Arithmetic density and congruences of $t$-core partitions

TL;DR

This paper investigates the divisibility properties of $t$-core partition functions, establishing almost always divisibility by powers of primes and providing algorithms and proofs for various congruences, extending previous results in the field.

Contribution

It generalizes divisibility and congruence results for $t$-core partition functions, including new algorithms and proofs for prime and composite $t$, building on Radu and Seller's approach.

Findings

$a_{3^{eta} m}(n)$ is almost always divisible by powers of 2 and 3.

$a_{t}(n)$ is almost always divisible by powers of primes $p_i \\geq 5$.

Provided algorithms and alternate proofs for congruences modulo 3 and 5.

Abstract

A partition of $n$ is called a $t$-core partition if none of its hook number is divisible by $t.$ In 2019, Hirschhorn and Sellers \cite{Hirs2019} obtained a parity result for $3$-core partition function $a_3(n)$. Recently, both authors \cite{MeherJindal2022} proved density results for $a_3(n)$, wherein we proved that $a_3(n)$ is almost always divisible by arbitrary power of $2$ and $3.$ In this article, we prove that for a non-negative integer $\alpha,$ $a_{3^{\alpha} m}(n)$ is almost always divisible by arbitrary power of $2$ and $3.$ Further, we prove that $a_{t}(n)$ is almost always divisible by arbitrary power of $p_i^j,$ where $j$ is a fixed positive integer and $t= p_1^{a_1}p_2^{a_2}\ldots p_m^{a_m}$ with primes $p_i \geq 5.$ Furthermore, by employing Radu and Seller's approach, we obtain an algorithm and we give alternate proofs of several congruences modulo $3$ and $5$ for $a_{p}(n)$, where $p$ is prime number. Our results also generalizes the results in \cite{radu2011a}.

Equations

$$\sum _{n=0}^{\infty }{a}_t(n)q^n=\prod _{n=1}^{\infty }\frac{(1-q^{nt}{)}^t}{(1-q^n)}=\frac{(q^t;q^t{)}_{\infty }^t}{(q;q{)}_{\infty }}$$

$$a_p(p^{j}n-{\delta }_p)\equiv 0(\mathrm{mod}{p}^j)$$

$$a_{5^j}(5^{j}n-{\delta }_{5,j})$$

$$a_{7^j}(7^{j}n-{\delta }_{7,j})$$

$$a_{11^j}(11^{j}n-{\delta }_{11,j})$$

$$a_3(n)=\{\begin{array}{ll}1(\mathrm{mod}2)& \text{if }n=3m^2+2m\text{ for some integer }m,\\ 0(\mathrm{mod}2)& \text{otherwise.}\end{array}$$

$$\{n\in \mathbb{N}:a_{3^{\alpha }m}(n)\equiv 0(\mathrm{mod}{2}^j)\}$$

$$\{n\in \mathbb{N}:a_{3^{\alpha }m}(n)\equiv 0(\mathrm{mod}{3}^j)\}$$

$$\left\{0<n<x:a_t(n)\equiv 0(\mathrm{mod}{p}_i^j)\right\}$$

$$\underset{X\to \infty }{\lim }\frac{\#\left\{0<n\le X:a_p(n)\equiv 0(\mathrm{mod}{p}^j)\right\}}{X}=1.$$

$$\kappa :=$$

$$\hat{p}:=$$

$$A_t:=$$

$${ϵ}_2:=$$

$${ϵ}_p:=$$

$$P(t):=\left\{t^{{}^{\prime }}:\exists [s{]}_{24m}\text{ such that }{t}^{{}^{\prime }}\equiv ts+\frac{(s-1)(p^2-1)}{24}(\mathrm{mod}m)\right\}$$

$$P(t):=\left\{t^{{}^{\prime }}:\exists [s{]}_{24m}\text{ such that }{t}^{{}^{\prime }}\equiv ts+(s-1)\hat{p}(\mathrm{mod}m)\right\}$$

$$P(t):=\left\{t^{{}^{\prime }}:\exists [s{]}_{48p}\text{ such that }{t}^{{}^{\prime }}\equiv ts+(s-1)\hat{p}(\mathrm{mod}48p)\right\}$$

$$P(t)=\left\{t^{{}^{\prime }}:\left(\frac{24t-1}{24}\right)=\left(\frac{24t^{{}^{\prime }}-1}{24}\right),t\equiv t^{{}^{\prime }}(\mathrm{mod}2),0\le t^{{}^{\prime }}\le 2p-1\right\}.$$

$$P(t):=\left\{t^{{}^{\prime }}:\exists [s{]}_{24pq}\text{ such that }{t}^{{}^{\prime }}\equiv ts+(s-1)\hat{p}(\mathrm{mod}pq)\right\}$$

$$a_5(15n+6,10,12,13)\equiv 0(\mathrm{mod}3),$$

$$a_7(21n+3,8,11,15,17,18)\equiv 0(\mathrm{mod}3),$$

$$a_{11}(33n+3,11,12,20,24,26,27,29,30,32)\equiv 0(\mathrm{mod}3),$$

$$a_{13}(39n+3,7,9,10,15,16,18,22,28,31,33,36)\equiv 0(\mathrm{mod}3),$$

$$a_{17}(51n+10,14,16,19,20,23,25,26,28,34,35,38,41,46,47,49)\equiv 0(\mathrm{mod}3),$$

$$a_{19}(57n+7,14,16,17,19,22,25,26,31,35,37,38,41,44,50,52,55,56)\equiv 0(\mathrm{mod}3),$$

$$a_{23}(69n+3,9,16,22,27,30,31,33,34,36,42,43,46,48,51,52,58,60,61,63,64,64,67)$$

$$\equiv 0(\mathrm{mod}3).$$

$$a_7(35n+4,17,22,24,29,32)\equiv 0(\mathrm{mod}5),$$

$$a_{11}(55n+4,7,9,11,18,21,22,26,29,31,32,33,37,42,43,44,48,51,53,54)$$

$$\equiv 0(\mathrm{mod}5),$$

$$a_{17}(85n+4,7,9,14,24,37,47,52,54,57,64,69,72,74,77,82)\equiv 0(\mathrm{mod}5).$$

$$a_5(8n+3)\equiv 0(\mathrm{mod}3),$$

$$a_5(49n+6,13,20,27,34,41)\equiv 0(\mathrm{mod}3),$$

$$a_7(25n+3,8,13,18)\equiv 0(\mathrm{mod}3),$$

$$a_{11}(44n+9,21,29,33,37)\equiv 0(\mathrm{mod}3),$$

$$a_{13}(12n+3)\equiv 0(\mathrm{mod}3),$$

$$a_{13}(64n+25)\equiv 0(\mathrm{mod}3),$$

$$a_{19}(76n+3,7,19,31,35,55,63,71,75)\equiv 0(\mathrm{mod}3).$$

$$a_5(5n+4)$$

$$a_7(7n+5)$$

$$a_{11}(11n+6)$$

Full text

2010 Mathematics Subject Classification: Primary 05A17, 11P83, Secondary 11F11

Keywords: $t$-core partitions; Eta-quotients; Congruence; modular forms; arithmetic density.

