DIVISIBILITY OF THE PARTITION FUNCTION $\text {PDO}_t(n)$ BY POWERS OF $2$ AND $3$

Abstract

Lin introduced the partition function $\text {PDO}_t(n)$, which counts the total number of tagged parts over all the partitions of n with designated summands in which all parts are odd. Lin also proved some congruences modulo 3 and 9 for $\text {PDO}_t(n)$, and conjectured certain congruences modulo $3^{k+2}$ for $k\geq 0$. He proved the conjecture for $k=0$ and $k=1$ [‘The number of tagged parts over the partitions with designated summands’, J. Number Theory 184 (2018), 216–234]. We prove the conjecture for $k=2$. We also study the lacunarity of $\text {PDO}_t(n)$ modulo arbitrary powers of 2 and 3. Using nilpotency of Hecke operators, we prove that there exists an infinite family of congruences modulo any power of 2 satisfied by $\text {PDO}_t(n)$.

1 Introduction and statement of results

A partition of a positive integer n is a nonincreasing sequence of positive integers, called parts, whose sum is n. In [1], Andrews, Lewis and Lovejoy investigated the partition function $\text {PD}(n)$ which counts the number of partitions of n with designated summands. A partition of n with designated summands is obtained from an ordinary partition of n by tagging exactly one of each part size. For example, $\text {PD}(4)=10$ with the relevant partitions being $4', 3'+1', 2'+2, 2+2', 2'+1'+1, 2'+1+1', 1' +1+1+1, 1+1' +1+1, 1+1+1' +1, 1+1+1+1'$ . They also studied another partition function $\text {PDO}(n)$ which counts the number of partitions of n with designated summands in which all parts are odd. From the above example, $\text {PDO}(4)=5$ . Later, many authors have studied these two partition functions (see for example [2–4, 17]).

Recently, Lin [9] introduced two new partition functions $\text {PD}_t(n)$ and $\text {PDO}_t(n)$ related to partitions with designated summands. The partition function $\text {PD}_t(n)$ counts the total number of tagged parts over all the partitions of n with designated summands. For instance, $\text {PD}_t(4)=13$ . The other partition function $\text {PDO}_t(n)$ counts the total number of tagged parts over all the partitions of n with designated summands in which all parts are odd. For example, $\text {PDO}_t(4)=6$ . Lin found the generating functions of $\text {PD}_t(n)$ and $\text {PDO}_t(n)$ . The generating function of $\text {PDO}_t(n)$ is given by

(1.1) $$ \begin{align} G(q):=\sum_{n=0}^{\infty}\text{PDO}_t(n)q^n=\frac{qf_2f_3^2f_{12}^2}{f_1^2f_6}, \end{align} $$

where $f_k=\displaystyle (q^k; q^k)_{\infty }$ and $(a; q)_{\infty }:= \prod _{j=0}^{\infty }(1-aq^j)$ .

Lin also established many congruences modulo small powers of 3 satisfied by $\text {PD}_t(n)$ and $\text {PDO}_t(n)$ . For example, he proved the following Ramanujan-type congruences modulo $9$ and $27$ satisfied by $\text {PDO}_t(n)$ : for $n\geq 0$ ,

$$ \begin{align*} \text{PDO}_t(8n)&\equiv \text{PDO}_t(12n)\equiv \text{PDO}_t(12n+8)\equiv 0 \pmod{9},\\ \text{PDO}_t(24n)&\equiv \text{PDO}_t(36n)\equiv \text{PDO}_t(36n+24)\equiv 0 \pmod{27}. \end{align*} $$

He further conjectured the following congruences.

Conjecture 1.1 [9, Conjecture 6.1]

For $k,n\geq 0$ ,

$$ \begin{align*} \mathrm{PDO}_t(8\cdot 3^k n)\equiv 0 \pmod{3^{k+2}}, \\ \mathrm{PDO}_t(12\cdot 3^k n)\equiv 0\pmod{3^{k+2}}. \end{align*} $$

Lin proved Conjecture 1.1 for $k=0,1$ using basic q-series techniques. We prove the following theorem which establishes Conjecture 1.1 for $k=2$ .

