1 Introduction

In 2007, Andrews and Paule [1] introduced a new class of combinatorial objects called broken \(k\)-diamond partitions. Let \(\Delta _k(n)\) denote the number of broken \(k\)-diamond partitions of \(n\), and they proved that

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

In [1], they proved a congruence for \(\Delta _1(n)\) that for all \(n\ge 0\), \(\Delta _1(2n+1)\equiv 0\pmod {3}\). They also conjectured several other congruences modulo 2 satisfied by certain \(\Delta _k(n)\). Since then, mathematicians have provided numerous additional congruences satisfied by \(\Delta _k(n)\) for small integers \(k\). For example, Hirschhorn and Sellers [3] proved some parity results for \(\Delta _1(n)\) and \(\Delta _2(n)\):

$$\begin{aligned} \Delta _1(4n+2)&\equiv 0\pmod {2}\\ \Delta _1(4n+3)&\equiv 0\pmod {2}\\ \Delta _2(10n+2)&\equiv 0\pmod {2}\\ \Delta _2(10n+6)&\equiv 0\pmod {2}. \end{aligned}$$

After that, Chan [2] provided a different proof of the above results and obtained some new congruences for \(\Delta _2(n)\) modulo 5. Paule and Radu [6] extended the results of \(\Delta _2(n)\) modulo 5, and made four conjectures about \(\Delta _3(n)\) modulo 7 and \(\Delta _5(n)\) modulo 11. Two of those conjectures were proved by Xiong [10] in 2011, and the other two were proved by Jameson [4] recently. See more results of congruences for \(\Delta _k(n)\) in Radu and Sellers [7–9], Yao [11], etc.

In 2010, Radu and Sellers [7] provided many beautiful congruences for \(\Delta _k(n)\) modulo 2 when \(k=2,3,5,6,8,9,11\). Among them when \(k=8\), they proved that for all \(n\ge 0\),

$$\begin{aligned} \Delta _8(34n+r)\equiv ~0\pmod {2} \end{aligned}$$
(2)

when \(r\in \{11,15,17,19,25,27,29,33\}\). In our article, by using properties of modular forms, we have obtained many more congruences for \(\Delta _8(n)\) modulo 2. In fact, we have completely determined the behavior of \(\Delta _8(2n+1)\) modulo 2. It can be characterized in the following theorem:

Theorem 1

The broken 8-diamond partition function is defined by

$$\begin{aligned} \sum _{n=0}^{\infty }\Delta _8(n)q^n:=\prod _{n=1}^{\infty }\frac{(1-q^{2n})(1-q^{17n})}{(1-q^n)^3 (1-q^{34n})}, \end{aligned}$$
(3)

and then we have

$$\begin{aligned} \sum _{n=0}^{\infty }\Delta _8(2n+1)q^n\equiv \prod _{n=1}^{\infty }(1-q^n)^3+q^2\prod _{n=1}^{\infty }(1-q^{17n})^3\pmod {2}. \end{aligned}$$
(4)

With the help of the well-known identity [5, thm 1.60]

$$\begin{aligned} q\cdot \prod _{n=1}^{\infty }(1-q^{8n})^3=\sum _{n=0}^{\infty }(-1)^n (2n+1)q^{(2n+1)^2}, \end{aligned}$$
(5)

equation (4) can be changed to

$$\begin{aligned} \sum _{n=0}^{\infty }\Delta _8(2n+1)q^{8n+1}\equiv \sum _{n=0}^{\infty }q^{(2n+1)^2}+\sum _{n=0}^{\infty }q^{17(2n+1)^2}\pmod {2}. \end{aligned}$$
(6)

That means \(\Delta _8(2n+1)\equiv 1\pmod {2}\) if and only if \(8n+1\) is a square or \(17\) times a square. So we obtain a corollary to judge whether a congruence for \(\Delta _8(n)\) holds or not:

Corollary 1

Let \(A,B\) be two nonnegative integers, \(B<A\), then the congruence

$$\begin{aligned} \Delta _8(2An+2B+1)\equiv ~0\pmod {2} \end{aligned}$$
(7)

holds for all \(n\ge 0\) if and only if \(8B+1\) is a quadratic nonresidue mod \(8A\) and \(136B+17\) is a quadratic nonresidue mod \(136A\).

