Families of Ramanujan-Type Congruences Modulo 4 for the Number of Divisors

Abstract

In this paper, we explore Ramanujan-type congruences modulo 4 for the function ${\sigma }_0\left(n\right)$, counting the positive divisors of n. We consider relations of the form ${\sigma }_0\left(8(\alpha n+\beta )+r\right)\equiv 0(mod4),$ with $(\alpha ,\beta )\in {\mathbb{N}}^2$ and $r\in \{1,3,5,7\}$. In this context, some conjectures are made and some Ramanujan-type congruences involving overpartitions are obtained.

1. Introduction

Recall [1] that an overpartition of the positive integer n is an ordinary partition of n where the first occurrence of parts of each size may be overlined. Let $\overline{p}\left(n\right)$ denote the number of overpartitions of n. For example, the overpartitions of the integer 3 are:

$$3,\overline{3},2+1,\overline{2}+1,2+\overline{1},\overline{2}+\overline{1},1+1+1\mathrm{and}\overline{1}+1+1.$$

We see that $\overline{p}\left(3\right)=8$. It is well-known that the generating function of $\overline{p}\left(n\right)$ is given by

$$\sum _{n=0}^{\infty }\overline{p}\left(n\right)q^n=\frac{{(-q;q)}_{\infty }}{{(q,q)}_{\infty }}.$$

Here and throughout this paper, we use the following customary q-series notation:

$$\begin{array}{cc}\hfill & {(a;q)}_n=\left\{\begin{array}{cc}1,\hfill & \mathrm{for}n=0,\hfill \\ (1-a)(1-aq)\cdots (1-a{q}^{n-1}),\hfill & \mathrm{for}n>0;\hfill \end{array}\right.\hfill \\ \hfill & {(a;q)}_{\infty }=\underset{n\to \infty }{lim}{(a;q)}_n.\hfill \end{array}$$

Many congruences for the number of overpartitions have been discovered in the recent years by authors such as Chen [2], Chen, Hou, Sun and Zhang [3], Chern and Dastidar [4], Dou and Lin [5], Fortin, Jacob and Mathieu [6], Hirschhorn and Sellers [7], Kim [8,9], Lovejoy and Osburn [10], Mahlburg [11], Xia [12], Xiong [13] and Yao and Xia [14].

Fortin, Jacob and Mathieu [6] founded in 2003 the first Ramanujan-type congruences modulo power of 2 for $\overline{p}\left(n\right)$ and for all n that cannot be written as a sum of s or less squares, they obtained that

$$\overline{p}\left(n\right)\equiv 0(mod{2}^{s+1}).$$

(1)

This result is meaningful only for $s<4$ since, by Lagrange’s four-square theorem, all numbers can be written as a sum of four squares. A complete characterization of Ramanujan-type congruences modulo 16 for the overpartition function $\overline{p}\left(n\right)$ was provided in 2019 using the function ${\sigma }_0\left(n\right)$ that counts the positive divisors of n [15]. By the proofs of Theorems 1.3 and 1.4 in [15], we easily deduce the following result.

Theorem 1.

Let $r\in \{3,5\}$ be a fixed integer. For all $n⩾0$, we have

$$\overline{p}(8n+r)\equiv 0(mod16)\iff {\sigma }_0(8n+r)\equiv 0(mod4).$$

In this paper, apart from $\overline{p}\left(n\right)$, we consider the overpartition function $\overline{p_o}\left(n\right)$ that counts the overpartitions of n into odd parts. The generating function for the number of overpartitions into odd parts is given by

$$\sum _{n=0}^{\infty }\overline{p_o}\left(n\right)q^n=\frac{{(-q;q^2)}_{\infty }}{{(q;q^2)}_{\infty }}.$$

(2)

The expression of the generating function for $\overline{p_o}\left(n\right)$ was first used by Lebesgue [16] in 1840 in the following series-product identity

$$\sum _{n=0}^{\infty }\frac{{(-1;q)}_n{q}^{n(n+1)/2}}{{(q;q)}_n}=\frac{{(-q;q^2)}_{\infty }}{{(q;q^2)}_{\infty }}.$$

Although authors such as Bessenrodt [17], Santos and Sills [18] utilized more recently the generating function (2) for $\overline{p_o}\left(n\right)$, none of them connected their works to overpartitions into odd parts.

