# On Generalizations of a Conjecture of Kang and Park

Ryota Inagaki  University of California, Berkeley<EMAIL_ADDRESS>and  Ryan Tamura  University of California, Berkeley<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

Let $\Delta_{d}^{(a,-)}(n)=q_{d}^{(a)}(n)-Q_{d}^{(a,-)}(n)$ where
$q_{d}^{(a)}(n)$ counts the number of partitions of $n$ into parts with
difference at least $d$ and size at least $a$, and $Q_{d}^{(a,-)}(n)$ counts
the number of partitions into parts $\equiv\pm a\pmod{d+3}$ excluding the
$d+3-a$ part. Motivated by generalizing a conjecture of Kang and Park, Duncan,
Khunger, Swisher, and the second author conjectured that
$\Delta_{d}^{(3,-)}(n)\geq 0$ for all $d\geq 1$ and $n\geq 1$ and were able to
prove this when $d\geq 31$ is divisible by $3$. They were also able to
conjecture an analog for higher values of $a$ that the modified difference
function $\Delta_{d}^{(a,-,-)}(n)=q_{d}^{(a)}(n)-Q_{d}^{(a,-,-)}(n)\geq 0$
where $Q_{d}^{(a,-,-)}(n)$ counts the number of partitions into parts
$\equiv\pm a\pmod{d+3}$ excluding the $a$ and $d+3-a$ parts and proved it for
infinitely many classes of $n$ and $d$.

We prove that $\Delta_{d}^{(3,-)}(n)\geq 0$ for all but finitely many $d$. We
also provide a proof of the generalized conjecture for all but finitely many
$d$ for fixed $a$ and strengthen the results of Duncan et.al. We provide a
conditional linear lower bound on d for the generalized conjecture by using a
variant of Alder’s conjecture. Additionally, we obtain asymptotic evidence
that this modification holds for sufficiently large n.

2010 _Mathematics Subject Classification_. 05A17, 11P82, 11P84, 11F37

_Key words and phrases_. partitions, Rogers-Ramanujan identities, Alder’s
conjecture.

##  1\. Introduction

A partition of a positive integer $n$ is a non-increasing sequence of positive
integers, called parts, that sum to $n$. Let $p(n\mid\mbox{condition})$ be the
number of partitions of $n$ satisfying a certain condition. Euler famously
proved that the number of partitions of a positive integer $n$ into odd parts
equals the number of partitions of $n$ into distinct parts. Two other famous
partition identities are those of Rogers and Ramanujan. The first Rogers-
Ramanujan identity states that the number of partitions of $n$ with parts
having difference at least $2$ is equal to the number of partitions of $n$
with parts congruent to $\pm 1\pmod{5}$ and the second Rogers-Ramanujan
identity states the number of partitions of $n$ with parts at least $2$ and
difference at least $2$ is equal to the number of partitions of $n$ with parts
congruent to $\pm 2\pmod{5}$. These identities are encapsulated by the
$q-$series

$\displaystyle\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q\mbox{;}q)_{n}}$
$\displaystyle=\frac{1}{(q\mbox{;}q^{5})_{\infty}(q^{4}\mbox{;}q^{5})_{\infty}},$
$\displaystyle\sum_{n=0}^{\infty}\frac{q^{n(n+1)}}{(q\mbox{;}q)_{n}}$
$\displaystyle=\frac{1}{(q^{2}\mbox{;}q^{5})_{\infty}(q^{3}\mbox{;}q^{5})_{\infty}},$

where $(a\mbox{;}q)_{0}=1$, and
$(a\mbox{;}q)_{n}=\prod_{k=0}^{n-1}(1-aq^{k})$, where $n=\infty$ is allowed.

Motivated in generalizing the Rogers-Ramanujuan identities, Schur discovered
the number of partitions with parts having difference at least $3$ which have
no consecutive multiples of $3$ as parts is equal to the number of partitions
with parts congruent to $\pm 1\pmod{6}$.

Remarkably Alder [2] and Lehmer [12] showed there are no other partition
identities similar to those of Euler, Rogers-Ramanujan, and Schur. In $1956$,
Alder [1] conjectured a generalization of a related family of partition
identities. Alder’s conjecture states that the number of partitions with parts
that differ by at least $d$ is greater than or equal to the number of
partitions with parts congruent to $\pm 1\pmod{d+3}$. Note that this
conjecture generalizes the Euler, Rogers-Ramanujan, and Schur identities.
Alder’s conjecture was proven by Andrews [5] for $n\geq 1$ and
$d=2^{r}-1,r\geq 4$ in $1971$. In $2004$ and $2008$, Yee [15, 16] proved the
conjecture for $n\geq 1$, $d\geq 32$ and $d=7$. In 2011, the remaining cases
of Alder’s conjecture were proven by Alfes, Jameson, and Lemke Oliver [3] by
using the asymptotic methods of Meinardus [13], [14].

In $2020$, Kang and Park [11] investigated how to construct an analog of
Alder’s conjecture that incorporates the second Rogers-Ramanujan identity.
Kang and Park compared the partition functions

$\displaystyle q_{d}^{(a)}(n):=p(n\mid\text{ parts}\geq a\text{ and parts
differ by at least }d),$ $\displaystyle Q_{d}^{(a)}(n):=p(n\mid\text{
parts}\equiv\pm a\ \left(\mathrm{mod}\ d+3\right)),$

by utilizing the difference function

$\displaystyle\Delta_{d}^{(a)}(n)$
$\displaystyle:=q_{d}^{(a)}(n)-Q_{d}^{(a)}(n).$

In their attempts to create an analog of Alder’s conjecture for the second
Rogers-Ramanujan identity, Kang and Park found that

$\Delta_{d}^{(2)}(n)<0\text{ for some choices of }d,n\geq 1.$

However, by employing a minor modification of $Q_{d}^{(a)}(n)$ by defining for
$d,a,n\geq 1$,

$\displaystyle Q_{d}^{(a,-)}\left(n\right)$ $\displaystyle:=p(n\mid\text{
parts}\equiv\pm a\ \left(\mathrm{mod}\ d+3\right),\text{ excluding the part
}d+3-a),$ $\displaystyle\Delta_{d}^{(a,-)}\left(n\right)$
$\displaystyle:=q_{d}^{(a)}\left(n\right)-Q_{d}^{(a,-)}\left(n\right),$

they presented the following conjecture.

######  Conjecture 1.1 (Kang, Park [11], 2020).

For all $d$, $n\geq 1$,

$\Delta_{d}^{(2,-)}\left(n\right)\geq 0.$

Kang and Park’s conjecture was proven for all but finitely many $d$ by Duncan,
et al. by employing a modification of Alder’s conjecture and the results of
Andrews [5] and Yee [16].

######  Theorem 1.2 (Duncan, et al. [8], 2021).

For all $d\geq 62$ and $n\geq 1$,

$\Delta_{d}^{(2,-)}(n)\geq 0.$

Motivated by Kang and Park’s conjecture, Duncan et al. attempted to find a
generalization for higher values of $a$. Remarkably, they observed that the
removal of the $d+3-a$ as a part appeared to be sufficient when $a=3$.

######  Conjecture 1.3 (Duncan, et al. [8], 2021).

For all $d,n\geq 1$,

$\Delta_{d}^{(3,-)}(n)\geq 0.$

By employing the methods of Duncan, et al. [8], we prove Conjecture 1.3 for
all but finitely many $d$.

######  Theorem 1.4.

For $n\geq 1$ and $d=1,2,91,92,93$ or $d\geq 187$,

$\Delta_{d}^{(3,-)}(n)\geq 0.$

When $a\geq 4$, it is not the case that $\Delta_{d}^{(a,-)}(n)$ is non-
negative for all $d,n\geq 1$. By considering the functions

$\displaystyle Q_{d}^{(a,-,-)}(n)$ $\displaystyle:=p(n\mid\text{
parts}\equiv\pm a\ \left(\mathrm{mod}\ d+3\right),\text{excluding the parts
}a\text{ and }d+3-a),$ $\displaystyle\Delta_{d}^{(a,-,-)}\left(n\right)$
$\displaystyle:=q_{d}^{(a)}(n)-Q_{d}^{(a,-,-)}\left(n\right),$

Duncan, et al. conjectured the following analog of Kang and Park’s conjecture.

######  Conjecture 1.5 (Duncan, et al. [8], 2021).

For $d,a,n\geq$1 with $1\leq a\leq d+2$,

$\Delta_{d}^{(a,-,-)}(n)\geq 0.$

They were able to prove infinitely many cases of Conjecture 1.5 by employing
the methods of Yee [16] and Andrews [5].

Throughout this paper, we let $h_{d}^{(a)}$ and $h_{n}^{(a)}$ denote the least
non-negative residues of $-d$ and $-n$ modulo $a$ respectively. Note that
$\frac{d+h_{d}^{(a)}}{a}=\lceil\frac{d}{a}\rceil$ and
$\frac{n+h_{n}^{(a)}}{a}=\lceil\frac{n}{a}\rceil$.

We prove a strengthening of [8, Theorem 1.6].

######  Theorem 1.6.

For $a\geq 4$ and $\lceil\frac{d}{a}\rceil=31$ or $\lceil\frac{d}{a}\rceil\geq
63$ such that $h_{d}^{(a)}\leq 3$, $d\not\equiv-3\pmod{a}$, and $n\geq 1$,

$\Delta_{d}^{(a,-)}(n)\geq 0.$

Moreover, when $d\equiv-3\pmod{a}$, $\Delta_{d}^{(a,-)}(n)\geq 0$ for all
$n\neq d+3+a$.

In particular, for $a=4$, $121\leq d\leq 124$ or $d\geq 249$, and $n\geq 1$,
$n\neq d+7$ when $d\equiv 1\pmod{4}$,

$\Delta_{d}^{(4,-)}(n)=q_{d}^{(4)}\left(n\right)-Q_{d}^{(4,-)}\left(n\right)\geq
0.$

######  Remark 1.7.

This is a strengthening of [8, Theorem 1.6] since

$\Delta_{d}^{(a,-,-)}(n)\geq\Delta_{d}^{(a,-)}(n)\geq\Delta_{d}^{(a)}(n)\text{
for all }d,a,n\geq 1.$

By developing a new method in comparing these partition functions, we also
prove Conjecture 1.5 for all but finitely many $d$ for fixed $a$.

######  Theorem 1.8.

For $d,n\geq 1$ and $a\geq 5$ such that $\lceil\frac{d}{a}\rceil\geq
2^{a+3}-1$,

$\Delta_{d}^{(a,-,-)}(n)\geq 0.$

######  Remark 1.9.

The cases of Conjecture 1.5 for $1\leq a\leq 3$ have already been proven for
all but finitely many $d$ by Alder’s conjecture, Theorems 1.2 and 1.4. Note
when $a=4$ and $n=d+7$ that Conjecture 1.5 is trivially verified. Hence, with
Theorem 1.6, the case of Conjecture 1.5 for $a=4$ have been proven for all but
finitely many $d$. Hence, we’ll focus on proving Theorem 1.8.

It is a natural question of determining for fixed $a\geq 1$ which $d,n\geq 1$
are sufficient to allow

$\Delta_{d}^{(a,-)}(n)\geq 0.$

Towards answering this question, we present the following Alder-type
inequality.

