1 Introduction

A partition of a positive integer n is a nonincreasing sequence of positive integers whose sum is n. An overpartition of n is a partition of n in which the first occurrence of a number may be overlined. Let \(\overline{p}(n)\) denote the number of overpartitions of n, and we assume that \(\overline{p}(0) = 1\). From [6], we know that the generating function for \(\overline{p}(n)\) is given by

$$\begin{aligned} \sum _{n=0}^\infty \overline{p}(n)q^n=\frac{(-q;q)_\infty }{(q;q)_\infty }, \end{aligned}$$
(1.1)

where

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

The arithmetic properties of \(\overline{p}(n)\) were widely studied in the literature. Fortin, Jacob and Mathieu [9], and Hirschhorn and Sellers [10] established the 2-, 3-, and 4-dissections of the generating function for \(\overline{p}(n)\), from which some congruences modulo 4 and 8 are obtained. In particular, they obtained the following three Ramanujan type identities:

$$\begin{aligned} \sum _{n=0}^\infty \overline{p}(2n+1)q^n&=2\frac{(q^2;q^2)_\infty ^2(q^8;q^8)_\infty ^2}{(q;q)_\infty ^4(q^4;q^4)_\infty },\end{aligned}$$
(1.2)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(4n+3)q^n&=8\frac{(q^2;q^2)_\infty (q^4;q^4)_\infty ^6}{(q;q)_\infty ^8},\end{aligned}$$
(1.3)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8n+7)q^n&=64\frac{(q^2;q^2)_\infty ^{22}}{(q;q)_\infty ^{23}}. \end{aligned}$$
(1.4)

Hirschhorn and Sellers [10] also conjectured that if p is an odd prime and r is a quadratic nonresidue modulo p,

$$\begin{aligned} \overline{p}(pn+r)\equiv \left\{ \begin{array}{ll}0 \quad (\mathrm{mod}\,\, 4) &{}\quad \text { if } p \equiv \pm 3 \quad (\mathrm{mod}\,\, {8}),\\ 0 \quad (\mathrm{mod}\,\, 8) &{}\quad \text { if }p \equiv \pm 1\quad (\mathrm{mod}\,\, {8}). \end{array}\right. \end{aligned}$$

The above conjecture later was confirmed by Kim [14]. Mahlburg [16] conjectured that for all positive integers k, \(\overline{p}(n)\equiv 0 \quad (\mathrm{mod}\,\, {2^k})\) holds for a set of integers of arithmetic density 1 and proved the case \(k=6\). In [13], Kim confirmed the case \(k=7\), and the conjecture is still open. Recently, Ramanujan type congruences modulo 16 and 32 have been considered by several authors, see [5, 20, 21], for example. For congruences modulo 5 for \(\overline{p}(n)\), we refer the reader to [3, 4, 8, 15, 17, 18]. For modulo powers of 3, see [11, 19, 21].

The aim of this paper is to derive infinite families of congruences for \(\overline{p}(n)\) modulo powers of 2. Here we list our main results in the following theorems.

Theorem 1.1

For \(\alpha \ge 0\) and \(n\ge 0\), we have

$$\begin{aligned} \overline{p}(8\cdot 5^{8\alpha +2}n+31\cdot 5^{8\alpha +1})&\equiv 0\quad (\mathrm{mod}\,\,{2^5}),\end{aligned}$$
(1.5)
$$\begin{aligned} \overline{p}(8\cdot 5^{8\alpha +2}n+39\cdot 5^{8\alpha +1})&\equiv 0\quad (\mathrm{mod}\,\,{2^5}),\end{aligned}$$
(1.6)
$$\begin{aligned} \overline{p}(8\cdot 5^{8\alpha +4}n+31\cdot 5^{8\alpha +3})&\equiv 0\quad (\mathrm{mod}\,\,{2^6}),\end{aligned}$$
(1.7)
$$\begin{aligned} \overline{p}(8\cdot 5^{8\alpha +4}n+39\cdot 5^{8\alpha +3})&\equiv 0\quad (\mathrm{mod}\,\,{2^6}),\end{aligned}$$
(1.8)
$$\begin{aligned} \overline{p}\left( 8\cdot 5^{8\alpha +8}n+(8i+7)\cdot 5^{8\alpha +7}\right)&\equiv 0\quad (\mathrm{mod}\,\, {2^6}), \end{aligned}$$
(1.9)

where \(i=0,2,3,4\).

