# Arithmetic Properties Modulo Powers of $2$ for Overpartition $k$-Tuples with
Odd Parts

Manjil P. Saikia  Mathematical and Physical Sciences division, School of Arts
and Sciences, Ahmedabad University, Ahmedabad 380009, Gujarat, India
<EMAIL_ADDRESS>,  Abhishek Sarma  Department of Mathematical Sciences,
Tezpur University, Napaam, Tezpur, India, PIN-784028<EMAIL_ADDRESS>and  James A. Sellers  Department of Mathematics and Statistics, University of
Minnesota Duluth, Duluth, MN 55812, USA<EMAIL_ADDRESS>
###### Abstract.

Recently, Drema and Saikia (2023) proved several congruences modulo powers of
$2$ and $3$ for overpartition triples with odd parts. We extend their list
substantially. We prove several congruences modulo powers of $2$ for
overpartition $k$-tuples with odd parts, along with a few infinite families of
congruences for overpartition triples with odd parts and for overpartition
$k$-tuples with odd parts (for $k=4$ and for odd $k$).

###### Key words and phrases:

integer partitions, Ramanujan-type congruences.

###### 2020 Mathematics Subject Classification:

11P81, 11P83.

##  1\. Introduction

A partition of a positive integer $n$ is a finite non-increasing sequence of
positive integers $\lambda=(\lambda_{1},\lambda_{2},\ldots,\lambda_{k})$ such
that the parts $\lambda_{i}$ sum up to $n$. For instance, $4,3+1,2+2,2+1+1$
and $1+1+1+1$ are the five partitions of $4$. The number of partitions of $n$
is denoted by $p(n)$, and its generating function found by Euler is given by

$\sum_{n\geq 0}p(n)q^{n}=\frac{1}{(q;q)_{\infty}},$

where

$(a;q)_{\infty}:=\prod_{i\geq 0}(1-aq^{i}),\quad|q|<1.$

Throughout the paper, we will use the notation
$f_{k}:=(q^{k};q^{k})_{\infty}$.

Among the many beautiful properties satisfied by the partition function,
undoubtedly some of the most beautiful are the congruences modulo primes that
the partition function satisfies, as discovered by Ramanujan. Several
mathematicians have since studied the arithmetic properties of the partition
function, as well as other generalized classes of partitions. Recently, Drema
and Saikia [DS23] studied congruences modulo small powers of $2$ and $3$
satisfied by the function which counts overpartition triples with odd parts.
The aim of the present paper is to extend their list and to also look at some
more general cases.

An overpartition of a nonnegative integer $n$ is a non-increasing sequence of
natural numbers whose sum is $n$, where the first occurrence (or equivalently,
the last occurrence) of a number may be overlined. The eight overpartitions of
3 are

$3,\bar{3},2+1,\bar{2}+1,2+\bar{1},\bar{2}+\bar{1},1+1+1,~{}\text{and}~{}\bar{1}+1+1.$

The number of overpartitions of $n$ is denoted by $\overline{p}(n)$ and its
generating function is given by

$\sum_{n\geq 0}\overline{p}(n)q^{n}=\frac{f_{2}}{f_{1}^{2}}.$

An overpartition triple $\xi$ of $n$ is a triplet $(\xi_{1},\xi_{2},\xi_{3})$
of overpartitions such that the sum of all the parts is $n$. An overpartition
triple into odd parts is a triplet of overpartitions
$(\xi_{1},\xi_{2},\xi_{3})$ such that the parts of all the overpartitions
$\xi_{1},\xi_{2}$ and $\xi_{3}$ are restricted to odd integers. Let
$\overline{OPT}(n)$ denote the number of such overpartition triples of $n$
with odd parts. Then the generating function for $\overline{OPT}(n)$ is given
by

(1) $\sum_{n\geq
0}\overline{OPT}(n)q^{n}=\frac{f_{2}^{9}}{f_{1}^{6}f_{4}^{3}}=\prod_{i\geq
0}\left(\frac{1+q^{2i+1}}{1-q^{2i+1}}\right)^{3}.$

Before proceeding further, we mention (without commentary) the following easy
consequence of the binomial theorem which will be used below.

######  Lemma 1.

For a prime $p$, and positive integers $k$ and $l$, we have

$\displaystyle f_{k}^{p^{l}}\equiv f_{pk}^{p^{l-1}}\pmod{p^{l}}.$

It can be shown easily [DS23, Eq. (43)] that

$\overline{OPT}(2n+1)\equiv 0\pmod{2},$

for all $n\geq 0$.

Indeed, working modulo 2, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(n)q^{n}\equiv\prod_{i\geq
0}\left(\frac{1+q^{2i+1}}{1-q^{2i+1}}\right)^{3}\equiv\prod_{i\geq
0}\left(\frac{1-q^{2i+1}}{1-q^{2i+1}}\right)^{3}\equiv 1\pmod{2}.$

Hence, for $n\geq 1$, we have

$\displaystyle\overline{OPT}(n)\equiv 0\pmod{2}.$

The last congruence is reminiscent of the fact that

$\overline{p}(n)\equiv 0\pmod{2},\quad\text{for all}~{}n>0.$

In this paper we prove several infinite families of Ramanujan-type congruences
modulo powers of $2$ for the $\overline{OPT}(n)$ function (Theorems 1, 2, 3
and 5.)

As can be seen, there is nothing special about the definition of an
overpartition triple, and it can be easily extended to an overpartition
$k$-tuple, as was done by the third author with Keister and Vary [KSV09]. An
overpartition $k$-tuple of $n$ is a $k$-tuple of overpartitions
$\xi_{1},\xi_{2},\ldots,\xi_{k})$ such that the sum of the parts of
$\xi_{i}$’s equals $n$. The generating function for the number of
overpartition $k$-tuples of $n$, denoted by $\overline{p}_{k}(n)$ is given by

$\sum_{n\geq 0}\overline{p}_{k}(n)q^{n}=\frac{f_{2}^{k}}{f_{1}^{2k}}.$

We can similarly define an overpartition $k$-tuple of $n$ with odd parts to be
an overpartition $k$-tuple $(\xi_{1},\xi_{2},\ldots,\xi_{k})$ of $n$ where all
parts of $\xi_{i}$’s are odd. The generating function for the number of
overpartition $k$-tuples of $n$ with odd parts, denoted by
$\overline{OPT}_{k}(n)$ is given by

$\sum_{n\geq
0}\overline{OPT}_{k}(n)q^{n}=\frac{f_{2}^{3k}}{f_{1}^{2k}f_{4}^{k}}.$

The case $k=3$ here corresponds to overpartition tuples with odd parts. The
case $k=2$ has also been studied, the interested reader can look at the work
of Lin [Lin12]. The study of the arithmetic properties of the case $k=1$ was
initiated by the third author and Hirschhorn [HS06], who used the notation
$\overline{p}_{o}(n)$ for $\overline{OPT}_{1}(n)$, which we will use from now
on.

In this paper we prove several Ramanujan-type congruences modulo powers of $2$
for the functions $\overline{OPT}_{2k+1}(n)$ (for $k\geq 1$) (Theorems 6, 7, 8
and 9). We also give two infinite families of congruences modulo powers of $2$
for the $\overline{OPT}_{4}(n)$ (Theorem 10) and $\overline{OPT}_{2k+1}(n)$
(Theorem 11) functions.

The paper is organized as follows: in Section 2 we state our main results (and
prove three of them quickly), and in Section 3 we state some known results
which will be used for our proofs. Theorems 1, 2 and 3 are proved in Section
4. Theorem 5 is then proved in Section 5, Theorem 9 is proved in Section 6 and
Theorems 10 and 11 are proved in Section 7. We end the paper with some
concluding remarks in Section 8, which also includes Conjecture 1, a
counterpart of Theorem 9.