######  Conjecture 1.10.

For $d,n\geq 12$ such that $n\geq d+2$,

$q_{d}^{(1)}(n)-Q_{d-4}^{(1,-)}(n)\geq 0.$

######  Remark 1.11.

We have learned that Armstrong, et.al [7] have proven Conjecture 1.10 for
$n\geq 1$ and $d\geq 105$. They were also able to obtain a generalization of
Conjecture 1.10, showing for $N\geq 2$, $d\geq\max\\{63,46N-79\\}$, and $n\geq
d+2$, that

$q_{d}^{(1)}(n)\geq Q_{d-N}^{(1,-)}\left(n\right).$

Assuming Conjecture 1.10, we are able to prove a strengthening of Conjecture
1.5 which allows $a$ as a part. This conditional result remarkably reduces the
exponential lower bound of $d$ in Theorem 1.8 to a linear one.

######  Theorem 1.12.

Suppose that Conjecture 1.10 holds for the prescribed bounds. Then for $a\geq
1$, $\lceil\frac{d}{a}\rceil\geq 12$, and $n\geq 5\lceil\frac{d}{a}\rceil+1$,

$\Delta_{d}^{(a,-)}(n)\geq 0.$

Moreover, suppose the bounds on $a$ and $d$ are as above, and
$1\leq\lceil\frac{n}{a}\rceil\leq 5\lceil\frac{d}{a}\rceil$,
$d\not\equiv-3\pmod{a}$, unconditionally we have

$\Delta_{d}^{(a,-)}(n)\geq 0.$

If $d\equiv-3\pmod{a}$, then unconditionally $\Delta_{d}^{(a,-)}(n)\geq 0$ for
all $1\leq\lceil\frac{n}{a}\rceil\leq 5\lceil\frac{d}{a}\rceil$ and $n\neq
d+3+a$.

######  Remark 1.13.

The choice of the upper bound $1\leq\lceil\frac{n}{a}\rceil\leq
5\lceil\frac{d}{a}\rceil$ for the unconditional component of Theorem 1.12
comes from applying the function $\mathcal{G}_{d}^{(1)}(n)$ as defined by Yee
[16].

We now outline the rest of this paper. In Section 2, we present a modification
of [5, Theorem 3] and establish some other important lemmas. In Section 3, we
prove a modification of Alder’s conjecture, which forms the critical component
of the proofs of Theorems 1.4 and 1.6. In Section 4, we prove Theorems 1.4 and
1.6 by reducing to our result in Section 3. In Section 5, we prove Theorem 1.8
by using the partition counting functions from Yee [16]. In Section 6, we
conditionally prove Theorem 1.12 by reducing to the case of $a=1$. We also
prove Conjecture 1.10 for small $n$ to derive our unconditional result in
Theorem 1.12. Finally, in Section 7, we obtain asymptotic evidence that
Proposition 3.1 and Conjecture 1.10 holds for large $n$ and $d\geq 10$ and
$d\geq 12$ respectively. We conclude with potential avenues for resolving more
small $d$ cases of Conjectures 1.3, 1.5 and extending Theorem 1.6.

##  2\. Preliminaries

In this section, we first establish generating functions for our partition
counting functions. We then introduce several lemmas which our results are
based on.

We find by employing combinatorial methods that the generating functions for
$q_{d}^{(a)}(n)$ and $Q_{d}^{(a)}(n)$ for $1\leq a\leq d+2$ are

$\displaystyle\sum_{n=0}^{\infty}q_{d}^{(a)}(n)q^{n}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{q^{d{k\choose
2}+ka}}{(q\mbox{;}q)_{k}},$
$\displaystyle\sum_{n=0}^{\infty}Q_{d}^{(a)}(n)q^{n}$
$\displaystyle=\frac{1}{(q^{d+3-a}\mbox{;}q^{d+3})_{\infty}(q^{a}\mbox{;}q^{d+3})_{\infty}}.$

As described in [8], we find by removing the $d+3-a$ term that the generating
function for $Q_{d}^{(a,-)}\left(n\right)$ is

$\displaystyle\sum_{n=0}^{\infty}Q_{d}^{(a,-)}(n)q^{n}=\begin{cases}\frac{1}{(q^{2d+6-a}\mbox{;}q^{d+3})_{\infty}}&\text{
for }a=\frac{d+3}{2},\\\
\frac{1}{(q^{2d+6-a}\mbox{;}q^{d+3})_{\infty}(q^{a}\mbox{;}q^{d+3})_{\infty}}&\text{
otherwise.}\end{cases}$

Similarly, we find that the generating function of
$Q_{d}^{(a,-,-)}\left(n\right)$ to be

$\displaystyle\sum_{n=0}^{\infty}Q_{d}^{(a,-,-)}(n)q^{n}=\begin{cases}\frac{1}{(q^{2d+6-a}\mbox{;}q^{d+3})_{\infty}}&\text{
for }a=\frac{d+3}{2},\\\
\frac{1}{(q^{2d+6-a}\mbox{;}q^{d+3})_{\infty}(q^{d+3+a}\mbox{;}q^{d+3})_{\infty}}&\text{
otherwise.}\end{cases}$

We employ the following notation: for a set of positive integers $R$, we
define

$\rho(R\mbox{;}n):=p(n\mid\text{parts from the set }R).$

Suppose that $\lambda$ is a partition. We let $\lambda\vdash n$ to denote that
$\lambda$ is a partition of $n$. For fixed positive integers $d$ and $r$,
define as in Andrews [5] the partition function $\rho(T_{r,d}\mbox{;}n)$ with
set of parts

$T_{r,d}=\\{x\in\mathbb{N}\mid x\equiv 1,d+2,\cdots,d+2^{r-1}\pmod{2d}\\}.$

The generating function of $\rho(T_{r,d}\mbox{;}n)$ is

$\sum_{n=0}^{\infty}\rho(T_{r,d}\mbox{;}n)q^{n}=\frac{1}{(q^{1}\mbox{;}q^{2d})_{\infty}(q^{d+2}\mbox{;}q^{2d})_{\infty}\cdots(q^{d+2^{r-1}}\mbox{;}q^{2d})_{\infty}}.$

Suppose that $d\geq 1$. Throughout this paper, we define $r_{d}$ to be the
largest positive integer $r$ such that $d\geq 2^{r}-1$.

We also use the partition function $\mathcal{G}_{d}^{(1)}(n)$ considered by
Yee [16], which is defined by

(2.1)
$\sum_{n=0}^{\infty}\mathcal{G}_{d}^{(1)}(n)q^{n}:=\frac{(-q^{d+2^{r_{d}-1}}\mbox{;}q^{2d})_{\infty}}{(q^{1}\mbox{;}q^{2d})_{\infty}(q^{d+2}\mbox{;}q^{2d})_{\infty}\cdots(q^{d+2^{r_{d}-2}}\mbox{;}q^{2d})_{\infty}}.$

From (2.1), we find that $\mathcal{G}_{d}^{(1)}(n)$ counts partitions with
distinct parts from the set $\\{x\in\mathbb{N}\mid x\equiv
d+2^{r_{d}-1}\pmod{2d}\\}$ and unrestricted parts from the set

$T_{r_{d}-1,d}=\\{y\in\mathbb{N}\mid y\equiv
1,d+2,\cdots,d+2^{r_{d}-2}\pmod{2d}\\}.$

We now state several lemmas that will allow us to prove our main results. We
first present a comparison theorem of Andrews [5, Theorem 3].

######  Theorem 2.1 (Andrews [5]).

Let $S=\\{x_{i}\\}_{i=1}^{\infty}$ and $T=\\{y_{i}\\}_{i=1}^{\infty}$ be two
strictly increasing sequences of positive integers such that $y_{1}=1$ and
$x_{i}\geq y_{i}$ for all $i$. Then for all $n\geq 1$,

$\rho(T\mbox{;}n)\geq\rho(S\mbox{;}n).$

We present the following modification of Theorem 2.1, which will be
extensively employed throughout the rest of this paper.

######  Lemma 2.2.

Let $S=\\{x_{i}\\}$ and $T=\\{y_{i}\\}$ be two strictly increasing sequences
of positive integers such that $y_{1}=a$, $a$ divides each $y_{i}$, and
$x_{i}\geq y_{i}$ for all $i$. Then for all $n\geq 1$,

$\rho(T\mbox{;}n+h_{n}^{(a)})\geq\rho(S\mbox{;}n).$

###### Proof..

We construct an injection $\varphi:X\to Y$ with $X$ and $Y$ being the sets of
partitions counted by $\rho(S\mbox{;}n)$ and $\rho(T\mbox{;}n+h_{n}^{(a)})$
respectively. Suppose $\lambda\in X$. Let $p_{i}$ and $q_{i}$ denote the
multiplicity of $x_{i}$ (wrt $y_{i}$) occurs as a part of $\lambda$ (wrt
$\varphi(\lambda)$).

We define an associated sum of differences for $\lambda$ by

(2.2) $\alpha(\lambda):=\sum_{i\geq 1}p_{i}(x_{i}-y_{i}).$

We observe that $\alpha(\lambda)$ is non-negative since $x_{i}\geq y_{i}$ for
all $i.$ Since $n=\sum_{i\geq 1}p_{i}y_{i}+\alpha$ and $a$ divides
$\sum_{i\geq 1}p_{i}y_{i}$, we have that $a$ divides $\alpha+h_{n}^{(a)}$.

Using (2.2), we define

$q_{i}=\begin{cases}p_{1}+\frac{h_{n}^{(a)}+\alpha(\lambda)}{a},i=1\\\
p_{i},i\geq 2.\end{cases}$

Note

$\displaystyle\sum_{i\geq 1}q_{i}y_{i}$
$\displaystyle=\left(p_{1}+\frac{\alpha(\lambda)+h_{n}^{(a)}}{a}\right)a+\sum_{i\geq
2}p_{i}y_{i}=p_{1}x_{1}+\sum_{i\geq 2}p_{i}x_{i}+h_{n}^{(a)}=n+h_{n}^{(a)},$

hence $\varphi$ is well defined.

We now show that $\varphi$ is injective. Suppose that
$\lambda,\lambda^{\prime}\in X$, such that
$\varphi(\lambda)=\varphi(\lambda^{\prime})$. Let $p_{i}$ and $p_{i}^{\prime}$
denote the multiplicity of $x_{i}$ occurring as parts of $\lambda$ and
$\lambda^{\prime}$ respectively. Observe from construction of $\varphi$ that
we must have $p_{i}=p_{i}^{\prime}$ for all $i\geq 2$. Note that this implies
that

$\alpha(\lambda)=\sum_{i\geq 1}p_{i}(x_{i}-y_{i})=\sum_{i\geq
1}p_{i}^{\prime}(x_{i}-y_{i})^{\prime}=\alpha(\lambda^{\prime}),$

thus $p_{1}=p_{1}^{\prime}$. Hence, we have $\lambda=\lambda^{\prime},$
implying that $\varphi$ is injective.

∎

We also employ the following lemmas of Duncan et al. [8, Lemmas 2.4 and 2.5]
which allow us to use our modification of Alder’s conjecture as described in
Section 3.

######  Lemma 2.3 (Duncan et al. [8]).

For all $d,a\geq 1$ and $n\geq d+2a$,

$q_{d}^{(a)}(n)\geq
q_{\lceil\frac{d}{a}\rceil}^{(1)}\left(\lceil\frac{n}{a}\rceil\right).$

######  Lemma 2.4 (Duncan et al. [8]).

For $d,a,n\geq 1$ such that $a$ divides $d+3$,

$\displaystyle Q_{d}^{(a,-)}\left(an\right)$
$\displaystyle=Q_{\frac{d+3}{a}-3}^{(1,-)}\left(n\right),$ $\displaystyle
Q_{d}^{(a,-,-)}\left(an\right)$
$\displaystyle=Q_{\frac{d+3}{a}-3}^{(1,-,-)}\left(n\right).$

We now present a combined result of Andrews [5], which will be employed in the
proofs of Theorems 1.4 and 1.6.