Theorem 1.2

For \(\alpha \ge 0\) and \(n\ge 0\), we have

$$\begin{aligned} \overline{p}(8\cdot 7^{4\alpha +2}n+(8i+5)7^{4\alpha +1})\equiv 0\quad (\mathrm{mod}\,\, {2^5}), \end{aligned}$$
(1.10)

where \(i=3,4,6\), and

$$\begin{aligned} \overline{p}(8\cdot 7^{4\alpha +4}n+(8i+5)7^{4\alpha +3})\equiv 0\quad (\mathrm{mod}\,\, {2^5}), \end{aligned}$$
(1.11)

where \(i=0,1,3,4,5,6\).

Theorem 1.3

For \(\alpha \ge 0\) and \(n\ge 0\), we have

$$\begin{aligned} \overline{p}(8\cdot 3^{4\alpha +4}n+5\cdot 3^{4\alpha +3})\equiv 0\quad (\mathrm{mod}\,\, {2^8}),\end{aligned}$$
(1.12)
$$\begin{aligned} \overline{p}(8\cdot 3^{4\alpha +4}n+13\cdot 3^{4\alpha +3})\equiv 0\quad (\mathrm{mod}\,\, {2^8}). \end{aligned}$$
(1.13)

2 Preliminaries

Let f(a, b) be Ramanujan’s general theta function given by

$$\begin{aligned} f(a,b):=\sum _{n=-\infty }^{\infty }a^{\frac{n(n+1)}{2}}b^{\frac{n(n-1)}{2}},\quad |ab|<1. \end{aligned}$$

Jacobi’s triple product identity can be stated in Ramanujan’s notation as follows:

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

Thus,

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

In order to prove our results, we need the following lemmas.

Lemma 2.1

[7] For any odd prime p,

$$\begin{aligned} \psi (q)=\sum _{k=0}^{\frac{p-3}{2}}q^{\frac{k^{2}+k}{2}}f\left( q^{\frac{p^{2}+(2k+1)p}{2}},q^{\frac{p^{2}-(2k+1)p}{2}}\right) +q^{\frac{p^{2}-1}{8}}\psi (q^{p^{2}}). \end{aligned}$$

Furthermore, we claim that for \(0\le k\le (p-3)/2\),

$$\begin{aligned} \frac{k^{2}+k}{2} \not \equiv \frac{p^{2}-1}{8} \quad (\mathrm{mod}\,\, p). \end{aligned}$$

In particular, setting \(p=5,7\) in Lemma 2.1, we have

$$\begin{aligned} \psi (q)&=f(q^{10},q^{15})+qf(q^5,q^{20})+q^3\psi (q^{25}),\end{aligned}$$
(2.1)
$$\begin{aligned} \psi (q)&=f(q^{21},q^{28})+qf(q^{14},q^{35})+q^3f(q^7,q^{42})+q^6\psi (q^{49}). \end{aligned}$$
(2.2)

For convenience, we rewrite (2.1) as the following simple form

$$\begin{aligned} \psi (q)&=A_0+qA_1+q^3A_3, \end{aligned}$$
(2.3)

where \(A_0=f(q^{10},q^{15}), A_1=f(q^5,q^{20}), A_3=\psi (q^{25})\).

Lemma 2.2

[1, p. 26, (1.6.7)]

$$\begin{aligned} \psi ^2(q)-q\psi ^2(q^5)=f(q,q^4)f(q^2,q^3). \end{aligned}$$

From (2.3) and Lemma 2.2, we see that

$$\begin{aligned} \psi ^2(q^5)-q^5\psi ^2(q^{25})=A_0A_1. \end{aligned}$$
(2.4)

Lemma 2.3

For integer \(n\ge 1\), we have

$$\begin{aligned} (q^n;q^n)_\infty ^4&\equiv (q^{2n};q^{2n})_\infty ^2\quad (\mathrm{mod}\,\, 4),\\ (q^n;q^n)_\infty ^8&\equiv (q^{2n};q^{2n})_\infty ^4\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

3 Proof of Theorem 1.1

To prove Theorem 1.1, we first need to establish following lemma.