Many congruences for the number of overpartitions into odd parts have been discovered lately [19,20]. It appears that the first Ramanujan-type congruences modulo power of 2 for $\overline{p_o}\left(n\right)$ was found in 2006 by Hirschhorn and Sellers [20]. Very recently, Theorem 1 in [21], we introduced a complete characterization of Ramanujan-type congruences modulo 8 for the overpartition function $\overline{p_o}\left(n\right)$ considering again the divisor function ${\sigma }_0\left(n\right)$. By the proof of Theorem 1 in [21], we easily deduce the following result.

Theorem 2.

Let $r\in \{1,3\}$ be a fixed integer. For all $n⩾0$, we have

$$\overline{p_o}(8n+r)\equiv 0(mod8)\iff {\sigma }_0(8n+r)\equiv 0(mod4).$$

Theorems 1 and 2 may be viewed as steps towards classifying all Ramanujan-type congruences for overpartitions, particularly because the divisibility properties of multiplicative functions are more directly accessible with elementary methods than those of functions defined in terms of partitions. Recall that a multiplicative function is an arithmetic function $f\left(n\right)$ of a positive integer n with the property that $f\left(1\right)$ = 1 and $f\left(ab\right)=f\left(a\right)f\left(b\right)$ whenever a and b are coprime.

In this paper, motivated by Theorems 1 and 2, we consider $r\in \{1,3,5,7\}$ to be a fixed integer and investigate pairs $(\alpha ,\beta )$ of positive integers for which the following statement is true:

$$\mathrm{For}\mathrm{all}n⩾0,{\sigma }_0\left(8(\alpha n+\beta )+r\right)\equiv 0(mod4).$$

(3)

There is a substantial amount of numerical evidence to conjecture the following.

Conjecture 1.

If the statement (3) is true, then there is an odd prime p such that α is divisible by $p^2$ and $8\beta +r$ is divisible by p.

Since a multiplicative function is defined by its values at prime powers, this conjecture boils down to understanding how the divisibility properties of the divisor function ${\sigma }_0\left(n\right)$ at prime powers intersect with arithmetic progressions.

If the statement (3) is true for $(\alpha ,\beta )$, then the statement (3) is true for any pair $(k\alpha ,b\alpha +\beta )$, with $k\in \mathbb{N}$ and $b\in \{0,1,\dots ,k-1\}$. To prove this fact, it is enough to replace n by $kn+b$ in (3). This makes us not very attracted to cases where $\alpha$ is not a square of an odd prime.

Definition 1.

For each odd prime p, we define ${\mathcal{B}}_{r,p}$ to be the set of nonnegative integers $\beta <p^2$ such that

$${\sigma }_0\left(8(p^{2}n+\beta )+r\right)\equiv 0(mod4),$$

for all nonnegative integers n.

Assuming Conjecture 1, we state the following.

Conjecture 2.

For each odd prime p, we have

$$|{\mathcal{B}}_{1,p}|=\left\{\begin{array}{cc}p-1,\hfill & \mathit{if}p-1\mathit{is}\mathit{cubefree},\hfill \\ (p-1)/2,\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Conjecture 3.

Let $r\in \{3,5,7\}$ be a fixed integer. For each odd prime p, we have

$$|{\mathcal{B}}_{r,p}|=\left\{\begin{array}{cc}(p-1)/2,\hfill & \mathit{if}p\equiv r(mod8),\hfill \\ p-1,\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Conjecture 4.

Let $r\in \{1,3,5,7\}$ be a fixed integer. Then,

$$\bigcup _{p\mathit{odd}\mathit{prime}}{\mathcal{B}}_{r,p}=\{n\in \mathbb{N}:{\sigma }_0(8n+r)\equiv 0(mod4)\}\backslash \left\{\begin{array}{cc}\left\{3\right\},\hfill & \mathrm{if}r=3,\hfill \\ \varnothing ,\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Assuming the last conjecture, we remark that there is not an odd prime p such that

$${\sigma }_0(8p^{2}n+27)\equiv 0(mod4),$$

for all nonnegative integers n.

In this paper, we consider some special cases of our conjectures and present a strategy for proving them. These special cases together with our Theorems 1 and 2 allow us to easily obtain some Ramanujan-type congruences for the overpartition functions $\overline{p}\left(n\right)$ and $\overline{p_o}\left(n\right)$. Somewhat unrelated to our topics, we will show that these congruences are precursors of stronger congruences. In fact, these stronger congruences were discovered considering few Ramanujan-type congruences modulo 4 for the divisor function ${\sigma }_0\left(n\right)$.

