mathmode, boxsize=0.9em

Strict Log-Subadditivity for Overpartition Rank

Helen W.J. Zhang1,2 and Ying Zhong1

1School of Mathematics

Hunan University

Changsha 410082, P. R. China

2Hunan Provincial Key Laboratory of

Intelligent Information Processing and Applied Mathematics

Changsha 410082, P. R. China

Emails<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Bessenrodt and Ono initially found the strict log-subadditivity of partition
function $p(n)$, that is, $p(a+b)<p(a)p(b)$ for $a,b>1$ and $a+b>9$. Many
other important statistics of partitions are proved to enjoy similar
properties. Lovejoy introduced the overpartition rank as an analog of Dyson’s
rank for partitions from the $q$-series perspective. Let
${\overline{N}}(a,c,n)$ denote the number of overpartitions with rank
congruent to $a$ modulo $c$. Ciolan computed the asymptotic formula of
${\overline{N}}(a,c,n)$ and showed that
${\overline{N}}(a,c,n)>{\overline{N}}(b,c,n)$ for $c\geq 7$ and $n$ large
enough. In this paper, we derive an upper bound and a lower bound of
${\overline{N}}(a,c,n)$ for each $c\geq 3$ by using the asymptotics of Ciolan.
Consequently, we establish the strict log-subadditivity of
${\overline{N}}(a,c,n)$ analogous to the partition function $p(n)$.

Mathematics Subject Classification: 05A20, 11P82, 11P83
Keywords: Rank, Overpartition, Strict Log-Subadditivity

##  1 Introduction

Strict log-subadditivity phenomena of partition statistics oriented from the
work of Bessenrodt and Ono [4]. Recall that a partition of a positive integer
$n$ is a sequence of non-increasing positive integers whose sum equals $n$.
Let $p(n)$ denote the number of partitions of $n$. Bessenrodt and Ono showed
that $p(n)$ satisfies the strict log-subadditivity result, that is, for
$a,b>1$ and $a+b>9$

$p(a+b)<p(a)p(b).$

Since then different partition functions are proven to enjoy the strict log-
subadditivity as $p(n)$, for instance, the overpartition function [18], the
$k$-regular partition function [3] and the spt-function [10, 12].

Strict log-subadditivity of other statistics concerned with partitions comes
into sight. Dyson [13] introduced the rank of a partition to interpret
Ramanujan’s congruences for $p(n)$ combinatorially. This is defined as the
largest part of the partition minus the number of parts. Let $N(a,c,n)$ denote
the number of partitions of $n$ with rank congruent to $a$ modulo $c$. Hou and
Jagadeesan [17] showed that

$N(r,3,a+b)<N(r,3,a)N(r,3,b)$

when $r=0$ (resp. $r=1,2$) and $a+b\geq 12$ (resp. $11,12$). They also
conjectured the general strict log-subadditivity: for $0\leq r<t$ and $t\geq
2$,

$N(r,t,a+b)<N(r,t,a)N(r,t,b)$ (1.1)

holds for sufficiently large $a$ and $b$. This conjecture is confirmed by
Males [20]. Gomez and Zhu [15] proved that (1.1) holds for $t=2$ and $a,b\geq
12$. The strict log-subadditivity for other statistics are referred to [2, 7,
16].

In this paper, we show the strict log-subadditivity rank of overpartitions.
Recall that an overpartition of a nonnegative integer $n$ is a partition of
$n$ where the first occurrence of each distinct part may be overlined. Let
$\ell(\lambda)$ denote the largest part of $\lambda$, $\\#(\lambda)$ denote
the number of parts of $\lambda$. Lovejoy [19] defined the rank of an
overpaprtition $\lambda$ as

$rank(\lambda)=\ell(\lambda)-\\#(\lambda).$

Let $\overline{N}(a,c,n)$ be the number of overpartitions of $n$ with rank
congruent to $a$ modulo $c$. Our main result as stated as follows.

######  Theorem 1.1.

Given any residue $a\pmod{c}$, we have

$\displaystyle{\overline{N}}(a,c,n_{1}+n_{2})<{\overline{N}}(a,c,n_{1}){\overline{N}}(a,c,n_{2})$
(1.2)

if $3\leq c\leq 5$ and $n_{1},n_{2}\geq 9$ or if $c\geq 6$ and
$n_{1},n_{2}\geq M_{c}$, where

$M_{c}=1.691\times
10^{13}c^{20}\exp\left(\frac{16c^{2}(2c^{2}+\pi)}{\pi^{2}}\right).$

This paper is organized as follows. In Section 2, we state the asymptotic
behavior of ${\overline{N}}(a,c,n)$ due to Ciolan. In Section 3, we estimate
the upper bounds for both main terms and error terms in the asymptotic
formula. Based on the estimates, we finally prove Theorem 1.1 in Section 4.

##  2 Asymptotic formula of Ciolan

Our approach to the strict log-subadditivity of overpartition rank relies on
the asymptotic behavior of generating function of ${\overline{N}}(a,c,n)$. In
this section, we recall the asymptotic formula due to Ciolan [8, 9].

We begin with some notations that we need. Let

$\omega_{h,k}:=\exp(\pi is(h,k)),$

where the Dedekind sum $s(h,k)$ is defined by

$s(h,k):=\sum_{u~{}(\mathrm{mod}~{}k)}\left(\left(\frac{u}{k}\right)\right)\left(\left(\frac{hu}{k}\right)\right).$

Here $((\cdot))$ is the sawtooth function defined by

$((x)):=\begin{cases}x-\lfloor
x\rfloor-\frac{1}{2},&\text{if}~{}~{}x\in\mathbb{R}\setminus\mathbb{Z},\\\
0,&\text{if}~{}~{}x\in\mathbb{Z}.\end{cases}$

In what follows, $0\leq h<k$ are coprime integers, and
$h^{\prime}\in\mathbb{Z}$ is defined by the congruence $hh^{\prime}\equiv 1$
(mod $k$). Further, let $0<a<c$ are coprime positive integers with $c>2$,
$c_{1}:=\frac{c}{(c,k)}$ and $k_{1}:=\frac{k}{(c,k)}$. Let the integer $0\leq
l<c_{1}$ be the solution to $l\equiv ak_{1}$ (mod $c_{1}$).

We then introduce several Kloosterman sums. Here and throughout we write
$\sum_{h}^{{}^{\prime}}$ to denote summation over the integers $0\leq h<k$
that are coprime to $k$.