##  2\. Main Results

The first set of results we prove in this paper are as follows.

######  Theorem 1.

For all $n\geq 0,\alpha\geq 0$, we have

(2) $\overline{OPT}(2^{\alpha}(4n+3))\equiv 0\pmod{4}.$

######  Theorem 2.

For all $n\geq 0,\alpha\geq 0$, we have

(3) $\overline{OPT}(2^{\alpha}(8n+5))\equiv 0\pmod{8}.$

######  Theorem 3.

For all $n\geq 0,\alpha\geq 0$, we have

(4) $\overline{OPT}(2^{\alpha}(16n+14))\equiv 0\pmod{16}.$

######  Remark 4.

Theorem 3 is an extension of a result of Drema and Saikia [DS23, Eq. (18)] who
proved that, for all $n\geq 0,$

$\overline{OPT}(16n+14)\equiv 0\pmod{8}.$

Theorems 1, 2 and 3 are proved in Section 4.

Many mathematicians working in this area, including Drema and Saikia [DS23,
Theorems 2 and 3] have studied infinite families of congruences. We present
two such infinite families in the following theorem.

######  Theorem 5.

If $p\geq 3$ is a prime, then for all $n\geq 0$, $\alpha\geq 0$ and
$\delta\geq 0$, we have

(5) $\displaystyle\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2\delta+1}(pn+t)+3^{2\alpha}\cdot p^{2(\delta+1)})\equiv 0\pmod{4},$

where $t\in\\{1,2,\ldots,p-1\\}$,

If $p>3$ is a prime satisfying $\left(\frac{-2}{p}\right)=-1$, then for all
$n\geq 0$, $\alpha\geq 0$ and $\delta\geq 0$, we have

(6) $\displaystyle\overline{OPT}(8\cdot 3^{2\alpha}\cdot
p^{2\delta+1}(pn+r)+3^{2\alpha}\cdot p^{2(\delta+1)})\equiv 0\pmod{8},$

where $r\in\\{1,2,\ldots,p-1\\}$.

We prove the above theorem in Section 5.

We now state some results for overpartition $k$-tuples with odd parts. The
first such result is reminiscent of a result proved by the third author with
Keister and Vary [KSV09, Theorem 8].

######  Theorem 6.

Let $k=(2^{m})r$ with $m>0$ and $r$ odd. Then for all $n\geq 0$ we have

$\overline{OPT}_{k}(n)\equiv 0\pmod{2^{m+1}}.$

The proof is very short and we can give it immediately using the following
lemma from Keister, Sellers and Vary [KSV09, Lemma 7].

######  Lemma 2.

Let $m$ be a nonnegative integer, then for all $1\leq n\leq 2^{m}$, we have

$\binom{2^{m}}{n}2^{n}\equiv 0\pmod{2^{m+1}}.$

###### Proof of Theorem 6.

We have the following

$\sum_{n\geq
0}\overline{OPT}_{k}(n)q^{n}=\prod_{i=1}^{\infty}\left(\frac{1+q^{2i+1}}{1-q^{2i+1}}\right)^{k}=\left[\prod_{i=1}^{\infty}\left(\frac{1+q^{2i+1}}{1-q^{2i+1}}\right)^{2^{m}}\right]^{r}=\left[\prod_{i=1}^{\infty}\left(1+\frac{2q^{2i+1}}{1-q^{2i+1}}\right)^{2^{m}}\right]^{r}.$

Now, using the binomial theorem and Lemma 2 we can conclude

$\sum_{n\geq 0}\overline{OPT}_{k}(n)q^{n}\equiv 1\pmod{2^{m+1}}.$

This completes the proof. ∎

We now state three results for the general cases of
$\overline{OPT}_{2k+1}(n)$.

######  Theorem 7.

For all $n\geq 0$ and $k\geq 1$ we have

$\overline{OPT}_{2k+1}(n)\equiv\overline{p}_{o}(n)\pmod{4}.$

###### Proof.

We have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{2k+1}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{3(2k+1)}}{f_{1}^{2(2k+1)}f_{4}^{2k+1}}$
$\displaystyle\equiv\frac{f_{2}^{6k}f_{2}^{3}}{f_{2}^{2k}f_{2}^{4k}f_{1}^{2}f_{4}}\pmod{4}$
$\displaystyle\equiv\frac{f_{2}^{3}}{f_{1}^{2}f_{4}}\pmod{4}$
$\displaystyle\equiv\sum_{n\geq 0}\overline{p}_{o}(n)q^{n}\pmod{4}.$

∎

We can use Theorem 7 to prove an infinite family of Ramanujan–like congruences
modulo 4 for the functions $\overline{OPT}_{2k+1}.$

######  Theorem 8.

For all $n\geq 0$, $k\geq 0$, $p\geq 5$ prime, and all quadratic nonresidues
$r$ modulo $p$ with $1\leq r\leq p-1$ we have

$\overline{OPT}_{2k+1}(2pn+R)\equiv 0\pmod{4}$

where

$R=\begin{cases}r,&\text{if $r$ is odd,}\\\ p+r,&\text{if $r$ is
even.}\end{cases}$

###### Proof.

From the work of Hirschhorn and the third author [HS06, Theorem 1.1], we know
that, for all $n\geq 1,$ $\overline{p}_{o}(n)\equiv 0\pmod{4}$ if and only if
$n$ is neither a square nor twice a square. So we simply need to show that
$2pn+R$ as defined above is never a square and never twice a square. Note that
$2pn+R$ is always odd by definition, so we see immediately that $2pn+R$ can
never be twice a square. Next, we see that $2pn+R\equiv r\pmod{p}$ again from
the definition of $R.$ And since $r$ is defined to be a quadratic nonresidue
modulo $p$, we know that $r$ cannot be congruent to a square modulo $p.$ Thus,
$2pn+R$ cannot equal a square. This concludes the proof. ∎

######  Theorem 9.

For all $n\geq 0$ and $k\geq 0$, we have

$\displaystyle\overline{OPT}_{2k+1}(8n+1)$ $\displaystyle\equiv 0\pmod{2},$
$\displaystyle\overline{OPT}_{2k+1}(8n+2)$ $\displaystyle\equiv 0\pmod{2},$
$\displaystyle\overline{OPT}_{2k+1}(8n+3)$ $\displaystyle\equiv 0\pmod{4},$
$\displaystyle\overline{OPT}_{2k+1}(8n+4)$ $\displaystyle\equiv 0\pmod{2},$
$\displaystyle\overline{OPT}_{2k+1}(8n+5)$ $\displaystyle\equiv 0\pmod{8},$
$\displaystyle\overline{OPT}_{2k+1}(8n+6)$ $\displaystyle\equiv 0\pmod{4},$
$\displaystyle\overline{OPT}_{2k+1}(8n+7)$ $\displaystyle\equiv 0\pmod{16}.$

Theorem 9 is proved in Section 6. The approach used to prove Theorem 9 is very
similar to the one that was taken recently by the third author [Sel24] to
prove a conjecture of the first author [Sai23].

We will also prove the following infinite family of congruences similar to
(6).

######  Theorem 10.

If $p\geq 3$ is a prime, then for all $n\geq 0$, $\alpha\geq 0$ and
$\delta\geq 0$, we have

(7) $\displaystyle\overline{OPT}_{4}(8\cdot 3^{2\alpha}\cdot
p^{2\delta+1}(pn+t)+2\cdot 3^{2\alpha}\cdot
p^{2(\delta+1)}+3^{2\alpha}-1)\equiv 0\pmod{16},$

where $t\in\\{1,2,\ldots,p-1\\}$.

Lastly, the following result generalizes (6) for $\overline{OPT}_{k}(n)$,
where $k$ is odd.