Theorem 1.2. For all $n\geq 0$ ,

(1.2) $$ \begin{align} \kern5pt\mathrm{PDO}_t(72n)&\equiv 0\pmod{81}, \end{align} $$

(1.3) $$ \begin{align} \mathrm{PDO}_t(108n)&\equiv 0\pmod{81}. \end{align} $$

In addition to the study of Ramanujan-type congruences, it is an interesting problem to study the distribution of the partition function modulo positive integers M. To be precise, given an integral power series $F(q):=\sum _{n=0}^{\infty }a(n)q^n$ and $0\leq r<M$ , we define

$$ \begin{align*} \delta_r(F, M; X):=\frac{\#\{n\leq X: a(n)\equiv r \pmod{M}\}}{X}. \end{align*} $$

An integral power series F is called lacunary modulo M if

$$ \begin{align*} \lim _{X\rightarrow \infty}\delta_0(F, M; X)=1, \end{align*} $$

that is, almost all of the coefficients of F are divisible by M.

It is a well-known fact that modular forms with integer Fourier coefficients are lacunary modulo any positive integer. Recently, in [5, Theorem 1.1], Cotron et al. extended this fact to integral weight eta-quotients modulo arbitrary powers of primes under certain strong conditions. In [9], Lin remarked that the generating function of $\text {PDO}_t(n)$ is a modular form. However, this observation is not quite correct because $\text {PDO}_t(n)$ is not holomorphic at the cusp 1. Also, the generating function of $\text {PDO}_t(n)$ does not satisfy the conditions of [5, Theorem 1.1]. Therefore, it is an interesting problem to study the lacunarity of $G(q)=\sum _{n=0}^{\infty }\text {PDO}_t(n)q^n$ modulo arbitrary powers of primes. In the following theorem, we prove that $G(q)$ is lacunary modulo arbitrary powers of $2$ and $3$ .

Theorem 1.3. For any positive integer k,

(1.4) $$ \begin{align} \lim _{X\rightarrow \infty}\delta_0(G, 2^k; X)=1, \end{align} $$

(1.5) $$ \begin{align} \lim _{X\rightarrow \infty}\delta_0(G, 3^k; X)=1. \end{align} $$

Serre observed and Tate proved that the action of Hecke algebras on spaces of modular forms of level 1 modulo 2 is locally nilpotent (see for example [14–16]). Ono and Taguchi [13] showed that this phenomenon generalises to higher levels. We observe that, for any positive integer k, the generating function of $\text {PDO}_t(n)$ is congruent to an eta-quotient modulo $2^k$ , and the eta-quotient is a modular form whose level is in Ono and Taguchi’s list. This allows us to use a result of Ono and Taguchi to prove the following congruence for $\text {PDO}_t(n)$ .

Theorem 1.4. Let n be a nonnegative integer. Then there is an integer $s \geq 0$ such that for every $u \geq 1$ and distinct primes $p_1, \ldots , p_{s+u}$ coprime to $6$ ,

$$ \begin{align*} \mathrm{PDO}_t(p_1\cdots p_{s+u}\cdot n )\equiv 0\pmod{2^{u}} \end{align*} $$

whenever n is coprime to $p_1, \ldots , p_{s+u}$ .

The rest of this paper is organised as follows. In Section 2, we recall some basic properties of modular forms and $\eta $ -quotients. In Section 3, we prove Theorem 1.2 using standard dissection of q-series. The proofs of Theorems 1.3 and 1.4 rely on properties of modular forms and we prove these theorems in Sections 4 and 5, respectively. Finding a proof of Conjecture 1.1 for $k\geq 3$ using standard dissection of q-series looks difficult. However, it might be possible to prove Conjecture 1.1 using modular forms.

2 Preliminaries

In this section, we recall some definitions and basic facts on modular forms and eta-quotients. For more details, see for example [8, 12].