2. Some Special Cases

This section is devoted to the presentation of the proof strategy of some special cases of Conjectures 2 and 3 listed bellow. We will rely on the fact that the divisor function ${\sigma }_0\left(n\right)$ is a multiplicative function.

Theorem 3.

(i) 

${\mathcal{B}}_{1,3}=\{4,7\}$;

(ii) 

${\mathcal{B}}_{1,5}=\{8,13,18,23\}$.

Theorem 4.

(i) 

${\mathcal{B}}_{3,3}=\left\{6\right\}$;

(ii) 

${\mathcal{B}}_{3,5}=\{4,14,19,24\}$.

Theorem 5.

(i) 

${\mathcal{B}}_{5,3}=\{2,8\}$;

(ii) 

${\mathcal{B}}_{5,5}=\{10,20\}$.

To proof these identities, the following steps have to be performed.

Step 1. The first step in all our proofs is to verify that for each $\beta \in {\mathcal{B}}_{r,p}$, $(8\beta +r)/p\in \mathbb{N}.$

Step 2. For each $\beta \in {\mathcal{B}}_{r,p}$, we prove that $gcd\left(p,8pn+(8\beta +r)/p\right)=1$, for all $n⩾0$.

Step 3. For each $\beta \in {\mathcal{B}}_{r,p}$, we prove that $8pn+(8\beta +r)/p$ is not a square, for all $n⩾0$. Thus, for each $\beta \in {\mathcal{B}}_{r,p}$, we deduce that

$${\sigma }_0(8p^{2}n+8\beta +r)={\sigma }_0\left(p\right){\sigma }_0\left(8pn+\frac{8\beta +r}{p}\right)\equiv 0(mod4).$$

Step 4. For each $\beta \in \{0,1,2,\dots ,p^2-1\}\backslash {\mathcal{B}}_{r,p}$, we show that there is an integer n such that

$${\sigma }_0(8p^{2}n+8\beta +r)\not\equiv 0(mod4).$$

Now, we provide full details for the proofs of Theorems 3–5.

Proof of Theorem 3.

 

(i).

Step 1. We have $(8\times 4+1)/3=11$ and $(8\times 7+1)/3=19$.

Step 2. For all $n⩾0$, it is clear that $gcd(3,24n+11)=1$ and $gcd(3,24n+19)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $24n+11$ is a square. Thus, we deduce that $24n+11={(2k+1)}^2$ or $12n+5=2k^2+2k$. This identity is not possible, because $12n+5$ is odd and $2k^2+2k$ is even. It is clear that $24n+11$ cannot be a square. Similarly, it can be proved that $24n+19$ is not a square. For all $n⩾0$, we deduce that

$${\sigma }_0\left(8(9n+4)+1\right)={\sigma }_0(72n+33)={\sigma }_0\left(3\right){\sigma }_0(24n+11)\equiv 0(mod4)$$

and

$${\sigma }_0\left(8(9n+7)+1\right)={\sigma }_0(72n+57)={\sigma }_0\left(3\right){\sigma }_0(24n+19)\equiv 0(mod4).$$

Step 4. Considering that

$$\begin{array}{cc}\hfill & {\sigma }_0\left(8(9\times 1+0)+1\right)\equiv {\sigma }_0\left(8(9\times 2+1)+1\right)\equiv {\sigma }_0\left(8(9\times 0+2)+1\right)\hfill \\ \hfill & \equiv {\sigma }_0\left(8(9\times 1+3)+1\right)\equiv {\sigma }_0\left(8(9\times 0+5)+1\right)\equiv {\sigma }_0\left(8(9\times 2+6)+1\right)\hfill \\ \hfill & \equiv {\sigma }_0\left(8(9\times 1+8)+1\right)\equiv 2(mod4),\hfill \end{array}$$

the proof is finished.

(ii).

Step 1. We have $(8\times 8+1)/5=13$, $(8\times 13+1)/5=21$, $(8\times 18+1)/5=29$ and $(8\times 23+1)/5=37$.