######  Theorem 11.

If $p\geq 3$ is a prime, then for all $n\geq 0$, $\alpha\geq 0$, $\delta\geq
0$ and $k\geq 1$, we have

$\displaystyle\overline{OPT}_{2k+1}(2\cdot 3^{2\alpha}\cdot
p^{2\delta+1}(pn+t)+3^{2\alpha}\cdot p^{2(\delta+1)})\equiv 0\pmod{4},$

where $t\in\\{1,2,\ldots,p-1\\}$.

Theorems 10 and 11 are proved in Section 7.

##  3\. Preliminaries

Ramanujan’s general theta function $f(a,b)$ is defined by

$\displaystyle
f(a,b)=\sum_{k=-\infty}^{\infty}a^{\frac{k(k+1)}{2}}b^{\frac{k(k-1)}{2}},\quad|ab|<1.$

Two special cases of $f(a,b)$ are given by

(8) $\displaystyle\phi(q)$
$\displaystyle:=f(q,q)=\sum_{k=-\infty}^{\infty}q^{k^{2}}=1+2\sum_{n\geq
1}q^{n^{2}},$

and

$\displaystyle\psi(q):=f(q,q^{3})=\sum_{k=0}^{\infty}q^{k(k+1)/2}.$

We recall the following results.

######  Lemma 3.

[Ber91, p. 40, Entries 25(i) and 25(ii)] We have

(9) $\displaystyle\phi(q)=\phi(q^{4})+2q\psi(q^{8}).$

######  Theorem 12.

[HS06, Theorem 2.4] We have

$\displaystyle\sum_{n\geq
0}\overline{p}_{o}(n)q^{n}=\phi(q)\phi(q^{2})\phi(q^{4})^{2}\phi(q^{8})^{4}\cdots$

Some known 2-, 3-, 5-dissections (see for example [BD22, Lemmas 2, 3 and 4])
are stated in the following lemma, which will be used subsequently.

######  Lemma 4.

We have

(10) $\displaystyle f_{1}^{2}$
$\displaystyle=\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}},$
(11) $\displaystyle\frac{1}{f_{1}^{2}}$
$\displaystyle=\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}},$
(12) $\displaystyle f_{1}^{4}$
$\displaystyle=\frac{f_{4}^{10}}{f_{2}^{2}f_{8}^{4}}-4q\frac{f_{2}^{2}f_{8}^{4}}{f_{4}^{2}},$
(13) $\displaystyle\frac{1}{f_{1}^{4}}$
$\displaystyle=\frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}}+4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}},$
(14) $\displaystyle f_{1}f_{2}$
$\displaystyle=\frac{f_{6}f_{9}^{4}}{f_{3}f_{18}^{2}}-qf_{9}f_{18}-2q^{2}\frac{f_{3}f_{18}^{4}}{f_{6}f_{9}^{2}},$
(15) $\displaystyle f_{1}^{3}$
$\displaystyle=\frac{f_{6}f_{9}^{6}}{f_{3}f_{18}^{3}}-3qf_{9}^{3}+4q^{3}\frac{f_{3}^{2}f_{18}^{6}}{f_{6}^{2}f_{9}^{3}},.$

The next lemma gives a $p$-dissection of $f_{1}$.

######  Lemma 5.

[CG13, Theorem 2.2] For a prime $p>3$, we have

$\displaystyle f_{1}=(-1)^{\frac{\pm
p-1}{6}}q^{\frac{p^{2}-1}{24}}f_{p^{2}}+\sum_{k\neq\frac{\pm
p-1}{6},k=-\frac{p-1}{2}}^{\frac{p-1}{2}}(-1)^{k}q^{\frac{3k^{2}+k}{2}}f\Bigg{(}-q^{\frac{3p^{2}+(6k+1)p}{2}},-q^{\frac{3p^{2}-(6k+1)p}{2}}\Bigg{)},$

where

$\dfrac{\pm p-1}{6}=$
$\begin{cases}\dfrac{p-1}{6},&if~{}p\equiv~{}1~{}\pmod{6},\\\
\dfrac{-p-1}{6},&if~{}p\equiv~{}-1~{}\pmod{6}.\end{cases}$

Furthermore, for $\dfrac{-(p-1)}{2}\leq k\leq\dfrac{p-1}{2}$ and
$k\neq\dfrac{\pm p-1}{6}$,

$\displaystyle\dfrac{3k^{2}+k}{2}\not\equiv\dfrac{p^{2}-1}{24}\pmod{p}.$

The following lemma gives a $p$-dissection of $f_{1}^{3}$.

######  Lemma 6.

[AB16, Lemma 2.3] If $p\geq 3$ is a prime, then

(16) $\displaystyle
f_{1}^{3}=\sum_{k\neq\frac{p-1}{2},k=0}^{p-1}(-1)^{k}q^{\frac{k(k+1)}{2}}\sum_{n=0}^{\infty}(-1)^{n}(2pn+2k+1)q^{pn\cdot\frac{pn+2k+1}{2}}+p(-1)^{\frac{p-1}{2}}q^{\frac{p^{2}-1}{8}}f_{p^{2}}^{3}.$

Furthermore, if $k\neq\dfrac{p-1}{2},0\leq k\leq p-1$, then

$\displaystyle\frac{k^{2}+k}{2}\not\equiv\frac{p^{2}-1}{8}\pmod{p}.$

##  4\. Proofs of Theorems 1, 2 and 3

We now provide proofs of Theorems 1–3 via elementary generating function
manipulations.

###### Proof of Theorem 1.

From (1), we recall

(17) $\displaystyle\sum_{n\geq 0}\overline{OPT}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{9}}{f_{1}^{6}f_{4}^{3}}=\frac{f_{2}^{9}}{f_{4}^{3}}\Bigg{(}\frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}}+4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}\Bigg{)}\Bigg{(}\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}\Bigg{)},$

where we have employed (11) and (13). Now, working modulo $4$, we have

(18) $\displaystyle\sum_{n\geq 0}\overline{OPT}(n)q^{n}$
$\displaystyle\equiv\frac{f_{2}^{7}}{f_{4}^{3}}\Bigg{(}\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}\Bigg{)}\pmod{4}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

(19) $\displaystyle\sum_{n\geq 0}\overline{OPT}(2n+1)q^{n}$
$\displaystyle\equiv 2\frac{f_{1}^{2}f_{8}^{2}}{f_{2}f_{4}}\equiv
2f_{2}^{6}\pmod{4}.$

Extracting the terms involving $q^{2n+1}$, we have

(20) $\displaystyle\overline{OPT}(4n+3)\equiv 0\pmod{4}.$

This proves (2) for the case $\alpha=0$.

Now, extracting the terms involving $q^{2n}$ from (18) and replacing $q^{2}$
by $q$, we have

(21) $\displaystyle\sum_{n\geq 0}\overline{OPT}(2n)q^{n}$
$\displaystyle\equiv\frac{f_{1}^{2}f_{4}^{5}}{f_{2}^{3}f_{8}^{2}}\pmod{4}.$

From equations (1) and (21) we can conclude via elementary simplifications
that, for all $n\geq 0$,

(22) $\overline{OPT}(2n)\equiv\overline{OPT}(n)\pmod{4}.$

Finally, combining equations (22) with (20), we have (2). ∎

###### Proof of Theorem 2.

Extracting the terms involving $q^{2n+1}$ from (17), dividing both sides by
$q$ and replacing $q^{2}$ by $q$ and working modulo 8, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(2n+1)q^{n}$ $\displaystyle\equiv
2\frac{f_{2}f_{8}^{2}}{f_{4}f_{1}^{2}}+4f_{4}f_{8}\equiv
2\frac{f_{2}f_{8}^{2}}{f_{4}}\Bigg{(}\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}\Bigg{)}+4f_{4}f_{8}\pmod{8},$

thanks to (11). Extracting the terms involving $q^{2n}$ and replacing $q^{2}$
by $q$, we have

(23) $\displaystyle\sum_{n\geq 0}\overline{OPT}(4n+1)q^{n}$
$\displaystyle\equiv
2\frac{f_{4}^{7}}{f_{1}^{4}f_{2}f_{8}^{2}}+4f_{2}f_{4}\equiv
2\frac{f_{4}^{7}}{f_{2}f_{8}^{2}}\Bigg{(}\frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}}+4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}\Bigg{)}+4f_{2}f_{4}\pmod{8},$

where in the last step, we have employed (13).