######  Theorem 2.5 (Andrews [5], 1971).

For $d=2^{r}-1,r\geq 4$, and $n\geq 1$,

$q_{d}^{(1)}\left(n\right)\geq\rho(T_{r,d}\mbox{;}n).$

###### Proof..

The result follows by combining the proofs of [5, Theorems 1 and 4]. ∎

We conclude this section with a result from Yee [16] which will be used in our
modification of Alder’s conjecture.

######  Lemma 2.6 (Yee [16], 2008).

For $d\geq 31,r\geq 1,d\neq 2^{r}-1$, and $n\geq 4d+2^{r_{d}}$,

$q_{d}^{(1)}\left(n\right)\geq\mathcal{G}_{d}^{(1)}(n).$

###### Proof..

Combine both [16, Lemmas 2.2 and 2.7] to obtain the result. ∎

##  3\. A modification of Alder’s conjecture

In this section, we use the work of Andrews [5] and Yee [16] to prove a
modification of Alder’s conjecture. We use this modification to give simple
proofs of Theorems 1.4 and 1.6.

######  Proposition 3.1.

For $d=31$ or $d\geq 63$ and $n\geq d+2$,

$q_{d}^{(1)}(n)\geq Q_{d-3}^{(1,-)}(n).$

We prove Proposition 3.1 in two cases based on the form and size of $d$ and
$n$. We will use throughout the paper the notation $\lambda^{x}$ to denote
that $\lambda$ appears as a part $x$ times in a partition.

######  Lemma 3.2.

For $d+2\leq n\leq 5d$ and $d\geq 31$,

$q_{d}^{(1)}\left(n\right)\geq Q_{d-3}^{(1,-)}\left(n\right).$

###### Proof..

We define

$S_{d}=\\{x\in\mathbb{N}\mid x\equiv\pm 1\pmod{d}\\}\setminus{\\{d-1\\}}$

so that $\rho(S_{d}\mbox{;}n)=Q_{d-3}^{(1,-)}\left(n\right)$. We prove Lemma
3.2 based on relating the size of the parts in $S_{d}$ and the size of $n$.

We note that $q_{d}^{(1)}(n)$ is a weakly increasing function of $n$ since for
any partition of $n$ counted by $q_{d}^{(1)}(n)$, we can add $1$ to its
largest part to create a partition of $n+1$ counted by $q_{d}^{(1)}(n+1)$. In
a similar fashion, $Q_{d-3}^{(1,-)}(n)$ is weakly increasing since for any
partition of $n$ counted by $Q_{d-3}^{(1,-)}(n)$, we can adjoin $1$ as a part
to create a partition of $n+1$ counted by $Q_{d-3}^{(1,-)}(n+1)$.

Notice that for $d+2\leq n\leq 2d-2$ that we have
$q_{d}^{(1)}\left(d+2\right)=2$ with partitions $(d+1,1),(d+2)$. Note
$Q_{d-3}^{(1,-)}\left(2d-2\right)=2$ with partitions
$(d+1,1^{d-3}),(1^{2d-2})$, hence the inequality holds.

We now consider the interval $2d-1\leq n\leq 4d-1$. We notice that
$q_{d}^{(1)}\left(2d-1\right)\geq 16$ since the partitions $(2d-1),(2d-1-i,i)$
with $1\leq i\leq 15$ are counted by $q_{d}^{(1)}\left(2d-1\right)$ due to
$d\geq 31$. Note that at $n=4d-1$, we have
$Q_{d-3}^{(1,-)}\left(4d-1\right)=12$ partitions since there is one partition
with largest part for each element in $\\{4d-1$, $3d+1,3d-1\\}$, two with
largest part $2d+1$, three with largest part for each element in
$\\{2d-1,d+1\\}$, and one with largest part $1$. Hence for all $2d-1\leq n\leq
4d-1$ the inequality holds.

We now verify for all $4d\leq n\leq 5d$ that the inequality holds. Notice that
we have the lower bound $q_{d}^{(1)}\left(4d\right)\geq 47$ since the
partitions $(4d),(4d-i,i)$ with $1\leq i\leq 46$ are counted by
$q_{d}^{(1)}\left(4d\right)$ due to $d\geq 31$. We observe that
$Q_{d-3}^{(1,-)}\left(5d\right)=26$ since there is one partition with largest
part for each element in $\\{5d-1,4d+1\\}$, two partitions with largest part
$4d-1$, three partitions with largest part $3d+1$, four partitions with
largest part $3d-1$, five partitions with largest part for each element in
$\\{2d+1,2d-1\\}$, four partitions with largest part $d+1$, and one partition
with largest part $1$. Hence, we obtain $q_{d}^{(1)}\left(n\right)\geq
Q_{d-3}^{(1,-)}\left(n\right)$ for $4d\leq n\leq 5d$.

∎

######  Lemma 3.3.

For $d=31$ or $d\geq 63$ and $n\geq 4d+2^{r_{d}}$,

$q_{d}^{(1)}(n)\geq Q_{d-3}^{(1,-)}\left(n\right).$

###### Proof..

We prove Lemma 3.3 by showing the following inequalities,

$q_{d}^{(1)}(n)\geq\mathcal{G}_{d}^{(1)}(n)\geq
Q_{d-3}^{(1,-)}\left(n\right).$

In the case when $d\neq 2^{s}-1$, recall that from Lemma 2.6 that
$q_{d}^{(1)}\left(n\right)\geq\mathcal{G}_{d}^{(1)}(n)$. From (2.1),
$\mathcal{G}_{d}^{(1)}(n)\geq\rho(T_{r_{d}-1,d}\mbox{;}n)$. Since $r_{d}\geq
6$, we have $\rho(T_{r_{d}-1,d}\mbox{;}n)\geq\rho(T_{5,d}\mbox{;}n)$ from
Theorem 2.1.

In the case when $d=2^{s}-1$ with $s\geq 5$, from Theorem 2.5, we have
$q_{d}^{(1)}(n)\geq\rho(T_{5,d}\mbox{;}n)$. Hence, in both cases it suffices
to show $\rho(T_{5,d}\mbox{;}n)\geq Q_{d-3}^{(1,-)}\left(n\right)$

Let $S$ and $T$ denote the sets of partitions counted by
$Q_{d-3}^{(1,-)}\left(n\right)$ and $\rho(T_{5,d}\mbox{;}n)$ respectively. We
set $x_{i}\in S_{d}$ and $y_{i}\in T_{5,d}$ to denote the associated $ith$
smallest element of $S_{d}$ and $T_{5,d}$. Using Table 1, note that the only
$i$ where $x_{i}<y_{i}$ is when $i=2$.

We will construct an injection $\varphi:S\to T$. Let $\lambda\vdash n$ be an
element in $S$ and $p_{i}$ and $q_{i}$ denote the number of times $x_{i}$ (wrt
$y_{i}$) occurs as a part of $\lambda$ (wrt $\varphi(\lambda)$). Set

$\alpha:=\sum_{i\neq 2}p_{i}(x_{i}-y_{i})$

to be the difference sum.

Let $S_{1}$ denote the subset of $S$ where the partitions satisfy the
constraint $p_{1}+\alpha\geq p_{2}$. Note that if $\lambda\vdash n$ has
$p_{2}=1$ that $\lambda\in S_{1}$ since $n\geq d+2$. We define the function
$\varphi_{1}:S_{1}\to T$ as follows:

I: $p_{1}+\alpha\geq p_{2}$. We set

$q_{i}=\begin{cases}-p_{2}+p_{1}+\alpha,i=1\\\ p_{i},i\geq 2\\\ \end{cases}$

Table 1. Values of $x_{i}\in S_{d},y_{i}\in T_{5,d}$ for $i=10\alpha+\bar{i}$ where $\bar{i}$ and $\alpha$ are respectively the remainder and quotient from Euclidean division of $i$ by $10$.  $i$ | $x_{i}$ | $y_{i}$
---|---|---
$1$ | $1$ | $1$
$10\alpha+1:\mbox{ ($i\neq 1$})$ | $\left(5\alpha+1\right)d-1$ | $4d\alpha+1$
$10\alpha+2$ | $\left(5\alpha+1\right)d+1$ | $4d\alpha+d+2$
$10\alpha+3$ | $\left(5\alpha+2\right)d-1$ | $4d\alpha+d+4$
$10\alpha+4$ | $\left(5\alpha+2\right)d+1$ | $4d\alpha+d+8$
$10\alpha+5$ | $\left(5\alpha+3\right)d-1$ | $4d\alpha+d+16$
$10\alpha+6$ | $\left(5\alpha+3\right)d+1$ | $4d\alpha+2d+1$
$10\alpha+7$ | $\left(5\alpha+4\right)d-1$ | $4d\alpha+3d+2$
$10\alpha+8$ | $\left(5\alpha+4\right)d+1$ | $4d\alpha+3d+4$
$10\alpha+9$ | $\left(5\alpha+5\right)d-1$ | $4d\alpha+3d+8$
$10\alpha:\alpha>0$ | $\left(5\alpha\right)d+1$ | $4d\alpha-d+16$

Observed that if $\lambda\in S_{1}$ that we have $\varphi_{1}(\lambda)\vdash
n$ since

$\displaystyle\sum_{i\geq 1}q_{i}y_{i}$
$\displaystyle=-p_{2}+p_{1}+\alpha+p_{2}(d+2)+\sum_{i\geq 3}p_{i}y_{i}$
$\displaystyle=p_{1}+p_{2}(d+1)+\sum_{i\geq 3}p_{i}x_{i}=n.$

Hence $\varphi_{1}$ is well-defined.

Let $S_{2}$ denote the set of partitions of $S$ with $p_{2}>p_{1}+\alpha$. We
define $S_{(2,\beta)}\subset S_{2}$ to be the set of partitions which
additionally satisfy $p_{1}+p_{6}=\beta(d-2)+\bar{p}$ with
$\bar{p}\in\\{0,\cdots,d-3\\}$ and $\beta\in\mathbb{Z}_{\geq 0}$. Observe by
construction that $S_{2}$ is the disjoint union
$\bigcup_{\beta\in\mathbb{Z}_{\geq 0}}S_{(2,\beta)}$.

We define for fixed $\beta\geq 0$,

$\epsilon=\epsilon(\lambda)=\begin{cases}0\ \text{if }p_{2}\text{ is even}\\\
1\ \text{if }p_{2}\text{ is odd}.\par\end{cases}$

For each $\beta$, we define the function $\varphi_{(2,\beta)}:S_{(2,\beta)}\to
T$ as follows:

II: $p_{1}+\alpha<p_{2}$ and $\lambda\in S_{(2,\beta)}$. We set

$q_{i}=\begin{cases}\frac{p_{2}-2\beta-3\epsilon}{2}+\alpha+p_{1}-2\beta,i=1\\\
2\beta+\epsilon,i=2\\\ p_{i},3\leq i\leq 5\\\
\frac{p_{2}-2\beta-\epsilon}{2}+p_{6},i=6\\\ p_{i},i\geq 7.\end{cases}$

We now show that $\varphi_{2,\beta}$ is well-defined. If $\lambda\in
S_{2,\beta}$, notice

$\displaystyle\sum_{i\geq 1}q_{i}y_{i}$
$\displaystyle=\left(\frac{p_{2}-2\beta-3\epsilon}{2}+\alpha+p_{1}-2\beta\right)+(2\beta+\epsilon)(d+2)$
$\displaystyle+$
$\displaystyle\left(\frac{p_{2}-2\beta-\epsilon}{2}+p_{6}\right)(2d+1)+\sum_{i\neq
1,2,6}p_{i}y_{i}$
$\displaystyle=p_{1}+\frac{p_{2}-2\beta-3\epsilon}{2}-2\beta+(2\beta+\epsilon)(d+2)$
$\displaystyle+$
$\displaystyle\left(\frac{p_{2}-2\beta-\epsilon}{2}\right)(2d+1)+\sum_{i\geq
3}p_{i}x_{i}$
$\displaystyle=p_{1}+p_{2}-2\beta-\epsilon+(2\beta+\epsilon)(d+1)+d(p_{2}-2\beta-\epsilon)+\sum_{i\geq
3}p_{i}x_{i}$ $\displaystyle=p_{1}+p_{2}(d+1)+\sum_{i\geq 3}p_{i}x_{i}=n.$

Hence $\varphi_{2,\beta}(\lambda)\vdash n$.