Step 2. For all $n⩾0$, it is clear that $gcd(5,40n+13)=1$, $gcd(5,40n+21)=1$, $gcd(5,40n+29)=1$ and $gcd(5,40n+37)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $40n+13$ is a square. Thus, we deduce that $40n+13={(2k+1)}^2$ or $10n+3=k^2+k$. This identity is not possible, because $10n+3$ is odd and $k^2+k$ is even. It is clear that $40n+13$ cannot be a square. Similarly, it can be proved that $40n+21$, $40n+29$ and $40n+37$ are not squares. For $\beta \in {\mathcal{B}}_{1,5}$ and $n⩾0$, we deduce that

$${\sigma }_0(200n+8\beta +1)={\sigma }_0\left(5\right){\sigma }_0\left(40n+\frac{8\beta +1}{5}\right)\equiv 0(mod4).$$

Step 4. For $\beta \in \{0,1,\dots ,24\}\backslash \left\{{\mathcal{B}}_{1,5}\cup \{4,7,16,20,22\}\right\}$, it is not difficult to check that ${\sigma }_0\left(8(25\times 0+\beta )+1\right)$ is not congruent to 0 mod 4. In addition, for $\beta \in \{4,7,20\}$, we have ${\sigma }_0\left(8(25\times 1+\beta )+1\right)\not\equiv 0(mod4)$. For $\beta \in \{16,22\}$, we see that ${\sigma }_0\left(8(25\times 2+\beta )+1\right)$ is not congruent to 0 mod 4. The proof is finished.    □

Proof of Theorem 4.

 

(i).

Step 1. We have $(8\times 6+3)/3=17$.

Step 2. For all $n⩾0$, it is clear that $gcd(3,24n+17)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $24n+17$ is a square. Thus, we deduce that $24n+17={(2k+1)}^2$ or $3n+2=k(k+1)/2$. On the other hand,

$$\frac{k(k+1)}{2}\equiv \left\{\begin{array}{cc}1(mod3),\hfill & \mathrm{if}k\equiv 1(mod3)\hfill \\ 0(mod3),\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

It is clear that $24n+17$ cannot be a square. For all $n⩾0$, we deduce that

$${\sigma }_0\left(8(9n+6)+3\right)={\sigma }_0(72n+51)={\sigma }_0\left(3\right){\sigma }_0(24n+17)\equiv 0(mod4).$$

Step 4. Taking into account that

$$\begin{array}{cc}\hfill & {\sigma }_0\left(8(9\times 0+0)+3\right)\equiv {\sigma }_0\left(8(9\times 0+1)+3\right)\equiv {\sigma }_0\left(8(9\times 0+2)+3\right)\hfill \\ \hfill & \equiv {\sigma }_0\left(8(9\times 1+3)+3\right)\equiv {\sigma }_0\left(8(9\times 1+4)+3\right)\equiv {\sigma }_0\left(8(9\times 0+5)+3\right)\hfill \\ \hfill & \equiv {\sigma }_0\left(8(9\times 0+7)+3\right)\equiv {\sigma }_0\left(8(9\times 0+8)+3\right)\equiv 2(mod4),\hfill \end{array}$$

the proof is finished.

(ii).

Step 1. We have $(8\times 4+3)/5=7$, $(8\times 14+3)/5=23$, $(8\times 19+3)/5=31$ and $(8\times 24+4)/5=39$.

Step 2. For all $n⩾0$, it is clear that $gcd(5,40n+7)=1$, $gcd(5,40n+23)=1$, $gcd(5,40n+31)=1$ and $gcd(5,40n+39)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $40n+7$ is a square. Thus, we deduce that $40n+7={(2k+1)}^2$ or $20n+3=2k^2+2k$. This identity is not possible, because $20n+3$ is odd and $2k^2+2k$ is even. It is clear that $20n+3$ cannot be a square. Similarly, it can be proved that $40n+23$, $40n+31$ and $40n+39$ are not squares. For $\beta \in {\mathcal{B}}_{3,5}$ and $n⩾0$, we deduce that

$${\sigma }_0(200n+8\beta +3)={\sigma }_0\left(5\right){\sigma }_0\left(40n+\frac{8\beta +3}{5}\right)\equiv 0(mod4).$$

Step 4. For $\beta \in \{0,1,\dots ,24\}\backslash \left\{{\mathcal{B}}_{3,5}\cup \{3,6,11,15,23\}\right\}$, it is not difficult to check that ${\sigma }_0\left(8(25\times 0+\beta )+3\right)$ is not congruent to 0 mod 4. In addition, for $\beta \in \{3,6,23\}$, we have ${\sigma }_0\left(8(25\times 1+\beta )+3\right)\not\equiv 0(mod4)$. For $\beta \in \{11,15\}$, we see that ${\sigma }_0\left(8(25\times 2+\beta )+3\right)$ is not congruent to 0 mod 4. The proof is finished.    □

Proof of Theorem 5.