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+5)q^{n}$ $\displaystyle\equiv
8\frac{f_{2}^{9}f_{4}^{2}}{f_{1}^{11}}\pmod{8}.$

Hence,

(24) $\displaystyle\overline{OPT}(8n+5)\equiv 0\pmod{8}.$

This proves (3) for $\alpha=0$.

Now, extracting the terms involving $q^{2n}$ from (17), replacing $q^{2}$ by
$q$ and working modulo 8, we have

(25) $\displaystyle\sum_{n\geq 0}\overline{OPT}(2n)q^{n}$
$\displaystyle\equiv\frac{f_{2}^{11}f_{4}}{f_{1}^{10}f_{8}^{2}}\equiv\frac{f_{2}^{7}f_{4}}{f_{1}^{2}f_{8}^{2}}\equiv\frac{f_{2}^{7}f_{4}}{f_{8}^{2}}\Bigg{(}\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}\Bigg{)}\pmod{8}$

via (11). Taking the odd dissection from the above, we obtain

$\displaystyle\sum_{n\geq 0}\overline{OPT}(4n+2)q^{n}$ $\displaystyle\equiv$
$\displaystyle
2\frac{f_{1}^{7}f_{2}}{f_{4}^{2}}\cdot\frac{f_{2}^{2}f_{8}^{2}}{f_{1}^{5}f_{4}}\equiv
2\frac{f_{1}^{2}f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\pmod{8}$ $\displaystyle\equiv$
$\displaystyle
2\left(\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\right)\frac{f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\pmod{8}$

by (10).

Now we take the even dissection of the above to obtain

$\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+2)q^{n}$ $\displaystyle\equiv$
$\displaystyle
2\frac{f_{1}f_{4}^{5}}{f_{2}^{2}f_{8}^{2}}\cdot\frac{f_{1}^{3}f_{4}^{2}}{f_{2}^{3}}\equiv
2\frac{f_{1}^{4}f_{4}^{7}}{f_{2}^{5}f_{8}^{2}}\equiv
2\frac{f_{2}^{2}f_{4}^{7}}{f_{2}^{5}f_{8}^{2}}\equiv
2\frac{f_{4}^{7}}{f_{2}^{3}f_{8}^{2}}\pmod{8}.$

This last expression is an even function of $q$. Thus, for all $n\geq 0$, we
know

(26) $\overline{OPT}(8(2n+1)+2)=\overline{OPT}(16n+10)\equiv 0\pmod{8}.$

This proves the $\alpha=1$ case of (3).

Extracting the terms involving even powers of $q$ from (25), we have

(27) $\displaystyle\sum_{n\geq 0}\overline{OPT}(4n)q^{n}$
$\displaystyle\equiv\frac{f_{1}^{2}f_{2}f_{4}^{3}}{f_{8}^{2}}\equiv\frac{f_{2}f_{4}^{3}}{f_{8}^{2}}\Bigg{(}\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\Bigg{)}\pmod{8}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+4)q^{n}$
$\displaystyle\equiv-2\frac{f_{1}^{2}f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\equiv-2\frac{f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\Bigg{(}\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\Bigg{)}\pmod{8}.$

Extracting the terms involving $q^{2n}$ from (17) and replacing $q^{2}$ by
$q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(16n+4)q^{n}$
$\displaystyle\equiv-2\frac{f_{1}^{4}f_{4}^{7}}{f_{2}^{5}f_{8}^{2}}\equiv-2\frac{f_{4}^{7}}{f_{2}^{5}f_{8}^{2}}\Bigg{(}\frac{f_{4}^{10}}{f_{2}^{2}f_{8}^{4}}-4q\frac{f_{2}^{2}f_{8}^{4}}{f_{4}^{2}}\Bigg{)}\pmod{8},$

where in the last step, we have employed (12).

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(32n+20)q^{n}$ $\displaystyle\equiv
8\frac{f_{2}^{5}f_{4}^{2}}{f_{1}^{3}}\pmod{8}.$

Hence, for all $n\geq 0,$

(28) $\displaystyle\overline{OPT}(32n+20)\equiv 0\pmod{8}.$

This proves (3) for $\alpha=2$.

Lastly, from equation (27) we have

(29) $\displaystyle\sum_{n\geq
0}\overline{OPT}(8n)q^{2n}\equiv\frac{f_{2}^{2}f_{4}^{3}f_{8}^{5}}{f_{8}^{2}f_{4}^{2}f_{16}^{2}}\pmod{8}$

which means

(30) $\displaystyle\sum_{n\geq
0}\overline{OPT}(8n)q^{n}\equiv\frac{f_{1}^{2}f_{2}f_{4}^{3}}{f_{8}^{2}}\pmod{8},$

and thus we can conclude that, for all $n\geq 0$

(31) $\overline{OPT}(8n)\equiv\overline{OPT}(4n)\pmod{8}$

thanks to (27). Combining equation (31) with (24), (26), and (28), we have
(3). ∎

###### Proof of Theorem 3.

From (17), we have

$\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{9}}{f_{1}^{6}f_{4}^{3}}\equiv\frac{f_{1}^{16}f_{2}}{f_{1}^{6}f_{4}^{3}}\equiv\frac{f_{2}}{f_{4}^{3}}\cdot
f_{1}^{2}\cdot\left(f_{1}^{4}\right)^{2}$
$\displaystyle\equiv\frac{f_{2}}{f_{4}^{3}}\left(\frac{f_{4}^{10}}{f_{2}^{2}f_{8}^{4}}-4q\frac{f_{2}^{2}f_{8}^{4}}{f_{4}^{2}}\right)^{2}\left(\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\right)\pmod{16}.$

Extracting the terms involving even powers of $q$, we arrive at

(32) $\displaystyle\sum_{n\geq 0}\overline{OPT}(2n)q^{n}$
$\displaystyle\equiv\frac{f_{2}^{15}}{f_{1}^{2}f_{4}^{3}f_{8}^{2}}\equiv\frac{f_{2}^{15}}{f_{4}^{3}f_{8}^{2}}\Bigg{(}\frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}}+2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}\Bigg{)}\pmod{16}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(4n+2)q^{n}$ $\displaystyle\equiv
2\frac{f_{1}^{2}f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\equiv
2\frac{f_{2}^{3}f_{8}^{2}}{f_{4}^{3}}\Bigg{(}\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\Bigg{)}\pmod{16}.$

Extracting the terms involving odd powers of $q$, we have

(33) $\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+6)q^{n}$
$\displaystyle\equiv-4\frac{f_{4}f_{8}^{2}}{f_{2}}\pmod{16}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

(34) $\displaystyle\overline{OPT}(16n+14)$ $\displaystyle\equiv 0\pmod{16}.$

This proves (4) for $\alpha=0$.