Lemma 3.1

For \(\alpha \ge 0\) and \(n\ge 0\), we have

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}\left( 8\cdot 5^{8\alpha }n+3\cdot 5^{8\alpha }\right) q^{n}&\equiv 8\psi ^3(q)\quad (\mathrm{mod}\,\, {64}),\end{aligned}$$
(3.1)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +1}n+7\cdot 5^{8\alpha +1})q^n&\equiv 24q\psi ^3(q^{5})-16\psi (q^{5})\psi ^2(q)\quad (\mathrm{mod}\,\,{64}),\end{aligned}$$
(3.2)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +3}n+7\cdot 5^{8\alpha +3})q^n&\equiv 40q\psi ^3(q^{5})+32\psi (q^{5})\psi ^2(q)\quad (\mathrm{mod}\,\, {64}),\end{aligned}$$
(3.3)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}\left( 8\cdot 5^{8\alpha +7}n+7\cdot 5^{8\alpha +7}\right) q^{n}&\equiv 8q\psi ^3(q^5)\quad (\mathrm{mod}\,\, {64}). \end{aligned}$$
(3.4)

Proof

From (1.3), we have

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(4n+3)q^n&\equiv 8\frac{(q^2;q^2)_\infty (q^4;q^4)_\infty ^6}{(q^2;q^2)_\infty ^4}\quad (\mathrm{mod}\,\, {64}), \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8n+3)q^n&\equiv 8\psi ^3(q) \quad (\mathrm{mod}\,\, {64}). \end{aligned}$$
(3.5)

Define a(n) as follows:

$$\begin{aligned} \sum _{n=0}^{\infty }a(n)q^n=\psi ^3(q). \end{aligned}$$
(3.6)

Thus,

$$\begin{aligned} \overline{p}(8n+3)\equiv 8a(n)\quad (\mathrm{mod}\,\, {64}). \end{aligned}$$
(3.7)

Applying (2.3), we have

$$\begin{aligned} \sum _{n=0}^{\infty }a(n)q^n =&A^3_0+3qA^2_0A_1+3q^2A_0A^2_1+q^3A^3_1+3q^3A^2_0A_3\\&+6q^4A_0A_1A_3+3q^5A^2_1A_3+3q^6A_0A^2_3+3q^7A_1A^2_3+q^9A^3_3. \end{aligned}$$

Applying (2.4), it can be seen that

$$\begin{aligned} \sum _{n=0}^{\infty }a(5n+4)q^{5n+4}&= q^9\psi ^3(q^{25})+6q^4\psi (q^{25})f(q^5,q^{20})f(q^{10},q^{15})\\&=q^9\psi ^3(q^{25})+6q^4\psi (q^{25})\left( \psi ^2(q^5)-q^5\psi ^2(q^{25})\right) \\&=-5q^9\psi ^3(q^{25})+6q^4\psi (q^{25})\psi ^2(q^5), \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^{\infty }a(5n+4)q^{n}\equiv 3q\psi ^3(q^{5})-2\psi (q^{5})\psi ^2(q)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$
(3.8)

Using (2.3) and (2.4) again, it follows that

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5(5n+1)+4\right) q^{5n+1}&=\sum _{n=0}^{\infty }a\left( 25n+9\right) q^{5n+1}\\&\equiv 3q\psi ^3(q^5)-2\psi (q^5)\left( q^6A^2_3+2qA_0A_1\right) \\&\equiv 3q\psi ^3(q^5)-2\psi (q^5)\left( q^6\psi ^2(q^{25})+ 2q\psi ^2(q^5)-2q^6\psi ^2(q^{25})\right) \\&\equiv -q\psi ^3(q^5)+2q^6\psi ^2(q^{25})\psi (q^5)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

That is,

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 25n+9\right) q^{n}&\equiv -\psi ^3(q)+2q\psi ^2(q^{5})\psi (q)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