2.1 Spaces of modular forms

We first define the matrix groups

$$ \begin{align*} \text{SL}_2(\mathbb{Z}) & :=\left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix}: a, b, c, d \in \mathbb{Z}, ad-bc=1 \right\},\\ \Gamma_{0}(N) & :=\left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \text{SL}_2(\mathbb{Z}) : c\equiv 0\pmod N \right\}, \\ \Gamma_{1}(N) & :=\left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \Gamma_0(N) : a\equiv d\equiv 1\pmod N \right\}, \end{align*} $$

and

$$ \begin{align*}\Gamma(N) & :=\left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \text{SL}_2(\mathbb{Z}) : a\equiv d\equiv 1\pmod N ~\text{and}~ b\equiv c\equiv 0\pmod N\right\}, \end{align*} $$

where N is a positive integer. A subgroup $\Gamma $ of $\text {SL}_2(\mathbb {Z})$ is called a congruence subgroup if $\Gamma (N)\subseteq \Gamma $ for some N. The smallest N such that $\Gamma (N)\subseteq \Gamma $ is called the level of $\Gamma $ . For example, $\Gamma _0(N)$ and $\Gamma _1(N)$ are congruence subgroups of level N.

Let $\mathbb {H}:=\{z\in \mathbb {C}: \text {Im}(z)>0\}$ be the upper half of the complex plane. The group

$$ \begin{align*}\text{GL}_2^{+}(\mathbb{R})=\left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix}: a, b, c, d\in \mathbb{R}~\text{and}~ad-bc>0\right\}\end{align*} $$

acts on $\mathbb {H}$ by

$$ \begin{align*}\begin{bmatrix} a & b \\ c & d \end{bmatrix} z=\displaystyle \frac{az+b}{cz+d}.\end{align*} $$

We identify $\infty $ with ${1}/{0}$ and define

$$ \begin{align*}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \displaystyle\frac{r}{s}=\displaystyle \frac{ar+bs}{cr+ds},\end{align*} $$

where ${r}/{s}\in \mathbb {Q}\cup \{\infty \}$ . This gives an action of $\text {GL}_2^{+}(\mathbb {R})$ on the extended upper half-plane $\mathbb {H}^{\ast }=\mathbb {H}\cup \mathbb {Q}\cup \{\infty \}$ . Suppose that $\Gamma $ is a congruence subgroup of $\text {SL}_2(\mathbb {Z})$ . A cusp of $\Gamma $ is an equivalence class in $\mathbb {P}^1=\mathbb {Q}\cup \{\infty \}$ under the action of $\Gamma $ .

The group $\text {GL}_2^{+}(\mathbb {R})$ also acts on functions $f: \mathbb {H}\rightarrow \mathbb {C}$ . In particular, suppose that $\gamma =[\begin {smallmatrix} a & b \\ c & d \end {smallmatrix}]\in \text {GL}_2^{+}(\mathbb {R})$ . If $f(z)$ is a meromorphic function on $\mathbb {H}$ and $\ell $ is an integer, then define the slash operator $|_{\ell }$ by

$$ \begin{align*}(f|_{\ell}\gamma)(z):=(\text{det}~{\gamma})^{\ell/2}(cz+d)^{-\ell}f(\gamma z).\end{align*} $$

Definition 2.1. Let $\Gamma $ be a congruence subgroup of level N. A holomorphic function $f: \mathbb {H}\rightarrow \mathbb {C}$ is called a modular form with integer weight $\ell $ on $\Gamma $ if it satisfies the following conditions:

1. (1) $f( ({az+b})/({cz+d}))=(cz+d)^{\ell }f(z)$ for all $z\in \mathbb {H}$ and all $[\begin {smallmatrix} a & b \\ c & d \end {smallmatrix}] \in \Gamma $ ;

2. (2) if $\gamma \in \text {SL}_2(\mathbb {Z})$ , then $(f|_{\ell }\gamma )(z)$ has a Fourier expansion of the form

   $$ \begin{align*}(f|_{\ell}\gamma)(z)=\displaystyle\sum_{n\geq 0}a_{\gamma}(n)q_N^n,\end{align*} $$

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

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

Definition 2.2 [12, Definition 1.15]

If $\chi $ is a Dirichlet character modulo N, then we say that a modular form $f\in M_{\ell }(\Gamma _1(N))$ has Nebentypus character $\chi $ if

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

for all $z\in \mathbb {H}$ and all $[\begin {smallmatrix} a & b \\ c & d \end {smallmatrix}] \in \Gamma _0(N)$ . The space of such modular forms is denoted by $M_{\ell }(\Gamma _0(N), \chi )$ .