In the case when $\epsilon=0$, note $p_{2}-2\beta\geq 0$ since
$p_{2}-\frac{2p_{2}}{d-2}\geq 0$ if $d\geq 4$. Additionally,
$\alpha+p_{1}-2\beta\geq 0$ if $d\geq 4$, thus $q_{1},q_{6}\geq 0$.

In the case when $\epsilon=1$, note if $p_{2}=3$ we must have $\beta=0$, since
$p_{2}>p_{1}+\alpha\geq p_{1}+p_{6}\geq\beta(d-2)$, $d\geq 31$, and
$\lambda\in\varphi_{2,\beta}$. For $p_{2}\geq 5$, note $p_{2}\geq
3+\frac{2p_{2}}{d-2}$ if $d\geq 7$, implying $p_{2}-3-2\beta\geq 0$. Observe
from the definition of $\beta$ that $\alpha+p_{1}-2\beta\geq 0$ for $d\geq 4$.
Thus $q_{1},q_{6}\geq 0$. Hence, we obtain that $\varphi_{2,\beta}$ is well
defined.

We define $\varphi:S\to T$ to be the function defined piecewise from
$\varphi_{1},\varphi_{(2,\beta)}$ as above. In order to show that $\varphi$ is
injective, it suffices to show that $\varphi_{1},\varphi_{(2,\beta)}$ are
injective and that the images of distinct cases are disjoint.

Injectivity of $\varphi_{1}$ follows in the same manner as the function
$\varphi$ present in Lemma 2.2. We now show for fixed $\beta$ that
$\varphi_{(2,\beta)}$ is injective on its domain $S_{(2,\beta)}$. Suppose that
$\lambda,\lambda^{\prime}\in S_{(2,\beta)}$ are partitions such that
$\varphi_{(2,\beta)}(\lambda)=\varphi_{(2,\beta)}(\lambda^{\prime})$. Let
$p_{i}$ and $p_{i}^{\prime}$ denote the multiplicity numbers of $x_{i}$
occurring as a part of $\lambda$ and $\lambda^{\prime}$ respectively.
Similarly, let $\bar{p},\bar{p}^{\prime}$ denote the remainders when
$p_{1}+p_{6},p_{1}^{\prime}+p_{6}^{\prime}$ are divided by $d-2$ respectively.

Observe from the definition of $\varphi_{(2,\beta)}$ that we may immediately
have $p_{i}=p_{i}^{\prime}$ for $i\neq 1,2,6$. Similarly, we must have
$\epsilon(\lambda)=\epsilon(\lambda^{\prime})$ otherwise $q_{2}$ and
$q_{2}^{\prime}$ will have opposite parity. Hence, we will assume that $p_{i}$
for $i\neq 1,2,6$ and $\epsilon(\lambda)$, $\epsilon(\lambda^{\prime})$ are
zero. We obtain from the definition of $\varphi_{(2,\beta)}$ and using that
$\beta$ is fixed the following system of equations

$\displaystyle\frac{p_{2}}{2}+dp_{6}+p_{1}$
$\displaystyle=\frac{p_{2}^{\prime}}{2}+dp_{6}^{\prime}+p_{1}^{\prime}$
$\displaystyle\frac{p_{2}}{2}+p_{6}$
$\displaystyle=\frac{p_{2}^{\prime}}{2}+p_{6}^{\prime}$ $\displaystyle
p_{1}+(d-1)p_{6}$ $\displaystyle=p_{1}^{\prime}+(d-1)p_{6}^{\prime}.$

Assume without loss of generality that $p_{1}\geq p_{1}^{\prime}$. Observe
that since $\lambda,\lambda^{\prime}\in S_{(2,\beta)}$, we have
$(p_{1}-p_{1}^{\prime})+(p_{6}-p_{6}^{\prime})=\bar{p}-\bar{p}^{\prime}<d-2$.
We note that this yields $\mid\bar{p}-\bar{p}^{\prime}\mid<d-2$ since
$0\leq\bar{p},\bar{p}^{\prime}<d-2$. Using this and the third equation above,
we have

(3.1)
$(p_{1}-p_{1}^{\prime})=(\bar{p}-\bar{p}^{\prime})+(p_{6}^{\prime}-p_{6})=(d-1)(p_{6}^{\prime}-p_{6}).$

From (3.1), we have that $\bar{p}=\bar{p}^{\prime}$ since
$d-2\nmid(\bar{p}-\bar{p}^{\prime})$ if $\bar{p}-\bar{p}^{\prime}\neq 0$. This
implies that $(d-2)(p_{6}^{\prime}-p_{6})=0$, which yields that
$p_{6}=p_{6}^{\prime}$ since $d\geq 31$. Via the three equations above, we
obtain that $p_{1}=p_{1}^{\prime}$ and $p_{2}=p_{2}^{\prime}$, thus
$\lambda=\lambda^{\prime}$. Hence $\varphi_{2,\beta}$ is an injection.

We now show that images of the subfunctions forming $\varphi$ are disjoint. We
observe from construction that if $\beta\neq\beta^{\prime}$ that
$\text{im}\varphi_{(2,\beta)}\cap\text{im}\varphi_{(2,\beta^{\prime})}=\varnothing$.
Hence it suffices to show
$\text{im}\varphi_{1}\cap\text{im}\varphi_{(2,\beta)}=\varnothing$ for all
$\beta\in\mathbb{Z}_{\geq 0}$. Let $\lambda\in S_{1}$, $\lambda^{\prime}\in
S_{(2,\beta)}$, and suppose that
$\varphi_{1}(\lambda)=\varphi_{(2,\beta)}(\lambda^{\prime})$. We can assume
from the construction of the function $\varphi_{2,\beta}$ that
$p_{i}=p_{i}^{\prime}=0$ for all $i\neq 1,2,6$.

We observe that we obtain the system of equations

$\displaystyle p_{1}+dp_{6}-p_{2}$
$\displaystyle=p_{1}^{\prime}+dp_{6}^{\prime}+\frac{p_{2}^{\prime}-2\beta-3\epsilon}{2}-2\beta$
$\displaystyle p_{2}$ $\displaystyle=2\beta+\epsilon$ $\displaystyle p_{6}$
$\displaystyle=p_{6}^{\prime}+\frac{p_{2}^{\prime}-2\beta-\epsilon}{2}.$

Using these three equations and $p_{1}^{\prime}<p_{2}^{\prime}$, we obtain

(3.2)
$p_{1}=-d(p_{6}-p_{6}^{\prime})+p_{1}^{\prime}+\frac{p_{2}^{\prime}-2\beta-3\epsilon}{2}+\epsilon<\frac{-d+1}{2}(p_{2}^{\prime}-2\beta-\epsilon)+p_{2}^{\prime}.$

Note from (3.2) that it suffices to show

(3.3) $\frac{-d+1}{2}(p_{2}^{\prime}-2\beta-\epsilon)+p_{2}^{\prime}\leq 0.$

From (3.3), $(d-1)\beta\leq\frac{d-1}{d-2}p_{2}^{\prime}$ and $\epsilon\leq
1$, it suffices to show that
$\frac{d-1}{2}(p_{2}^{\prime}-1)-\frac{d-1}{d-2}p_{2}^{\prime}\geq
p_{2}^{\prime}.$ We notice that $p_{2}^{\prime}-1\geq\frac{p_{2}^{\prime}}{2}$
since $p_{2}^{\prime}\geq 2$. Hence it suffices to show that

$\displaystyle\frac{d-1}{4}\geq 1+\frac{d-1}{d-2},$

which is true for $d\geq 10$. This yields that $p_{1}<0$ which is a
contradiction. ∎

###### Proof of Proposition 3.1..

Combine Lemmas 3.2 and 3.3 to obtain the result. ∎

##  4\. Proofs of Theorems 1.4 and 1.6

In this section, we provide proofs of Theorems 1.4 and 1.6 using the lemmas
established in Section 2 as well as Proposition 3.1.

###  4.1. Proof of Theorem 1.4

We first prove Theorem 1.4 for $d=1$ by employing the Glaisher bijection [9].

######  Proposition 4.1.

For $n\geq 1$ and $d=1$,

$\Delta_{d}^{(3,-)}(n)\geq 0.$

###### Proof..

Let $S$ and $T$ denote the sets of partitions counted by
$Q_{1}^{(3,-)}\left(n\right)$ and $q_{1}^{(3)}\left(n\right)$ respectively.
Observe that the parts of partitions in $S$ are congruent to $\pm 3\pmod{4}$
and strictly greater than $1$. Thus we have that each part is odd and is
greater than or equal to $3$. Suppose $\lambda\in S$ and let $\lambda_{i}$
denote its $ith$ largest part.

We define an injection $\varphi:S\to T$ by modifying the Glaisher bijection
[9]. Let $p_{i}$ denote the multiplicity of $\lambda_{i}$ as a part of
$\lambda$. We write $p_{i}=2^{a_{1}(i)}+\cdots+2^{a_{j}(i)}$ (with
$a_{1}(i)<a_{2}(i)<\cdots<a_{j}(i)$) in its binary expansion. Define $\varphi$
part-wise by

$\varphi(\lambda_{i})=(2^{a_{1}(i)}\lambda_{i},\cdots,2^{a_{j}(i)}\lambda_{i}).$

We observe that $\varphi$ replaces $\lambda_{i}$ with the distinct parts
$2^{a_{1}(i)}\lambda_{i},\cdots,2^{a_{j}(i)}\lambda_{i}$. Since
$\lambda_{i}\geq 3$, the parts of $\varphi(\lambda)$ are greater than or equal
to $3$. Note if $\lambda_{i}\neq\lambda_{j}$ that $2^{a}\lambda_{i}\neq
2^{b}\lambda_{j}$ for any positive integers $a,b$. Hence, the parts of
$\varphi(\lambda)$ are distinct. Observe that $\varphi$ is injective by the
same reasoning in [4]. ∎

We now provide a proof of the $d=2$ case for Theorem 1.4.