1. Introduction

A partition $\alpha=(\alpha_{1},\alpha_{2},\cdots,\alpha_{s})$ of $n$ is a non-increasing sequence of positive integers whose sum is $n$ and the positive integers $\alpha_{i}$ are called parts of the partitions. A partition $\alpha$ of $n$ can be represented by the Young diagram $[\alpha]$ (also known as the Ferrers graph) which consists of the $s$ number of rows such that the $i^{th}$ row has $\alpha_{i}$ number of dots $\bullet$ and all the rows start from the same column. An illustration of the Young diagram for $\alpha=(\alpha_{1},\alpha_{2},\cdots,\alpha_{s})$ is as follows.

$[\alpha]$:=

$\bullet$ $\bullet$ \cdots$$\cdots$$\cdots   $\bullet$ $\alpha_{1}$ dots

$\bullet$ $\bullet$ \cdots$$\cdots   $\bullet$ $\alpha_{2}$ dots

$\vdots$

⋮

$\bullet$ $\bullet$ $\cdots$   $\bullet$ $\alpha_{s}$ dots

For $1\leq i\leq s$ and $1\leq j\leq\alpha_{i}$, the dot of $[\alpha]$ which lies in the $i^{th}$ row and $j^{th}$ column is denoted by $(i,j)^{th}$-dot of $[\alpha]$. Let $\alpha_{j}^{{}^{\prime}}$ denotes the number of dots in the $j^{th}$ column. The hook number $H_{i,j}$ of $(i,j)^{th}$-dot is defined by $\alpha_{i}+\alpha_{j}^{{}^{\prime}}-i-j+1$. In other words, $H_{i,j}=1+h_{0}$ where $h_{0}$ is the sum of the number of dots lying to the right of the $(i,j)^{th}$-dot in the $i^{th}$ row, and the number of dots lying below the $(i,j)^{th}$-dot in the $j^{th}$ column. Given a partition $\alpha$ of $n$, we say that it is a $t$-core partition if none of its hook numbers is divisible by $t$.

Example. The Young diagram of the partition $\alpha=(5,3,2)$ of $10$ is

[TABLE]

where the superscripts on each dot represents its hook number. It can be easily observed that $\alpha$ is a $t$-core partition of $10$ for $t=5$ and $t\geqslant 8$.

For $n\geq 0$, let $a_{t}(n)$ denote the number of partitions of $n$ that are $t$-core partitions. The generating function for $a_{t}(n)$ is given by

[TABLE]

where $(a;q)_{n}=(1-a)(1-aq)(1-aq^{2})\cdots(1-aq^{n})$ and $(a;q)_{\infty}=\lim\limits_{n\to\infty}(a;q)_{n}$.

In [4, Corollary 1], Garvan, Kim and Stanton proved the congruence

[TABLE]

where $p\in\{5,7,11\}$, $n$ and $j$ are positive integers and $\delta_{p}=\frac{p^{2}-1}{24}$. In [5, Proposition 3], Granville and Ono proved similar congruences, namely

[TABLE]

where $n$ and $j$ are positive integers and $\delta_{p,j}\equiv\frac{1}{24}\pmod{p^{j}}$ for $p\in\{5,7,11\}$.

Recently, Hirschhorn and Sellers [6] proved that, for all $n\geq 0$,

[TABLE]

In [8], the authors proved that the set $\{n\in\mathbb{N}:a_{3}(n)\equiv 0\pmod{p^{j}}\}$ has arithmetic density $1$ for $p\in\{2,3\}$. In this article, we study the arithmetic densities of the partition function $a_{t}(n)$ modulo arbitrary powers of $2$ and $3$ when $t=3^{\alpha}m$, where $\alpha\geq 0$, $m\geq 1$ are integers with $\gcd(m,6)=1$, and modulo arbitrary prime powers $p_{i}^{j}$ when $t=p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{m}^{a_{m}},$ where $p_{i}\geq 5$ are prime numbers. Precisely, we prove the following results.

Theorem 1.1**.**

Let $j\geq 1$, $\alpha\geq 0$ and $m\geq 1$ be integers with $\gcd(m,6)=1$. Then the set

[TABLE]

has arithmetic density $1$.

Theorem 1.2**.**

Let $j\geq 1$, $\alpha\geq 0$ and $m\geq 1$ be integers with $\gcd(m,6)=1$. Then the set

[TABLE]

has arithmetic density $1.$

Theorem 1.3**.**

For a positive integer $m,$ let $a_{1},a_{2},\ldots,a_{m}$ be non negative integers. Let $t=p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{m}^{a_{m}},$ where $p_{i}\geq 5$’s are prime numbers. Then for every positive integer $j,$ the set

[TABLE]

has arithmetic density $1.$

As a consequence of the above theorem, we obtain the following result.

Corollary 1.4**.**

Let $j$ be a positive integer and $p\geq 5$ be a prime number. Then $a_{p}(n)$ is almost always is divisible by $p^{j},$ namely

[TABLE]

In the remaining part of this section, we give an algorithm for $p$-core partitions for primes $p\geq 3$ and we obtain several congruences for $p$-core partitions modulo 3 and modulo 5 for $5\leq p\leq 23$ using this algorithm. We use the technique given by Radu and Sellers [11]. Before proceeding further, we define some notation. For an integer $m$, a prime $p\geqslant 3$ and $t\in\{0,1,\dots,m-1\}$, we set

[TABLE]

We note that \hat{p}\in\mbox{\mathbb{Z}} for $p\geq 5$. Also, it is immediate that A_{t}\in\mbox{\mathbb{Z}} for each $t\in\{0,1,\dots,m-1\}$.