2.2 Modularity of eta-quotients

The relevant modular forms in this paper are those that arise from eta-quotients. Recall that the Dedekind eta-function $\eta (z)$ is defined by

$$ \begin{align*} \eta(z):=q^{1/24}(q;q)_{\infty}=q^{1/24}\prod_{n=1}^{\infty}(1-q^n), \end{align*} $$

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

$$ \begin{align*} f(z)=\prod_{\delta\mid N}\eta(\delta z)^{r_{\delta}}, \end{align*} $$

where N is a positive integer and $r_{\delta }$ is an integer.

We now recall two theorems from [12, page 18] on modularity of eta-quotients. We will use these two results to verify modularity of certain eta-quotients appearing in the proofs of our main results.

Theorem 2.3 [12, Theorem 1.64]

If $f(z)=\prod _{\delta \mid N}\eta (\delta z)^{r_{\delta }}$ is an eta-quotient such that $\ell =\tfrac 12\sum _{\delta \mid N}r_{\delta }\in \mathbb {Z}$ ,

$$ \begin{align*}\sum_{\delta\mid N} \delta r_{\delta}\equiv 0 \pmod{24}\end{align*} $$

and

$$ \begin{align*}\sum_{\delta\mid N} \frac{N}{\delta}r_{\delta}\equiv 0 \pmod{24},\end{align*} $$

then $f(z)$ satisfies

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

for every $[\begin {smallmatrix} a & b \\ c & d \end {smallmatrix}] \in \Gamma _0(N)$ . Here, the character $\chi $ is defined by $\chi (d):=\big(\frac{(-1)^\ell s}{d}\big)$ , where $s:= \prod _{\delta \mid N}\delta ^{r_{\delta }}$ .

Suppose that f is an eta-quotient satisfying the conditions of Theorem 2.3 and that the associated weight $\ell $ is a positive integer. If $f(z)$ is holomorphic at all of the cusps of $\Gamma _0(N)$ , then $f(z)\in M_{\ell }(\Gamma _0(N), \chi )$ . The following theorem gives the necessary criterion for determining orders of an eta-quotient at cusps.

Theorem 2.4 [12, Theorem 1.65]

Let $c, d$ and N be positive integers with $d\mid N$ 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 ${c}/{d}$ is

$$ \begin{align*}\frac{N}{24}\sum_{\delta\mid N}\frac{\gcd(d,\delta)^2r_{\delta}}{\gcd(d,{N}/{d})d\delta}.\end{align*} $$

Finally, we recall the definition of Hecke operators. Let m be a positive integer and $f(z) = \sum _{n=0}^{\infty } a(n)q^n \in M_{\ell }(\Gamma _0(N),\chi )$ . Then, the action of the Hecke operator $T_m$ on $f(z)$ is defined by

$$ \begin{align*} f(z)|T_m := \sum_{n=0}^{\infty} \bigg(\sum_{d\mid \gcd(n,m)}\chi(d)d^{\ell-1}a\bigg(\frac{nm}{d^2}\bigg)\bigg)q^n. \end{align*} $$

In particular, if $m=p$ is prime,

$$ \begin{align*} f(z)|T_p := \sum_{n=0}^{\infty} \bigg(a(pn)+\chi(p)p^{\ell-1}a\bigg(\frac{n}{p}\bigg)\bigg)q^n. \end{align*} $$

We adopt the convention that $a(n)=0$ unless n is a nonnegative integer.

3 Proof of Theorem 1.2

To prove Theorem 1.2, we need the following lemma.