Based on the above corollary, we can derive many new congruences for \(\Delta _8(n)\) modulo \(2\). Following the notation in [7], we write

$$\begin{aligned} f(tn+r_1,r_2,\ldots ,r_m)\equiv 0\pmod {2} \end{aligned}$$

to mean that, for each \(i\in \{1,2,\ldots ,m\}\),

$$\begin{aligned} f(tn+r_i)\equiv 0\pmod {2}. \end{aligned}$$

Example 1

The following congruences hold for all \(n\ge 0\):

$$\begin{aligned} \begin{array}{ll} \displaystyle \Delta _8(18n+9,15)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(26n+9,11,15,19,23,25)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(34n+11,15,17,19,25,27,29,33)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(36n+9,15,27,33)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(38n+9,11,17,23,25,27,29,33,37)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(42n+17,19,25,29,35,37)&{}\equiv 0\pmod {2}\\ \displaystyle \Delta _8(50n+17,27,37,47)&{}\equiv 0\pmod {2}. \end{array} \end{aligned}$$

We can also directly obtain infinite families of congruences for \(\Delta _8(n)\) modulo 2.

Corollary 2

Let \(p\) be a odd prime, \(p\ne 17\), and \(t>0\) and \(\alpha \ge 0\) are integers with \((p,t)=1\), and \(t\cdot p^{2\alpha +1}\equiv 1\pmod {8}\). Then the following congruence holds for all \(n\ge 0\):

$$\begin{aligned} \Delta _8\left( 2p^{2\alpha +2}n+\frac{1}{4}(p^{2\alpha +1}t-1)+1\right) \equiv 0\pmod {2}. \end{aligned}$$
(8)

We will prove these theorems and corollaries in Sects. 3 and 4. Note that after examining small congruences for \(\Delta _8(n)\) modulo 2 by computer, we find that Corollary 1 has covered all congruences in the form \(\Delta _8(An+B)\equiv 0\pmod {2}\) for \(A\le 100\).

2 Preliminaries

In this section, we will introduce some notations of modular forms. The Dedekind’s eta function \(\eta (z)\) is defined as

$$\begin{aligned} \eta (z):=q^{1/24}\prod _{n=1}^{\infty }(1-q^n), \end{aligned}$$

where \(q=2\pi i z\), \(z\in \mathcal {H}\), \(\mathcal {H}\) is the upper half complex plane. A function \(f(z)\) is called an eta-quotient if it can be written as

$$\begin{aligned} f(z)=\prod _{\delta |N}\eta (\delta z)^{r_{\delta }}, \end{aligned}$$

where \(\delta \) and \(N\) are positive integers and \(r_\delta \) is an integer corresponding to \(\delta \). Let \(M_k(\varGamma _0(N),\chi )\) (resp. \(S_k(\varGamma _0(N),\chi )\)) denote the set of all holomorphic modular forms (resp. cusp forms) with respect to \(\varGamma _0(N)\) with weight \(k\) and character \(\chi \); the following theorem helps us to determine when an eta-quotient is of a modular form:

Theorem 2

[5, thm 1.64] If \(f(z)=\prod _{\delta |N}\eta (\delta z)^{r_{\delta }}\) is an eta-quotient with \(k=\frac{1}{2}\sum _{\delta | N}r_{\delta }\), with the additional properties that

$$\begin{aligned} \sum _{\delta | N}\delta r_{\delta }\equiv 0\pmod {24} \end{aligned}$$

and

$$\begin{aligned} \sum _{\delta | N}\frac{N}{\delta } r_{\delta }\equiv 0\pmod {24}, \end{aligned}$$

then \(f(z)\) satisfies

$$\begin{aligned} f\left( \frac{az+b}{cz+d}\right) =\chi (d)(cz+d)^k f(z) \end{aligned}$$

for every \(\left( \begin{array}{ccc} a&{}b\\ c&{}d \end{array}\right) \in \varGamma _0(N)\). Here the character \(\chi \) is defined by \(\chi (d):=\left( \frac{(-1)^k\cdot s}{d}\right) \), where \(s:=\prod _{\delta | N}\delta ^{r_{\delta }}\). Moreover, if \(f(z)\) is holomorphic (resp. vanishes) at all of the cusps of \(\varGamma _0(N)\), then \(f(z)\in M_k(\varGamma _0(N),\chi )\) (resp. \(S_k(\varGamma _0(N),\chi )\)).