Theorem 1.5**.**

Let $p\geq 3$ be a prime and let $u$ be an integer. For a positive integer $g$, let $e_{1},e_{2},\cdots,e_{g}$ be non-negative integers. Let $m=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{g}^{e_{g}}$ where $p_{i}$’s are prime numbers. Let $t\in\{0,1,\ldots,m-1\}$ such that $A_{t}$ divides $2^{\epsilon_{2}}p^{\epsilon_{p}}p_{1}p_{2}\cdots p_{g}$. Define

[TABLE]

where $[s]_{24m}$ is the residue class of $s$ in $\mathbb{Z}_{24m}$. If the congruence $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $0\leq n\leq\left\lfloor\frac{2^{\epsilon_{2}}(p+1)^{\epsilon_{p}}(p-1)(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)}{24}-\frac{(p^{2}-1)}{24m}\right\rfloor$, then $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $n\geq 0$.

Let $m=pp_{1}p_{2}\cdots p_{g}$ be a square-free integer where $p$ and $p_{i}$’s are prime numbers. Since $p|m$, we have $\epsilon_{p}=0$. If $p=3$, then $3\nmid\kappa$ gives $9|A_{t}$ which implies that $A_{t}\nmid 2^{\epsilon_{2}}pp_{1}p_{2}\cdots p_{g}$ for each $t\in\{0,1,\ldots,m-1\}$. If $p>3$, then $A_{t}|2^{\epsilon_{2}}pp_{1}p_{2}\cdots p_{g}$ for each $t\in\{0,1,\ldots,m-1\}$. Thus we have the following corollary to Theorem 1.5.

Corollary 1.6**.**

Let $p\geq 5$ be a prime and let $u$ be an integer. For a positive integer $g$, let $m=pp_{1}p_{2}\cdots p_{g}$ where $p$ and $p_{i}$’s are distinct prime numbers. Let $t\in\{0,1,\ldots,m-1\}$. Define

[TABLE]

where $[s]_{24m}$ is the residue class of $s$ in $\mathbb{Z}_{24m}$. If the congruence $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $0\leq n\leq\left\lfloor\frac{2^{\epsilon_{2}}(p^{2}-1)(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)}{24}-\frac{(p^{2}-1)}{24m}\right\rfloor\leq 2^{\epsilon_{2}}\hat{p}(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)-1$, then $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $n\geq 0$.

In order to further simplify, we next consider the particular case of a square-free integer $m$ of the type $m=pq$ in Corollary 1.6. When we consider $q=2$, we obtain Corollary 1.7 and when we consider $q\geq 3$, we deduce Corollary 1.8.

Corollary 1.7**.**

Let $p\geq 5$ be a prime and let $u$ be an integer. Let $t\in\{0,1,\ldots,2p-1\}$. Define

[TABLE]

where $[s]_{48p}$ is the residue class of $s$ in $\mathbb{Z}_{48p}$. If the congruence $a_{p}(2pn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $0\leq n\leq 2^{\frac{1-(-1)^{\frac{p-1}{2}}}{2}}3\hat{p}-1$, then $a_{p}(2pn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $n\geq 0$.

In [12, Theorem 1.4], Radu and Sellers obatined several congruences of the form $a_{p}(2pn+t)\equiv 0\pmod{3}$ for $5\leq p\leq 23$ which can also be obtained from the above corollary by performing a finite check for certain initial values of $n$. They used a similar approach as in this paper to obtain their results. In fact, they simplified the set $P(t)$ in case of $m=2p$ as

[TABLE]

Also, they connected there results with broken $k$-diamond partitions. Our result Theorem 1.5 is generalizes their work.

Corollary 1.8**.**

Let $p,q$ be two distinct primes with $p\geq 5$ and $q\geq 3$ and let $u$ be an integer. Let $t\in\{0,1,\ldots,pq-1\}$. Define

[TABLE]

where $[s]_{24pq}$ is the residue class of $s$ in $\mathbb{Z}_{24pq}$. If the congruence $a_{p}(pqn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $0\leq n\leq\frac{(p^{2}-1)(q+1)}{24}=\hat{p}(q+1)$, then $a_{p}(pqn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P(t)$ and $n\geq 0$.

We obtain Theorems 1.9 and 1.10 as applications of Corollary 1.8.

Theorem 1.9**.**

For all $n\geq 0$, we have

[TABLE]

Theorem 1.10**.**

For all $n\geq 0$, we have

[TABLE]

We also obtain a number of congruences for $a_{p}(n)$ modulo $3$ for $p\in\{5,7,11,13,19\}$ in the following result using Theorem 1.5.

Theorem 1.11**.**

For all $n\geq 0$, we have

[TABLE]

Alternate proofs of the congruences obtained in Theorems 1.9, 1.10 and 1.11 can be found in Garvan [3] and Chen [2]. Further Theorem 1.5 can be used to obtain an alternate proof of the congruences

[TABLE]

for $n\geq 0$, which coincide with the congruences (1.2) with $j=1$.

Our paper is of two folds. Firstly, we prove density results for the parity of the partition function $a_{3^{\alpha}m}(n)$ in Theorems 1.1 and 1.2 using modular form techniques and Serre’s result. Secondly, we provide an algorithm using Radu and Seller’s method and we also obtain some algebraic results in Theorems 1.9 and 1.11 which supports our analytical results.

2. Preliminaries

We recall some basic facts and definition on modular forms. For more details, one can see [7], [9]. We start with some matrix groups. We define

[TABLE]

For a positive integer $N$, we define

[TABLE]

and

[TABLE]

A subgroup of $\Gamma=\mathrm{SL_{2}}(\mathbb{Z})$ is called a congruence subgroup if it contains $\Gamma(N)$ for some $N$ and the smallest $N$ with this property is called its level. Note that $\Gamma_{0}(N)$ and $\Gamma_{1}(N)$ are congruence subgroup of level $N,$ whereas $\mathrm{SL_{2}}(\mathbb{Z})$ and $\Gamma_{\infty}$ are congruence subgroups of level $1.$ The index of $\Gamma_{0}(N)$ in $\Gamma$ is

[TABLE]

where $p$ runs over the prime divisors of $N$.