If $c|k$, let

$A_{a,c,k}(n,m):=(-1)^{k_{1}+1}\tan\left(\frac{\pi
a}{c}\right){\sum_{h}}^{\prime}\frac{\omega_{h,k}^{2}}{\omega_{h,k/2}}\cdot\cot\left(\frac{\pi
ah^{\prime}}{c}\right)\cdot e^{-\frac{2\pi ih^{\prime}a^{2}k_{1}}{c}}\cdot
e^{\frac{2\pi i}{k}(nh+mh^{\prime})},$

and

$B_{a,c,k}(n,m):=-\frac{1}{\sqrt{2}}\tan\left(\frac{\pi
a}{c}\right){\sum_{h}}^{\prime}\frac{\omega_{h,k}^{2}}{\omega_{2h,k}}\cdot\frac{1}{\sin\left(\frac{\pi
ah^{\prime}}{c}\right)}\cdot e^{-\frac{2\pi ih^{\prime}a^{2}k_{1}}{c}}\cdot
e^{\frac{2\pi i}{k}(nh+mh^{\prime})}.$

If $c\nmid k$ and $0<\frac{l}{c_{1}}\leq\frac{1}{4}$, let

$D_{a,c,k}(n,m):=\frac{1}{\sqrt{2}}\tan\left(\frac{\pi
a}{c}\right){\sum_{h}}^{\prime}\frac{\omega_{h,k}^{2}}{\omega_{2h,k}}\cdot
e^{\frac{2\pi i}{k}(nh+mh^{\prime})},$

and if $c\nmid k$ and $\frac{3}{4}<\frac{l}{c_{1}}<1$, let

$D_{a,c,k}(n,m):=-\frac{1}{\sqrt{2}}\tan\left(\frac{\pi
a}{c}\right){\sum_{h}}^{\prime}\frac{\omega_{h,k}^{2}}{\omega_{2h,k}}\cdot
e^{\frac{2\pi i}{k}(nh+mh^{\prime})}.$

Finally, if $c\nmid k$, set

$\delta_{c,k,r}:=\begin{cases}\frac{1}{16}-\frac{l}{2c_{1}}+\frac{l^{2}}{c_{1}^{2}}-r\frac{l}{c_{1}},&\text{if}~{}~{}0<\frac{l}{c_{1}}\leq\frac{1}{4},\\\\[3.0pt]
0,&\text{if}~{}~{}\frac{1}{4}<\frac{l}{c_{1}}\leq\frac{3}{4},\\\\[3.0pt]
\frac{1}{16}-\frac{3l}{2c_{1}}+\frac{l^{2}}{c_{1}^{2}}+\frac{1}{2}-r\left(1-\frac{l}{c_{1}}\right),&\text{if}~{}~{}\frac{3}{4}<\frac{l}{c_{1}}<1,\end{cases}$

and

$m_{a,c,k,r}:=\begin{cases}-\frac{1}{2c_{1}^{2}}(2(ak_{1}-l)^{2}+c_{1}(ak_{1}-l)+2rc_{1}(ak_{1}-l)),&\text{if}~{}~{}0<\frac{l}{c_{1}}\leq\frac{1}{4},\\\\[3.0pt]
0,&\text{if}~{}~{}\frac{1}{4}<\frac{l}{c_{1}}\leq\frac{3}{4},\\\\[3.0pt]
-\frac{1}{2c_{1}^{2}}(2(ak_{1}-l)^{2}+3c_{1}(ak_{1}-l)-2rc_{1}(ak_{1}-l)-c_{1}^{2}(2r-1)),&\text{if}~{}~{}\frac{3}{4}<\frac{l}{c_{1}}<1.\end{cases}$

Let ${\overline{N}}(m,n)$ denote the number of overpartitions of $n$ with rank
$m$. We adopt the notation $\mathcal{O}(u;q)$ in [19] to denote the generating
function of ${\overline{N}}(m,n)$, that is,

$\displaystyle\mathcal{O}(u;q):=\sum_{n\geq
0}\sum_{m}{\overline{N}}(m,n)u^{m}q^{n}.$ (2.1)

If $0<a<c$ are coprime positive integers, and if by $\zeta_{n}=e^{\frac{2\pi
i}{n}}$ we denote the primitive $n$-th root of unity, we set

$\displaystyle\mathcal{O}\left(\frac{a}{c};q\right):=\mathcal{O}\left(\zeta_{c}^{a};q\right)=1+\sum_{n\geq
1}A\left(\frac{a}{c};n\right)q^{n}.$ (2.2)

Ciolan obtained the following asymptotic formula for
$A\left(\frac{a}{c};n\right)$.

######  Theorem 2.1.

[8, Theorem 1] If $0<a<c$ are coprime positive integers with $c>2$, and
$\varepsilon>0$ is arbitrary, then

$\displaystyle A\left(\frac{a}{c};n\right)=$ $\displaystyle
i\sqrt{\frac{2}{n}}\sum_{\begin{subarray}{c}1\leq k\leq\sqrt{n}\\\\[2.0pt]
c|k,~{}2\nmid
k\end{subarray}}\frac{B_{a,c,k}(-n,0)}{\sqrt{k}}\cdot\sinh\left(\frac{\pi\sqrt{n}}{k}\right)$
$\displaystyle+2\sqrt{\frac{2}{n}}\sum_{\begin{subarray}{c}1\leq
k\leq\sqrt{n}\\\\[2.0pt] c|k,~{}2\nmid k,~{}c_{1}\neq 4\\\\[2.0pt] r\geq
0,~{}\delta_{c,k,r}>0\end{subarray}}\frac{D_{a,c,k}(-n,m_{a,c,k,r})}{\sqrt{k}}\cdot\sinh\left(\frac{4\pi\sqrt{\delta_{c,k,r}n}}{k}\right)+O(n^{\varepsilon}).$
(2.3)

To obtain the inequality of ${\overline{N}}(a,c,n)$, Ciolan [9] bounded the
main term and the error term in above theorem. Here we list these bounds
respectively.

  * $\bullet$

Bounds in the main terms: Ciolan [9, § 4.2] gave upper bounds for Kloosterman
sums and hyperbolic sine function in (2.1).