Extracting the terms involving $q^{2n}$ from (32) and replacing $q^{2}$ by
$q$, we have

(35) $\displaystyle\sum_{n\geq 0}\overline{OPT}(4n)q^{n}$
$\displaystyle\equiv\frac{f_{1}^{10}f_{4}^{3}}{f_{2}^{3}f_{8}^{2}}\equiv\frac{f_{4}^{3}}{f_{2}^{3}f_{8}^{2}}\cdot(f_{1}^{4})^{2}\cdot
f_{1}^{2}$
$\displaystyle\equiv\frac{f_{4}^{3}}{f_{2}^{3}f_{8}^{2}}\left(\frac{f_{4}^{10}}{f_{2}^{2}f_{8}^{4}}-4q\frac{f_{2}^{2}f_{8}^{4}}{f_{4}^{2}}\right)^{2}\left(\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\right)\pmod{16}.$

Extracting the terms involving odd powers of $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+4)q^{n}$
$\displaystyle\equiv-8\frac{f_{2}^{9}f_{4}^{3}}{f_{1}^{2}f_{8}^{2}}-2\frac{f_{2}^{23}f_{8}^{2}}{f_{1}^{6}f_{4}^{11}}$
$\displaystyle\equiv-8\frac{f_{2}^{9}f_{4}^{3}}{f_{2}f_{8}^{2}}-2\frac{f_{2}^{23}f_{8}^{2}}{f_{2}^{4}f_{4}^{11}}f_{1}^{2}$
$\displaystyle\equiv-8\frac{f_{2}^{8}f_{4}^{3}}{f_{8}^{2}}-2\frac{f_{2}^{23}f_{8}^{2}}{f_{2}^{4}f_{4}^{11}}\Bigg{(}\frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{2}f_{16}^{2}}{f_{8}}\Bigg{)}\pmod{16}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}(16n+12)q^{n}$ $\displaystyle\equiv
4\frac{f_{1}^{20}f_{4}f_{8}^{2}}{f_{2}^{11}}\equiv
4\frac{f_{4}f_{8}^{2}}{f_{2}}\pmod{16}.$

Extracting the terms involving $q^{2n+1}$, dividing both sides by $q$ and
replacing $q^{2}$ by $q$, we have

(36) $\displaystyle\overline{OPT}(32n+28)\equiv 0\pmod{16}.$

This proves (4) for $\alpha=1$.

From equation (35) with a little simplification we obtain

$\sum_{n\geq
0}\overline{OPT}(8n)q^{2n}\equiv\frac{f_{4}^{21}}{f_{2}^{6}f_{8}^{5}f_{16}^{2}}\pmod{16}.$

Further simplifying and using (35) yields

(37) $\overline{OPT}(8n)\equiv\overline{OPT}(4n)\pmod{16}.$

Combining equations (37) with (34) and (36) we have (4). ∎

##  5\. Proof of Theorem 5

###### Proof of (5).

From (19), we recall

(38) $\displaystyle\sum_{n\geq 0}\overline{OPT}(2n+1)q^{n}$
$\displaystyle\equiv 2f_{2}^{6}\equiv 2f_{4}^{3}\pmod{4}.$

Employing (15) in the last step, extracting the terms involving $q^{3n+4}$,
dividing both sides by $q^{4}$ and replacing $q^{3}$ by $q$, we arrive at

$\displaystyle\sum_{n\geq 0}\overline{OPT}(6n+9)q^{n}$ $\displaystyle\equiv
2f_{12}^{3}\pmod{4}.$

Further extracting the terms involving $q^{3n}$ and replacing $q^{3}$ by $q$,
we have

(39) $\displaystyle\sum_{n\geq 0}\overline{OPT}(18n+9)q^{n}$
$\displaystyle\equiv 2f_{4}^{3}\pmod{4}.$

From (38) and (39), we have

$\displaystyle\overline{OPT}(2n+1)$ $\displaystyle\equiv\overline{OPT}(18n+9)$
$\displaystyle\equiv\overline{OPT}(9(2n+1))$ $\displaystyle\qquad\vdots$
$\displaystyle\equiv\overline{OPT}(2\cdot 3^{2\alpha}n+3^{2\alpha})\pmod{4}$

for any $\alpha>0$. Hence,

(40) $\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(2\cdot
3^{2\alpha}n+3^{2\alpha})q^{n}\equiv 2f_{4}^{3}\pmod{4}.$

Next, with the help of mathematical induction, we prove that if $p\geq 3$ is a
prime, then for $\delta\geq 0$, we have

(41) $\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2\delta}n+3^{2\alpha}\cdot p^{2\delta})q^{n}\equiv 2\cdot
p^{\delta}(-1)^{\delta\left(\frac{p-1}{2}\right)}f_{4}^{3}\pmod{4},$

Clearly the case $\delta=0$ is true by (40).

Using the $p$-dissection of $f_{1}^{3}$ stated in (16), we obtain

$\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2\delta}n+3^{2\alpha}\cdot p^{2\delta})q^{n}$ $\displaystyle\equiv
2p^{\delta}(-1)^{\delta\left(\frac{p-1}{2}\right)}\Bigg{[}\sum_{k\neq\frac{p-1}{2},k=0}^{p-1}(-1)^{k}q^{4\frac{k(k+1)}{2}}\sum_{0}^{\infty}(-1)^{n}(2pn+2k+1)q^{4pn\cdot\frac{pn+2k+1}{2}}$
$\displaystyle\qquad+p(-1)^{\frac{p-1}{2}}q^{4\frac{p^{2}-1}{8}}f_{4p^{2}}^{3}\Bigg{]}\pmod{4},$

Extracting the terms involving $q^{pn+\frac{p^{2}-1}{2}}$, dividing both sides
by $q^{\frac{p^{2}-1}{2}}$ and replacing $q^{p}$ by $q$, we arrive at

(42) $\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2\delta+1}n+3^{2\alpha}\cdot p^{2(\delta+1)})q^{n}\equiv 2\cdot
p^{\delta+1}(-1)^{(\delta+1)\left(\frac{p-1}{2}\right)}f_{4p}^{3}\pmod{4},$

Extracting the terms involving $q^{pn}$ and replacing $q^{p}$ by $q$, we
obtain

$\displaystyle\sum_{n=0}^{\infty}\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2(\delta+1)}n+3^{2\alpha}\cdot p^{2(\delta+1)})q^{n}\equiv 2\cdot
p^{\delta+1}(-1)^{(\delta+1)\left(\frac{p-1}{2}\right)}f_{4}^{3}\pmod{4},$

which is the $\delta+1$ case of (5). This completes the proof of (41).

Now, comparing the coefficients of $q^{pn+t}$ for $t\in\\{1,2,\cdots,p-1\\}$
from (42), we have

$\displaystyle\overline{OPT}(2\cdot 3^{2\alpha}\cdot
p^{2\delta+1}(pn+t)+3^{2\alpha}\cdot p^{2(\delta+1)})q^{n}\equiv 0\pmod{4},$

which completes the proof of (5). ∎

###### Proof of (6).

From $\eqref{4n+18}$, we recall

(43) $\displaystyle\sum_{n\geq 0}\overline{OPT}(4n+1)q^{n}$
$\displaystyle\equiv
2\frac{f_{4}^{7}}{f_{2}f_{8}^{2}}\Bigg{(}\frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}}+4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}\Bigg{)}+4f_{2}f_{4}\pmod{8}.$

Extracting the even powered terms of $q$, we obtain

$\displaystyle\sum_{n\geq 0}\overline{OPT}(8n+1)q^{n}$ $\displaystyle\equiv
2\frac{f_{2}^{7}}{f_{1}f_{4}^{2}}\cdot\frac{f_{2}^{14}}{f_{1}^{14}f_{4}^{4}}+4f_{1}f_{2}$
(44) $\displaystyle\equiv 6f_{1}f_{2}$ $\displaystyle\equiv
6\left(\frac{f_{6}f_{9}^{4}}{f_{3}f_{18}^{2}}-qf_{9}f_{18}-2q^{2}\frac{f_{3}f_{18}^{4}}{f_{6}f_{9}^{2}}\right)\pmod{8},$

where in the last step, we have employed (14).
Extracting the terms involving $q^{3n+1}$, dividing both sides by $q$ and
replacing $q^{3}$ by $q$, we obtain