######  Proposition 4.2.

For $n\geq 1$ and $d=2$,

$\Delta_{d}^{(3,-)}(n)\geq 0.$

###### Proof..

Recall that the second Rogers-Ramanujan identity yields that for all $n\geq
1$,

(4.1)
$\sum_{n=0}^{\infty}q_{2}^{(2)}(n)q^{n}=\sum_{n=0}^{\infty}\frac{q^{n(n+1)}}{(q\mbox{;}q)_{n}}=\frac{1}{(q^{2}\mbox{;}q^{5})_{\infty}(q^{3}\mbox{;}q^{5})_{\infty}}=\sum_{n=0}^{\infty}Q_{2}^{(2)}(n)q^{n}.$

Multiplying by the factor $(1-q^{2})$ on both sides of (4.1) yields

$\displaystyle(1-q^{3})\sum_{n=0}^{\infty}q_{2}^{(2)}(n)q^{n}$
$\displaystyle=\frac{(1-q^{2})}{(q^{2}\mbox{;}q^{5})_{\infty}(q^{3}\mbox{;}q^{5})_{\infty}}=\sum_{n=0}^{\infty}Q_{2}^{(3,-)}\left(n\right)q^{n}.$

Note that when $n=1,2$, it’s clear that
$q_{2}^{(3)}(n)=Q_{2}^{(3,-)}\left(n\right)=0$. By setting $m=n+2$, it
suffices to show for $m\geq 3$ the inequality,

$q_{2}^{(2)}(m)-q_{2}^{(2)}(m-2)\leq q_{2}^{(3)}(m).$

Let $q_{2}^{(2)}(m)^{*}$ denote the set of partitions of $m$ counted by
$q_{2}^{(2)}(m)$ with the additional property that they contain a part of size
$2$. Note that $q_{2}^{(2)}(m)-q_{2}^{(2)}(m)^{*}=q_{2}^{(3)}(m)$ by
construction. Thus it suffices to show that $q_{2}^{(2)}(m)^{*}\leq
q_{2}^{(2)}(m-2)$.

Let $X$ and $Y$ denote the set of partitions counted by $q_{2}^{(2)}(m)^{*}$
and $q_{2}^{(2)}(m-2)$ respectively. We also let
$\lambda=(\lambda_{1},\cdots,\lambda_{i-1},2)\vdash m$ be a partition counted
by $q_{2}^{(2)}(m)^{*}$. Assuming that $X$ is non-empty, we define the
function $\varphi:X\to Y$,

$\varphi(\lambda)=(\lambda_{1},\cdots,\lambda_{i-1}).$

It is clear from construction that $\varphi(\lambda)\in Y$. We now show that
$\varphi$ is injective. Note that if $\lambda,\lambda^{\prime}\in X$ have
different lengths, then $\varphi(\lambda)\neq\varphi(\lambda^{\prime})$ since
$\varphi$ subtracts the length of the partitions by $1$. Hence we may assume
that $\lambda,\lambda^{\prime}$ have the same length. Since $\varphi$ only
removes the last part of partitions of $X$, we immediately must have
$\lambda=\lambda^{\prime}$ if $\varphi(\lambda)=\varphi(\lambda^{\prime})$.
This yields $q_{2}^{(2)}(m)^{*}\leq q_{2}^{(2)}(m-2)$, which completes the
proof. ∎

###### Proof of Theorem 1.4..

Let $91\leq d\leq 93$ and $d\geq 187$. The case when $n=1,2$ is trivial since
$q_{d}^{(3)}\left(n\right)=Q_{d}^{(3,-)}\left(n\right)=0$. We observe for
$d\geq 1$ and $3\leq n\leq d+5$ that

$\Delta_{d}^{(3,-)}(n)\geq 0$

since $3$ is the only possible part of partition counted by
$Q_{d}^{(3,-)}\left(n\right)$ in the interval. Thus
$Q_{d}^{(3,-)}\left(n\right)\leq 1\leq q_{d}^{(3)}(n)$. In the case when
$n=d+6$, we have $q_{d}^{(3)}(d+6)=2\geq Q_{d}^{(3,-)}\left(d+6\right)$. Now
let $n\geq d+7$. To prove Theorem 1.4 for this range of $d$ and $n$, it
suffices to show the inequality chain

(4.2) $q_{d}^{(3)}(n)\geq
q_{\frac{d+h_{d}^{(3)}}{3}}^{(1)}\left(\frac{n+h_{n}^{(3)}}{3}\right)\geq
Q_{\frac{d+h_{d}^{(3)}}{3}-3}^{(1,-)}\left(\frac{n+h_{n}^{(3)}}{3}\right)=Q_{d+h_{d}^{(3)}-3}^{(3,-)}\left(n+h_{n}^{(3)}\right)\geq
Q_{d}^{(3,-)}\left(n\right).$

The first inequality in (4.2) is justified by Lemma 2.3 for $n\geq d+6$. For
$d$ such that $d+h_{d}^{(3)}=93$ or $d+h_{d}^{(3)}\geq 189$, the second
inequality follows from Proposition 3.1. The equality follows from applying
Lemma 2.4. The final inequality follows by applying Lemma 2.2 with
$\rho(T\mbox{;}n+h_{n}^{(3)})=Q_{d+h_{d}^{(3)}-3}^{(3,-)}\left(n+h_{n}^{(3)}\right)$
and $\rho(S\mbox{;}n)=Q_{d}^{(3,-)}\left(n\right)$. Note that the application
of Lemma 2.2 is justified since $h_{d}^{(3)}\leq 3$.

∎

###  4.2. The Proof of Theorem 1.6

###### Proof of Theorem 1.6..

The case when $1\leq n\leq a-1$ is trivial since
$q_{d}^{(a)}(n)=Q_{d}^{(a,-)}\left(n\right)=0$. We observe for
$d+h_{d}^{(a)}\geq 31a$ and $a\leq n\leq d+(a+2)$ that $a$ is the only
possible part of a partition counted by $Q_{d}^{(a,-)}\left(n\right)$, hence
$Q_{d}^{(a,-)}\left(n\right)\leq 1\leq q_{d}^{(a)}(n)$. Note for $d+(3+a)\leq
n\leq d+2a-1$ that the only potential partitions counted by
$Q_{d}^{(a,-)}\left(n\right)$ are $(a^{\lambda}),(d+3+a)$ with $\lambda$ such
that $d+(3+a)\leq\lambda a\leq d+2a$. Hence $Q_{d}^{(a,-)}\left(n\right)\leq
2$. Note that $Q_{d}^{(a,-)}\left(n\right)=2$ when $n=d+3+a$ and $a$ divides
$d+3+a$. Observe that this happens only when $d\equiv-3\pmod{a}$. Clearly
$q_{d}^{(a)}(n)\geq 1$ on this interval. Thus for $d+h_{d}^{(a)}\geq 31a$,
$1\leq n\leq d+2a$, $n\neq d+3+a$ when $d\equiv-3\pmod{a}$ ,

$\Delta_{d}^{(a,-)}(n)\geq 0.$

Now let $n\geq d+2a$. In order to prove Theorem 1.6, we derive the following
inequality chain

(4.3) $q_{d}^{(a)}(n)\geq
q_{\frac{d+h_{d}^{(a)}}{a}}^{(1)}\left(\frac{n+h_{n}^{(a)}}{a}\right)\geq
Q_{\frac{d+h_{d}^{(a)}}{a}-3}^{(1,-)}\left(\frac{n+h_{n}^{(a)}}{a}\right)=Q_{d+h_{d}^{(a)}-3}^{(a,-)}\left(n+h_{n}^{(a)}\right)\geq
Q_{d}^{(a,-)}\left(n\right).$

We utilize the same argument present in our proof of Theorem 1.4. The first
inequality in (4.3) is justified by Lemma 2.3 for $n\geq d+2a$. For $d$ such
that $d+h_{d}^{(a)}=31a$ or $d+h_{d}^{(a)}\geq 63a$, the second inequality
follows from Proposition 3.1. The equality is a result of applying Lemma 2.4.
The final inequality follows by applying Lemma 2.2 with
$\rho(T\mbox{;}n+h_{n}^{(a)})=Q_{d+h_{d}^{(a)}-3}^{(a,-)}(n+h_{n}^{(a)})$ and
$\rho(S\mbox{;}n)=Q_{d}^{(a,-)}(n)$. The application of Lemma 2.2 holds since
$h_{d}^{(a)}\leq 3$.

We end by remarking that $h_{d}^{(4)}$ is always less than or equal to 3.
Thus, for $n\geq 1$ and $\lceil\frac{d}{4}\rceil=31$ or
$\lceil\frac{d}{4}\rceil\geq 63$, and when $d\equiv 1\pmod{4}$ that $n\neq
d+7$,

$q_{d}^{(4)}\left(n\right)\geq Q_{d}^{(4,-)}\left(n\right).$

∎

##  5\. On the generalized Kang-Park conjecture

In this section, we provide a proof of Theorem 1.8. This allows us to give an
extension of [8, Theorem 1.6].

We define $r_{d,a}$ to be the largest non-negative integer $r$ such that
$2^{r_{d,a}}-1\leq\frac{d+h_{d}^{(a)}}{a}$. Recall that
$\mathcal{G}_{\frac{d+h_{d}^{(a)}}{a}}^{(1)}(\frac{n+h_{n}^{(a)}}{a})$ counts
the number partitions of $\frac{n+h_{n}^{(a)}}{a}$ with the set of parts

$\displaystyle\\{\lambda_{i}\equiv
1,\frac{d+h_{d}^{(a)}}{a}+2,\cdots,\frac{d+h_{d}^{(a)}}{a}+2^{r_{d,a}-2},\frac{d+h_{d}^{(a)}}{a}+2^{r_{d,a}-1}\pmod{2(\frac{d+h_{d}^{(a)}}{a})}\\},$

where parts congruent to
$\frac{d+h_{d}^{(a)}}{a}+2^{r_{d,a}-1}\pmod{2(\frac{d+h_{d}^{(a)}}{a})}$ are
distinct.

For the intermediate partition functions in this section, we define for fixed
$\ell,a,d\geq 1$ the set $T^{\ell}_{a,d}$ to be

$T^{\ell}_{a,d}:=\\{\lambda_{i}\equiv\ell,(d+2)\ell,\cdots,(d+2^{a-1})\ell\pmod{2d\ell}\\}.$

We will use $T^{\ell}_{a,d}$ to compare $\rho(T^{1}_{a,d}\mbox{;}n)$ and
$\mathcal{G}^{(1)}_{d}(n)$ with other partition functions of similar forms and
$Q_{d}^{(a,-,-)}\left(n\right)$.

We will prove Theorem 1.8 in three cases based on the form and size of $d$ and
$n$. In addressing all three cases, we use the following result.

######  Lemma 5.1.

Let $d$, $a$, and $n$ be positive integers such that $a\geq 5$,
$d+h_{d}^{(a)}\geq 2^{a+3}a-a$. Then,

$\rho(T_{a,d+h_{d}^{(a)}}^{a};n+h_{n}^{(a)})\geq
Q_{d}^{(a,-,-)}\left(n\right).$

###### Proof..

Let $S_{d}$ denote the set of allowed parts of partitions counted by the
functions $Q_{d}^{(a,-,-)}\left(n\right)$. We set $x_{i}\in S_{d}$ and
$y_{i}\in T_{a,d+h_{d}^{(a)}}^{a}$ to denote the $ith$ part of their
respective sets in ascending order. To compare the values of $x_{i},y_{i}$, we
use that $d+h_{d}^{(a)}\leq d+a$, $d+h_{d}^{(a)}\geq a2^{a+3}-a$, and
$a\leq\frac{d}{15}$. Note that by using Table 2, we find that a sufficient
bound for $d$ to ensure $x_{i}\geq y_{i}$ is

(5.1) $d\geq 2^{a-1}a+2a-3.$

Table 2.  Values of $x_{i}$, $y_{i}$ over $j$ for $a\geq 5$ where $j$ and
$\alpha$ are integers such that $j\in\\{2,3,...,a\\}$ and $i=a\ell+j$ for some
$\ell\in\mathbb{N}$.