Applying (2.1) and (3.8), it follows that

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^2(5n+4)+9\right) q^{n}&=\sum _{n=0}^{\infty }a\left( 5^3n+\frac{7\cdot 5^3-3}{8}\right) q^{n}\nonumber \\&\equiv 5q\psi ^3(q^5)+2\psi (q^5)\psi ^2(q)+2\psi ^2(q)\psi (q^5)\quad (\mathrm{mod}\,\, 8)\nonumber \\&=5q\psi ^3(q^5)+4\psi (q^5)\psi ^2(q). \end{aligned}$$
(3.9)

Using (2.3) and (2.4), we deduce that

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^3(5n+1)+\frac{7\cdot 5^3-3}{8}\right) q^{5n+1}&=\sum _{n=0}^{\infty }a\left( 5^4n+\frac{3\cdot (5^4-1)}{8}\right) q^{5n+1}\\&\equiv 5q\psi ^3(q^5)+4\psi (q^5)\left( q^6A^2_3+2qA_0A_1\right) \\&\equiv 5q\psi ^3(q^5)+4q^6\psi ^2(q^{25})\psi (q^5) \;\, (\mathrm{mod}\,\, 8), \end{aligned}$$

namely,

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^4n+\frac{3\cdot (5^4-1)}{8}\right) q^{n}&\equiv 5\psi ^3(q)+4q\psi ^2(q^{5})\psi (q)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

Based on (2.1) and (3.8), we have

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^4(5n+4)+\frac{3\cdot (5^4-1)}{8}\right) q^{n}&=\sum _{n=0}^{\infty }a\left( 5^5n+\frac{7\cdot 5^5-3}{8}\right) q^{n}\nonumber \\&\equiv -25q\psi ^3(q^5)-10\psi ^2(q)\psi (q^5)+4\psi ^2(q)\psi (q^5)\nonumber \\&\equiv -q\psi ^3(q^5)+2\psi ^2(q)\psi (q^5)\quad (\mathrm{mod}\,\, 8) . \end{aligned}$$
(3.10)

Similarly,

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^5(5n+1)+\frac{7\cdot 5^5-3}{8}\right) q^{5n+1}&=\sum _{n=0}^{\infty }a\left( 5^6n+\frac{3\cdot (5^6-1)}{8}\right) q^{5n+1}\\&\equiv -q\psi ^3(q^5)+2\psi (q^5)\!\left( q^6A^2_3+2qA_0A_1\!\right) \\&\equiv 3q\psi ^3(q^5){-}2q^6\psi (q^5)\psi ^2(q^{25})\;\, (\mathrm{mod}\,\, 8), \end{aligned}$$

so it follows that

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^6n+\frac{3\cdot (5^6-1)}{8}\right) q^{n} \equiv 3\psi ^3(q)-2q\psi (q)\psi ^2(q^{5})\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

From (2.1) and (3.8), it can be seen that

$$\begin{aligned}&\sum _{n=0}^{\infty }a\left( 5^6(5n+4)+\frac{3\cdot (5^6-1)}{8}\right) q^{n}\nonumber \\&\quad =\sum _{n=0}^{\infty }a\left( 5^7n+\frac{7\cdot 5^7-3}{8}\right) q^{n}\nonumber \\&\quad \equiv -15q\psi ^3(q^5)-6\psi (q^5)\psi ^2(q)-2\psi (q^5)\psi ^2(q)\nonumber \\&\quad \equiv q\psi ^3(q^5)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$
(3.11)

Then we have

$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^7(5n+1)+\frac{7\cdot 5^7-3}{8}\right) q^{n}&=\sum _{n=0}^{\infty }a\left( 5^8n+\frac{3\cdot (5^8-1)}{8}\right) q^{n}\\&\equiv \psi ^3(q)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

So we have the following useful relation:

$$\begin{aligned} a\left( 5^8n+\frac{3\cdot (5^8-1)}{8}\right) \equiv a(n)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$

By induction, we have

$$\begin{aligned} a\left( 5^{8\alpha }n+\frac{3\cdot (5^{8\alpha }-1)}{8}\right) \equiv a(n)\quad (\mathrm{mod}\,\, 8). \end{aligned}$$
(3.12)