$\displaystyle\left|B_{j,c,k}(-n,0)\right|\leq\cot\left(\frac{\pi}{2c}\right)\frac{2k\left(1+\log\left(\frac{c-1}{2}\right)\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)},$
(2.4)
$\displaystyle\left|D_{j,c,k}(-n,m_{j,c,k,r})\right|\leq\frac{k}{\sqrt{2}}\cot\left(\frac{\pi}{2c}\right),$
(2.5)
$\displaystyle\sinh\left(\frac{4\pi\sqrt{\delta_{c,k,r}n}}{k}\right)\leq\sinh\left(\pi\sqrt{n}\left(1-\frac{4}{c}\right)\right).$
(2.6)

  * $\bullet$

Bounds in the error terms: the error term $O(n^{\varepsilon})$ is constituted
of $S_{1},S_{2},\ldots,S_{8}$, the error terms of
$\sum_{1},\sum_{2},\ldots,\sum_{8}$ respectively. Here we strictly follows the
notations of Ciolan in [9, § 4.3]. Let ${\overline{p}}(n)$ denote the number
of overpartitions of $n$. Then

$\begin{split}&S_{1}<4C_{3}e^{2\pi}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\qquad
S_{2}<4C_{1}e^{2\pi}\sqrt{2}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\\\\[6.0pt]
&S_{3}<2C_{4}e^{2\pi}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\qquad
S_{4}<C_{5}e^{2\pi}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\\\\[6.0pt]
&S_{5}<C_{2}e^{2\pi}\sqrt{2}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\quad
S_{6}<C_{2}e^{2\pi}\frac{1}{\sqrt{2}}\cot\left(\frac{\pi}{2c}\right)\sum_{k}k^{-\frac{1}{2}},\\\\[6.0pt]
&S_{7}<\frac{n^{\frac{3}{4}}\log\left(\frac{n}{4}\right)}{2\pi\left(1-\frac{\pi^{2}}{24}\right)\sin\left(\frac{\pi}{c}\right)},\qquad\qquad
S_{8}<\frac{n^{\frac{3}{4}}\log\left(\frac{n}{4}\right)}{2\pi\left(1-\frac{\pi^{2}}{24}\right)\sin\left(\frac{\pi}{c}\right)},\end{split}$
(2.7)

where

$\displaystyle C_{1}=\sum_{r\geq
1}{\overline{p}}(r)\left(e^{-\frac{(16r-1)\pi}{16}}+e^{-\frac{(16r+7)\pi}{16}}\right),\quad
C_{2}=2\sum_{r\geq 1}{\overline{p}}(r)e^{-\frac{(c^{2}-8)\pi r}{16c^{2}}},$
(2.8) $\displaystyle C_{3}=\sum_{r\geq 1}{\overline{p}}(r)e^{-\pi r},\quad
C_{4}=\sum_{r\geq 1}{\overline{p}}(r)e^{-\frac{\pi r}{2c^{2}}},\quad
C_{5}=\sum_{r\geq 1}{\overline{p}}(r)e^{-\frac{\pi(2r+1)}{8}}.$ (2.9)

To our need, we adopt the following notations. The errors arising from
integrating over the remaining parts of the interval are denoted by
$S_{2err},S_{5err},S_{6err}$, and errors introduced by integrating along a
smaller arc are denoted by $I_{2err},I_{5err},I_{6err}$. Ciolan [9, § 4.3]
provided the following bounds on each of those pieces:

$\displaystyle S_{2err}$
$\displaystyle<\sqrt{2}e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)n^{-\frac{1}{2}}\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}\sum_{k}k^{-\frac{1}{2}},$
$\displaystyle S_{5err}$
$\displaystyle<2\sqrt{2}e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)n^{-\frac{1}{2}}\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}\sum_{k}k^{-\frac{1}{2}},$
(2.10) $\displaystyle S_{6err}$
$\displaystyle<\sqrt{2}e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)n^{-\frac{1}{2}}\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}\sum_{k}k^{-\frac{1}{2}}$

and

$\displaystyle I_{2err}$
$\displaystyle<2\sqrt{2}\left(\frac{4}{3}+2^{\frac{5}{4}}\right)e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}n^{\frac{1}{4}},$
$\displaystyle I_{5err}$
$\displaystyle<4\sqrt{2}\left(\frac{4}{3}+2^{\frac{5}{4}}\right)e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}n^{\frac{1}{4}},$
(2.11) $\displaystyle I_{6err}$
$\displaystyle<2\sqrt{2}\left(\frac{4}{3}+2^{\frac{5}{4}}\right)e^{2\pi+\frac{\pi}{8}}\cot\left(\frac{\pi}{2c}\right)\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}n^{\frac{1}{4}}.$

##  3 Bounds for ${\overline{N}}(a,c,n)$

In this section, we show that ${\overline{N}}(a,c,n)$ can be bounded by a
constant multiple of ${\overline{p}}(n)$ for each $c\geq 3$.

######  Theorem 3.1.

We have

$\displaystyle
0.0019~{}{\overline{p}}(n)<{\overline{N}}(a,3,n)<0.6648~{}{\overline{p}}(n),~{}~{}n\geq
2089,$ (3.1) $\displaystyle
0.0091~{}{\overline{p}}(n)<{\overline{N}}(a,4,n)<0.4909~{}{\overline{p}}(n),~{}~{}n\geq
272,$ (3.2) $\displaystyle
0.0103~{}{\overline{p}}(n)<{\overline{N}}(a,5,n)<0.3897~{}{\overline{p}}(n),~{}~{}n\geq
449.$ (3.3)

For $c\geq 6$ we define the constant

$M_{c}:=1.691\times
10^{13}c^{20}\exp\left(\frac{16c^{2}(2c^{2}+\pi)}{\pi^{2}}\right),$ (3.4)

we have

$\displaystyle\left(\frac{1}{2c}\right){\overline{p}}(n)<{\overline{N}}(a,c,n)<\left(\frac{3}{2c}\right){\overline{p}}(n),~{}~{}n\geq
M_{c}.$ (3.5)

Our main mission of this section is to prove the above theorem. We sketch our
strategy here. Setting $u=\zeta_{c}^{a}$ in (2.1), we have

$\displaystyle\mathcal{O}\left(\frac{a}{c};q\right)=\sum_{n\geq
0}\sum_{m}{\overline{N}}(m,n)\zeta_{c}^{am}q^{n}.$ (3.6)

Comparing the coefficient of $q^{n}$ in (2.2), (3.6) and with the
orthogonality of the roots of unity, we obtain

${\overline{N}}(a,c,n)=\frac{1}{c}{\overline{p}}(n)+\frac{1}{c}\sum_{1\leq
j\leq c-1}\zeta_{c}^{-aj}A\left(\frac{j}{c};n\right).$ (3.7)

The second summation in right-hand side of (3.7) is denoted by
${\overline{R}}(a,c,n)$, that is,

$\displaystyle{\overline{N}}(a,c,n)=\frac{1}{c}{\overline{p}}(n)+{\overline{R}}(a,c,n).$
(3.8)

Due to (3.8), we need to estimate ${\overline{R}}(a,c,n)$ and
${\overline{p}}(n)$ separately. For ${\overline{R}}(a,c,n)$, we need to give a
bound to the main terms as well as the error terms in view of Theorem 2.1.
These are completed in Theorem 3.3 and Theorem 3.5 respectively. After then,
we give an upper bound and lower bound for ${\overline{p}}(n)$ in Theorem 3.6
based on the asymptotic formula of Engel [14].

