1 Introduction

Andrews and Paule [1] introduced the concept of broken k-diamond partitions and showed that the generating function for \(\Delta _k(n)\), the number of broken k-diamond partitions of n, is

$$\begin{aligned} \sum _{n\ge 0}\Delta _k(n)q^n =\frac{(q^2;q^2)_\infty (q^{2k+1};q^{2k+1})_\infty }{(q;q)_\infty ^3(q^{2(2k+1)};q^{2(2k+1)})_\infty }. \end{aligned}$$
(1.1)

The following congruences were proved by Chan [2] and again by Radu [4]:

$$\begin{aligned} \Delta _2(25n+14)\equiv 0\pmod 5 \end{aligned}$$
(1.2)

and

$$\begin{aligned} \Delta _2(25n+24)\equiv 0\pmod 5. \end{aligned}$$
(1.3)

Indeed, Chan generalised these to

$$\begin{aligned} \Delta _2\left( 5^{\alpha +1}n+\frac{11\times 5^\alpha +1}{4}\right) \equiv 0\pmod 5 \end{aligned}$$
(1.4)

and

$$\begin{aligned} \Delta _2\left( 5^{\alpha +1}n+\frac{19\times 5^\alpha +1}{4}\right) \equiv 0\pmod 5. \end{aligned}$$
(1.5)

The object of this note is to give as simple a proof as I can of (1.2)–(1.5).

2 Proofs

We start by noting that the 5–dissection of \(\displaystyle \psi (q)=\sum _{n\ge 0}q^{(n^2+n)/2}\) is

$$\begin{aligned} \psi (q)&=1+q+q^3+q^6+q^{10}+q^{15}+q^{21}+q^{28}+q^{36}+q^{45}+q^{55}+q^{66}+q^{75}+\ \cdots \nonumber \\&=(1+q^{10}+q^{15}+q^{45}+q^{55}+\ \cdots )+q(1+q^5+q^{20}+q^{35}+q^{65}+\ \cdots \ )\nonumber \\&\quad +\, q^3(1+q^{25}+q^{75}+\ \cdots \ )\nonumber \\&=a+qb+q^3c, \end{aligned}$$
(2.1)

where

$$\begin{aligned} a=(-q^{10},-q^{15},q^{25};q^{25})_\infty ,\ \ \ b=(-q^5,-q^{20},q^{25};q^{25})_\infty ,\ \ \ c=\psi (q^{25}). \end{aligned}$$
(2.2)

Note that by [3, (34.1.21)]

$$\begin{aligned} ab+q^5c^2=\psi (q^5)^2. \end{aligned}$$
(2.3)

We have, modulo 5,

$$\begin{aligned} \sum _{n\ge 0}\Delta _2(n)q^n&=\frac{(q^2;q^2)_\infty (q^5;q^5)_\infty }{(q;q)_\infty ^3(q^{10};q^{10})_\infty } \equiv \frac{(q^2;q^2)_\infty (q;q)_\infty ^5}{(q;q)_\infty ^3(q^2;q^2)_\infty ^5} =\frac{(q;q)_\infty ^2}{(q^2;q^2)_\infty ^4}\nonumber \\&=\frac{1}{\psi (q)^2} =\frac{\psi (q)^3}{\psi (q)^5}\equiv \frac{\psi (q)^3}{\psi (q^5)}. \end{aligned}$$
(2.4)

Alternatively,

$$\begin{aligned}&\sum _{n\ge 0}\Delta _2(n)q^n=\frac{(q^2;q^2)_\infty (q^5;q^5)_\infty }{(q;q)_\infty ^3(q^{10};q^{10})_\infty }=\frac{\psi (q)^3}{\psi (q^5)}\frac{(q^{10};q^{10})_\infty }{(q^2;q^2)_\infty ^5}\equiv \frac{\psi (q)^3}{\psi (q^5)},\nonumber \\ \end{aligned}$$
(2.5)

or, again, by [3, (34.1.23)],

$$\begin{aligned} \sum _{n\ge 0}\Delta _2(n)q^n&=\frac{(q^2;q^2)_\infty (q^5;q^5)_\infty }{(q;q)_\infty ^3(q^{10};q^{10})_\infty }\nonumber \\&=\frac{1}{\psi (q)^2-5q\psi (q^5)^2}\equiv \frac{1}{\psi (q)^2}\equiv \frac{\psi (q)^3}{\psi (q^5)}. \end{aligned}$$
(2.6)

Thus, we have

$$\begin{aligned}&\sum _{n\ge 0}\Delta _2(n)q^n\equiv \frac{\psi (q)^3}{\psi (q^5)}=\frac{(a+qb+q^3c)^3}{\psi (q^5)}\nonumber \\&\quad =\frac{a^3+3qa^2b+3q^2ab^2+q^3b^3+3q^3a^2c+6q^4abc +3q^5b^2c+3q^6ac^2+3q^7bc^2+q^9c^3}{\psi (q^5)}.\nonumber \\ \end{aligned}$$
(2.7)

It follows that

$$\begin{aligned} \sum _{n\ge 0}\Delta _2(5n+4)q^{5n}&\equiv \frac{abc+q^5c^3}{\psi (q^5)}=\frac{c(ab+q^5c^2)}{\psi (q^5)}\nonumber \\&=\frac{\psi (q^{25})\psi (q^5)^2}{\psi (q^5)}=\psi (q^{25})\psi (q^5) \end{aligned}$$
(2.8)

and

$$\begin{aligned} \sum _{n\ge 0}\Delta _2(5n+4)q^n\equiv \psi (q^5)\psi (q)=\psi (q^5)(a+qb+q^3\psi (q^{25})). \end{aligned}$$
(2.9)

It is now an easy induction (replace n by \(5n+3\)) to deduce that for \(\alpha \ge 1\),

$$\begin{aligned}&\sum _{n\ge 0}\Delta _2\left( 5^\alpha n+\frac{3\times 5^\alpha +1}{4}\right) q^n\equiv \psi (q^5)\psi (q)\nonumber \\&\quad \quad =\psi (q^5)(a+qb+q^3\psi (q^{25})). \end{aligned}$$
(2.10)

Since there are no terms on the right in which the power of q is congruent to 2 or 4 modulo 5, we have that for \(\alpha \ge 1\),

$$\begin{aligned} \Delta _2\left( 5^\alpha (5n+2)+\frac{3\times 5^\alpha +1}{4}\right) \equiv 0 \end{aligned}$$
(2.11)

and

$$\begin{aligned} \Delta _2\left( 5^\alpha (5n+4)+\frac{3\times 5^\alpha +1}{4}\right) \equiv 0, \end{aligned}$$
(2.12)

as claimed.