$i$ | $x_{i}$ | $y_{i}$
---|---|---
$a\ell+1:\mbox{ ($a\ell+1$ is even)}$ | $\left(\frac{a}{2}\alpha+\frac{3}{2}\right)(d+3)-a$ | $2(d+h_{d}^{(a)})\alpha+a$
$a\ell+1:\mbox{ ($a\ell+1$ is odd)}$ | $\left(\frac{a}{2}\alpha+1\right)(d+3)+a$ | $2(d+h_{d}^{(a)})\alpha+a$
$a\ell+j:\mbox{ ($a\ell+j$ is even)}$ | $\left(\frac{a}{2}\alpha+\frac{j+2}{2}\right)(d+3)-a$ | $2(d+h_{d}^{(a)})\alpha+(d+h_{d}^{(a)})+2^{j-1}a$
$a\ell+j:\mbox{ ($a\ell+j$ is odd)}$ | $\left(\frac{a}{2}\alpha+\frac{j+1}{2}\right)(d+3)+a$ | $2(d+h_{d}^{(a)})\alpha+(d+h_{d}^{(a)})+2^{j-1}a$

By our assumption on $d$, inequality (5.1) is satisfied, hence $x_{i}\geq
y_{i}$. Since $x_{i}\geq y_{i}$ for all positive $i$, we can apply Lemma 2.2
with
$\rho(T\mbox{;}n+h_{n}^{(a)})=\rho(T_{a,d+h_{d}^{(a)}}^{a}\mbox{;}n+h_{n}^{(a)})$
and $\rho(S\mbox{;}n)=Q_{d}^{(a,-,-)}\left(n\right)$ to obtain the result.

∎

Now we prove Theorem 1.8. For brevity we denote $m=\frac{n+h_{n}^{(a)}}{a}$
and $k=\frac{d+h_{d}^{(a)}}{a}$.

###### Proof of Theorem 1.8..

In the case when $1\leq n\leq d+2+a$, one can check that
$Q_{d}^{(a,-,-)}\left(n\right)=0$, hence $q_{d}^{(a)}(n)\geq
Q_{d}^{(a,-,-)}\left(n\right)$. When $d+3+a\leq n\leq d+2a$, note that
$Q_{d}^{(a,-,-)}\left(n\right)\leq 1\leq q_{d}^{(a)}(n)$, since
$Q_{d}^{(a,-,-)}\left(n\right)=1$ only when $n=d+a+3$.

We now consider the case when $n>d+2a$. For these $n$ we consider the
following inequality chain:

(5.2) $\begin{split}q_{d}^{(a)}\left(n\right)\geq
q_{k}^{(1)}\left(m\right)\geq\rho(T_{a,k}^{1}\mbox{;}m)=\rho(T_{d+h_{d},a}^{a}\mbox{;}n+h_{n}^{(a)})\geq
Q_{d}^{(a,-,-)}\left(n\right).\end{split}$

We note that the first inequality in (5.2) holds by Lemma 2.3. The last
inequality holds by Lemma 5.1. The equality follows from the natural bijection
of multiplying the parts by $a$. Therefore we reduce to showing that

(5.3) $q_{k}^{(1)}\left(m\right)\geq\rho(T_{a,k}^{1}\mbox{;}m)$

for the following three cases.

Case 1: Let $m\geq 4k+2^{r_{k}}$ and $k\neq 2^{s}-1$. We prove Inequality
(5.3) by showing

(5.4)
$q_{k}^{(1)}\left(m\right)\geq\mathcal{G}_{k}^{(1)}(m)\geq\rho(T^{1}_{a,k}\mbox{;}m).$

The first inequality of (5.4) is a result from Lemma 2.6. The second
inequality follows from $k=\frac{d+h_{d}^{(a)}}{a}\geq 2^{a+3}-1$ and that
there are more parts allowed for partitions counted by
$\mathcal{G}_{k}^{(1)}(m)$ than $\rho(T^{1}_{a,k}\mbox{;}m)$.

Case 2: Let
$\frac{n+h_{n}^{(a)}}{a}\geq\frac{4(d+h_{d}^{(a)})}{a}+2^{r_{d,a}}$ and
$\frac{d+h_{d}^{(a)}}{a}=2^{r_{d,a}}-1$. We prove (5.3) by showing

(5.5)
$q_{k}^{(1)}\left(m\right)\geq\rho(T_{r_{d,a},k}^{1}\mbox{;}m)\geq\rho(T^{1}_{a,k}\mbox{;}m).$

We observe that it suffices to prove the inequality when $r_{d,a}=a+3$ since
this yields the minimum number of congruence classes for
$\rho(T^{1}_{r_{d,a},k}\mbox{;}m)$. Observe that the first inequality of (5.5)
follows from Theorem 2.5. The second inequality follows from
$k=\frac{d+h_{d}^{(a)}}{a}\geq 2^{a+3}-1$ and that there are more parts
allowed for partitions counted by $\rho(T^{1}_{r_{d,a},k}\mbox{;}m)$ than
$\rho(T_{a,k}^{1}\mbox{;}m)$.

Case 3: Let $m\leq 4k+2^{r_{k}}$. It suffices to show for $k\geq 2^{a+3}-1$
and $1\leq m\leq 5k+1$ the inequality (5.3).

We again use that $q_{k}^{(1)}(m)$ is a weakly increasing function. We also
observe that $\rho(T^{1}_{a,k}\mbox{;}m)$ is weakly increasing since we can
add $1$ to be an additional part of a partition of $m$ to obtain a partition
for $m+1$.

We prove (5.3) for the interval $1\leq m\leq 2k+6$. Note that both functions
on the interval $1\leq m\leq k+1$ return one, hence we suppose that $k+2\leq
m\leq 2k$. We notice that $\rho(T^{1}_{a,k}\mbox{;}k+2^{i})=i+1$ for $1\leq
i\leq a-1$ and is constant on the intervals $2^{i}+k\leq m\leq k+2^{i+1}-1$.
We note that $q_{k}^{(1)}\left(2^{i}+k\right)\geq i+1$ for $i\geq 1$ since the
partitions $(k+2^{i}),(k+2^{i}-\alpha,\alpha)$ with $\alpha\leq 2^{i-1}$ are
counted by $q_{k}^{(1)}(2^{i}+k)$.

Note that in the interval $k+2^{a-1}\leq m\leq 2k$ that we have
$\rho(T^{1}_{a,k}\mbox{;}m)=a$. Hence, we consider the value of the functions
on the interval $2k+1\leq m\leq 2k+6$. We have that the partitions
$(2k+1-\alpha,\alpha)$ with $\alpha\leq\lfloor\frac{k+1}{2}\rfloor$ are
counted by $q_{k}^{(1)}\left(2k+1\right)$, yielding
$q_{k}^{(1)}\left(2k+1\right)\geq 1+\lfloor\frac{k+1}{2}\rfloor\geq
2^{a+2}+1$. We have $\rho(T^{1}_{a,k}\mbox{;}2k+6)=a+3$, from counting the
number of partitions with largest parts $2k+1$,
$k+2^{a-1},k+2^{a-2},...,k+2,1$. Therefore, for $2k+1\leq m\leq 2k+6$,
$q_{k}^{(1)}\left(m\right)\geq\rho(T^{1}_{a,k}\mbox{;}m)$

We now show (5.3) for the interval $2k+6\leq m\leq 3k+1$. Since
$q_{k}^{(1)}(m)$ is weakly increasing, we will find a lower bound for
$q_{k}^{(1)}\left(2k+6\right)$. The partitions of $2k+6$ of the form
$(2k+6-i,i)$ where $i\in\\{1,...,\lfloor\frac{k}{2}+3\rfloor\\}$ and the
trivial partition $(2k+6)$ are counted by $q_{k}^{(1)}\left(2k+6\right)$. From
this, we obtain the lower bound
$q_{k}^{(1)}\left(2k+6\right)\geq\lfloor\frac{k}{2}+3\rfloor+1$. By $k\geq
2^{a+3}-1$, we find that

$\displaystyle
q_{k}^{(1)}(2k+6)\geq\lfloor\frac{2^{a+3}-1}{2}+3\rfloor+1=2^{a+2}+3.$

Since $\rho(T^{1}_{a,k}\mbox{;}m)$ is weakly increasing, we will provide an
upper bound for $\rho(T^{1}_{a,k}\mbox{;}3k+1)$. The largest part that could
be in a partition counted by $\rho(T^{1}_{a,k}\mbox{;}3k+1)$ is $2k+1$. We
observe that $(2k+1,1^{k})$ is the only partition of $3k+1$ counted by
$\rho(T\mbox{;}3k+1)$ that includes $2k+1$ since $3k+1-2k-1=k$.

We now consider partitions counted by $\rho(T^{1}_{a,k}\mbox{;}3k+1)$ with
largest part $k+2^{\ell}$ where $\ell\in\\{1,2,...,a-1\\}$. For each $\ell$,
we notice that $3k+1-k-2^{\ell}=2k+1-2^{\ell}$. Therefore, any partition whose
largest part is of the form $k+2^{\ell}$ can have at most one other element of
the form $k+2^{w}$ with $w\in\\{1,....,\ell\\}$. Thus, for each fixed
$\ell\in\\{1,...,a-1\\}$, there are at most $(\ell+1)$ partitions of $3k+1$
such that $k+2^{\ell}$ is the largest part. Finally, there is only one
partition of $3k+1$ with largest part of 1. Therefore, we obtain that

$\displaystyle\rho(T^{1}_{a,k}\mbox{;}3k+1)\leq\frac{(a+2)(a-1)}{2}+1+1=1+\frac{a(a+1)}{2}.$

Note that since $2^{a+2}+3\geq\frac{a(a+1)}{2}+1$ for $a\geq 1$, we have that
(5.3) holds for this interval.

We now prove (5.3) for $3k+1\leq m\leq 4k+1$. Since $q_{k}^{(1)}(m)$ is weakly
increasing, we will find a lower bound for $q_{k}^{(1)}\left(3k+1\right)$. The
partitions of $3k+1$ of the form $(3k+1-i,i)$ for
$i\in\\{1,...,\lfloor\frac{2k+1}{2}\rfloor\\}$ and the trivial partition
$(3k+1)$ are counted by $q_{k}^{(1)}\left(3k+1\right)$. Thus, we have that

$\displaystyle
q_{k}^{(1)}(3k+1)\geq\lfloor\frac{2k+1}{2}\rfloor+1\geq\lfloor\frac{2\cdot(2^{a+3}-1)+1}{2}\rfloor+1=2^{a+3}.$

We now provide an upper bound for $\rho(T^{1}_{a,k}\mbox{;}m)$ in $3k+1\leq
m\leq 4k+1$ by obtaining an upper bound for $\rho(T^{1}_{a,k}\mbox{;}4k+1)$.
By repeating the same procedure by considering partitions with fixed largest
part, there is one partition with largest part $4k+1$, there are at most $a-1$
partitions with largest part of the form $3k+2^{\ell}$ with
$\ell\in\\{1,2,...,a-1\\}$, at most $a$ partitions with largest part $2k+1$,
and for each $h\in\\{1,\cdots,a-1\\}$ at most $(h+1)^{2}$ partitions with
largest part $k+2^{h}$ by considering the partitions with one, two, and three
parts that are not equal to $1$. Finally, there is only one partition of
$4k+1$ whose largest part is 1. Using this method, we have

$\displaystyle\rho(T^{1}_{a,k}\mbox{;}4k+1)\leq
1+(a-1)+a+(2^{2}+3^{2}+...+a^{2})+1=2a+\frac{a(a+1)(2a+1)}{6}.$

Thus, we have that (5.3) holds for $3k+1\leq m\leq 4k+1$ since $k\geq
2^{a+3}-1$.