From the view of Theorem 2.1, we have the following estimates to bound the
main terms of ${\overline{R}}(a,c,n)$.

######  Lemma 3.2.

For integer $c>2$, then we have

$\displaystyle\left|B_{j,c,k}(-n,0)\right|\leq 0.3444kc^{2},$ (3.9)
$\displaystyle\left|D_{j,c,k}(-n,m_{j,c,k,r})\right|\leq 0.4503kc,$ (3.10)
$\displaystyle\sinh\left(\frac{4\pi\sqrt{\delta_{c,k,r}n}}{k}\right)\leq\frac{1}{2}e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}.$
(3.11)

###### Proof.

We shall estimate (3.9). Using the fact in [6, P. 159],

$\displaystyle\log x$ $\displaystyle\leq a(x^{1/a}-1),~{}~{}~{}~{}a,x>0,$
(3.12)

we have

$\displaystyle\frac{1+\log\left(\frac{c-1}{2}\right)}{\pi\left(1-\frac{\pi^{2}}{24}\right)}<0.2704c.$
(3.13)

Substituting (3.13) and

$\displaystyle\cot\left(\frac{\pi}{2c}\right)\leq\frac{2c}{\pi}<0.6367c$
(3.14)

into (2.4), we get the first inequality (3.9). By means of the estimates in
(2.5) and (3.14), we arrive at the second inequality (3.10). The third
inequality (3.11) immediately follows from the definition of hyperbolic sine
function. This completes the proof.

We can give a bound for the main term of ${\overline{R}}(a,c,n)$.

######  Theorem 3.3.

A bound for the main term of ${\overline{R}}(a,c,n)$ is

$0.1624e^{\frac{\pi\sqrt{n}}{c}}n^{\frac{1}{4}}c+(0.0266c+0.2123)e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}n^{\frac{1}{4}}c.$

###### Proof.

Immediately, the main term of (2.1) gives the main term of
${\overline{R}}(a,c,n)$

$\displaystyle\frac{1}{c}\sum_{1\leq j\leq
c-1}\zeta_{c}^{-aj}i\sqrt{\frac{2}{n}}\sum_{\begin{subarray}{c}1\leq
k\leq\sqrt{n}\\\\[2.0pt] c|k,2\nmid
k\end{subarray}}\frac{B_{j,c,k}(-n,0)}{\sqrt{k}}\cdot\sinh\left(\frac{\pi\sqrt{n}}{k}\right)$
$\displaystyle\quad+\frac{1}{c}\sum_{1\leq j\leq
c-1}\zeta_{c}^{-aj}2\sqrt{\frac{2}{n}}\sum_{\begin{subarray}{c}1\leq
k\leq\sqrt{n}\\\\[2.0pt] c|k,2\nmid k,c_{1}\neq 4\\\\[2.0pt] r\geq
0,\delta_{c,k,r}>0\end{subarray}}\frac{D_{j,c,k}(-n,m_{j,c,k,r})}{\sqrt{k}}\cdot\sinh\left(\frac{4\pi\sqrt{\delta_{c,k,r}n}}{k}\right).$
(3.15)

Let $G_{1}$ and $G_{2}$ denote the summations of (3), respectively. By the
bounds in (3.9), we have

$\displaystyle\left|G_{1}\right|\leq\sqrt{\frac{2}{n}}\sinh\left(\frac{\pi\sqrt{n}}{c}\right)\sum_{\begin{subarray}{c}1\leq
k\leq\sqrt{n}\\\\[2.0pt] c|k\end{subarray}}0.3444k^{\frac{1}{2}}c^{2}\leq
0.1624e^{\frac{\pi\sqrt{n}}{c}}n^{\frac{1}{4}}c,$ (3.16)

where the last inequality is due to

$\sum_{\begin{subarray}{c}1\leq k\leq\sqrt{n}\\\\[2.0pt]
c|k\end{subarray}}k^{\frac{1}{2}}\leq c^{\frac{1}{2}}\sum_{1\leq
m\leq\lfloor\frac{N}{c}\rfloor}m^{\frac{1}{2}}\leq
c^{\frac{1}{2}}\int_{1}^{\lfloor\frac{N}{c}\rfloor}x^{\frac{1}{2}}dx\leq\frac{2}{3c}n^{\frac{3}{4}}.$

Now we consider the bound of $|G_{2}|$. For fixed $k$, it is easy to see that

$\sum_{\begin{subarray}{c}r\geq 0\\\\[2.0pt] c|k,2\nmid k,c_{1}\neq
4\\\\[2.0pt] \delta_{c,k,r}>0\end{subarray}}1\leq\frac{c+8}{16}.$ (3.17)

Then by (3.10), (3.11) and (3.17), we get

$\displaystyle\left|G_{2}\right|$ $\displaystyle\leq
0.4503c\sqrt{\frac{2}{n}}e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}\frac{c+8}{16}\sum_{\begin{subarray}{c}1\leq
k\leq\sqrt{n}\\\\[2.0pt] c\nmid k\end{subarray}}k^{\frac{1}{2}}.$

Since

$\sum_{\begin{subarray}{c}1\leq k\leq\sqrt{n}\\\\[2.0pt] c\nmid
k\end{subarray}}k^{\frac{1}{2}}\leq\int_{1}^{\sqrt{n}}x^{\frac{1}{2}}dx\leq\frac{2}{3}n^{\frac{3}{4}},$

we have

$\displaystyle\left|G_{2}\right|$ $\displaystyle\leq
0.4503c\sqrt{\frac{2}{n}}e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}\cdot\frac{c+8}{16}\cdot\frac{2}{3}n^{\frac{3}{4}}$
$\displaystyle\leq(0.0266c+0.2123)e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}n^{\frac{1}{4}}c.$
(3.18)

Plugging (3.16) and (3) into (3), we complete the proof.

Next we shall give a bound for the error terms of ${\overline{R}}(a,c,n)$. To
this end, we need to analyse the terms in (2.7), (2.10) and (2.11) more
detailedly.