Using (3.6), (3.8), (3.9), (3.11) and (3.12), we deduce that

$$\begin{aligned} \sum _{n=0}^\infty a\left( 5^{8\alpha }n+\frac{3\cdot (5^{8\alpha }-1)}{8}\right) q^n&\equiv \psi ^3(q)\quad (\mathrm{mod}\,\, 8),\end{aligned}$$
(3.13)
$$\begin{aligned} \sum _{n=0}^{\infty }a\left( 5^{8\alpha +1}n+\frac{7\cdot 5^{8\alpha +1}-3}{8}\right) q^{n}&\equiv 3q\psi ^3(q^{5})-2\psi (q^{5})\psi ^2(q)\quad (\mathrm{mod}\,\, 8),\end{aligned}$$
(3.14)
$$\begin{aligned} \sum _{n=0}^\infty a \left( 5^{8\alpha +3}n+\frac{7\cdot 5^{8\alpha +3}-3}{8}\right) q^n&\equiv 5q\psi ^3(q^5)+4\psi (q^5)\psi ^2(q)\quad (\mathrm{mod}\,\, 8),\end{aligned}$$
(3.15)
$$\begin{aligned} \sum _{n=0}^\infty a \left( 5^{8\alpha +7}n+\frac{7\cdot 5^{8\alpha +7}-3}{8}\right) q^n&\equiv q\psi ^3(q^5)\;\,(\mathrm{mod}\,\, 8). \end{aligned}$$
(3.16)

Using the above relations and (3.7), we can easily get the desired results. \(\square \)

Proof of Theorem 1.1

Applying (2.3) and (3.2), we deduce that

$$\begin{aligned}&\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +1}n+7\cdot 5^{8\alpha +1})q^n\\&\equiv -40q\psi ^3(q^{5})-16\psi (q^{5})\\&\quad \times \left( A_0^2+q^2A_1^2+q^6A^2_3+2qA_0A_1+2q^3A_0A_3+2q^4A_1A_3\right) \quad (\mathrm{mod}\, {64}). \end{aligned}$$

Thus, it follows that

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +1}(5n+3)+7\cdot 5^{8\alpha +1})q^{5n+3}&=\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +2}n+31\cdot 5^{8\alpha +1})q^{5n+3}\\&\equiv -16\psi (q^{5})(2q^3A_0A_3)\quad (\mathrm{mod}\,\, {64}), \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +1}(5n+4)+7\cdot 5^{8\alpha +1})q^{5n+4}&=\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +2}n+39\cdot 5^{8\alpha +1})q^{5n+4}\\&\equiv -16\psi (q^{5})(2q^4A_1A_3)\quad (\mathrm{mod}\,\, {64}). \end{aligned}$$

This yields the first two congruences of the theorem. In addition, applying (2.3) and (3.3), we deduce that

$$\begin{aligned}&\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +3}n+7\cdot 5^{8\alpha +3})q^n\\&\equiv 40q\psi ^3(q^{5})+32\psi (q^{5})\\&\quad \times \left( A_0^2+q^2A_1^2+q^6A^2_3+2qA_0A_1+2q^3A_0A_3+2q^4A_1A_3\right) \quad (\mathrm{mod}\,\, {64}). \end{aligned}$$

Hence, we obtain

$$\begin{aligned}&\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +3}(5n+3)+7\cdot 5^{8\alpha +3})q^{5n+3}\nonumber \\&\quad =\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +4}n+31\cdot 5^{8\alpha +3})q^{5n+3} \nonumber \\&\quad \equiv 32q^3\psi (q^{5})(2A_0A_3)\nonumber \\&\quad \equiv 0\quad (\mathrm{mod}\,\, {64}), \end{aligned}$$
(3.17)
$$\begin{aligned}&\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +3}(5n+4)+7\cdot 5^{8\alpha +3})q^{5n+4}\nonumber \\&\quad =\sum _{n=0}^{\infty }\overline{p}(8\cdot 5^{8\alpha +4}n+39\cdot 5^{8\alpha +3})q^{5n+4}\nonumber \\&\quad \equiv 32q^4\psi (q^{5})(2A_1A_3)\nonumber \\&\quad \equiv 0\quad (\mathrm{mod}\,\, {64}). \end{aligned}$$
(3.18)