Lemma 3.1. The following identities hold:

$$ \begin{align*} \frac{f_3}{{f}^3_1}&=\frac{{f}^6_4{f}^3_6}{{f}^9_2{f}^2_{12}}+3q\frac{{f}^2_4f_6{f}^2_{12}}{{f}^7_2},\\ \frac{1}{{f}^4_1}&=\frac{{f}^{14}_4}{{f}^{14}_2{f}^4_8}+4q\frac{{f}^2_4{f}^4_8}{{f}^{10}_2},\\ \frac{f_2}{{f}^2_1}&=\frac{{f}^4_6{f}^6_9}{{f}^8_3{f}^3_{18}}+2q\frac{{f}^3_6{f}^3_9}{{f}^7_3}+4q^2\frac{{f}^2_6{f}^3_{18}}{{f}^6_3},\\ f_1f_2&=\frac{f_6{f}^4_9}{f_3{f}^2_{18}}-qf_9f_{18}-2q^2\frac{f_3{f}^4_{18}}{f_6{f}^2_9},\\ {f}^3_1&=\frac{f_6{f}^6_9}{f_3{f}^3_{18}}-3q{f}^3_9+4q^3\frac{{f}^2_3{f}^6_{18}}{{f}^2_6{f}^3_9}. \end{align*} $$

Proof. For the proof of first identity, see [10]. The second identity is proved in [6]. The remaining three identities of the lemma are proved in [7].

Proof of Theorem 1.2

By (1.1), we have

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(n)q^n=q\frac{f_2{f}^2_3{f}^2_{12}}{{f}^2_1f_6}. \end{align*} $$

Substituting the 3-dissection formula for ${f_2}/{{f}^2_1}$ from Lemma 3.1,

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(n)q^n=q\frac{{f}^2_3{f}^2_{12}}{f_6}\bigg( \frac{{f}^4_6{f}^6_9}{{f}^8_3{f}^3_{18}}+2q\frac{{f}^3_6{f}^3_9}{{f}^7_3}+4q^2\frac{{f}^2_6{f}^3_{18}}{{f}^6_3}\bigg). \end{align*} $$

Extracting those terms of the form $q^{3n}$ on both sides of this equation and replacing $q^3$ by q, we find that

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(3n)q^n=4q\frac{f_2{f}^2_4{f}^3_6}{{f}^4_1}. \end{align*} $$

Substituting the 2-dissection formula for ${1}/{{f}^4_1}$ from Lemma 3.1 yields

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(3n)q^n=4qf_2{f}^2_4{f}^3_6\bigg(\frac{{f}^{14}_4}{{f}^{14}_2{f}^4_8}+4q\frac{{f}^2_4{f}^4_8}{{f}^{10}_2} \bigg). \end{align*} $$

Extracting those terms of the form $q^{2n}$ on both sides of this equation and replacing $q^2$ by q,

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(6n)q^n=16q{f}^4_2{f}^4_4\bigg( \frac{f_3}{{f}^3_1}\bigg)^3. \end{align*} $$

Substituting the 2-dissection formula for ${f_3}/{{f}^3_1}$ from Lemma 3.1 yields

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(6n)q^n & =16q\frac{{f}^{22}_4{f}^9_6}{{f}^{23}_2{f}^6_{12}}+16\cdot9q^2\frac{{f}^{18}_4{f}^7_6}{{f}^{21}_2{f}^2_{12}}+16\cdot27q^3\frac{{f}^{14}_4{f}^5_6{f}^2_{12}}{{f}^{19}_2} \\ &\quad +16\cdot27q^4\frac{{f}^{10}_4{f}^3_6{f}^6_{12}}{{f}^{17}_2}. \end{align*} $$

Extracting those terms of the form $q^{2n}$ on both sides of this equation and replacing $q^2$ by q,

(3.1) $$ \begin{align} \sum_{n=0}^{\infty}\text{PDO}_t(12n)q^n=16\cdot9q\frac{{f}^{18}_2{f}^7_3}{{f}^{21}_1{f}^2_6}+16\cdot27q^2\frac{{f}^{10}_2{f}^3_3{f}^6_6}{{f}^{17}_1}. \end{align} $$