 

(i).

Step 1. We have $(8\times 2+5)/3=7$ and $(8\times 8+5)/3=23$.

Step 2. For all $n⩾0$, it is clear that $gcd(3,24n+7)=1$ and $gcd(3,24n+23)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $24n+7$ is a square. Thus, we deduce that $24n+7={(2k+1)}^2$ or $12n+3=2k^2+2k$. This identity is not possible, because $12n+3$ is odd and $2k^2+2k$ is even. It is clear that $24n+7$ cannot be a square. Similarly, it can be proved that $24n+23$ is not a square. For all $n⩾0$, we deduce that

$${\sigma }_0\left(8(9n+2)+5\right)={\sigma }_0(72n+21)={\sigma }_0\left(3\right){\sigma }_0(24n+7)\equiv 0(mod4)$$

and

$${\sigma }_0\left(8(9n+8)+5\right)={\sigma }_0(72n+69)={\sigma }_0\left(3\right){\sigma }_0(24n+23)\equiv 0(mod4).$$

Step 4. For $\beta \in \{0,1,\dots ,8\}\backslash {\mathcal{B}}_{5,3}$, it is not difficult to check that ${\sigma }_0\left(8(9\times 0+\beta )+5\right)$ is congruent to 2 mod 4. The proof is finished.

(ii).

Step 1. We have $(8·10+5)/5=17$ and $(8·20+5)/5=33$.

Step 2. For all $n⩾0$, it is clear that $gcd(5,40n+17)=1$ and $gcd(5,40n+33)=1$.

Step 3. We suppose that there is an integer $n⩾0$ such that $40n+17$ is a square. Thus, we deduce that $40n+17={(2k+1)}^2$ or $5n+2=k(k+1)/2$. On the other hand,

$$\frac{k(k+1)}{2}\equiv \left\{\begin{array}{cc}3(mod5),\hfill & \mathrm{if}k\equiv 2(mod5)\hfill \\ 1(mod5),\hfill & \mathrm{if}k\equiv \{1,3\}(mod5)\hfill \\ 0(mod5),\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

It is clear that $40n+17$ cannot be a square. Similarly, we suppose that there is an integer $n⩾0$ such that $40n+33$ is a square. Thus, we deduce that $40n+33={(2k+1)}^2$ or $5n+4=k(k+1)/2$. Because $k(k+1)/2\not\equiv 4mod5$, this identity is not possible. For $\beta \in {\mathcal{B}}_{5,5}$ and $n⩾0$, we deduce that

$${\sigma }_0(200n+8\beta +5)={\sigma }_0\left(5\right){\sigma }_0\left(40n+\frac{8\beta +5}{5}\right)\equiv 0(mod4).$$

Step 4. For $\beta \in \{0,1,\dots ,24\}\backslash \left\{{\mathcal{B}}_{5,5}\cup \{2,8,9,11,15,16,17,23\}\right\}$, it is not difficult to check that ${\sigma }_0\left(8(25\times 0+\beta )+5\right)$ is congruent to 2 mod 4. In addition, for $\beta \in \{8,9,11,15,16,23\}$, we have ${\sigma }_0\left(8(25\times 1+\beta )+5\right)\equiv 2(mod4)$. For $\beta \in \{2,17\}$, we see that ${\sigma }_0\left(8(25\times 2+\beta )+5\right)$ is congruent to 2 mod 4. The proof is finished.    □

It seems that the approach outlined in Steps 1, 2 and 4 can be easily automated. Unfortunately, we cannot say the same about Step 3 because we do not have a criterion which establishes the parity of $(8\beta +r)/p$. Is the number $(8\beta +r)/p$ always odd? When $(8\beta +r)/p$ is an odd number, we need to investigate identities of the form

$$8pn+\frac{8\beta +r}{p}-1=4k(k+1).$$

When $(8\beta +r)/p$ is an even number, we need to investigate identities of the form

$$8pn+\frac{8\beta +r}{p}=4k^2.$$

Can the investigation of these identities be automated? We do not have an answer to this question yet.