From (3.4), we see that for \(i=0,2,3,4\),

$$\begin{aligned} \overline{p}\left( 8\cdot 5^{8\alpha +7}(5n+i)+7\cdot 5^{8\alpha +7}\right) \equiv 0\quad (\mathrm{mod}\,\,{64}). \end{aligned}$$

Therefore, we finish the proof. \(\square \)

4 Proof of Theorem 1.2

Lemma 4.1

For \(\alpha \ge 0\) and \(n\ge 0\), we have

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 7^{4\alpha }n+3\cdot 7^{4\alpha })q^n&\equiv 8\psi ^3(q)\quad (\mathrm{mod}\,\, {32}),\nonumber \\ \sum _{n=0}^{\infty }\overline{p}(8\cdot 7^{4\alpha +1}n+5\cdot 7^{4\alpha +1})q^n&\equiv 16f_1f_{14}+8q^2\psi ^3(q^7)\quad (\mathrm{mod}\,\, {32}),\end{aligned}$$
(4.1)
$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 7^{4\alpha +2}n+3\cdot 7^{4\alpha +2})q^n&\equiv 16f_2f_{7}+8\psi ^3(q)\quad (\mathrm{mod}\,\, {32}),\nonumber \\ \sum _{n=0}^{\infty }\overline{p}(8\cdot 7^{4\alpha +3}n+5\cdot 7^{4\alpha +3})q^n&\equiv 8q^2\psi ^3(q^7)\quad (\mathrm{mod}\,\, {32}), \end{aligned}$$
(4.2)

where \( f_n:=(q^n;q^n)_\infty .\)

Proof

From (3.5), we have

$$\begin{aligned} \overline{p}(8n+3)\equiv 8a(n)\quad (\mathrm{mod}\,\, {32}). \end{aligned}$$
(4.3)

Recalling the generating function (3.6) of a(n) and the following fact

$$\begin{aligned} f(q,q^6)f(q^2,q^5)f(q^3,q^4)=\frac{f_2f^4_7}{f_1f_{14}}, \end{aligned}$$

it is not hard to see that

$$\begin{aligned} \sum _{n=0}^{\infty }a(7n+4)q^n&\equiv 2\frac{f_2f^4_7}{f_1f_{14}}+q^2\psi ^3(q^7)\quad (\mathrm{mod}\,\, 4)\nonumber \\&\equiv 2f_1f_{14}+q^2\psi ^3(q^7)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$
(4.4)

From [2, p. 303, Entry 17(v)], we have the 7-dissection

$$\begin{aligned} f_1=f_{49}\frac{f(-q^{14},-q^{35})}{f(-q^{7},-q^{42})}-qf_{49}\frac{f(-q^{21},-q^{28})}{f(-q^{14},-q^{35})}\nonumber \\ -q^2f_{49}+q^5f_{49}\frac{f(-q^{7},-q^{42})}{f(-q^{21},-q^{28})}. \end{aligned}$$
(4.5)

Thanks to (4.5), we obtain that

$$\begin{aligned} \sum _{n=0}^{\infty }a(7(7n+2)+4)q^n=\sum _{n=0}^{\infty }a(49n+18)q^n\equiv -2f_7f_2+\psi ^3(q)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$
(4.6)

Then in view of (4.4) and (4.5), it can be seen that

$$\begin{aligned} \sum _{n=0}^{\infty }a(49(7n+4)+18)q^n&=\sum _{n=0}^{\infty }a(7^3n+214)q^n\nonumber \\&\equiv -2f_1f_{14}+2f_1f_{14}+q^2\psi ^3(q^7)\quad (\mathrm{mod}\,\, 4)\nonumber \\&=q^2\psi ^3(q^7), \end{aligned}$$
(4.7)

and

$$\begin{aligned} \sum _{n=0}^{\infty }a(7^3(7n+2)+214)q^n&=\sum _{n=0}^{\infty }a(7^4n+900)q^n\equiv \psi ^3(q)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$
(4.8)

Thus, from (3.6) and (4.8), we see that

$$\begin{aligned} a(7^4n+900)\equiv a(n)\quad (\mathrm{mod}\, 4). \end{aligned}$$
(4.9)

Using (3.6), (4.4), (4.6), (4.7), (4.9) and (4.3), by induction, it is easy to establish the desired results. \(\square \)

We are now in a position to prove Theorem 1.2.