And the orders of an eta-quotient at cusps are determined by

Theorem 3

[5, thm 1.65] Let \(c\), \(d\), and \(N\) be the positive integers with \(d|N\) and \(\gcd (c,d)=1\). If \(f(z)\) is an eta-quotient satisfying the conditions of Theorem 2.1 for \(N\), then the order of vanishing of \(f(z)\) at the cusp \(\frac{c}{d}\) is

$$\begin{aligned} \frac{N}{24}\sum _{\delta |N}\frac{gcd(d,\delta )^2 r_\delta }{gcd(d,\frac{N}{d})d\delta }. \end{aligned}$$

If \(d\) is a positive integer and \(f(q)=\sum _{n=0}^{\infty }a(n)q^n\) is a formal power series, we define the operator \(U(d)\) by

$$\begin{aligned} f(q)|U(d):=\sum _{n=0}^{\infty }a(dn)q^n. \end{aligned}$$

Proposition 2.22 in [5] shows that if \(d|N\), \(f(z)\in M_k(\varGamma _0(N),\chi )\), then \(f(z)|U(d)\in M_k(\varGamma _0(N),\chi )\). Also note that the \(U(d)\) operator has the property

$$\begin{aligned} \left[ \left( \,\sum _{n=0}^{\infty }a(n)q^n\right) \left( \,\sum _{n=0}^{\infty }b(n)q^{dn}\right) \right] |U(d)=\left( \,\sum _{n=0}^{\infty }a(dn)q^n\right) \left( \,\sum _{n=0}^{\infty }b(n)q^n\right) . \end{aligned}$$
(9)

For the purposes of studying congruences, we introduce the Sturm’s Theorem, to show that every holomorphic modular form modulo \(M\) is determined by its “first few” coefficients. Let \(f(q)=\sum _{n=0}^{\infty }a(n)q^n\) be a formal power series with \(a(n)\in \mathbb {Z}\) and \(M\) be a positive integer, and define the order of \(f\) modulo \(M\) by

$$\begin{aligned} \mathrm{ord}_M (f):=\min \{n:a(n)\not \equiv 0\pmod {M}\}, \end{aligned}$$

then Sturm’s Theorem can be stated as

Theorem 4

[5, thm 2.58] Suppose that \(f(z),g(z)\in M_k(\varGamma _0(N),\chi )\bigcap \mathbb {Z}[[q]]\) and \(M\) is prime. If

$$\begin{aligned} \mathrm{ord}_M(f(z)-g(z))\ge 1+\frac{kN}{12}\prod _{p|N}\left( 1+\frac{1}{p}\right) , \end{aligned}$$

where the product is over all prime divisors \(p\) of \(N\), then \(f(z)\equiv g(z)\pmod {M}\).

With the above theorems, we can now start our proof of Theorem 1. We will frequently use the following congruence during the proof:

$$\begin{aligned} \prod _{n=1}^{\infty }(1-q^n)^2 \equiv \prod _{n=1}^{\infty }(1-q^{2n})\pmod {2}. \end{aligned}$$
(10)

3 Proof of Theorem 1

Proof

In this section, we will prove Eq. (4). Define eta-quotients \(F(z)\) and \(G(z)\) as

$$\begin{aligned} F(z)&:=\frac{\eta (z)^3\eta (2z)^{43}\eta (17z)}{\eta (34z)}\\&=q^3 \prod _{n=1}^{\infty }\frac{(1-q^n)^3(1-q^{2n})^{43}(1-q^{17n})}{(1-q^{34n})} \end{aligned}$$

and

$$\begin{aligned} G(z)&:=\eta (z)^{44}\eta (2z)^2+\eta (z)^{43}\eta (2z)\eta (17z)\eta (34z)\\&=q^2\prod _{n=1}^{\infty }(1-q^n)^{44}(1-q^{2n})^2\\&\quad +q^4\prod _{n=1}^{\infty }(1-q^n)^{43}(1-q^{2n})(1-q^{17n})(1-q^{34n}). \end{aligned}$$