3. Some Ramanujan-Type Congruences

Let $a\left(n\right)$ be a sequence of integers defined by

$$\begin{array}{c}\hfill \sum _{n=0}^{\infty }a\left(n\right)q^n=\prod _{\delta |M}{(q^{\delta };q^{\delta })}_{\infty }^{r_{\delta }},\end{array}$$

(4)

where M is a positive integer and $r_{\delta }$ are integers. Based on the ideas of Rademacher [22], Newman [23,24] and Kolberg [25], Radu [26] developed in 2009 an algorithm to verify the congruences

$$a(mn+t)\equiv 0(modu),$$

for any given m, t and u, and for all $n⩾0$.

In 2015, Radu [27] constructed an algorithm, called the Ramanujan–Kolberg algorithm, to derive identities on the generating functions of $a(mn+t)$ using modular functions for ${\Gamma }_0\left(N\right)$. A description of the Ramanujan–Kolberg algorithm can be found in Paule and Radu [28]. Recently, Smoot [29] provided a successful Mathematica implementation of Radu’s algorithm. This package is called RaduRK.

In this section, we use the RaduRK package to obtain some Ramanujan-type congruences for the overpartition functions $\overline{p}\left(n\right)$ and $\overline{p_o}\left(n\right)$. According to Theorems 2 and 3, we can write the following result.

Corollary 1.

For $n\equiv \{4,7\}(mod9)$ or $n\equiv \{8,13,18,23\}(mod25)$, we have

$$\overline{p_o}(8n+1)\equiv 0(mod8).$$

Upon reflection, one expects that there might be a stronger result.

Theorem 6.

(i) 

For all $n\equiv \{4,7\}(mod9)$, we have

$$\overline{p_o}(8n+1)\equiv 0(mod24).$$

(ii) 

For all $n\equiv \{8,13,18,23\}(mod25)$, we have

$$\overline{p_o}(8n+1)\equiv 0(mod32).$$

Proof. 

The generating function for $\overline{p_o}\left(n\right)$ can be written as

$$\frac{{(q^2;q^2)}_{\infty }^3}{{(q;q)}_{\infty }^2{(q^4;q^4)}_{\infty }}.$$

This can be described by setting $M=4$ and $r_1=-2$, $r_2=3$, $r_4=-1$.

(i) Considering the RaduRK program with

$$\mathtt{RK}[\mathtt{12},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{72},\mathtt{33}]$$

and

$$\mathtt{RK}[\mathtt{12},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{72},\mathtt{57}],$$

we deduce that

$$\sum _{n=0}^{\infty }\overline{p_o}(72n+33)q^n\equiv 0(mod24)$$

and

$$\sum _{n=0}^{\infty }\overline{p_o}(72n+57)q^n\equiv 0(mod24).$$

(ii) To obtain the second congruence identity, we consider the RaduRK program with

$$\mathtt{RK}[\mathtt{2},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{200},\mathtt{65}]$$

and

$$\mathtt{RK}[\mathtt{2},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{200},\mathtt{105}].$$

We deduce that

$$\begin{array}{cc}\hfill & \left(\sum _{n=0}^{\infty }\overline{p_o}(200n+65)q^n\right)\left(\sum _{n=0}^{\infty }\overline{p_o}(200n+185)q^n\right)\equiv 0(mod{2}^{10})\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \left(\sum _{n=0}^{\infty }\overline{p_o}(200n+105)q^n\right)\left(\sum _{n=0}^{\infty }\overline{p_o}(200n+145)q^n\right)\equiv 0(mod{2}^{10}).\hfill \end{array}$$

Having

$$\begin{array}{cc}\hfill & \overline{p_o}\left(65\right)=2^5\times 16851,\hfill \\ \hfill & \overline{p_o}(200+105)=2^5\times 6293025198351,\hfill \\ \hfill & \overline{p_o}\left(145\right)=2^5\times 64201703,\hfill \\ \hfill & \overline{p_o}\left(185\right)=2^5\times 1713260289,\hfill \end{array}$$

for $\alpha \in \{65,105,145,185\}$, we notice that

$$\sum _{n=0}^{\infty }\overline{p_o}(200n+\alpha )q^n\not\equiv 0(mod{2}^6)$$

and

$$\sum _{n=0}^{\infty }\overline{p_o}(200n+\alpha )q^n\equiv 0(mod{2}^5).$$

This concludes the proof.   □

According to Theorems 1, 2 and 4, we can write the following result.

Corollary 2.