Let $\mathbb{H}$ denote the upper half of the complex plane $\mathbb{C}$. The group

[TABLE]

acts on $\mathbb{H}$ by $\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}z=\frac{az+b}{cz+d}.$ We identify $\infty$ with $\frac{1}{0}$ and define $\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}\frac{r}{s}=\frac{ar+bs}{cr+ds},$ where $\frac{r}{s}\in\mathbb{Q}\cup\{\infty\}$. This gives an action of $\mathrm{GL_{2}^{+}}(\mathbb{R})$ on the extended half plane $\mathbb{H}^{*}=\mathbb{H}\cup\mathbb{Q}\cup\{\infty\}$. Suppose that $\Gamma$ is a congruence subgroup of $\mathrm{SL_{2}}(\mathbb{Z})$. A cusp of $\Gamma$ is an equivalence class in $\mathbb{P}^{1}=\mathbb{Q}\cup\{\infty\}$ under the action of $\Gamma$.

The group $\mathrm{GL_{2}^{+}}(\mathbb{R})$ also acts on functions $g:\mathbb{H}\rightarrow\mathbb{C}$. In particular, suppose that $\gamma=\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}\in\mathrm{GL_{2}^{+}}(\mathbb{R})$. If $g(z)$ is a meromorphic function on $\mathbb{H}$ and $k$ is an integer, then define the slash operator $|_{k}$ by

[TABLE]

Definition 2.1**.**

Let $\Gamma$ be a congruence subgroup of level $N$. A holomorphic function $g:\mathbb{H}\rightarrow\mathbb{C}$ is called a modular form with integer weight $k$ on $\Gamma$ if the following hold:

1. $(1)$

We have

[TABLE]

for all $z\in\mathbb{H}$ and $\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}\in\Gamma$. 2. $(2)$

If $\gamma\in SL_{2}(\mathbb{Z})$, then $(g|_{k}\gamma)(z)$ has a Fourier expnasion of the form

[TABLE]

where $q_{N}:=e^{2\pi iz/N}$.

For a positive integer $k$, the complex vector space of modular forms of weight $k$ with respect to a congruence subgroup $\Gamma$ is denoted by $M_{k}(\Gamma)$.

Definition 2.2**.**

[9, Definition 1.15]** If $\chi$ is a Dirichlet character modulo $N$, then we say that a modular form $g\in M_{k}(\Gamma_{1}(N))$ has Nebentypus character $\chi$ if

[TABLE]

for all $z\in\mathbb{H}$ and $\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}\in\Gamma_{0}(N)$. The space of such modular forms is denoted by $M_{k}(\Gamma_{0}(N),\chi)$.

The relevant modular forms for the results obtained in this article arise from eta-quotients. Recall that the Dedekind eta-function $\eta(z)$ is defined by

[TABLE]

where $q:=e^{2\pi iz}$ and $z\in\mathbb{H}$. A function $g(z)$ is called an eta-quotient if it is of the form

[TABLE]

where $N$ and $r_{\delta}$ are integers with $N>0$.

Theorem 2.3**.**

[9, Theorem 1.64]** If $g(z)=\prod\limits_{\delta|N}\eta(\delta z)^{r_{\delta}}$ is an eta-quotient such that $k=\frac{1}{2}$ $\sum_{\delta|N}r_{\delta}\in\mathbb{Z}$,

[TABLE]

then $g(z)$ satisfies

[TABLE]

for each $\begin{bmatrix}a&&b\\ c&&d\end{bmatrix}\in\Gamma_{0}(N)$. Here the character $\chi$ is defined by $\chi(d):=\left(\frac{(-1)^{k}s}{d}\right)$ where $s=\prod_{\delta|N}\delta^{r_{\delta}}$.

Theorem 2.4**.**

[9, Theorem 1.65]** Let $c,d$ and $N$ be positive integers with $d|N$ and $\gcd(c,d)=1$. If $f$ is an eta-quotient satisfying the conditions of Theorem 2.3 for $N$, then the order of vanishing of $f(z)$ at the cusp $\frac{c}{d}$ is

[TABLE]

Suppose that $g(z)$ is an eta-quotient satisfying the conditions of Theorem 2.3 and that the associated weight $k$ is a positive integer. If $g(z)$ is holomorphic at all of the cusps of $\Gamma_{0}(N)$, then $g(z)\in M_{k}(\Gamma_{0}(N),\chi)$. Theorem 2.4 gives the necessary criterion for determining orders of an eta-quotient at cusps. In the proofs of our results, we use Theorems 2.3 and 2.4 to prove that $g(z)\in M_{k}(\Gamma_{0}(N),\chi)$ for certain eta-quotients $g(z)$ we consider in the sequel.

We shall now mention a result of Serre [14, P. 43] which will be used later.

Theorem 2.5**.**

Let $g(z)\in M_{k}(\Gamma_{0}(N),\chi)$ has Fourier expansion

[TABLE]

Then for a positive integer $r$, there is a constant $\alpha>0$ such that

[TABLE]

Equivalently

[TABLE]

We finally recall the definition of Hecke operators and a few relavent results. Let $m$ be a positive integer and $g(z)=\sum\limits_{n=0}^{\infty}a(n)q^{n}\in M_{k}(\Gamma_{0}(N),\chi)$. Then the action of Hecke operator $T_{m}$ on $f(z)$ is defined by

[TABLE]

In particular, if $m=p$ is a prime, we have

[TABLE]

We note that $a(n)=0$ unless $n$ is a non-negative integer.

3. Proof of Theorem 1.1

We put $t=3^{\alpha}m$ in $\eqref{1e1}$ to obtain

[TABLE]

We define

[TABLE]

Note that for any prime $p$ and positive integer $j,$ we have

[TABLE]

Using the above formula, we get

[TABLE]

We define

[TABLE]

Using (3.3), we have

[TABLE]

From (3.1) and (3.5), we obtain

[TABLE]

Next, we prove that $B_{\alpha,m,j}(z)$ is a modular form. Applying Theorem 2.3, we first estimate the level of eta quotient $B_{\alpha,m,j}(z)$ . The level of $B_{\alpha,m,j}(z)$ is $N=2^{4}3^{\alpha+1}mM,$ where $M$ is the smallest positive integer which satisfies

[TABLE]

Therefore $M=4$ and the level of $B_{\alpha,m,j}(z)$ is $N=2^{6}3^{\alpha+1}m$. The cusps of $\Gamma_{0}(2^{6}3^{\alpha+1}m)$ are given by fractions $\frac{c}{d}$ where $d|2^{6}3^{\alpha+1}m$ and $\gcd(c,d)=1.$ By using Theorem 2.4, we have that $B_{\alpha,m,j}$ is holomorphic at a cusp $\frac{c}{d}$ if and only if

[TABLE]

where $G_{1}=\frac{\gcd^{2}(d,2^{3}3^{\alpha+1}m)}{\gcd^{2}(d,2^{4}3^{\alpha+1}m)}$ and $G_{2}=\frac{\gcd^{2}(d,24)}{\gcd^{2}(d,2^{4}3^{\alpha+1}m)}.$ Let $d$ be a divisor of $2^{6}3^{\alpha+1}m$. We can write $d=2^{r_{1}}3^{r_{2}}t$ where $0\leq r_{1}\leq 6$, $0\leq r_{2}\leq\alpha+1$ and $t|m$. We now consider the following two cases depending on $r_{1}$.