$\displaystyle\sum_{n\geq 0}\overline{OPT}(24n+9)q^{n}$
$\displaystyle\equiv-6f_{3}f_{6}\pmod{8}.$

Extracting the terms involving $q^{3n}$ and replacing $q^{3}$ by q, we obtain

(45) $\displaystyle\sum_{n\geq 0}\overline{OPT}(72n+9)q^{n}$
$\displaystyle\equiv-6f_{1}f_{2}\pmod{8}.$

From (45) and (44), we arrive at

$\displaystyle\overline{OPT}(8n+1)$
$\displaystyle\equiv-\overline{OPT}(72n+9)$
$\displaystyle\equiv-\overline{OPT}(8(3^{2}n+1)+1)$
$\displaystyle\equiv\overline{OPT}(72(3^{2}n+1)+9)$
$\displaystyle\equiv\overline{OPT}(8(3^{4}n+3^{2}\cdot 1+1)+1)$
$\displaystyle\qquad\vdots$
$\displaystyle\equiv(-1)^{\alpha}\overline{OPT}\left(8\left(3^{2\alpha}n+\frac{3^{2\alpha}-1}{8}\right)+1\right)$
$\displaystyle\equiv(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}n+3^{2\alpha})\pmod{8}$

for any $\alpha>0$. Hence,

(46) $\displaystyle\sum_{n\geq 0}(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}n+3^{2\alpha})q^{n}\equiv 6f_{1}f_{2}\pmod{8}$

for any $\alpha>0$. Next, with the aid of mathematical induction, we prove
that, if $p$ is a prime satisfying $\left(\frac{-2}{p}\right)=-1$, then for
$\delta\geq 0$,

(47) $\displaystyle\sum_{n\geq 0}(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}\cdot p^{2\delta}n+3^{2\alpha}\cdot
p^{2\delta})q^{n}\equiv(-1)^{\pm\delta\left(\frac{p-1}{3}\right)}6f_{1}f_{2}\pmod{8}.$

Clearly, the case $\delta=0$ is true as (46) holds.

Suppose that the result holds true for some $\delta>0$. Then, by using the
$p$-dissection of $f_{1}$ stated in Lemma 5, we have

$\displaystyle\sum_{n\geq 0}(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}\cdot p^{2\delta}n+3^{2\alpha}\cdot p^{2\delta})q^{n}$
$\displaystyle\equiv
6\cdot(-1)^{\pm\delta\left(\frac{p-1}{3}\right)}\Bigg{[}\sum_{k\neq\frac{\pm
p-1}{6},k=-\frac{p-1}{2}}^{\frac{p-1}{2}}(-1)^{k}q^{\frac{3k^{2}+k}{2}}f\Bigg{(}-q^{\frac{3p^{2}+(6k+1)p}{2}},-q^{\frac{3p^{2}-(6k+1)p}{2}}\Bigg{)}+(-1)^{\frac{\pm
p-1}{6}}q^{\frac{p^{2}-1}{24}}f_{p^{2}}\Bigg{]}$ (48)
$\displaystyle\quad\times\Bigg{[}\sum_{k\neq\frac{\pm
p-1}{6},k=-\frac{p-1}{2}}^{\frac{p-1}{2}}(-1)^{k}q^{2\cdot\frac{3k^{2}+k}{2}}f\Bigg{(}-q^{2\cdot\frac{3p^{2}+(6k+1)p}{2}},-q^{2\cdot\frac{3p^{2}-(6k+1)p}{2}}\Bigg{)}+(-1)^{\frac{\pm
p-1}{6}}q^{2\cdot\frac{p^{2}-1}{24}}f_{2\cdot p^{2}}\Bigg{]}\pmod{8}.$

Now we consider the congruence

(49)
$\displaystyle\frac{3k^{2}+k}{2}+2\cdot\frac{3m^{2}+m}{2}\equiv\frac{p^{2}-1}{8}\pmod{p},$

where $-\frac{p-1}{2}\leq k,m\leq\frac{p-1}{2}$. Since the above congruence is
equivalent to solving the congruence

$\displaystyle(6k+1)^{2}+2(6m+1)^{2}\equiv 0\pmod{p},$

and $\left(\frac{-2}{p}\right)=-1$, it follows that (49) has the unique
solution $k=m=\frac{\pm p-1}{6}$. Therefore, extracting the terms involving
$q^{pn+\frac{p^{2}-1}{8}}$ from both sides of (5), dividing by
$q^{\frac{p^{2}-1}{8}}$ and replacing $q^{p}$ by $q$ we find that

(50) $\displaystyle\sum_{n\geq 0}(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}\cdot p^{2\delta+1}n+3^{2\alpha}\cdot
p^{2(\delta+1)})q^{n}\equiv(-1)^{\pm(\delta+1)\left(\frac{p-1}{3}\right)}6f_{p}f_{2p}\pmod{8}.$

Extracting the terms involving $q^{pn}$ and replacing $q^{p}$ by $q$, we
arrive at

$\displaystyle\sum_{n\geq 0}(-1)^{\alpha}\overline{OPT}(8\cdot
3^{2\alpha}\cdot p^{2(\delta+1)}n+3^{2\alpha}\cdot
p^{2(\delta+1)})q^{n}\equiv(-1)^{\pm(\delta+1)\left(\frac{p-1}{3}\right)}6f_{1}f_{2}\pmod{8},$

which is the $\delta+1$ case of (47). Hence (47) holds.

Now, equating the coefficients of $q^{pn+r}$ for $r\in\\{1,2,\cdots,p-1\\}$
from both sides of (50), we have (6). ∎

##  6\. Proof of Theorem 9

We first state and prove a result that will help us prove Theorem 9.

######  Lemma 7.

For all odd $t\geq 1$, we have

$\displaystyle(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{t}=\sum_{j=0}^{7}a_{t,j}q^{j}F_{t,j}(q^{8}),$

where $F_{t,j}(q^{8})$ is a function of $q^{8}$ whose power series
representation has integer coefficients, and the following divisibilities
hold:

$\displaystyle a_{t,1}$ $\displaystyle\equiv 0\pmod{2},$ $\displaystyle
a_{t,2}$ $\displaystyle\equiv 0\pmod{2},$ $\displaystyle a_{t,3}$
$\displaystyle\equiv 0\pmod{4},$ $\displaystyle a_{t,4}$ $\displaystyle\equiv
0\pmod{2},$ $\displaystyle a_{t,5}$ $\displaystyle\equiv 0\pmod{8},$
$\displaystyle a_{t,6}$ $\displaystyle\equiv 0\pmod{4},$ $\displaystyle
a_{t,7}$ $\displaystyle\equiv 0\pmod{16}.$

###### Proof.

We prove the above lemma with the help of induction on $t$. First, we prove
the result for $t=1$.