######  Lemma 3.4.

For integer $c>2$, then we have

$\displaystyle S_{1}$ $\displaystyle\leq 1496.9n^{\frac{1}{4}}c,\quad
S_{2}\leq 3111.36n^{\frac{1}{4}}c,\quad~{}~{}S_{3}\leq
1363.79\overline{C}_{4}n^{\frac{1}{4}}c,$ $\displaystyle S_{4}$
$\displaystyle\leq 82469.8n^{\frac{1}{4}}c,\quad S_{5}\leq
964.35\overline{C}_{2}n^{\frac{1}{4}}c,\quad S_{6}\leq
482.18\overline{C}_{2}n^{\frac{1}{4}}c,$ $\displaystyle S_{7}$
$\displaystyle\leq 0.9093n^{\frac{7}{8}}c,\quad~{}~{}S_{8}\leq
0.9093n^{\frac{7}{8}}c$

and

$\displaystyle S_{2err}$ $\displaystyle\leq 386.18n^{-\frac{1}{4}}c^{2},\quad
S_{5err}\leq 772.36n^{-\frac{1}{4}}c^{2},\quad S_{6err}\leq
386.18n^{-\frac{1}{4}}c^{2},$ $\displaystyle I_{2err}$ $\displaystyle\leq
1433.39n^{\frac{1}{4}}c^{2},\quad I_{5err}\leq
2866.78n^{\frac{1}{4}}c^{2},\quad~{}~{}~{}I_{6err}\leq
1433.39n^{\frac{1}{4}}c^{2},$

where

$\displaystyle\overline{C}_{2}=2\exp\left(\frac{32c^{2}(16c^{2}+(c^{2}-8)\pi)}{\pi^{2}(c^{2}-8)^{2}}\right),\quad\overline{C}_{4}=\exp\left(\frac{4c^{2}(2c^{2}+\pi)}{\pi^{2}}\right).$

###### Proof.

First, we give the estimations of $C_{1},C_{2},C_{3},C_{4},C_{5}$ in (2.8) and
(2.9). By the fact

$\displaystyle{\overline{p}}(n)<e^{\pi\sqrt{n}},$ (3.19)

it is easy to check that

$\displaystyle C_{1}\leq 0.8066,\quad C_{3}\leq 0.5488,\quad C_{5}\leq
120.942.$ (3.20)

To estimate $C_{2}$, we claim that

$\displaystyle\sum_{n\geq 0}{\overline{p}}(n)e^{-2\pi
ny}\leq\exp\left(\frac{2e^{-2\pi y}}{(1-e^{-2\pi y})^{2}}\right),$ (3.21)

where $y$ is a positive real number. Recall that the generating function of
${\overline{p}}(n)$ is given by Corteel and Lovejoy [11]

$F(q):=\sum_{n\geq 0}{\overline{p}}(n)q^{n}=\prod_{n\geq
1}\frac{(1+q^{n})}{(1-q^{n})},$ (3.22)

where $q=e^{2\pi iz}$ with $z=x+iy$. Taking logarithm of both sides of (3.22),
we have

$\displaystyle\log F(q)$ $\displaystyle=\sum_{n\geq
1}\left(\log(1+q^{n})-\log(1-q^{n})\right)$ $\displaystyle=\sum_{n\geq
1}\sum_{m\geq 1}\left(\frac{q^{nm}}{m}-\frac{(-1)^{m}q^{nm}}{m}\right)$
$\displaystyle<\frac{2q}{(1-q)^{2}}.$

Hence

$F(|q|)\leq\exp\left(\frac{2|q|}{(1-|q|)^{2}}\right),$

together with $|q|=e^{-2\pi y}$, which gives (3.21) as claimed. Since integer
$c>2$, $c^{2}-8>0$. Take $y=(c^{2}-8)/32c^{2}$ in (3.21) and we have

$\displaystyle C_{2}$ $\displaystyle\leq
2\exp\left(\frac{2e^{-\frac{(c^{2}-8)\pi}{16c^{2}}}}{\left(1-e^{\frac{(c^{2}-8)\pi}{16c^{2}}}\right)^{2}}\right).$

For $x>-1$, it can be easily derived from

$\frac{x}{1+x}<1-e^{-x}<x$

that

$\frac{e^{-x}}{(1-e^{-x})^{2}}<\frac{1+x}{x^{2}}.$ (3.23)

Therefore

$\displaystyle C_{2}\leq
2\exp\left(\frac{32c^{2}(16c^{2}+(c^{2}-8)\pi)}{\pi^{2}(c^{2}-8)^{2}}\right):=\overline{C}_{2}.$
(3.24)

By the same argument as above, one can derive

$C_{4}\leq\exp\left(\frac{4c^{2}(2c^{2}+\pi)}{\pi^{2}}\right):=\overline{C}_{4}$
(3.25)

from (3.21) and (3.23).

Since

$\displaystyle\sum_{k}k^{-\frac{1}{2}}=\int_{1}^{\sqrt{n}}x^{-\frac{1}{2}}dx\leq
2n^{\frac{1}{4}}.$ (3.26)

Substituting (3.14), (3.26) and the bounds for $C_{1},\ldots,C_{5}$ as given
in (3.20), (3.24) and (3.25) into (2.7), we finally derive estimates for
$S_{1},\ldots,S_{6}$ in Lemma 3.4.

As for $S_{7}$ and $S_{8}$, first we note that when
$x\in\left(0,\frac{\pi}{2}\right)$,

$\sin x\geq\frac{2x}{\pi}.$

Then we have

$\displaystyle\sin\left(\frac{\pi}{c}\right)\geq\frac{2}{c}.$ (3.27)

Meanwhile, setting $x=\frac{n}{4}$ and $a=8$ in (3.12), we have

$\log\left(\frac{n}{4}\right)\leq 8\left(\frac{n}{4}\right)^{\frac{1}{8}}.$
(3.28)

Thus $S_{7}$ and $S_{8}$ is bounded by $0.9093n^{\frac{7}{8}}c$ as we
substitute (3.27) and (3.28) into (2.7). The asymptotic estimates $S_{2err}$,
$S_{5err}$, $S_{6err}$, $I_{2err}$, $I_{5err}$ and $I_{6err}$ easily follow
from (3.13), (3.14) and (3.26). This completes the proof.

A bound for the error terms of ${\overline{R}}(a,c,n)$ immediately follows
from the above lemma.

######  Theorem 3.5.

The bound for the error term of ${\overline{R}}(a,c,n)$ is

$\displaystyle 1544.72n^{-\frac{1}{4}}c^{2}+87078.1n^{\frac{1}{4}}c$
$\displaystyle+5733.56n^{\frac{1}{4}}c^{2}+1.8186n^{\frac{7}{8}}c$
$\displaystyle\quad+1363.79\overline{C}_{4}n^{\frac{1}{4}}c+1446.53\overline{C}_{2}n^{\frac{1}{4}}c.$
(3.29)

###### Proof.

Adding the estimates in Lemma 3.4 together and applying to (2.1), we get the
bound for the error term of ${\overline{R}}(a,c,n)$ as desired.

The following bounds for ${\overline{p}}(n)$ serves as a connection between
${\overline{p}}(n)$ and ${\overline{R}}(a,c,n)$.