Case 1: Let $0\leq r_{1}\leq 3,$ $0\leq r_{2}\leq\alpha+1$. Then $G_{1}=1$ and $\frac{1}{3^{2\alpha}t^{2}}\leq G_{2}\leq 1$ which implies $L\geq 2\left(3^{\alpha}m+2^{j+1}\right)-2\cdot 3^{\alpha}m-2^{j}=3\cdot 2^{j}>0.$

Case 2: Let $4\leq r_{1}\leq 6$, $0\leq r_{2}\leq\alpha+1$. Then $G_{1}=\frac{1}{4}$ and $\frac{1}{43^{2\alpha}t^{2}}\leq G_{2}\leq\frac{1}{4}$. This gives $L=2\left(3^{\alpha}m+2^{j+1}\right)G_{1}-2\cdot 3^{\alpha}mG_{2}-2^{j}\geq\frac{3^{\alpha}m}{2}+2^{j}-\frac{3^{\alpha}m}{2}-2^{j}=0.$

Therefore, $B_{\alpha,m,j}(z)$ is holomorphic at every cusp $\frac{c}{d}.$ Using Theorem 2.3, we estimate that the weight of $B_{\alpha,m,j}(z)$ is $k=\frac{3^{\alpha}m-1}{2}+2^{j-1}$ which is a positive integer. The associated character for $B_{\alpha,m,j}(z)$ is

[TABLE]

Thus, $B_{\alpha,m,j}(z)\in M_{k}(\Gamma_{0}(N),\chi)$ where $k$, $N$ and $\chi$ are as above. Applying Theorem 2.5, we obtain that the Fourier coefficients of $B_{\alpha,m,j}(z)$ satisfies (2.1) which implies that the Fourier coefficient of $B_{\alpha,m,j}(z)$ are almost always divisible by $r=2^{j}$. Hence, from $\eqref{3e6}$, we conclude that $a_{3^{\alpha}m}(n)$ are almost always divisible by $2^{j}$. This completes the proof of Theorem 1.1. ∎

4. Proof of Theorem 1.2

We follow the same approach as in the proof of Theorem 1.1. Here we define

[TABLE]

and then using (3.2) for $p=3$, we get

[TABLE]

Next we consider the eta-quotient

[TABLE]

From (1.1), we obtain

[TABLE]

Therefore, from the above discussion we conclude that

[TABLE]

Hence, to prove Theorem 1.2, it is enough to prove that the Fourier coefficients of $D_{\alpha,m,j}(z)$ are almost always divisible by $r=3^{j}$. We first prove that $D_{\alpha,m,j}(z)$ is a modular form. Using Theorem 2.3, we find that the level of eta quotient $D_{\alpha,m,j}(z)$ is equal to $N=2^{3}3^{\alpha+2}mM,$ where $M$ is the smallest positive integer which satisfies

[TABLE]

Therefore $M=1$ and the level of $D_{\alpha,m,j}(z)$ is $N=2^{3}3^{\alpha+2}m.$ The cusps of $\Gamma_{0}(2^{3}3^{\alpha+2}m)$ are given by fractions $\frac{c}{d}$ where $d|2^{3}3^{\alpha+2}m$ and $\gcd(c,d)=1.$ By using Theorem 2.4, we have that $B_{\alpha,m,j}$ is holomorphic at a cusp $\frac{c}{d}$ if and only if

[TABLE]

where $G_{1}=\frac{\gcd^{2}(d,2^{3}3^{\alpha+1}m)}{\gcd^{2}(d,2^{3}3^{\alpha+2}m)}$ and $G_{2}=\frac{\gcd^{2}(d,24)}{\gcd^{2}(d,2^{3}3^{\alpha+2}m)}$. Let $d$ be a divisor of $2^{3}3^{\alpha+2}m$. We write $d=2^{r_{1}}3^{r_{2}}t$ where $0\leq r_{1}\leq 3,$ $0\leq r_{2}\leq\alpha+2$ and $t|m$. We now consider the following two cases depending on $r_{2}$.

Case 1: Let $0\leq r_{1}\leq 3,$ $0\leq r_{2}\leq\alpha+1$. Then $G_{1}=1$ and $\frac{1}{3^{2\alpha}t^{2}}\leq G_{2}\leq 1$. Therefore $L=3^{\alpha+1}m+3^{j+2}-3^{\alpha+1}mG_{2}-3^{j}\geq 3^{\alpha+1}m+3^{j+2}-3^{\alpha+1}m-3^{j}=2\cdot 3^{j}>0$.

Case 2: Let $0\leq r_{1}\leq 3,$ ${r_{2}}=\alpha+2$. Then $G_{1}=\frac{1}{9},$ $\frac{1}{3^{2\alpha+2}t^{2}}\leq G_{2}\leq\frac{1}{9}.$ Hence, we have $L=(3^{\alpha+1}m+3^{j+2})G_{1}-3^{\alpha+1}mG_{2}-3^{j}\geq 3^{\alpha-1}m+3^{j}-3^{\alpha-1}m-3^{j}=0.$

This proves that $D_{\alpha,m,j}(z)$ is holomorphic at every cusp $\frac{c}{d}$. Applying Theorem 2.3, we obtain that the weight of $D_{\alpha,m,j}(z)$ is $k=\frac{3^{\alpha}m-1}{2}+3^{j}$ which is a positive integer. The associated character for $D_{\alpha,m,j}(z)$ is

[TABLE]

Thus, $D_{\alpha,m,j}(z)\in M_{k}(\Gamma_{0}(N),\chi)$ where $k$, $N$ and $\chi$ are as above. Applying a deep result of Serre (Theorem 2.5), we obtain that the Fourier coefficients of $D_{\alpha,m,j}(z)$ satisfies (2.1). Hence the Fourier coefficients of $D_{\alpha,m,j}(z)$ are almost always divisible by $r=3^{j}$. This completes the proof. ∎