By multiple applications of Lemma 3, we have

$\displaystyle\phi(q)\phi(q^{2})\phi(q^{4})^{2}$
$\displaystyle=\left(2q\psi(q^{8})+\phi(q^{16})+2q^{4}\psi(q^{32})\right)$
$\displaystyle\quad\times\left(2q^{2}\psi(q^{16})+\phi(q^{32})+2q^{8}\psi(q^{64})\right)\times\left(2q^{4}\psi(q^{32})+\phi(q^{64})+2q^{16}\psi(q^{128})\right)^{2}.$

Now, we expand this expression in Mathematica and work modulo 16 to arrive at

$\displaystyle\phi(q)\phi(q^{2})\phi(q^{4})^{2}$
$\displaystyle\equiv\phi(q^{16})\phi(q^{32})\phi(q^{64})^{2}+2q\psi(q^{8})\phi(q^{32})\phi(q^{64})^{2}+2q^{2}\psi(q^{16})\phi(q^{16})\phi(q^{64})^{2}$
$\displaystyle\qquad+4q^{3}\psi(q^{8})\psi(q^{16})\phi(q^{64})^{2}+4q^{4}\psi(q^{32})\phi(q^{16})\phi(q^{32})\phi(q^{64})$
$\displaystyle\qquad+2q^{4}\psi(q^{32})\phi(q^{32})\phi(q^{64})^{2}+8q^{5}\psi(q^{8})\psi(q^{32})\phi(q^{32})\phi(q^{64})$
$\displaystyle\qquad+8q^{6}\psi(q^{16})\psi(q^{32})\phi(q^{16})\phi(q^{64})+4q^{6}\psi(q^{16})\psi(q^{32})\phi(q^{64})^{2}$
$\displaystyle\qquad+2q^{8}\psi(q^{64})\phi(q^{16})\phi(q^{64})^{2}+4q^{8}\psi(q^{32})^{2}\phi(q^{16})\phi(q^{32})+8q^{8}\psi(q^{32})^{2}\phi(q^{32})\phi(q^{64})$
$\displaystyle\qquad+8q^{9}\psi(q^{8})\psi(q^{32})^{2}\phi(q^{32})+4q^{9}\psi(q^{8})\psi(q^{64})\phi(q^{64})^{2}+8q^{10}\psi(q^{16})\psi(q^{32})^{2}\phi(q^{16})$
$\displaystyle\qquad+8q^{12}\psi(q^{32})^{3}\phi(q^{32})+8q^{12}\psi(q^{32})\psi(q^{64})\phi(q^{16})\phi(q^{64})+4q^{12}\psi(q^{32})\psi(q^{64})\phi(q^{64})^{2}$
$\displaystyle\qquad+8q^{16}\psi(q^{32})^{2}\psi(q^{64})\phi(q^{16})+4q^{16}\psi(q^{128})\phi(q^{16})\phi(q^{32})\phi(q^{64})$
$\displaystyle\qquad+8q^{17}\psi(q^{8})\psi(q^{128})\phi(q^{32})\phi(q^{64})+8q^{18}\psi(q^{16})\psi(q^{128})\phi(q^{16})\phi(q^{64})$
$\displaystyle\qquad+8q^{20}\psi(q^{32})\psi(q^{128})\phi(q^{16})\phi(q^{32})+8q^{20}\psi(q^{32})\psi(q^{128})\phi(q^{32})\phi(q^{64})$
$\displaystyle\qquad+8q^{24}\psi(q^{64})\psi(q^{128})\phi(q^{16})\phi(q^{64})+4q^{32}\psi(q^{128})^{2}\phi(q^{16})\phi(q^{32})$
$\displaystyle\qquad+8q^{33}\psi(q^{8})\psi(q^{128})^{2}\phi(q^{32})+8q^{34}\psi(q^{16})\psi(q^{128})^{2}\phi(q^{16})$
$\displaystyle\qquad+8q^{36}\psi(q^{32})\psi(q^{128})^{2}\phi(q^{32})+8q^{40}\psi(q^{64})\psi(q^{128})^{2}\phi(q^{16})\pmod{16},$

from which it is clear that the coefficients in the 8-dissection satisfy the
criteria in the statement of the lemma.

Next, we move to the induction step of the proof. Let the result be true for
some $t\geq 1$. Now,

$\displaystyle(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{t+2}$
$\displaystyle=(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{t}\times(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{2}$
$\displaystyle=\left(\sum_{j=0}^{7}a_{t,j}q^{j}F_{t,j}(q^{8})\right)(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{2}.$

Let us write

$\displaystyle\sum_{j=0}^{7}a_{t,j}q^{j}F_{t,j}(q^{8})$
$\displaystyle=b_{t,0}F_{t,0}(q^{8})+2b_{t,1}qF_{t,1}(q^{8})+2b_{t,2}q^{2}F_{t,2}(q^{8})+4b_{t,3}q^{3}F_{t,3}(q^{8})+2b_{t,4}q^{4}F_{t,4}(q^{8})$
$\displaystyle\quad+8b_{t,5}q^{5}F_{t,5}(q^{8})+4b_{t,6}q^{6}F_{t,6}(q^{8})+16b_{t,7}q^{7}F_{t,7}(q^{8}),$

for some integers $b_{t,i}$, $i=1,2,3,4,5,6,7$. Working modulo 16, we have

$\displaystyle\left(\sum_{j=0}^{7}a_{t,j}q^{j}F_{t,j}(q^{8})\right)(\phi(q)\phi(q^{2})\phi(q^{4})^{2})^{2}$
$\displaystyle\equiv
8q^{32}b_{t,0}F_{t,0}(q^{8})\psi(q^{128})^{2}\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{2}+8q^{20}b_{t,4}F_{t,4}(q^{8})\psi(q^{64})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{18}b_{t,2}F_{t,2}(q^{8})\psi(q^{64})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}+8q^{17}b_{t,1}F_{t,1}(q^{8})\psi(q^{64})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+4q^{16}b_{t,0}F_{t,0}(q^{8})\psi(q^{64})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}+8q^{16}b_{t,0}F_{t,0}(q^{8})\psi(q^{128})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{3}$
$\displaystyle\quad+8q^{12}b_{t,4}F_{t,4}(q^{8})\psi(q^{32})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{12}b_{t,4}F_{t,4}(q^{8})\psi(q^{64})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{10}b_{t,2}F_{t,2}(q^{8})\psi(q^{32})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{10}b_{t,0}F_{t,0}(q^{8})\psi(q^{16})\psi(q^{64})\phi(q^{16})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{10}b_{t,2}F_{t,2}(q^{8})\psi(q^{64})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}+8q^{9}b_{t,1}F_{t,1}(q^{8})\psi(q^{32})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{9}b_{t,1}F_{t,1}(q^{8})\psi(q^{64})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}+4q^{8}b_{t,0}F_{t,0}(q^{8})\psi(q^{32})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{8}b_{t,4}F_{t,4}(q^{8})\psi(q^{16})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}+4q^{8}b_{t,0}F_{t,0}(q^{8})\psi(q^{64})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{8}b_{t,0}F_{t,0}(q^{8})\psi(q^{32})^{2}\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{2}+8q^{8}b_{t,4}F_{t,4}(q^{8})\psi(q^{32})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{6}b_{t,2}F_{t,2}(q^{8})\psi(q^{16})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}+8q^{6}b_{t,2}F_{t,2}(q^{8})\psi(q^{32})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{6}b_{t,4}F_{t,4}(q^{8})\psi(q^{16})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}+4q^{6}b_{t,6}F_{t,6}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{6}b_{t,4}F_{t,4}(q^{8})\psi(q^{8})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{5}b_{t,1}F_{t,1}(q^{8})\psi(q^{16})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{5}b_{t,1}F_{t,1}(q^{8})\psi(q^{32})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{5}b_{,5}F_{t,5}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{5}b_{t,0}F_{t,0}(q^{8})\psi(q^{8})\psi(q^{32})\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{5}b_{t,4}F_{t,4}(q^{8})\psi(q^{8})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+4q^{4}b_{t,0}F_{t,0}(q^{8})\psi(q^{16})^{2}\phi(q^{16})^{2}\phi(q^{64})^{4}+4q^{4}b_{t,0}F_{t,0}(q^{8})\phi(q^{16})\phi(q^{32})^{2}\psi(q^{32})\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{4}b_{t,0}F_{t,0}(q^{8})\psi(q^{32})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{3}+8q^{4}b_{t,2}F_{t,2}(q^{8})\psi(q^{16})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}$
$\displaystyle\quad+2q^{4}b_{t,4}F_{t,4}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{4}b_{t,2}F_{t,2}(q^{8})\psi(q^{8})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{3}b_{t,1}F_{t,1}(q^{8})\psi(q^{16})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}+4q^{3}b_{t,3}F_{t,3}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+8q^{3}b_{t,1}F_{t,1}(q^{8})\psi(q^{8})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{3}b_{t,2}F_{t,2}(q^{8})\psi(q^{8})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+4q^{2}b_{t,0}F_{t,0}(q^{8})\psi(q^{16})\phi(q^{16})^{2}\phi(q^{32})\phi(q^{64})^{4}+2q^{2}b_{t,2}F_{t,2}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+4q^{2}b_{t.0}F_{t,0}(q^{8})\psi(q^{8})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+8q^{2}b_{t,1}F_{t,1}(q^{8})\psi(q^{8})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+2qb_{t,1}F_{t,1}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}+4qb_{t,0}F_{t,0}(q^{8})\psi(q^{8})\phi(q^{16})\phi(q^{32})^{2}\phi(q^{64})^{4}$
$\displaystyle\quad+b_{t,0}F_{t,0}(q^{8})\phi(q^{16})^{2}\phi(q^{32})^{2}\phi(q^{64})^{4}\pmod{16},$