By the binomial theorem,

$$ \begin{align*} \frac{{f}^{10}_2}{{f}^{17}_1}\equiv\frac{{f}^{9}_2}{{f}^{18}_1}f_1f_2\equiv\frac{{f}^{3}_6}{{f}^{6}_3}f_1f_2 \pmod{3}. \end{align*} $$

Substituting the 3-dissection formula for $f_1f_2$ from Lemma 3.1,

(3.2) $$ \begin{align} \frac{{f}^{10}_2}{{f}^{17}_1}\equiv\frac{{f}^{4}_6{f}^4_9}{{f}^7_3{f}^2_{18}}-q\frac{{f}^3_6f_9f_{18}}{{f}^6_3}-2q^2\frac{{f}^2_6{f}^4_{18}}{{f}^5_3{f}^2_9} \pmod{3}. \end{align} $$

Again, using the binomial theorem,

$$ \begin{align*} \frac{{f}^{18}_2}{{f}^{21}_1}\equiv\frac{{f}^{18}_2}{{f}^{27}_1}{f}^6_1\equiv\frac{{f}^{6}_6}{{f}^{9}_3}( {f}^3_1) ^2 \pmod{9}. \end{align*} $$

Substituting the 3-dissection formula for ${f}^3_1$ from Lemma 3.1,

(3.3) $$ \begin{align} \frac{{f}^{18}_2}{{f}^{21}_1}&\equiv\frac{{f}^8_6{f}^{12}_9}{{f}^{11}_3{f}^6_{18}}-6q\frac{{f}^7_6{f}^9_9}{{f}^{10}_3{f}^3_{18}}+9q^2\frac{{f}^6_6{f}^6_9}{{f}^9_3}+8q^3\frac{{f}^5_6{f}^3_9{f}^3_{18}}{{f}^{8}_3}-24q^4\frac{{f}^4_6{f}^6_{18}}{{f}^7_3}\notag \\ &\quad +16q^6\frac{{f}^2_6{f}^{12}_{18}}{{f}^5_3{f}^6_9} \pmod{9}. \end{align} $$

Substituting (3.2) and (3.3) in (3.1), and then extracting the terms of the form $q^{3n}$ on both sides, we find that

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(36n)q^{3n}\equiv16\cdot81q^3\frac{{f}^4_6{f}^6_9}{{f}^2_3}-16\cdot27q^3\frac{{f}^9_6f_9f_{18}}{{f}^3_3}\pmod{81}. \end{align*} $$

Replacing $q^3$ by q, and then substituting the 2-dissection formula for ${f_3}/{{f}^3_1}$ from Lemma 3.1,

(3.4) $$ \begin{align} \sum_{n=0}^{\infty}\text{PDO}_t(36n)q^n&\equiv-16\cdot27q\frac{{f}^9_2f_3f_6}{{f}^3_1} \pmod{81}\qquad\qquad\qquad\qquad \end{align} $$

(3.5) $$ \begin{align} &\qquad\ \ \,\kern1pt\qquad\qquad\equiv-16\cdot27q{f}^9_2f_6\frac{f_3}{{f}^3_1} \pmod{81}\nonumber\\ &\qquad\ \ \,\kern1pt\qquad\qquad\equiv-16\cdot27q\frac{{f}^6_4{f}^4_6}{{f}^2_{12}}-16\cdot81q^2{f}^2_2{f}^2_4{f}^2_6{f}^2_{12} \pmod{81}\nonumber\\ &\qquad\ \ \,\kern1pt\qquad\qquad\equiv -16\cdot27q\frac{{f}^6_4{f}^4_6}{{f}^2_{12}} \pmod{81}. \end{align} $$

Extracting the terms of the form $q^{2n}$ on both sides of (3.5),

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(72n)q^{2n}&\equiv0 \pmod{81}. \end{align*} $$

This completes the proof of (1.2).

We next prove (1.3). By the binomial theorem,

(3.6) $$ \begin{align} \frac{{f}^9_2}{{f}^3_1}\equiv\frac{{f}^3_6}{f_3} \pmod{3}. \end{align} $$

Combining (3.6) and (3.4),

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(36n)q^n\equiv-16\cdot27q{f}^4_6 \pmod{81}. \end{align*} $$

Extracting the terms of the form $q^{3n}$ on both sides yields

$$ \begin{align*} \sum_{n=0}^{\infty}\text{PDO}_t(108n)q^{3n}\equiv0 \pmod{81}. \end{align*} $$

This completes the proof of (1.3).