######  Theorem 3.6.

We have

$\frac{1}{8n}\left(1-\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}}\leq{\overline{p}}(n)\leq\frac{1}{8n}\left(1+\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}}.$
(3.30)

###### Proof.

Recall that Engel [14] provided an error term for the overpartition function

$\displaystyle{\overline{p}}(n)=\frac{1}{2\pi}\sum_{1\leq k\leq N\atop 2\nmid
k}\sqrt{k}\sum_{0\leq h\leq
k\atop(h,k)=1}\frac{\omega(h,k)^{2}}{\omega(2h,k)}e^{-\frac{2\pi
inh}{k}}\frac{\mathrm{d}}{\mathrm{d}n}\left(\frac{\sinh\left(\frac{\pi\sqrt{n}}{k}\right)}{\sqrt{n}}\right)+R_{2}(n,N),$

where

$\displaystyle|R_{2}(n,N)|\leq\frac{N^{\frac{5}{2}}}{\pi
n^{\frac{3}{2}}}\sinh\left(\frac{\pi\sqrt{n}}{N}\right).$

In particular, when $N=3$, we have

$\displaystyle{\overline{p}}(n)=\frac{1}{8n}\left[\left(1+\frac{1}{\pi\sqrt{n}}\right)e^{-\pi\sqrt{n}}+\left(1-\frac{1}{\pi\sqrt{n}}\right)e^{\pi\sqrt{n}}\right]+R_{2}(n,3),$
(3.31)

where

$\displaystyle|R_{2}(n,3)|\leq\frac{3^{\frac{5}{2}}}{\pi
n^{\frac{3}{2}}}\sinh\left(\frac{\pi\sqrt{n}}{3}\right)\leq\frac{3^{\frac{5}{2}}e^{\frac{\pi\sqrt{n}}{3}}}{2\pi
n^{\frac{3}{2}}}\leq\frac{\pi e^{\pi\sqrt{n}}}{16\pi n^{\frac{3}{2}}}.$ (3.32)

By (3.31), we get the following upper bound for ${\overline{p}}(n)$

$\displaystyle{\overline{p}}(n)\leq\frac{1}{8n}\left(1+\frac{1}{\pi\sqrt{n}}\right)e^{\pi\sqrt{n}}+\frac{\pi
e^{\pi\sqrt{n}}}{16\pi n^{\frac{3}{2}}}.$

The right hand side of above inequality can be rewritten as

$\displaystyle\frac{1}{8n}\left(1+\frac{1}{\pi\sqrt{n}}\right)e^{\pi\sqrt{n}}+\frac{\pi
e^{\pi\sqrt{n}}}{16\pi
n^{\frac{3}{2}}}=\frac{1}{8n}\left(1+\frac{2+\pi}{2\pi\sqrt{n}}\right)e^{\pi\sqrt{n}},$

which arrives at the right-hand side of (3.30). The left hand-side of (3.30)
can be verified easily. This completes the proof.

We are now ready to complete the proof of Theorem 3.1.

###### Proof of Theorem 3.1.

Combining Theorems 3.3 and 3.5, we have

$\displaystyle\left|{\overline{R}}(a,c,n)\right|\leq$
$\displaystyle\bigg{|}0.1624e^{\frac{\pi\sqrt{n}}{c}}n^{\frac{1}{4}}c+(0.0266c+0.2123)e^{\pi\sqrt{n}\left(1-\frac{4}{c}\right)}n^{\frac{1}{4}}c$
$\displaystyle\quad+1544.72n^{-\frac{1}{4}}c^{2}+87078.1n^{\frac{1}{4}}c+5733.56n^{\frac{1}{4}}c^{2}+1.8186n^{\frac{7}{8}}c$
$\displaystyle\quad+1363.79\overline{C}_{4}n^{\frac{1}{4}}c+1446.53\overline{C}_{2}n^{\frac{1}{4}}c\bigg{|}.$
(3.33)

We note that for $n\geq 2$,

$\frac{\sqrt{n}}{\sqrt{n}-1}\leq\frac{\sqrt{2}}{\sqrt{2}-1}\leq 3.415.$

By Theorem 3.6, the lower bound of ${\overline{p}}(n)^{-1}$ can be estimated
as

$\displaystyle{\overline{p}}(n)^{-1}\leq
8n\frac{\sqrt{n}}{\sqrt{n}-1}e^{-\pi\sqrt{n}}\leq 27.32ne^{-\pi\sqrt{n}}.$

Combining above inequality, we have for $c>2$

$\displaystyle\left|\frac{{\overline{R}}(a,c,n)}{{\overline{p}}(n)}\right|\leq$
$\displaystyle
4.44ce^{\left(\frac{1}{c}-1\right)\pi\sqrt{n}}n^{\frac{5}{4}}+(0.73c+5.81)ce^{-\frac{4}{c}\pi\sqrt{n}}n^{\frac{5}{4}}+e^{-\pi\sqrt{n}}\left[42201.8c^{2}n^{\frac{3}{4}}\right.$
$\displaystyle+156641c^{2}n^{\frac{5}{4}}+2.379\times
10^{6}cn^{\frac{5}{4}}+49.69cn^{\frac{15}{8}}+37258.7c\overline{C}_{4}n^{\frac{5}{4}}$
$\displaystyle\left.+39519.2c\overline{C}_{2}n^{\frac{5}{4}}\right].$ (3.34)

For $c=3$, recalling the definition of $C_{2}$ in (2.8), then using (3.19), we
compute that $C_{2}<4.5303\times 10^{52}$. Similarly, using (2.9), we have
$C_{4}<1.0535\times 10^{8}$. So replacing $\overline{C}_{2}$ and
$\overline{C}_{4}$ in (3) with $4.5303\times 10^{52}$ and $1.0535\times
10^{8}$, we obtain that

$\displaystyle\left|\frac{{\overline{R}}(a,3,n)}{{\overline{p}}(n)}\right|\leq$
$\displaystyle
13.32e^{\frac{-2\pi\sqrt{n}}{3}}n^{\frac{5}{4}}+24e^{-\frac{4}{3}\pi\sqrt{n}}n^{\frac{5}{4}}+e^{-\pi\sqrt{n}}\left(379816.2n^{\frac{3}{4}}\right.$
$\displaystyle\quad\quad\quad\left.+5.3711\times
10^{57}n^{\frac{5}{4}}+149.07n^{\frac{15}{8}}\right):=R_{3}.$ (3.35)

Plugging (3) into (3.8), we have

$\displaystyle{\overline{p}}(n)\left(\frac{1}{3}-R_{3}\right)<{\overline{N}}(a,3,n)<{\overline{p}}(n)\left(\frac{1}{3}+R_{3}\right).$
(3.36)

Note that $R_{3}$ is decreasing for $n>1$, then we have

$\displaystyle R_{3}<0.33142,~{}~{}n\geq 2089.$ (3.37)

Combining (3.36) and (3.37), we are led to (3.1).