Proof of Theorem 1.2

From (4.1) and (4.5), we deduce that

$$\begin{aligned} \overline{p}(8\cdot 7^{4\alpha +1}(7n+i)+5\cdot 7^{4\alpha +1})\equiv 0\quad (\mathrm{mod}\,\, {2^5}), \end{aligned}$$

where \(i=3,4,6\). In view of (4.2), we obtain

$$\begin{aligned} \overline{p}(8\cdot 7^{4\alpha +3}(7n+i)+5\cdot 7^{4\alpha +3})\equiv 0 \quad (\mathrm{mod}\,\, {2^5}), \end{aligned}$$

for \(i=0,1,3,4,5,6\). This completes the proof. \(\square \)

5 Proof of Theorem 1.3

From Lemma 2.1, we have

$$\begin{aligned} \psi (q)=B_0+B_1, \end{aligned}$$
(5.1)

where \(B_0=f(q^3,q^6),B_1=q\psi (q^9)\).

To prove Theorem 1.3, we need the following three lemmas.

Lemma 5.1

[12, Lemmas 2.1 and 2.2]

$$\begin{aligned}&\frac{1}{\psi (q)}=\frac{\psi (q^9)}{\psi ^4(q^3)}\left( B_0^2-B_0B_1+B_1^2\right) ,\end{aligned}$$
(5.2)
$$\begin{aligned}&B_0^3=\frac{\psi ^4(q^3)}{\psi (q^9)}-q^3\psi ^3(q^9). \end{aligned}$$
(5.3)

Lemma 5.2

Let c(n) be defined by

$$\begin{aligned} \sum _{n=0}^{\infty }c(n)q^n=\psi ^4(q). \end{aligned}$$

Then we have

$$\begin{aligned} \sum _{n=0}^{\infty }c(3n+1)q^{3n+1}\equiv q^4\psi ^4(q^9)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Proof

Using (5.1), we have

$$\begin{aligned} \sum _{n=0}^{\infty }c(n)q^{n}=(B_0+B_1)^4. \end{aligned}$$

Then

$$\begin{aligned} \sum _{n=0}^{\infty }c(3n+1)q^{3n+1}&=B^4_1+4B^3_0B_1\equiv q^4\psi ^4(q^9)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

\(\square \)

Lemma 5.3

Let d(n) be defined by

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

We have

$$\begin{aligned} \sum _{n=0}^{\infty }d(3n+2)q^{3n+2}&\equiv q^8\psi ^8(q^9)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Proof

Since

$$\begin{aligned} \sum _{n=0}^{\infty }d(n)q^{n}=(B_0+B_1)^8, \end{aligned}$$

we have

$$\begin{aligned} \sum _{n=0}^{\infty }d(3n+2)q^{3n+2}&=B^8_1+\frac{8!}{3!5!}B_1^5B_0^3+\frac{8!}{2!6!}B_1^2B_0^6\\&\equiv q^8\psi ^8(q^9)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

\(\square \)

Proof of Theorem 1.3

From (1.4), we see that

$$\begin{aligned} \frac{1}{64}\sum _{n=0}^{\infty }\overline{p}(8n+7)q^n\equiv \psi ^7(q) =(B_0+B_1)^7\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Setting

$$\begin{aligned} \frac{1}{64}\sum _{n=0}^{\infty }\overline{p}(8n+7)q^n=\sum _{n=0}^{\infty }f(n)q^n, \end{aligned}$$
(5.4)

we have

$$\begin{aligned} \sum _{n=0}^{\infty }f(3n+1)q^{3n+1}\equiv 7B_0^6B_1+35B_0^3B_1^4+B_1^7 \quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Applying (5.3), we see that

