n(n-1)}{2}}$ and $V_n=a^{\frac{(n-1)n}{2}}b^{\frac{n(n+1)}{2}}$. Taking $n=23$, $U_1=a=-q$ and $V_1=b=-q^{2}$ in (4.9), we have

$$\begin{aligned} f_1=&f\big (-q^{782}, -q^{805}\big ) - qf\big (-q^{851}, -q^{736}\big ) + q^5f\big (-q^{920}, -q^{667}\big ) \nonumber \\&- q^{12}f\big (-q^{989}, -q^{598}\big )+ q^{22}f_{529}- q^{35}f\big (-q^{1127}, -q^{460}\big ) \nonumber \\&+ q^{51}f\big (-q^{1196}, -q^{391}\big ) - q^{70}f\big (-q^{1265}, -q^{322}\big ) + q^{92}f\big (-q^{1334}, -q^{253}\big ) \nonumber \\&- q^{117}f\big (-q^{1403}, -q^{184}\big ) +q^{145 }f\big (-q^{1472}, -q^{115} \big )- q^{176}f\big (-q^{1541}, -q^{46}\big ) \nonumber \\&- q^{187}f\big (-q^{23}, -q^{1564}\big ) +q^{155}f\big (-q^{92}, -q^{1495}\big ) - q^{126}f\big (-q^{161}, -q^{1426}\big )\nonumber \\&+ q^{100}f\big (-q^{230}, -q^{1357}\big ) -q^{77}f\big (-q^{299}, -q^{1288}\big ) + q^{57}f\big (-q^{368}, -q^{1219}\big ) \nonumber \\&- q^{40}f\big (-q^{437}, -q^{1150}\big ) +q^{26}f\big (-q^{506}, -q^{1081}\big ) -q^{15}f\big (-q^{575}, -q^{1012}\big ) \nonumber \\&+ q^7f\big (-q^{644}, -q^{943}\big ) -q^2f\big (-q^{713}, -q^{874}\big ). \end{aligned}$$

(4.10)

Substituting (4.10) into (4.8), extracting the terms of the form $q^{23n}$ in both sides and then replacing $q^{23}$ by q, we deduce that

$$\begin{aligned} \sum _{n=0}^\infty \Delta _{11}(92n+1) q^n \equiv \sum _{n=0}^\infty \Delta _{11}(46n+1) q^n -q^2f_1^2f_{23}^2\ (\mathrm{mod}\ 4). \end{aligned}$$

(4.11)

It follows from (4.8) and (4.11) that for $n\ge 0$,

$$\begin{aligned} \Delta _{11}(92n+1)- \Delta _{11}(46n+1)\equiv \Delta _{11}(4n+1) -\Delta _{11}(2n+1) \ (\mathrm{mod}\ 4). \end{aligned}$$

(4.12)

Congruence (1.13) follows from (4.12) and mathematical induction. This completes the proof of Theorem 1.3.

5 Proof of Theorem 1.5

In this section, we prove Theorem 1.5.

By setting $k=2$ in (1.7),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(n)q^n=\frac{f_2f_5}{f_1^3f_{10}}. \end{aligned}$$

(5.1)

In view of (2.20) and (5.1),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(n)q^n\equiv \frac{f_5}{f_1f_{10} }\ (\mathrm{mod}\ 2). \end{aligned}$$

(5.2)

Xia and Yao [17] proved that

$$\begin{aligned} \frac{f_5}{f_1} =\frac{f_8f_{20 }^2 }{f_2^2f_{40}} +q \frac{ f_4^3f_{10}f_{40} }{f_2^3f_8f_{20} }. \end{aligned}$$

(5.3)

Thanks to (2.20) and (5.3),

$$\begin{aligned} \frac{f_5}{f_1} \equiv f_4 +q \frac{ f_{10}f_{20} }{f_2} \ ( \mathrm{mod}\ 2). \end{aligned}$$

(5.4)

By substituting (5.4) into (5.2),

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(n)q^n\equiv \frac{f_4}{f_{10}} +q \frac{ f_{20}}{f_2 } \ ( \mathrm{mod}\ 2), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(2n)q^n\equiv \frac{f_2 }{f_5} \ ( \mathrm{mod}\ 2) \end{aligned}$$

(5.5)

and

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(2n+1)q^n\equiv \frac{f_{10}}{f_1}\ ( \mathrm{mod}\ 2). \end{aligned}$$

(5.6)

Substituting (3.9) into (5.5), we obtain

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(2n)q^n\equiv \frac{f_{50} }{f_5}\left( R\big (q^{10}\big )-q^2-\frac{q^4}{R\big (q^{10}\big )}\right) \ ( \mathrm{mod}\ 2), \end{aligned}$$

which yields

$$\begin{aligned} \sum _{n=0}^\infty \Delta _2(10n+4)q^n\equiv \frac{f_{10}}{f_1}\ ( \mathrm{mod}\ 2). \end{aligned}$$

(5.7)

Congruence (1.14) follows from (5.6) and (5.7). This completes the proof of Theorem 1.5. $\square$