5. Proof of Theorem 1.3

Let $t=p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{m}^{a_{m}},$ where $p_{i}$’s are primes. From (1.1), we get

[TABLE]

Note that for any prime $p$ and positive integers $j,k$ we have

[TABLE]

For a positive integer $i,$ we define

[TABLE]

Using (5.2), we get

[TABLE]

Define

[TABLE]

On modulo $p_{i}^{j+1},$ we get

[TABLE]

Combining (5.1) and (5.3) together, we obtain

[TABLE]

Next, we prove that $B_{i,j,t}(z)$ is a modular form. Applying Theorem 2.3, we first estimate the level of eta quotient $B_{i,j,\ell}(z)$ . The level of $B_{i,j,t}(z)$ is $N=24p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{m}^{a_{m}}M,$ where $M$ is the smallest positive integer which satisfies

[TABLE]

Therefore $M=24$ and the level of $B_{i,j,t}(z)$ is $N=2^{6}3^{2}t$. The cusps of $\Gamma_{0}(2^{6}3^{2}t)$ are given by fractions $\frac{c}{d}$ where $d|2^{6}3^{2}t$ and $\gcd(c,d)=1.$ By using Theorem 2.4, we have that $B_{i,j,t}(z)$ is holomorphic at a cusp $\frac{c}{d}$ if and only if

[TABLE]

where $G_{1}=\frac{\gcd^{2}(d,24)}{\gcd^{2}(d,24t)}$ and $G_{2}=\frac{\gcd^{2}(d,24p_{i}^{a_{i}})}{\gcd^{2}(d,24t)}.$ Let $d$ be a divisor of $2^{6}3^{2}t$. We can write $d=2^{r_{1}}3^{r_{2}}p_{i}^{s}\ell$ where $0\leq r_{1}\leq 6$, $0\leq r_{2}\leq 2$, $0\leq s\leq a_{i}$ and $\ell|t$ but $p_{i}\nmid\ell$. It is immediate that $G_{1}=\frac{1}{p_{i}^{2s}\ell^{2}}$ and $G_{2}=\frac{1}{\ell^{2}}$. Therefore we have

[TABLE]

Note that $s\leq a_{i}$ implies $p^{2s-a_{i}}\leq p^{a_{i}}$. Thus $p^{j}(p_{i}^{a_{i}}-p_{i}^{2s-a_{i}})\geq 0$. Hence

[TABLE]

Therefore, $B_{i,j,t}(z)$ is holomorphic at every cusp $\frac{c}{d}.$ Using Theorem 2.3, we compute the weight of $B_{i,j,t}(z)$ is $k=\frac{t+\left(p_{i}^{a_{i}+j}-1\right)-p_{i}^{j}}{2}=\frac{t+p_{i}^{j}\left(p_{i}^{a_{i}}-1\right)-1}{2}$ which is a positive integer. The associated character for $B_{i,j,t}(z)$ is

[TABLE]

Thus $B_{i,j,t}(z)\in M_{k}(\Gamma_{0}(N),\chi)$ where $k$, $N$ and $\chi$ are as above. Applying Theorem 2.3, we obtain that the Fourier coefficients of $B_{i,j,t}(z)$ satisfies (2.1) which implies that the Fourier coefficient of $B_{i,j,t}(z)$ are almost always divisible by $p_{i}^{j}$. Hence, from $\eqref{eq704}$, we conclude that $a_{t}(n)$ are almost always divisible by $p_{i}^{j}$. This completes the proof of Theorem 1.3.

6. Proof of Theorems 1.5, 1.9, 1.10 and 1.11

6.1. An algorithmic approach by Radu and Sellers

We begin with recalling an algorithm developed by Radu and Sellers [11] that will be used to prove Theorem 1.5. Let $M$ be a positive integer and let $R(M)$ denote the set of integers sequences $r=(r_{\delta})_{\delta|M}$ indexed by the positive divisors of $M$. For $r\in R(M)$ and the positive divisors $1=\delta_{1}<\delta_{2}<\cdots<\delta_{i_{M}}=M$ of $M$, we set $r=(r_{\delta_{1}},r_{\delta_{2}},\dots,r_{\delta_{i_{M}}})$. We define $c_{r}(n)$ by

[TABLE]

Radu and Sellers [11] approach to prove congruences for $c_{r}(n)$ modulo a positive integer reduced the number of cases that we need to check as compared with the classical method which uses Sturm’s bound alone.

Let $m\geq 0$ and $s$ be integers. We denote by $[s]_{m}$ the residue class of $s$ in $\mathbb{Z}_{m}$ and we denote by $\mathbb{S}_{m}$ the set of squares in $\mathbb{Z}_{m}^{*}$. For $t\in\{0,1,\dots,m-1\}$ and $r\in R(M)$, the subset $P_{m,r}(t)\subseteq\{0,1,\dots,m-1\}$ is defined as

[TABLE]

Definition 6.1**.**

For positive integers $m$, $M$ and $N$, let $r=(r_{\delta})\in R(M)$ and $t\in\{0,1,\dots,m-1\}$. Let $\kappa=\kappa(m):=\gcd(m^{2}-1,24)$ and write

[TABLE]

where $s$ and $j$ are non-negative integers with $j$ odd. The set $\Delta^{*}$ is the collection of all tuples $(m,M,N,(r_{\delta}),t)$ satisfying the following conditions.

• (a)

Every prime divisor of $m$ is also a divisor of $N$.

• (b)

If $\delta|M,$ then $\delta|mN$ for every $\delta\geqslant 1$ such that $r_{\delta}\neq 0$.

• (c)

$\kappa N\sum\limits_{\delta|M}r_{\delta}mN/\delta\equiv 0\pmod{24}$.

• (d)

$\kappa N\sum\limits_{\delta|M}r_{\delta}\equiv 0\pmod{8}$.

• (e)

$\frac{24m}{\gcd(-24\kappa t-\kappa\sum\limits_{\delta|M}\delta r_{\delta},24m)}$divides $N$.

• (f)

If $2|m$, then either ($4|\kappa N$* and $8|sN$) or ($2|s$ and $8|(1-j)N$).*

For positive integers $m$, $M$ and $N$, $\gamma=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\in\Gamma$, $r\in R(M)$ and $a\in R(N)$, we define

[TABLE]

and

[TABLE]

The following lemma is given by Radu [10, Lemma $4.5$].