$$\begin{aligned} \sum _{n=0}^{\infty }f(3n+1)q^{3n+1}\equiv q^4\psi ^3(q^9)\psi ^4(q^3)+q^7\psi ^7(q^9)-q\frac{\psi ^8(q^3)}{\psi (q^9)}\quad (\mathrm{mod}\,\, 4), \end{aligned}$$
(5.5)

and

$$\begin{aligned} \sum _{n=0}^{\infty }f(3n+1)q^{n}\equiv q\psi ^3(q^3)\psi ^4(q)+q^2\psi ^7(q^3)-\frac{\psi ^8(q)}{\psi (q^3)}\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

From Lemmas 5.2 and 5.3, it follows that

$$\begin{aligned} \sum _{n=0}^{\infty }f(3(3n+2)+1)q^{3n+2}&\equiv q\psi ^3(q^3)\cdot q^4\psi ^4(q^9)+q^2\psi ^7(q^3)\\&-\frac{q^8\psi ^8(q^9)}{\psi (q^3)}\quad (\mathrm{mod}\,\, 4), \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^{\infty }f(9n+7)q^{n}\equiv q\psi ^3(q)\psi ^4(q^3)-\frac{q^2\psi ^8(q^3)}{\psi (q)}+\psi ^7(q)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Employing (5.1), (5.5) and Lemma 5.1, it can be seen that

$$\begin{aligned} \sum _{n=0}^{\infty }f(9(3n+1)+7)q^{3n+1}\equiv&q(B^3_0+B^3_1)\psi ^4(q^3)-q^2\psi ^8(q^3)\frac{q^2\psi ^3(q^9)}{\psi ^4(q^3)}\\&+q^4\psi ^3(q^9)\psi ^4(q^3){+}q^7\psi ^7(q^9)-q\frac{\psi ^8(q^3)}{\psi (q^9)}\quad (\mathrm{mod}\,\, 4)\\&=q^7\psi ^7(q^9), \end{aligned}$$

and

$$\begin{aligned} \sum _{n=0}^{\infty }f(27n+16)q^{n}&\equiv q^2\psi ^7(q^3)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$
(5.6)

Therefore,

$$\begin{aligned} \sum _{n=0}^{\infty }f(27(3n+2)+16)q^{n}=\sum _{n=0}^{\infty }f(3^4n+70)q^{n}&\equiv \psi ^7(q)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Thus, we have

$$\begin{aligned} f(n)\equiv f(3^4n+70)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Based on the above relation, by induction, we obtain that for \(\alpha \ge 0\),

$$\begin{aligned} \sum _{n=0}^{\infty }f\left( 3^{4\alpha } n+\frac{7\cdot 3^{4\alpha }-7}{8}\right) q^n\equiv \psi ^7(q)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$

Combining the above relation and (5.6), we find that

$$\begin{aligned} \sum _{n=0}^{\infty }f\left( 3^{4\alpha +3}n+\frac{5\cdot 3^{4\alpha +3}-7}{8}\right) q^n \equiv q^2\psi ^7(q^3)\quad (\mathrm{mod}\,\, 4). \end{aligned}$$
(5.7)

From (5.4) and (5.7), we see that

$$\begin{aligned} \sum _{n=0}^{\infty }\overline{p}(8\cdot 3^{4\alpha +3}n+5\cdot 3^{4\alpha +3})q^n \equiv q^2\psi ^7(q^3)\quad (\mathrm{mod}\,\, {2^8}). \end{aligned}$$
(5.8)

Since there are no terms on the right of (5.8) in which the powers of q are congruent to 0, 1 modulo 3, we have

$$\begin{aligned} \overline{p}(8\cdot 3^{4\alpha +3}(3n)+5\cdot 3^{4\alpha +3})\equiv 0\quad (\mathrm{mod}\,\, {2^8}),\\ \overline{p}(8\cdot 3^{4\alpha +3}(3n+1)+5\cdot 3^{4\alpha +3})\equiv 0\quad (\mathrm{mod}\,\, {2^8}). \end{aligned}$$

This completes the proof. \(\square \)