Similar to $c=3$, we first confirm the estimations of $C_{2}$ and $C_{4}$.
When $c=4$, $C_{2}<2.977\times 10^{13}$ and $C_{4}<1.4885\times 10^{13}$, for
$c=5$, we have $C_{2}<2.4221\times 10^{10}$ and $C_{4}<4.0102\times 10^{19}$.
Then we obtain

$\displaystyle\left|\frac{{\overline{R}}(a,4,n)}{{\overline{p}}(n)}\right|$
$\displaystyle\leq
17.76e^{\frac{-3\pi\sqrt{n}}{4}}n^{\frac{5}{4}}+e^{-\pi\sqrt{n}}\left(675228.9n^{\frac{3}{4}}\right.$
$\displaystyle\left.\qquad\qquad+6.9244\times
10^{18}n^{\frac{5}{4}}+198.76n^{\frac{15}{8}}\right):=R_{4},$ (3.38)

and

$\displaystyle\left|\frac{{\overline{R}}(a,5,n)}{{\overline{p}}(n)}\right|$
$\displaystyle\leq
69.5e^{\frac{-4\pi\sqrt{n}}{5}}n^{\frac{5}{4}}+e^{-\pi\sqrt{n}}\left(1.0551\times
10^{6}n^{\frac{3}{4}}\right.$ $\displaystyle\left.\qquad\qquad+7.4708\times
10^{24}n^{\frac{5}{4}}+248.45n^{\frac{15}{8}}\right):=R_{5}.$ (3.39)

Substituting (3) and (3) into (3.8), respectively, we have

$\displaystyle{\overline{p}}(n)\left(\frac{1}{4}-R_{4}\right)<{\overline{N}}(a,4,n)<{\overline{p}}(n)\left(\frac{1}{4}+R_{4}\right),$
(3.40)
$\displaystyle{\overline{p}}(n)\left(\frac{1}{5}-R_{5}\right)<{\overline{N}}(a,5,n)<{\overline{p}}(n)\left(\frac{1}{5}+R_{5}\right).$
(3.41)

Since

$\displaystyle R_{4}<0.24084,~{}~{}n\geq 272,$ (3.42) $\displaystyle
R_{5}<0.1897,~{}~{}~{}~{}n\geq 449.$ (3.43)

Applying the estimates (3.42) into (3.40) and (3.43) into (3.41), we reach
(3.2) and (3.3).

Next, we deal with the case $c\geq 6$. Here we combine all the terms in (3)
except the term of $49.69ce^{-\pi\sqrt{n}}n^{\frac{15}{8}}$ into an upper
bound. We take the sum of their coefficients, the highest order exponential,
and the highest power of $n$ from the all terms and put them together in one
term. For $c\geq 6$, $4/c<1-1/c$, so we have

$\displaystyle\left|\frac{{\overline{R}}(a,c,n)}{{\overline{p}}(n)}\right|\leq
37259c\exp\left(\frac{4c^{2}(2c^{2}+\pi)}{\pi^{2}}\right)e^{-\frac{4}{c}\pi\sqrt{n}}n^{\frac{5}{4}}+49.69ce^{-\pi\sqrt{n}}n^{\frac{15}{8}}:=R_{c}.$
(3.44)

In view of (3.8) and (3.44), we get the following bounds of
${\overline{N}}(a,c,n)$

$\displaystyle{\overline{p}}(n)\left(\frac{1}{c}-R_{c}\right)<{\overline{N}}(a,c,n)<{\overline{p}}(n)\left(\frac{1}{c}+R_{c}\right).$
(3.45)

We claim that

$\displaystyle R_{c}<\frac{1}{2c},~{}~{}n\geq M_{c},$ (3.46)

where $M_{c}$ is defined in (3.4).

To verify (3.46), we only need to show that

$\displaystyle
37259c\exp\left(\frac{4c^{2}(2c^{2}+\pi)}{\pi^{2}}\right)e^{-\frac{4}{c}\pi\sqrt{n}}n^{\frac{5}{4}}<\frac{1}{4c}$
(3.47)

and

$\displaystyle 49.69ce^{-\pi\sqrt{n}}n^{\frac{15}{8}}$
$\displaystyle<\frac{1}{4c},$ (3.48)

which are equivalent to prove that

$\frac{e^{\frac{4}{c}\pi\sqrt{n}}}{n^{\frac{5}{4}}}>4\times
37259c^{2}\exp\left(\frac{4c^{2}(2c^{2}+\pi)}{\pi^{2}}\right)$ (3.49)

and

$\frac{e^{\pi\sqrt{n}}}{n^{\frac{15}{8}}}>4\times 49.69c^{2}.$ (3.50)

In addiction, we recall the following inequality[21, P. 76]

$e^{x}>\left(1+\frac{x}{y}\right)^{y},~{}~{}~{}x,y>0.$

Hence, letting $x=\frac{4}{c}\pi\sqrt{n}$, $y=3$ and $x=\pi\sqrt{n}$, $y=4$
respectively, we obtain

$\displaystyle\frac{e^{\frac{4}{c}\pi\sqrt{n}}}{n^{\frac{5}{4}}}>\frac{1}{n^{\frac{5}{4}}}\left(1+\frac{4\pi\sqrt{n}}{3c}\right)^{3}>\frac{4^{3}\pi^{3}}{3^{3}c^{3}}n^{\frac{1}{4}}$
(3.51)

and

$\displaystyle\frac{e^{\pi\sqrt{n}}}{n^{\frac{15}{8}}}>\frac{1}{n^{\frac{15}{8}}}\left(1+\frac{\pi\sqrt{n}}{4}\right)^{4}>\frac{\pi^{4}}{4^{4}}n^{\frac{1}{8}}.$
(3.52)

By (3.51) and (3.52), we know that the inequalities (3.49) and (3.50) hold
when $n\geq M_{c}$ and $n\geq{M_{c}}^{\prime}$, respectively, where

$\displaystyle{M_{c}}^{\prime}$ $\displaystyle:=5.544\times 10^{21}c^{16}.$

Furthermore, we see that (3.47) and (3.48) hold when $n\geq M_{c}$ and
$n\geq{M_{c}}^{\prime}$, respectively. Therefore, (3.46) is verified for
$n\geq\max\\{M_{c},{M_{c}}^{\prime}\\}=M_{c}$. Plugging (3.46) into (3.45), we
complete the proof.

##  4 Strict log-subadditivity of ${\overline{N}}(a,c,n)$

In this section, we present a proof of Theorem 1.1 based on the intermediate
inequalities in the previous sections.