By Theorems 2 and 3, it is easy to verify that \(F(z),G(z)\in M_k(\varGamma _0(N),\chi )\), where \(k=23\), \(N=34*24\), and \(\chi (d)=\left( \frac{-1}{d}\right) \). Applying operator \(U(2)\) to \(F(z)\), and using Sturm’s Theorem, after checking the first \(23*8*18\) coefficients of \(F(z)|U(2)\) and \(G(z)\) by computer, we find that

$$\begin{aligned} F(z)|U(2)\equiv G(z)\pmod {2}. \end{aligned}$$
(11)

On the other hand, by the definition of \(\Delta _8(n)\) in Eq. (3) (we define \(\Delta _8(n)=0\) if \(n\) is not a nonnegative integer), using Eq. (10), we can see that

$$\begin{aligned} F(z)&\equiv q^3 \prod _{n=1}^{\infty }\frac{(1-q^{2n})(1-q^{17n})}{(1-q^n)^3(1-q^{34n})}(1-q^{2n})^{45}\\&\equiv \left( q^3\sum _{n=0}^{\infty }\Delta _8(n)q^n\right) \prod _{n=1}^{\infty }(1-q^{2n})^{45}\pmod {2}, \end{aligned}$$

so we have

$$\begin{aligned} F(z)|U(2)&\equiv \left[ \left( \,\sum _{n=0}^{\infty }\Delta _8(n-3)q^n\right) \left( \,\prod _{n=1}^{\infty }(1-q^{2n})^{45}\right) \right] |U(2)\nonumber \\&\equiv \left( \,\sum _{n=0}^{\infty }\Delta _8(2n-3)q^n\right) \left( \,\prod _{n=1}^{\infty }(1-q^{n})^{45}\right) \quad (\mathrm by Eq. 9 )\nonumber \\&\equiv \left( q^2\sum _{n=0}^{\infty }\Delta _8(2n+1)q^n\right) \left( \,\prod _{n=1}^{\infty }(1-q^{n})^{45}\right) \pmod {2}. \end{aligned}$$
(12)

Also note that

$$\begin{aligned} G(z)\equiv q^2\prod _{n=1}^{\infty }(1-q^n)^{48}+q^4\prod _{n=1}^{\infty }(1-q^n)^{45}(1-q^{17n})^3\pmod {2} \end{aligned}$$
(13)

Combining (11), (12), and (13) together, we get Eq. (4). \(\square \)

4 Proof of Corollaries 1 and 2

Since we have proved Theorem 1, those corollaries are quite simple. From Eq. (6), we know that if \(\Delta _8(2(An+B)+1)\equiv 0\pmod {2}\) for all \(n\ge 0\), then both \(8(An+B)+1\) and \(17*(8(An+B)+1)\) cannot be squares, which is equivalent to say that \(8B+1\) is a quadratic nonresidue mod \(8A\) and \(136B+17\) is a quadratic nonresidue mod \(136A\). For example, let \(A=9\), then \(8A=72\), and \(8B+1\in S:=\{1,9,17,25,33,41,49,57,65\}\). Among \(S\), only \(17,33,41,57,65\) are quadratic nonresidues mod 72, which means \(B\in \{2,4,5,7,8\}\). Similarly, \(136B+17\) is a quadratic nonresidue mod \(136A\) if and only if \(B\in \{0,3,4,6,7\}\). Thus, only \(B=4\) and \(B=7\) suit both conditions. Substituting \(A\) and \(B\) in \(\Delta _8(2(An+B)+1)\), we obtain the first result in Example 1.

To prove Corollary 2, write

$$\begin{aligned} 2p^{2\alpha +2}n+\frac{1}{4}(p^{2\alpha +1}t-1)+1=2\left( p^{2\alpha +2}n+\frac{1}{8}(p^{2\alpha +1}t-1)\right) +1. \end{aligned}$$

By Theorem 1, we need to show that

$$\begin{aligned} 8\left( p^{2\alpha +2}n+\frac{1}{8}(p^{2\alpha +1}t-1)\right) +1=p^{2\alpha +1}(8pn+t) \end{aligned}$$

is neither a square, nor 17 times a square. This is obvious because under our assumption, \(\mathrm{ord}_p\left( p^{2\alpha +1}(8pn+t)\right) =2\alpha +1\). Now we complete our proof.