Finally, we show that (5.3) holds for $4k+1\leq m\leq 5k+1$. Since
$q_{k}^{(1)}(m)$ is weakly increasing, we will find a lower bound for
$q_{k}^{(1)}\left(4k+1\right)$. The partitions of $4k+1$ of the form
$(4k+1-i,i)$ where $i\in\\{1,....,\lfloor\frac{3k+1}{2}\rfloor\\}$ and the
trivial partition $(4k+1)$ are counted by $q_{k}^{(1)}(4k+1)$. This yields the
lower bound,

$\displaystyle
q_{k}^{(1)}(4k+1)\geq\lfloor\frac{3k+1}{2}\rfloor+1\geq\lfloor\frac{3\cdot(2^{a+3}-1)+1}{2}\rfloor+1=3\cdot
2^{a+2}.$

We now provide an upper bound for $\rho(T^{1}_{a,k}\mbox{;}m)$ in $4k+1\leq
m\leq 5k+1$, which we’ll do by obtaining an upper bound for
$\rho(T^{1}_{a,k}\mbox{;}5k+1)$. The only partition with largest part $4k+1$
is $(4k+1,1^{k})$. By repeating the same procedure used in the interval
$2k+6\leq m\leq 3k+1$, we find that there are at most $a(a-1)$ partitions with
largest part of the form $3k+2^{\ell}$ with $\ell\in\\{1,2,...,a-1\\}$, at
most $1+\frac{a(a+1)}{2}$ partitions with largest part $2k+1$ by considering
various fixed second largest potential parts. We also find for each
$h\in\\{1,\cdots,a-1\\}$ at most $(h+1)^{3}$ partitions whose largest part is
$k+2^{h}$ by considering that there are at most four parts that are not equal
to $1$. Finally there is only one partition of $5k+1$ whose largest part is
$1$. We obtain by adding the upper bound

$\displaystyle\rho(T^{1}_{a,k}\mbox{;}5k+1)\leq
1+a(a-1)+1+\frac{a(a+1)}{2}+\frac{a^{2}(a+1)^{2}}{4}.$

Because $k\geq 2^{a+3}-1$, we have that (5.3) holds for this interval.

To summarize, we obtain bounds for $q_{k}^{(1)}\left(m\right)$ and
$\rho(T^{1}_{a,k}\mbox{;}m)$ displayed in Table 3. Notice that for $a\geq 5$,
this implies that $q_{k}^{(1)}\left(m\right)\geq\rho(T^{1}_{a,k}\mbox{;}m)$
for $2k+6\leq m\leq 5k+1$. This completes the proof of (5.3) for $1\leq m\leq
5k+1$

Table 3. Bounds on $q_{k}^{(1)}\left(m\right)$ and
$\rho(T_{a,k}^{1}\mbox{;}m)$ for intervals of $m$.

$m$ | $q_{k}^{(1)}\left(m\right)$ | $\rho(T^{1}_{a,k}\mbox{;}m)$
---|---|---
$2k+6\leq m\leq 3k+1$ | $\geq 2^{a+2}+3$ | $\leq 1+\frac{a(a+1)}{2}$
$3k+1\leq m\leq 4k+1$ | $\geq 2^{a+3}$ | $\leq 2a+\frac{a(a+1)(2a+1)}{6}$
$4k+1\leq m\leq 5k+1$ | $\geq 3\cdot 2^{a+2}$ | $\leq 2+a(a-1)+\frac{a(a+1)}{2}+\frac{a^{2}(a+1)^{2}}{4}.$

∎

##  6\. A strengthening of the generalized Kang-Park conjecture

In this section, we provide a proof of Theorem 1.12 by using Conjecture 1.10.
We also prove Conjecture 1.10 for $d+2\leq n\leq 5d$ to obtain that Theorem
1.12 holds unconditionally for the specified range of $n$ in Theorem 1.12.

######  Proof of conditional part of Theorem 1.12..

Via work done in the unconditional component of Theorem 1.12, we assume that
$n\geq d+2a$. We employ the following inequality chain:

$\displaystyle q_{d}^{(a)}(n)\geq
q_{\frac{d+h_{d}^{(a)}}{a}}^{(1)}\left(\frac{n+h_{n}^{(a)}}{a}\right)\geq
Q_{\frac{d+h_{d}^{(a)}}{a}-4}^{(1,-)}(\frac{n+h_{n}^{(a)}}{a})$
$\displaystyle=Q_{d+h_{d}^{(a)}-a-3}^{(a,-)}\left(n+h_{n}^{(a)}\right)\geq
Q_{d}^{(a,-)}\left(n\right).$

We note that the first inequality holds for $n\geq d+2a$ by Lemma 2.3. The
second inequality follows by assuming Conjecture 1.10. The equality follows
from the bijection by multiplying the parts of partitions counted by
$Q_{\frac{d+h_{d}^{(a)}}{a}-4}^{(1,-)}\left(\frac{n+h_{n}^{(a)}}{a}\right)$ by
$a$. We now show the last inequality.

Let $X$ and $Y$ be the set of partitions counted by
$Q_{d}^{(a,-)}\left(n\right)$ and
$Q_{d+h_{d}^{(a)}-3-a}^{(a,-)}\left(n+h_{n}^{(a)}\right)$ respectively. Let
$x_{i}$ and $y_{i}$ denote the allowed parts of partitions in $X$ and $Y$ with
indexing with respect to increasing size. Note that for all $i\geq 1$ that we
have $x_{i}\geq y_{i}$ since $h_{d}^{(a)}\leq a$. This allows us to apply
Lemma 2.2, yielding
$Q_{d+h_{d}^{(a)}-a-3}^{(a,-)}\left(n+h_{n}^{(a)}\right)\geq
Q_{d}^{(a,-)}\left(n\right)$. ∎

We now proceed with the unconditional part of Theorem 1.12. We first prove
Theorem 1.12 in the case when $1\leq n\leq d+2a$.

######  Lemma 6.1.

Let $a\geq 1$, $\lceil\frac{d}{a}\rceil\geq 12$, and $1\leq n\leq d+3a$. If
$d\not\equiv-3\pmod{a}$, we have

$\Delta_{d}^{(a,-)}(n)\geq 0.$

If $d\equiv-3\pmod{a}$ then $\Delta_{d}^{(a,-)}(n)\geq 0$ for all $1\leq n\leq
d+3a$ except when $n=d+3+a$.

###### Proof..

The case when $1\leq n\leq a-1$ is trivial since
$q_{d}^{(a)}(n)=Q_{d}^{(a,-)}\left(n\right)=0$. Observe for $a\leq n\leq
d+2+a$ that $q_{d}^{(a)}\left(n\right)\geq 1\geq Q_{d}^{(a,-)}\left(n\right)$.
Note for $d+3+a\leq n\leq d+2a$ that $q_{d}^{(a)}(n)\geq 1\geq
Q_{d}^{(a,-)}\left(n\right)$. If $d\not\equiv-3\pmod{a}$, we have
$Q_{d}^{(a,-)}\left(d+a+3\right)=1$, hence $q_{d}^{(a)}\left(d+a+3\right)\geq
1=Q_{d}^{(a,-)}\left(d+a+3\right)$.

If $d\equiv-3\pmod{a}$, we find that the only $n$ in the interval with
$Q_{d}^{(a,-)}\left(n\right)=2$ is $n=d+3+a$. Thus for any $d+3+a\leq n\leq
d+2a$ and $n\neq d+3+a$, we have $q_{d}^{(a)}(n)\geq 1\geq
Q_{d}^{(a,-)}\left(n\right)$.

We now consider the case when $d+2a\leq n\leq d+3a$. Observe that
$q_{d}^{(a)}(n)\geq 2$ with partitions $(n)$ and $(n-a,a)$ since $n\geq d+2a$.
Notice that $Q_{d}^{(a,-)}\left(n\right)\leq 2$ since the only partitions in
the interval are $a^{\lambda}$ and $(d+3+a,a)$. Thus $q_{d}^{(a)}(n)\geq 2\geq
Q_{d}^{(a,-)}\left(n\right)$. ∎

We now give an unconditional proof of the second statement of Theorem 1.12 for
the case where $\lceil\frac{d+2a}{a}\rceil\leq\lceil\frac{n}{a}\rceil\leq
5\lceil\frac{d}{a}\rceil$. For this we use the following lemma.

######  Lemma 6.2.

For integers $d\geq 12$ and $d+2\leq n\leq 5d$,

$q_{d}^{(1)}\left(n\right)\geq Q_{d-4}^{(1,-)}\left(n\right).$

###### Proof of Lemma 6.2..

Recall that $q_{d}^{(1)}\left(n\right)$ is weakly increasing since for any
partition of positive integer $n$ counted by the function
$q_{d}^{(1)}\left(n\right)$, one can add 1 to the largest part of the
partition to obtain a partition counted by $q_{d}^{(1)}\left(n+1\right)$.
Similarly, $Q_{d-4}^{(1,-)}(n)$ is weakly increasing since for any partition
of $n$ counted by $Q_{d-4}^{(1,-)}(n)$, we can adjoin $1$ as a part to create
a partition of $n+1$ counted by $Q_{d-4}^{(1,-)}(n+1)$.

We begin by showing that $q_{d}^{(1)}\left(n\right)\geq
Q_{d-4}^{(1,-)}\left(n\right)$ for $d+2\leq n\leq 2d-4$. Note that
$q_{d}^{(1)}\left(d+2\right)\geq 2$ since the partitions $(1,d+1)$ and $(d+2)$
are counted. Note that $Q_{d-4}^{(1,-)}\left(2d-4\right)=2$ with associated
partitions $(d,1^{d-4}),(1^{2d-4})$. Thus within the interval $d+2\leq n\leq
2d-4$ the inequality holds.

We now show $q_{d}^{(1)}\left(n\right)\geq Q_{d-4}^{(1,-)}\left(n\right)$ for
$2d-3\leq n\leq 3d-5$. We note that $Q_{d-4}^{(1,-)}\left(3d-5\right)=5$ with
associated partitions
$((2d-1),1^{d-4}),((2d-3),1^{d-2}),(d^{2},1^{d-5}),(d,1^{2d-5}),(1^{3d-5})$,
hence it suffices to show that $q_{d}^{(1)}(n)\geq 5$ in this interval. We
notice that the partitions of $2d-3$ in the set
$\\{(2d-3)\\}\cup\\{(2d-3-i,i):i\in\\{1,2,...,\lfloor\frac{d-3}{2}\rfloor\\}\\}$
are counted by $q_{d}^{(1)}(2d-3)$. Therefore, $q_{d}^{(1)}(2d-3)\geq
1+\lfloor\frac{d-3}{2}\rfloor$. Since $d\geq 12$, we have that
$q_{d}^{(1)}(2d-3)\geq 5$ as desired.

We now show that $q_{d}^{(1)}(n)\geq Q_{d-4}^{(1,-)}\left(n\right)$ for
$3d-4\leq n\leq 4d-6$. We note that $Q_{d-4}^{(1,-)}\left(4d-6\right)=11$
since there is one partition with largest part for each element in
$\\{3d-2,3d-4\\}$, two with largest part of $2d-1$, three with largest part
for each element in $\\{2d-3,d\\}$, and one partition of largest part $1$.
Hence, it suffices to show $q_{d}^{(1)}(n)\geq 11$ in the interval. We notice
that the partitions of $3d-4$ in the set
$\\{(3d-4)\\}\cup\\{(3d-4-i,i):i\in\\{1,2,...,d-2\\}\\}$ are counted by
$q_{d}^{(1)}(3d-4)$. Therefore, $q_{d}^{(1)}(3d-4)\geq d-1\geq 11$.