###### Proof of Theorem 1.1.

We begin with the case of $c=3$. Substituting (3.30) into (3.1), we obtain
that for $n\geq 2089$

$(0.0019)\frac{1}{8n}\left(1-\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}}<{\overline{N}}(a,3,n)<(0.6648)\frac{1}{8n}\left(1+\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}}.$
(4.1)

Letting $n_{2}=Cn_{1}$ for some $C\geq 1$, then by (4.1), we have

$\displaystyle{\overline{N}}(a,3;n_{1}){\overline{N}}(a,3;n_{2})>(0.0019)^{2}\frac{1}{64Cn_{1}^{2}}\left(1-\frac{1}{\sqrt{n_{1}}}\right)\left(1-\frac{1}{\sqrt{Cn_{1}}}\right)e^{\pi(\sqrt{n_{1}}+\sqrt{Cn_{1}})},$

and

$\displaystyle{\overline{N}}(a,3;n_{1}+n_{2})<(0.6648)\frac{1}{8(n_{1}+Cn_{1})}\left(1+\frac{1}{\sqrt{n_{1}+Cn_{1}}}\right)e^{\pi\sqrt{n_{1}+Cn_{1}}}.$

It is sufficient to show that

$T_{n_{1}}(C)>\log\left(V_{n_{1}}(C)\right)+\log\left(S_{n_{1}}(C)\right),$

where

$\displaystyle T_{n_{1}}(C)$
$\displaystyle:=\pi(\sqrt{n_{1}}+\sqrt{Cn_{1}})-\pi\sqrt{n_{1}+Cn_{1}},$
$\displaystyle V_{n_{1}}(C)$ $\displaystyle:=\frac{0.6648\times
8Cn_{1}}{(0.0019)^{2}(C+1)},$ $\displaystyle S_{n_{1}}(C)$
$\displaystyle:=\frac{1+\frac{1}{\sqrt{n_{1}+Cn_{1}}}}{\left(1-\frac{1}{\sqrt{n_{1}}}\right)\left(1-\frac{1}{\sqrt{Cn_{1}}}\right)}.$

As function of $C$, it can be shown that $T_{n_{1}}(C)$ is increasing and
$S_{n_{1}}(C)$ is decreasing for $C\geq 1$, and combined with

$V_{n_{1}}(C)<\frac{0.6648\times 8n_{1}}{(0.0019)^{2}},$

it suffices to show that

$\displaystyle T_{n_{1}}(1)$ $\displaystyle=2\pi\sqrt{n_{1}}-\pi\sqrt{2n_{1}}$
$\displaystyle>\log\left(\frac{0.6648\times
8n_{1}}{(0.0019)^{2}}\right)+\log\left(\frac{1+\frac{1}{\sqrt{2n_{1}}}}{\left(1-\frac{1}{\sqrt{n_{1}}}\right)^{2}}\right).$
(4.2)

By a short computation, we find that (4) holds for all $n_{1}\geq 109$. Hence,
if $n_{1},n_{2}\geq 2089$, we have

$\displaystyle{\overline{N}}(a,3,n_{1}+n_{2})<{\overline{N}}(a,3,n_{1}){\overline{N}}(a,3,n_{2}).$

It is routine to check that (1.2) is true for $c=3$ and $9\leq n\leq 2088$.

The proofs of $c=4,5$ are similar to that of $c=3$, and hence, they are
omitted.

We proceed to prove the case for $c\geq 6$. Plugging (3.30) into (3.5), we
obtain that

$\frac{1}{16nc}\left(1-\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}}<{\overline{N}}(a,c,n)<\frac{3}{16nc}\left(1+\frac{1}{\sqrt{n}}\right)e^{\pi\sqrt{n}},~{}~{}n\geq
M_{c},$ (4.3)

where $M_{c}$ is defined in (3.4).

Setting $n_{2}=Cn_{1}$ for some $C\geq 1$, one can easily find that for
$n_{1}\geq M_{c}$

$\displaystyle{\overline{N}}(a,c;n_{1}){\overline{N}}(a,c;n_{2})>\frac{1}{256Cn_{1}^{2}c^{2}}\left(1-\frac{1}{\sqrt{n_{1}}}\right)\left(1-\frac{1}{\sqrt{n_{2}}}\right)e^{\pi(\sqrt{n_{1}}+\sqrt{Cn_{1}})},$

and

$\displaystyle{\overline{N}}(a,c;n_{1}+n_{2})<\frac{3}{16(n_{1}+Cn_{1})c}\left(1+\frac{1}{\sqrt{n_{1}+Cn_{1}}}\right)e^{\pi\sqrt{n_{1}+Cn_{1}}}.$

It remains to show that for $n_{1}\geq M_{c}$

$T_{n_{1}}(C)>\log\left(W_{n_{1}}(C)\right)+\log\left(S_{n_{1}}(C)\right),$

where

$\displaystyle W_{n_{1}}(C)$ $\displaystyle:=(48c)\frac{Cn_{1}}{C+1}.$

Obviously,

$\displaystyle W_{n_{1}}(C)<48cn_{1}.$

Therefore, we only need show

$\displaystyle T_{n_{1}}(1)$ $\displaystyle=2\pi\sqrt{n_{1}}-\pi\sqrt{2n_{1}}$
$\displaystyle>\log\left(48cn_{1}\right)+\log\left(\frac{1+\frac{1}{\sqrt{2n_{1}}}}{\left(1-\frac{1}{\sqrt{n_{1}}}\right)^{2}}\right).$
(4.4)

If $n_{1}\geq 2$, we obtain

$\displaystyle\log\left(48cn_{1}\right)+\log\left(\frac{1+\frac{1}{\sqrt{2n_{1}}}}{\left(1-\frac{1}{\sqrt{n_{1}}}\right)^{2}}\right)<\log\left(48cn_{1}\right)+\log\left(\frac{1+\frac{1}{\sqrt{4}}}{\left(1-\frac{1}{\sqrt{2}}\right)^{2}}\right)<\log(840cn_{1}).$
(4.5)

Moreover, if $n_{1}\geq(840c)^{2}\geq(840\times 6)^{2}$, then

$T_{n_{1}}(1)=2\pi\sqrt{n_{1}}-\pi\sqrt{2n_{1}}>2\log(n_{1})\geq\log(n_{1})+2\log(840c)>\log(840cn_{1}).$
(4.6)

Combined with (4), (4.5) and (4.6), choosing $n_{1},n_{2}\geq(840c)^{2}$, we
complete the proof.

Acknowledgements

This work was supported by the National Natural Science Foundation of China
(Grant Nos. 12001182 and 12171487), the Fundamental Research Funds for the
Central Universities (Grant No. 531118010411) and Hunan Provincial Natural
Science Foundation of China (Grant No. 2021JJ40037).

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