Lemma 6.2**.**

Let $u$ be a positive integer, $(m,M,N,(r_{\delta}),t)\in\Delta^{*}$ and $a=(a_{\delta})\in R(N)$. Let $\{\gamma_{1},\gamma_{2},\dots,\gamma_{n}\}\subseteq\Gamma$ denote a complete set of representatives of the double cosets of $\Gamma_{0}(N)\backslash\Gamma/\Gamma_{\infty}$. Assume that $p_{m,r}(\gamma_{i})+p^{*}_{a}(\gamma_{i})\geq 0$ for all $1\leq i\leq n$. Let $t_{\min}=\min_{t^{{}^{\prime}}\in P_{m,r}(t)}t^{{}^{\prime}}$ and

[TABLE]

If the congruence $c_{r}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P_{m,r}(t)$ and $0\leq n\leq\lfloor\nu\rfloor$, then $c_{r}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P_{m,r}(t)$ and $n\geq 0$.

The next lemma is given by Wang [16, Lemma $4.3$]. This result gives the complete set of representatives of the double cosets in $\Gamma_{0}(N)\backslash\Gamma/\Gamma_{\infty}$ when $N$ or $\frac{N}{2}$ is a square-free integer.

Lemma 6.3**.**

If $N$ or $\frac{N}{2}$ is a square-free integer, then

[TABLE]

6.2. Proof of Theorem 1.5

For an integer $m$, a prime $p\geqslant 3$ and $t\in\{0,1,\dots,m-1\}$, we recall that

[TABLE]

We now prove the following three results specific to the proof of Theorem 1.5.

Lemma 6.4**.**

Let $p\geq 3$ be a prime number. For a positive integer $g$, let $e_{1},e_{2},\ldots,e_{g}$ be non-negative integers and let $p_{1},p_{2},\ldots,p_{g}$ be prime numbers. Let

[TABLE]

where $t\in\{0,1,\ldots,m-1\}$ is such that $A_{t}|N$. Then $(m,M,N,r,t)\in\Delta^{*}$.

Proof.

We first note that

[TABLE]

Therefore

[TABLE]

It is immediate that the conditions (a) and (b) in the definition of $\Delta^{*}$ are satisfied. Since $M=p$ and $r=(r_{1}=-1,r_{p}=p)$, we see that $\sum\limits_{\delta|M}r_{\delta}mN/\delta=r_{1}mN+r_{p}mN/p=0$ and therefore (c) is also satisfied. Next, we note that $\kappa N\sum\limits_{\delta|M}r_{\delta}=\kappa N(p-1)$ and $N(p-1)\equiv 0\pmod{4}$. We have $\kappa N(p-1)\equiv 0\pmod{8}$ if $\gcd(m,6)=1$ or $3$. Further, if $\gcd(m,6)=2$ or $6$, then $2|m$ gives $2^{1+\epsilon_{2}}|N$ and so it follows that $N(p-1)\equiv 0\pmod{8}$. Thus (d) holds. From the conditions on $t$, we see that (e) is also satisfied. For (f), we see that $\prod\limits_{\delta|M}\delta^{|r_{\delta}|}=p^{p}$ gives $s=0$ and $j=p^{p}$. As observed earlier in the justification for (c), we have $N(p-1)\equiv 0\pmod{8}$ if $2|m$. Thus if $2|m$, then $(1-p)|(1-j)$ implies $8|(1-j)N$. This completes the proof. ∎

Lemma 6.5**.**

Let $p\geq 3$ be a prime number. For a positive integer $m$, $M=p$, $r=(r_{1}=-1,r_{p}=p)$ and $\gamma=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\in\Gamma$, we have

[TABLE]

Proof.

We note that

[TABLE]

Since \gamma\in\Gamma=SL_{2}(\mbox{\mathbb{Z}}), we have $ad-bc=1$ which implies that $\gcd(a,c)=1$. Thus it follows that $\gcd(a+\kappa\lambda c,c)=1$. Therefore it is enough to prove that

[TABLE]

for each $\lambda\in\{0,1,\dots,m-1\}$.

Let $\lambda\in\{0,1,\dots,m-1\}$ be fixed. We consider the two cases $p|c$ and $p\nmid c$ separately.

Case 1: $p|c$. We set $c_{p}=\frac{c}{p}$. We observe that $\gcd(a,c)=1$ implies that $\gcd(a,c_{p})=1$ which in turn implies $\gcd(a+\kappa\lambda c,c_{p})=\gcd(a+\kappa\lambda pc_{p},c_{p})=1$. Hence we have

[TABLE]

Case 2: $p\nmid c$. In this case, we have

[TABLE]

If $p\nmid m$, then $\gcd(p(a+\kappa\lambda c),m)=\gcd(a+\kappa\lambda c,m)$ and thus $G=0$. We now assume that $p|m$. We set m_{p}=\frac{m}{p}\in\mbox{\mathbb{Z}}. Then

[TABLE]

Let $d=\gcd(a+\kappa\lambda c,m)$. Let ord${}_{p}(n)$ denote the highest exponent of $p$ dividing a positive integer $n$. It is clear that ord${}_{p}(d)\leq$ord${}_{p}(m)=$ord${}_{p}(m_{p})+1$. If ord${}_{p}(d)\leq$ord${}_{p}(m_{p})$, then $G=d^{2}(p^{2}-1)>0$. If ord${}_{p}(d)=$ord${}_{p}(m_{p})+1$, then $\gcd(a+\kappa\lambda c,m_{p})=\frac{d}{p}$ and therefore $G=p^{2}(\frac{d}{p})^{2}-d^{2}=0$. This completes the proof. ∎

Lemma 6.6**.**

Let $p\geq 3$ be a prime and let $u$ be an integer. Let $(m,M,N,r,t)$ be as defined in Lemma 6.4. Let $t_{\min}=\min_{t^{{}^{\prime}}\in P_{m,r}(t)}t^{{}^{\prime}}$. If the congruence $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P_{m,r}(t)$ and $0\leq n\leq\left\lfloor\frac{2^{\epsilon_{2}}(p+1)^{\epsilon_{p}}(p-1)(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)}{24}\right\rfloor$, then $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{u}$ holds for all $t^{{}^{\prime}}\in P_{m,r}(t)$ and $n\geq 0$.