For $n\equiv 6(mod9)$ or $n\equiv \{4,14,19,24\}(mod25)$, we have

$$\overline{p}(8n+3)\equiv 0(mod16)\mathrm{and}\overline{p_o}(8n+3)\equiv 0(mod8).$$

There are stronger results.

Theorem 7.

(i) 

For all $n\equiv 6(mod9)$, we have

$$\overline{p_o}(8n+3)\equiv 0(mod24).$$

(ii) 

For all $n\equiv \{4,14,19,24\}(mod25)$, we have

$$\overline{p_o}(8n+3)\equiv 0(mod64).$$

Proof. 

(i) To obtain the first congruence identity, we consider the RaduRK program with

$$\mathtt{RK}[\mathtt{4},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{72},\mathtt{51}]$$

and obtain

$$\sum _{n=0}^{\infty }\overline{p_o}(72n+51)q^n\equiv 0(mod24).$$

(ii) To obtain the second congruence identity, we consider again the RaduRK program with

$$\mathtt{RK}[\mathtt{2},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{200},\mathtt{35}]$$

and

$$\mathtt{RK}[\mathtt{2},\mathtt{4},\{-\mathtt{2},\mathtt{3},-\mathtt{1}\},\mathtt{200},\mathtt{155}].$$

These give us

$$\begin{array}{cc}\hfill & \left(\sum _{n=0}^{\infty }\overline{p_o}(200n+35)q^n\right)\left(\sum _{n=0}^{\infty }\overline{p_o}(200n+115)q^n\right)\equiv 0(mod{2}^{12})\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \left(\sum _{n=0}^{\infty }\overline{p_o}(200n+155)q^n\right)\left(\sum _{n=0}^{\infty }\overline{p_o}(200n+195)q^n\right)\equiv 0(mod{2}^{12}).\hfill \end{array}$$

Having

$$\begin{array}{cc}\hfill & \overline{p_o}\left(35\right)=2^6\times 113,\hfill \\ \hfill & \overline{p_o}\left(115\right)=2^6\times 2041219,\hfill \\ \hfill & \overline{p_o}(200+155)=2^6\times 59890735496633,\hfill \\ \hfill & \overline{p_o}\left(195\right)=2^6\times 1844065971,\hfill \end{array}$$

for $\alpha \in \{35,115,155,195\}$, we notice that

$$\sum _{n=0}^{\infty }\overline{p_o}(200n+\alpha )q^n\not\equiv 0(mod{2}^7)$$

and

$$\sum _{n=0}^{\infty }\overline{p_o}(200n+\alpha )q^n\equiv 0(mod{2}^6).$$

This concludes the proof.    □

Theorem 8.

For all $n\equiv \{19,24\}(mod25)$, we have

$$\overline{p}(8n+3)\equiv 0(mod160).$$

Proof. 

To obtain this congruence identity, we consider the RaduRK program with

$$\mathtt{RK}[\mathtt{2},\mathtt{2},\{-\mathtt{2},\mathtt{1}\},\mathtt{200},\mathtt{155}].$$

This gives us

$$\left(\sum _{n=0}^{\infty }\overline{p}(200n+155)q^n\right)\left(\sum _{n=0}^{\infty }\overline{p}(200n+195)q^n\right)\equiv 0(mod25600).$$

Having

$$\begin{array}{cc}\hfill & 25600=2^{10}\times 5^2,\hfill \\ \hfill & \overline{p}\left(155\right)=2^5\times 5\times 3^2\times 13\times 1693\times 2402791,\hfill \\ \hfill & \overline{p}\left(195\right)=2^5\times 5\times 3\times 6091\times 2417744023,\hfill \end{array}$$

for $\alpha \in \{155,195\}$, we notice that

$$\sum _{n=0}^{\infty }\overline{p}(200n+\alpha )q^n\not\equiv 0(mod{2}^6)$$

and

$$\sum _{n=0}^{\infty }\overline{p}(200n+\alpha )q^n\not\equiv 0(mod{5}^2).$$

Thus, for $\alpha \in \{155,195\}$, we deduce that

$$\sum _{n=0}^{\infty }\overline{p}(200n+\alpha )q^n\equiv 0(mod{2}^5·5).$$

This concludes the proof.    □

According to Theorems 1 and 5, we can write the following result.

Corollary 3.

For $n\equiv \{2,8\}(mod9)$ or $n\equiv \{10,20\}(mod25)$, we have

$$\overline{p}(8n+5)\equiv 0(mod16).$$

There are stronger results.