4 Proof of Theorem 1.3

Given a prime p, let

$$ \begin{align*} A_p(z) = \prod_{n=1}^{\infty} \frac{(1-q^{n})^p}{(1-q^{pn})} = \frac{\eta^p(z)}{\eta(pz)}. \end{align*} $$

Then, using the binomial theorem,

$$ \begin{align*} A_p^{p^k}(z) = \frac{\eta^{p^{k+1}}(z)}{\eta^{p^k}(pz)} \equiv 1 \pmod {p^{k+1}}. \end{align*} $$

Define $B_{p,k}(z)$ by

(4.1) $$ \begin{align} B_{p,k}(z) = \bigg(\frac{\eta(2z)\eta(3z)^2\eta(12z)^2}{\eta(z)^2\eta(6z)}\bigg)A_p^{p^k}(z). \end{align} $$

Modulo $p^{k+1}$ ,

(4.2) $$ \begin{align} B_{p,k}(z) \equiv \frac{\eta(2z)\eta(3z)^2\eta(12z)^2}{\eta(z)^2\eta(6z)} = \frac{qf_2f_3^2f_{12}^2}{f_1^2f_6}. \end{align} $$

Combining (1.1) and (4.2),

(4.3) $$ \begin{align} B_{p,k}(z) \equiv \sum_{n=0}^{\infty}\text{PDO}_t(n)q^n \pmod {p^{k+1}}. \end{align} $$

Proof of Theorem 1.3

We put $p=2$ in (4.1) to obtain

$$ \begin{align*} B_{2,k}(z) = \left(\frac{\eta(2z)\eta(3z)^2\eta(12z)^2}{\eta(z)^2\eta(6z)}\right)A_2^{2^k}(z)=\frac{\eta(2z)^{1-2^{k}}\eta(3z)^2\eta(12z)^2\eta(z)^{2^{k+1}-2}}{\eta(6z)}. \end{align*} $$

Now, $B_{2, k}$ is an eta-quotient with level $N=144$ . The cusps of $\Gamma _{0}(144)$ are represented by fractions ${c}/{d}$ , where $d\mid 144$ and $\gcd (c, d)=1$ (see for example [11, page 5]). By Theorem 2.4, $B_{2,k}(z)$ is holomorphic at a cusp ${c}/{d}$ if and only if

$$ \begin{align*} S&:=12\frac{\gcd(d,1)^2}{\gcd(d,12)^2} (2^{k+1}-2)+ 6\frac{\gcd(d,2)^2}{\gcd(d,12)^2}(1-2^k )+8\frac{\gcd(d,3)^2}{\gcd(d,12)^2} \\ & \quad -2\frac{\gcd(d,6)^2}{\gcd(d,12)^2}+2 \geq 0. \end{align*} $$

To find all the possible values of S, we prepared Table 1 using MATLAB. Using Table 1, we find that $S\geq 0$ for all $d\mid 144$ . Hence, $B_{2,k}(z)$ is holomorphic at every cusp ${c}/{d}$ . From Theorem 2.3, the weight of $B_{2,k}(z)$ is $\ell =2^{k-1}+1$ . Also, the associated character for $B_{2,k}(z)$ is given by

$$ \begin{align*} \chi_1=\bigg(\frac{(-1)^{2^{k-1}+1}2^{4- 2^{k}}3^{3}}{\bullet}\bigg). \end{align*} $$

Finally, by Theorem 2.3, $B_{2,k}(z) \in M_{2^{k-1}+1}(\Gamma _{0}(144), \chi _1)$ for $k\geq 1$ . Given any positive integer m, by a deep theorem of Serre [12, page 43], if $f(z)\in M_{\ell }(\Gamma _0(N), \chi )$ has the Fourier expansion

$$ \begin{align*} f(z)=\sum_{n=0}^{\infty}c(n)q^n\in \mathbb{Z}[[q]], \end{align*} $$

then there is a constant $\alpha>0$ such that

$$ \begin{align*} \# \{n\leq X: c(n)\not\equiv 0 \pmod{m} \}= \mathcal{O}\bigg(\frac{X}{(\log{}X)^{\alpha}}\bigg). \end{align*} $$