Proof.

It is enough to show that the assumptions of Lemma 6.2 are satisfied and that the upper bound in Lemma 6.2 is less than or equal to $\left\lfloor\frac{2^{\epsilon_{2}}(p+1)^{\epsilon_{p}}(p-1)(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)}{24}\right\rfloor$. For $\delta|N$, we set $\gamma_{\delta}=\begin{bmatrix}1&0\\ \delta&1\end{bmatrix}$. Since $\epsilon_{2}=0$ or $1$, $N$ or $\frac{N}{2}$ is a square-free integer. Thus Lemma 6.3 implies that $\{\gamma_{\delta}:\delta|N\}$ forms a complete set of double coset representatives of $\Gamma_{0}(N)\backslash\Gamma/\Gamma_{\infty}$. Lemma 6.5 implies that $p_{m,r}(\gamma_{\delta})\geq 0$ for each $\delta|N$. Therefore we take $a_{\delta}=0$ for each $\delta|N$, that is, $a=(0,0,\ldots,0)\in R(N)$ and hence $p_{m,r}(\gamma_{\delta})+p_{a}^{*}(\gamma_{\delta})\geq 0$ for each $\delta|N$. Since $t_{\min}\geq 0$, we have

[TABLE]

∎

Proof of Theorem 1.5.

We get from (1.1) that

[TABLE]

Let $(m,M,N,r,t)=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{g}^{e_{g}},p,2^{\epsilon_{2}}p^{\epsilon_{p}}p_{1}p_{2}\cdots p_{g},r=(r_{1}=-1,r_{p}=p),t)$ be such that $t\in\{0,1,\ldots,m-1\}$ and $A_{t}|N$. Then Lemma 6.4 implies $(m,M,N,r,t)\in\Delta^{*}$. Since $\sum_{\delta|M}\delta r_{\delta}=p^{2}-1$, we see that $P_{m,r}(t)=P(t)$. Now the assertion follows from Lemma 6.6. ∎

6.3. Proof of Theorem 1.9

We begin with the proof of the first congruence. We take $p=5$, $m=3\cdot 5=15$ and $t=6$. Using Sage [13], we find that $P(6)=\{6,10,12,13\}$. We compute that the upper bound in Corollary 1.8 is less than $4\hat{p}=4$. Using Mathematica, we verify that $a_{5}(15n+t)\equiv 0\pmod{3}$ for $t\in\{6,10,12,13\}$ and $0\leq n<4$. Now the first congruence follows immediately from Corollary 1.8.

A similar approach can be used to prove the other congruences. In particular, we set $B_{p}=4\hat{p}$ and we put the values of $p$, $m$, $t$, $P(t)$ and $B_{p}$ in the following table.

[TABLE]

For each $p$ and $m$ given in the above table, we verify, using Mathematica, that $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{3}$ for $t^{{}^{\prime}}\in P(t)$ and $0\leq n<B_{p}$ for respective values of $t$ and $B_{p}$. Now the congruences follows immediately from Corollary 1.8. This completes the proof.

6.4. Proof of Theorem 1.10

We start by proving the first congruence. We take $p=7$, $m=5\cdot 7=35$ and $t=4$. Using Sage [13], we find that $P(4)=\{4,17,22,24,29,32\}$. We compute that the upper bound in Corollary 1.8 is less than $6\hat{p}=12$. Using Mathematica, we verify that $a_{7}(35n+t)\equiv 0\pmod{5}$ for $t\in\{4,17,22,24,29,32\}$ and $0\leq n<12$. Now the first congruence follows immediately from Corollary 1.8.

A similar approach can be used to prove the other congruences. In particular, we set $B_{p}=6\hat{p}$ and we put the values of $p$, $m$, $t$, $P(t)$ and $B_{p}$ in the following table.

[TABLE]

For each $p$ and $m$ given in the above table, we verify, using Mathematica, that $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{5}$ for $t^{{}^{\prime}}\in P(t)$ and $0\leq n<B_{p}$ for respective values of $t$ and $B_{p}$. Now the congruences follows immediately from Corollary 1.8. This completes the proof.

6.5. Proof of Theorem 1.11

We first prove (1.4). We take $p=5$, $m=7^{2}$ and $t\in\{6,20\}$. Using Sage [13], we find that $P(6)=\{6,13,27\}$ and $P(20)=\{20,34,41\}$. We compute that the upper bound in Theorem 1.5 is less than $\lfloor\frac{6\cdot 4\cdot 8}{24}\rfloor=8$. Using Mathematica, we verify that $a_{5}(49n+t)\equiv 0\pmod{3}$ for $t\in\{6,13,20,27,34,41\}$ and $0\leq n<8$. Now (1.4) follows immediately from Theorem 1.5.

A similar method can be used to prove (1.3) and (1.5)-(1.9). In particular, we set $B_{p,m}=\left\lfloor\frac{2^{\epsilon_{2}}(p+1)^{\epsilon_{p}}(p-1)(p_{1}+1)(p_{2}+1)\cdots(p_{g}+1)}{24}-\frac{(p^{2}-1)}{24m}\right\rfloor$ and we put the values of $p$, $m$, $t$, $P(t)$ and $B_{p,m}$ in the following table.

[TABLE]

For each $p$ and $m$ given in the above table, we verify, using Mathematica, that $a_{p}(mn+t^{{}^{\prime}})\equiv 0\pmod{3}$ for $t^{{}^{\prime}}\in P(t)$ and $0\leq n\leq B_{p,m}$ for respective values of $t$ and $B_{p,m}$. Now the congruences follows immediately from Corollary 1.8. This completes the proof.

7. Conclusion

By Radu and Seller’s method, for a prime $p\geq 3$, we obtain an algorithm in Theorem 1.5 for congruences of the type $a_{p}(mn+t)\equiv 0\pmod{u}$ when $m=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{g}^{e_{g}}$ where $p_{i}$ are prime numbers and $e_{1},e_{2},\cdots,e_{g}$ are non-negative integers. Since we have obtained the density results for divisibility of $a_{3^{\alpha}m}(n)$ by powers of $2$ and $3$, and divisibility of $a_{t}(n)$, where $t=p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{m}^{a_{m}}$ with $p_{i}\geq 5$, by prime powers of $p_{i}^{j}$ in Theorems 1.1, 1.2 and 1.3, it will be interesting to see more algebraic results in the same direction.