Theorem 9.

For all $n\equiv 8(mod9)$, we have

$$\overline{p}(8n+5)\equiv 0(mod32).$$

Proof. 

To obtain this congruence identity, we consider the RaduRK program with

$$\mathtt{RK}[\mathtt{2},\mathtt{2},\{-\mathtt{2},\mathtt{1}\},\mathtt{72},\mathtt{69}].$$

This gives us

$$\sum _{n=0}^{\infty }\overline{p}(72n+69)q^n\equiv 0(mod32).$$

□

4. Open Problems and Concluding Remarks

In this paper, we show that each odd prime generates four families of Ramanujan-type congruences modulo 4 for the number of divisors. Assuming Conjecture 1, the algorithm for generating ${\mathcal{B}}_{r,p}$ is not difficult because $8\beta +r$ must be a multiple of the odd prime p. Related to the case $r=1$ of Conjecture 4, we remark that there is a substantial amount of numerical evidence to conjecture the following.

Conjecture 5.

If n is an integer that is not the difference between a triangular number and a square number, then

$${\sigma }_0(8n+1)\equiv 0(mod4).$$

We focused on the cases $(\alpha ,\beta )$, where $\alpha$ is the square of an odd prime. When $\alpha$ is a multiple of the square of an odd prime, we can derive other pairs $({\alpha }^{\prime },{\beta }^{\prime })$ for which the statement (3) is true. For example, considering ${\mathcal{B}}_{1,3}=\{4,7\}$, we easily deduce that the statement (3) is true if

$$\begin{array}{cc}\hfill (\alpha ,\beta )\in \{& (81,4),(81,7),(81,13),(81,16),(81,22),(81,25),\hfill \\ \hfill & (81,31),(81,34),(81,40),(81,43),(81,49),(81,52),\hfill \\ \hfill & (81,58),(81,61),(81,67),(81,70),(81,76),(81,79)\}.\hfill \end{array}$$

We remark that there are two pairs, $(81,37)$ and $(81,64)$, which cannot be derived from the pairs $(9,4)$ or $(9,7)$. In addition, we remark that

$${\sigma }_0\left(8(81n+37)+1\right)={\sigma }_0\left(27(24n+11)\right)\equiv 0(mod8)$$

and

$${\sigma }_0\left(8(81n+64)+1\right)={\sigma }_0\left(27(24n+19)\right)\equiv 0(mod8),$$

for all $n⩾0$. The proof of these congruences follows easily if we consider that

$$gcd(27,24n+11)\mathrm{and}gcd(27,24n+19)=1,$$

for all $n⩾0$. Moreover, $24n+11$ and $24n+19$ cannot be squares.

The study of congruences of the form

$${\sigma }_0(8n+r)\equiv 0(mod{2}^k),$$

with $r\in \{1,3,5,7\}$, can be a very appealing topic. In analogy with (3), we can consider the following statement:

$$\mathrm{For}\mathrm{all}n⩾0,{\sigma }_0\left(8(\alpha n+\beta )+r\right)\equiv 0(mod{2}^k).$$

(5)

There is a substantial amount of numerical evidence to state the following generalization of Conjecture 1.

Conjecture 6.

If the statement (5) is true, then there is a sequence of odd prime numbers, $p_1⩽p_2⩽\dots ⩽p_{k-1}$, such that α is divisible by ${(p_1{p}_2\cdots p_{k-1})}^2$ and $8\beta +r$ is divisible by $p_1{p}_2\cdots p_{k-1}$.

On the other hand, our investigations indicate that Conjecture 6 can be generalized if we consider congruences of the form

$${\sigma }_0(\alpha n+\beta )\equiv 0(mod{2}^k).$$

In analogy with (5), we can consider the following statement:

$$\mathrm{For}\mathrm{all}n⩾0,{\sigma }_0(\alpha n+\beta )\equiv 0(mod{2}^k).$$

(6)

We state the following generalization of Conjecture 6.

Conjecture 7.

If the statement (6) is true, then there is a sequence of prime numbers, $p_1⩽p_2⩽\dots ⩽p_{k-1}$, such that α is divisible by ${(p_1{p}_2\cdots p_{k-1})}^2$ and β is divisible by $p_1{p}_2\cdots p_{k-1}$.

Because ${\sigma }_0\left(n\right)$ is a multiplicative function, these conjectures motivate the question of identifying all Ramanujan-type congruences for multiplicative functions.