This yields

$$ \begin{align*} \lim _{X\rightarrow \infty}\delta_0(f, m; X)=\lim_{X\to\infty} \frac{\# \{n\leq X: c(n)\equiv 0 \pmod{m}\}}{X}&=1. \end{align*} $$

Since $B_{2,k}(z) \in M_{2^{k-1}+1}(\Gamma _{0}(144), \chi _1)$ , the Fourier coefficients of $B_{2,k}(z)$ are almost always divisible by $m=2^k$ . Now, using (4.3) completes the proof of (1.4).

Table 1 Data to find the values of $S$ .

We next prove (1.5). We put $p=3$ in (4.1) to obtain

$$ \begin{align*} B_{3,k}(z) = \bigg(\frac{\eta(2z)\eta(3z)^2\eta(12z)^2}{\eta(z)^2\eta(6z)}\bigg)\,A_3^{3^k}(z)=\frac{\eta(2z)\eta(3z)^{2-3^{k}}\eta(12z)^2\eta(z)^{3^{k+1}-2}}{\eta(6z)}. \end{align*} $$

Now, $B_{3, k}$ is an eta-quotient with $N=144$ . As before, the cusps of $\Gamma _{0}(144)$ are represented by fractions ${c}/{d}$ , where $d\mid 144$ and $\gcd (c, d)=1$ . By Theorem 2.4, $B_{3,k}(z)$ is holomorphic at a cusp ${c}/{d}$ if and only if

$$ \begin{align*} L:= &~12\frac{\gcd(d,1)^2}{\gcd(d,12)^2} (3^{k+1}-2)+4\frac{\gcd(d,3)^2}{\gcd(d,12)^2}(2-3^k) + 6\frac{\gcd(d,2)^2}{\gcd(d,12)^2}\\ &-2\frac{\gcd(d,6)^2}{\gcd(d,12)^2}+2 \geq 0. \end{align*} $$

From Table 1, $L\geq 0$ for all $d\mid 144$ . By Theorem 2.3, $B_{3,k}(z) \in M_{3^{k}+1}(\Gamma _{0}(144), \chi _2)$ , where $\chi_2$ is the associated Nebentypus character. Using the same reasoning and (4.3), we find that $\text {PDO}_t(n)$ is divisible by $3^k$ for almost all $n\geq 0$ . This completes the proof of (1.5).

5 Proof of Theorem 1.4

In this section, we prove Theorem 1.4 using nilpotency of Hecke operators. We apply a result of Ono and Taguchi [13] to the modular form $B_{2,k}(z)$ to deduce the infinite family of congruences.

Proof of Theorem 1.4

Taking $p=2$ in (4.3), we have

$$ \begin{align*} B_{2,k}(z) \equiv \sum_{n=0}^{\infty}\text{PDO}_t(n)q^{n} \pmod {2^{k+1}}. \end{align*} $$

Note that $B_{2,k}(z) \in M_{2^{k-1}+1}(\Gamma _{0}(144), \chi _1)$ . By [13, Theorem 1.3(3)], there is an integer $s\geq 0$ such that for any $u \geq 1$ ,

$$ \begin{align*} B_{2,k}(z)|T_{p_1}| T_{p_2}|\cdots| T_{p_{s+u}} \equiv 0 \pmod{2^u} \end{align*} $$

whenever $p_{1}, \ldots , p_{s+u}$ are coprime to $6$ . It follows from the definition of the Hecke operators that if $p_{1}, \ldots , p_{s+u}$ are distinct primes and if n is coprime to $p_{1} \cdots p_{s+u}$ , then

$$ \begin{align*} \text{PDO}_t(p_{1} \cdots p_{s+u}\cdot n) \equiv 0 \pmod{2^u}. \end{align*} $$

This completes the proof of the theorem.