We now show that $q_{d}^{(1)}\left(n\right)\geq Q_{d-4}^{(1,-)}\left(n\right)$
for $4d-5\leq n\leq 5d-8$. We find that $Q_{d-4}^{(1,-)}\left(5d-8\right)=20$
since there is one partition with largest part for each element in
$\\{4d-3,4d-5\\}$, two with largest part for each element in
$\\{3d-2,3d-4\\}$, five with largest part $2d-1$, four with largest part for
each element in $\\{2d-3,d\\}$, and one with largest part $1$. Hence, it
suffices to show $q_{d}^{(1)}(n)\geq 20$. We notice the partitions of $4d-5$
in the set
$\\{(4d-5)\\}\cup\\{(i,4d-5-i):i\in\\{1,2,...,\lfloor\frac{3d-5}{2}\rfloor\\}\\}$
are counted by $q_{d}^{(1)}(4d-5)$. Note that $q_{d}^{(1)}(4d-5)$ also counts
partitions of the form $(3d-6-j,j+d,1)$ with
$j\in\\{1,2,...,\lfloor\frac{d-6}{2}\rfloor\\}$ and partitions of form
$(3d-8-\ell,\ell+1+d,2)$ with
$\ell\in\\{1,2,...,\lfloor\frac{d-9}{2}\rfloor\\}$. Therefore,
$q_{d}^{(1)}(4d-5)\geq
1+\lfloor\frac{3d-5}{2}\rfloor+\lfloor\frac{d-6}{2}\rfloor+\lfloor\frac{d-9}{2}\rfloor$.
Since $d\geq 12$, we have the desired inequality.

Finally, we prove that $q_{d}^{(1)}\left(n\right)\geq
Q_{d-4}^{(1,-)}\left(n\right)$ for $5d-8\leq n\leq 5d$. We find that
$Q_{d-4}^{(1,-)}(5d)=36$ since there is one partition with largest part for
each element in $\\{5d-4,5d-6\\}$, two with largest part for each element in
$\\{4d-3,4d-5\\}$, five with largest part for each element in
$\\{3d-2,3d-4\\}$, eight with largest part $2d-1$, six with largest part
$2d-3$, five with largest part $d$, and one with largest part $1$. Hence it
suffices to show that $q_{d}^{(1)}(n)\geq 36$. We notice that the partitions
of $5d-8$ in the set $\\{(5d-8)\\}\cup\\{(5d-8-i,i):i\in\\{1,2,...,2d-4\\}\\}$
are counted by $q_{d}^{(1)}(5d-8)$. In addition, $q_{d}^{(1)}(5d-8)$ counts
the partitions of the form $(4d-9-j,d+j,1)$ for
$j\in\\{1,\cdots,\lfloor\frac{2d-9}{2}\rfloor\\}$. We also note
$q_{d}^{(1)}(5d-8)$ counts the partitions of the form
$(4d-11-\ell,d+1+\ell,2)$ for $\ell\in\\{1,\cdots,d-6\\}$ and
$(4d-13-r,d+2+r,3)$ for $r\in\\{1,\cdots,\lfloor\frac{2d-15}{2}\rfloor\\}$.
Therefore,

$q_{d}^{(1)}(5d-8)\geq
2d-4+\lfloor\frac{2d-9}{2}\rfloor+d-6+\lfloor\frac{2d-15}{2}\rfloor\geq 36$

using the assumption that $d\geq 12$.

∎

###### Proof of unconditional part of Theorem 1.12..

The case $1\leq n\leq d+3a$ with the additional condition $n\neq d+a-3$ when
$d\equiv-3\pmod{a}$ is addressed by Lemma 6.1.

For the case $n\geq d+3a$, we employ the same inequality chain as used in the
conditional part of Theorem 1.12. Recall that

$\displaystyle q_{d}^{(a)}(n)\geq
q_{\frac{d+h_{d}^{(a)}}{a}}^{(1)}\left(\frac{n+h_{n}^{(a)}}{a}\right)\geq
Q_{\frac{d+h_{d}^{(a)}}{a}-4}^{(1,-)}(\frac{n+h_{n}^{(a)}}{a})$
$\displaystyle=Q_{d+h_{d}^{(a)}-a-3}^{(a,-)}\left(n+h_{n}^{(a)}\right)\geq
Q_{d}^{(a,-)}\left(n\right).$

All inequalities and equalities except for the second one are justified by our
work in the conditional component of Theorem 1.12. The second inequality for
the cases $\lceil\frac{d}{a}\rceil+2\leq\lceil\frac{n}{a}\rceil\leq
5\lceil\frac{d}{a}\rceil$ is resolved by Lemma 6.2. ∎

##  7\. Potential methods and future directions

In this section, we describe potential methods in proving more cases of
Conjectures 1.3, 1.5 and extending the results of Theorem 1.6.

We observe by repeating the arguments presented in the proof of [8, Theorem
1.9] to obtain that for $a,n\geq 1$ and $d\geq 4$ such that $a<d+3$ and
$\mbox{gcd}(a,d+3)=1$,

$\lim_{n\to\infty}\Delta_{d}^{(a)}(n)=\lim_{n\to\infty}q_{d}^{(a)}(n)\left(1-\frac{Q_{d}^{(a)}(n)}{q_{d}^{(a)}(n)}\right)=+\infty.$

Observe Remark 1.7 indicates that
$\Delta_{d}^{(a,-)}(n)\geq\Delta_{d}^{(a)}(n)$, implying our desired result.

We note from Theorem 1.4 that the remaining cases of Conjecture 1.3 are $3\leq
d\leq 90,\ 94\leq d\leq 186$. The sub-cases when $d\geq 93$ are divisible by
$3$ are addressed by [8, Proposition 4.1].

We describe a potential method in resolving more cases of Conjecture 1.3 and
extending Theorem 1.6. First, one should derive computationally effective
asymptotic expressions for
$q_{\lceil\frac{d}{a}\rceil}^{(1)}(\lceil\frac{n}{a}\rceil)$ and
$Q_{\lceil\frac{d}{a}\rceil-3}^{(1,-)}\left(\lceil\frac{n}{a}\rceil\right)$ by
using the work of Alfes, et.al [3]. One can then use these expressions to show
$q_{\lceil\frac{d}{a}\rceil}^{(1)}(\lceil\frac{n}{a}\rceil)\geq
Q_{\lceil\frac{d}{a}\rceil-3}^{(1,-)}\left(\lceil\frac{n}{a}\rceil\right)$ for
suitable $\lceil\frac{n}{a}\rceil$ and use (4.2) and (4.3) to prove more small
$d$ cases of Conjecture 1.3 and extend Theorem 1.6.

From application of the work of Kang and Kim [10], it appears that this
approach will be feasible for $\lceil\frac{d}{a}\rceil\geq 10$. In particular,
one can apply [10, Theorem 1.1].

######  Theorem 7.1 (Kang-Kim [10], 2021).

Let $a$, $d$, $m$, $m_{1}$, $m_{2}$ be integers such that $0\leq
m_{1}<m_{2}<m$ and $a,d\geq 1$. For $\alpha_{d}$ the unique real root of
$x^{d}+x-1$ in the interval $(0,1)$ and

$A_{d}=\frac{d}{2}\log^{2}\alpha_{d}+\sum_{r=1}^{\infty}\frac{\alpha_{d}^{rd}}{r^{2}},$

we let $M_{d}=\lfloor\frac{\pi^{2}}{3A_{d}}\rfloor$. Then, it can be found
that

$\lim_{n\to\infty}q_{d}^{(a)}(n)-\rho(\\{p\in\mathbb{N}:p\equiv
m_{1}\pmod{m}\mbox{ or }p\equiv m_{2}\pmod{m}\\};n)$
$=\begin{cases}+\infty\mbox{, if $m>M_{d}$}\\\ -\infty\mbox{, if $m\leq
M_{d}$.}\\\ \end{cases}$

Calculations of $\alpha_{d},A_{d},M_{d}$ suggest that for $d\geq 10$, Theorem
7.1 implies that $q_{d}^{(1)}(n)-Q_{d-3}^{(1)}(n)\to\infty$ as $n\to\infty$.
This is further corroborated by computational data of verifying the inequality
for $d+2\leq n\leq 100000$.

Table 4.  Values of $d$, $\alpha_{d}$, $A_{d}$, $M_{d}$ as defined by Kang and Kim [10] over select $d\geq 3$.  $d$ |  $\alpha_{d}$ (approximate) |  $A_{d}$ (approximate) | $M_{d}$
---|---|---|---
3 | 0.682328 | 0.566433 | 5
4 | 0.724492 | 0.504981 | 6
5 | 0.754878 | 0.459731 | 7
6 | 0.778090 | 0.424486 | 7
7 | 0.796544 | 0.395966 | 8
8 | 0.811652 | 0.372243 | 8
9 | 0.824301 | 0.352090 | 9
10 | 0.835079 | 0.334683 | 9
11 | 0.844398 | 0.319446 | 10
12 | 0.852551 | 0.305958 | 10
20 | 0.893895 | 0.234874 | 14
100 | 0.966584 | 0.091456 | 35

We now consider the sub-cases of Conjecture 1.3 consisting of $3\leq d\leq
27$. Unfortunately, Table 4 and computational data suggest that the proposed
method fails for those values of $d$ since for $3\leq d\leq 27$, we have
$\lceil\frac{d}{3}\rceil\leq M_{\lceil\frac{d}{3}\rceil}$. This implies by
[10, Theorem 1.1], that

$\lim_{n\to\infty}\left(q_{d}^{(1)}(n)-Q_{d-3}^{(1,-)}\left(n\right)\right)=-\infty.$

To address this problematic case, one presumably could use the explicit
asymptotic expressions in Duncan, et al. [8] for $q_{d}^{(3)}(n)$ and
$Q_{d}^{(3,-)}(n)$ to find an $\Omega(d)$ such that $n>\Omega(d)$,

$\Delta_{d}^{(3,-)}(n)\geq 0$

to address the values of $d>3$ such that $\mathrm{gcd}(d,3)=1$. One presumably
could then employ a finite computation to show $\Delta_{d}^{(3,-)}(n)\geq 0$
for $n\leq\Omega(d)$ for these values of $d$.

Remarkably, Armstrong, et.al [7] have extended Theorems 1.8 and 1.12 with a
linear lower bound on $d$ by proving a vast generalization of Proposition 3.1.
In particular, they have shown that Theorem 1.8 holds for all $a,n\geq 1$ and
$\lceil\frac{d}{a}\rceil\geq 105$. However, their methods do not seem
applicable to reduce Conjecture 1.5 to a finite computation. Such a reduction,
unfortunately appears out of reach of current combinatorial and analytic
methods since Conjecture 1.10 fails for $1\leq d\leq 11$ via Theorem 7.1.

## Acknowledgments

We are immensely thankful for Holly Swisher in many useful conversations,
comments, and deeply encouraging feedback regarding this paper. We also thank
Robert J. Lemke Oliver for giving us his code which greatly aided our
calculations of $q_{d}^{(1)}(n)$ and $Q_{d}^{(1)}(n)$ for large $n$. We also
thank the anonymous reviewer for numerous helpful comments and suggestions
which greatly improved the exposition of this paper.

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