which shows that the coefficients of
$\left(\sum_{j=0}^{7}a_{t+2,j}q^{j}F_{t+2,j}(q^{8})\right)$ satisfy the
appropriate divisibilities. ∎

###### Proof of Theorem 9.

From Theorem 12 and the generating function for $\overline{OPT}_{k}(n)$ we
have, for any odd $t$,

$\displaystyle\sum_{n\geq
0}\overline{OPT}_{t}(n)q^{n}=\phi(q)^{t}\phi(q^{2})^{t}\phi(q^{4})^{2t}\phi(q^{8})^{4t}\cdots$

Since $\left(\prod_{i\geq 3}\phi(q^{2^{i}})\right)^{2^{i-1}\cdot t}$ is a
function of $q^{8}$, it is enough to do the 8-dissection of the first three
terms. Also, as the highest modulus involved in the theorem is 16 and the
other moduli are divisors of 16, we will prove our result modulo 16.

Hence,

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{t}(n)q^{n}$
$\displaystyle=\phi(q)^{t}\phi(q^{2})^{t}\phi(q^{4})^{2t}\phi(q^{8})^{4t}\cdots,$
$\displaystyle=\left(\sum_{j=0}^{7}a_{t,j}q^{j}F_{t,j}(q^{8})\right)\left(\prod_{i\geq
3}\phi(q^{2^{i}})\right)^{2^{i-1}\cdot t}.$

The result now follows easily from Lemma 7. ∎

##  7\. Proofs of Theorems 10 and 11

###### Proof of Theorem 10.

We have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{12}}{f_{1}^{8}f_{4}^{4}}\equiv\frac{f_{2}^{4}f_{1}^{8}}{f_{4}^{4}}\equiv\frac{f_{2}^{4}}{f_{4}^{4}}\cdot\left(f_{1}^{4}\right)^{2}\pmod{16}.$

Employing (13) in the last step and extracting the odd powered terms of $q$,
we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(2n+1)q^{n}$
$\displaystyle\equiv 8f_{32}\pmod{16}.$

Further, extracting the odd powered terms of $q$, we obtain

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(4n+1)q^{n}$
$\displaystyle\equiv 8f_{16}\pmod{16}.$

Again, extracting the terms involving $q^{2n+1}$, dividing both sides by $q$
and replacing $q^{2}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(8n+1)q^{n}$
$\displaystyle\equiv 8f_{2}^{3}\pmod{16}.$

Employing (15) in the last step, extracting the terms involving $q^{3n+2}$,
dividing both sides by $q^{2}$ and replacing $q^{3}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(24n+17)q^{n}$
$\displaystyle\equiv 8f_{6}^{3}\pmod{16}.$

Extracting the terms involving $q^{3n}$ and replacing $q^{3}$ by $q$, we have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{4}(72n+17)q^{n}$
$\displaystyle\equiv 8f_{2}^{3}\pmod{16}.$

Hence,

$\displaystyle\overline{OPT}_{4}(8n+1)\equiv\overline{OPT}_{4}(72n+17)\pmod{16}.$

The remainder of the proof is similar to the proof of (5) in Section 5. ∎

###### Proof of Theorem 11.

From the proof of Theorem 7, recall

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{2k+1}(n)q^{n}$
$\displaystyle\equiv\frac{f_{2}^{3}}{f_{1}^{2}f_{4}}\pmod{4}.$

Employing (11) in the last step, extracting the odd powered terms of $q$, we
have

$\displaystyle\sum_{n\geq 0}\overline{OPT}_{2k+1}(2n+1)q^{n}$
$\displaystyle\equiv 2\frac{f_{2}f_{8}^{2}}{f_{1}^{2}f_{4}}\equiv
2f_{4}^{3}\pmod{4}.$

The rest of the proof is exactly similar to the proof of (5) in Section 5. ∎

##  8\. Concluding Remarks

  1. (1)

Keister, Sellers and Vary [KSV09], and Chen [Che10] proved some other results
about overpartition $k$-tuples. It would be interesting to see if we can find
any analogous results for overpartition $k$-tuples with odd parts.

  2. (2)

Chen [Che14] and Kim [Kim11] have proved some congruences for overpartitions
with odd parts and overpartition pairs respectively, using the theory of
modular forms. It would be interesting to explore similar techniques for
overpartition $k$-tuples with odd parts.

  3. (3)

Based on numerical evidence, we conjecture the following counterpart of
Theorem 9.

######  Conjecture 1.

For all $i\geq 1$, $n\geq 0$ and odd $r$, we have

$\displaystyle\overline{OPT}_{2^{i}r}(8n+1)$ $\displaystyle\equiv
0\pmod{2^{i+1}},$ $\displaystyle\overline{OPT}_{2^{i}r}(8n+2)$
$\displaystyle\equiv 0\pmod{2^{2i+1}},$
$\displaystyle\overline{OPT}_{2^{i}r}(8n+3)$ $\displaystyle\equiv
0\pmod{2^{i+3}},$ $\displaystyle\overline{OPT}_{2^{i}r}(8n+4)$
$\displaystyle\equiv 0\pmod{2^{2i+4}},$
$\displaystyle\overline{OPT}_{2^{i}r}(8n+5)$ $\displaystyle\equiv
0\pmod{2^{i+2}},$ $\displaystyle\overline{OPT}_{2^{i}r}(8n+6)$
$\displaystyle\equiv 0\pmod{2^{2i+3}},$
$\displaystyle\overline{OPT}_{2^{i}r}(8n+7)$ $\displaystyle\equiv
0\pmod{2^{i+4}}.$

######  Remark 13.

We note that the first congruence of Conjecture 1 follows from Theorem 6. The
fifth congruence of Conjecture 1 follows from a result analogous to a result
of Keister, Vary and the third author [KSV09, Theorem 9]. We do not prove the
result here, the interested reader can verify the details. This means, only
five congruences of Conjecture 1 remain, and we do not have a unified proof of
them, like we have for the congruences in Theorem 2.

## Acknowledgements

The second author would like to thank Prof. Nayandeep Deka Baruah for his
encouragement and support. The second author was partially supported by an
institutional fellowship for doctoral research from Tezpur University, Assam,
India. He thanks the funding institution.

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