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\title{\bf On the subpartitions of\\ the ordinary partitions, II}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address 

\author{Byungchan Kim\thanks{This research has been supported by TJ Park Science Fellowship of POSCO TJ Park Foundation.}\\
\small School of Liberal Arts\\[-0.8ex]
\small Seoul National University of Science and Technology\\[-0.8ex] 
\small Seoul, Republic of  Korea\\
\small\tt bkim4@seoultech.ac.kr\\
\and
Eunmi Kim\\
\small Center for Applications of Mathematical Principles\\[-0.8ex]
\small National Institute for Mathematical Sciences\\[-0.8ex]
\small Daejeon, Republic of Korea\\
\small\tt ekim@nims.re.kr
}

\date{\dateline{Jun 30, 2013}{Oct 21, 2014}{Oct 30, 2014}\\
\small Mathematics Subject Classifications (2010): 05A17, 11P82}

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\begin{abstract} 
In this note, we provide a new proof for the number of partitions of $n$ having subpartitions of length $\ell$ with gap $d$. Moreover, by generalizing the definition of a subpartition, we show what is counted by $q$-expansion 
\[
\prod_{n=1}^{\infty} \frac{1}{1-q^n} \sum_{n=0}^{\infty} (-1)^n q^{(an^2 + bn)/2} 
\]
and how fast it grows. Moreover, we prove there is a special sign pattern for the coefficients of $q$-expansion
\[
\prod_{n=1}^{\infty} \frac{1}{1-q^n} \left( 1 - 2 \sum_{n=0}^{\infty} (-1)^n q^{(an^2 + bn)/2} \right).
\]

\bigskip\noindent \textbf{Keywords:} partition; subpartition; partial theta function.
\bigskip
\end{abstract}


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\section{Introduction}
Let $a_1 \ge a_2 \ge \cdots \ge a_{m}$ be an ordinary partition \cite{Andrews_Partition}. In a recent paper \cite{Kim_Sub}, the first author defines a subpartition of an ordinary partition as follows. Let us fix a positive integer $d$. Then, for a given partition, a subpartition with gap $d$ is defined as the longest sequence satisfying $a_1 > a_2 > \cdots > a_s$ and $a_s > a_{s+1}$, where $a_i - a_j \ge d$ for all $i < j \le s$. $a_{s+1}$ must be understood as a zero if it comes after the final part. This is a generalization of L. Kolitsch's Rogers-Ramanujan subpartition \cite{Kol}, which is the case $d=2$. We call the first condition involving $d$ a gap condition and the second condition $a_{s} > a_{s+1}$ a tail condition. For convenience, we define the subpartition of the empty partition as the empty partition. We define the length of the subpartition with gap $d$ as the number of parts in the subpartition. When the gap $d$ is clear in the context, we will write ``the subpartition'' instead of ``the subpartition with gap $d$''. In \cite{Kim_Sub}, the author uses subpartitions to find combinatorial proofs of entries in Ramanujan's lost notebook \cite{Rama}. Moreover, these subpartitions play a crucial role in obtaining an asymptotic formula for certain $q$-series involving partial theta functions \cite{J-Kim}. 



Define $p(n)$ to be the number of partitions of $n$ and $p(n,\ell,d)$ to be the number of partitions of $n$ having a subpartition of length $\ell$ with gap $d$. In \cite{Kim_Sub}, by finding a generating function via a case by case argument, the first author proved that

\begin{theorem} \label{main}
For all nonnegative integers $n$ and $\ell$ and a positive integer $d$, 
\[
 p(n,\ell,d) = p(n - S_{\ell,d} ) - p\left( n - S_{\ell+1,d} \right)
\]
where, for each nonnegative integer $k$,
\[
 S_{k,d}= \begin{cases}
          1 + (1+d) + (1 +2d) + \cdots + 1 + (k-1)d = \frac{dk^2 - (d-2)k}{2} \text{, if $k \neq 0$,} \\
          0 \text{, if $k = 0$.}
         \end{cases}
\]
\end{theorem}

\begin{example}
According to Theorem \ref{main}, there are $5$ partitions of $8$ having a subpartition of length $2$ with gap $2$ as $p(8,2,2) = p(8-4) - p(8-9) = p(4)-p(-1)=5-0=5$. Here are $5$ such partitions and the parts consisting of the subpartition are underlined:
\[
\underline{7}+\underline{1}, \;\; \underline{6} + \underline{2},\;\;  \underline{5} + \underline{3}, \;\; \underline{5}+ \underline{2} + 1, \;\; \text{and} \;\;  \underline{4}+\underline{2}+1+1.
\]
\end{example}

In this note, by employing a combinatorial argument, we give a simpler proof. 


Now we further generalize the notion of subpartitions as follows. We introduce a new parameter $t$ and replace the tail condition by $a_{s} - a_{s+1} \ge t$. The case $t=1$ is the original definition of a subpartition with gap $d$. Now we define $p(n,\ell,d,t)$ to be the number of partitions of $n$ having a subpartition of length $\ell$ with gap $d$ and tail condition $t$. Then, by employing essentially same argument, we can prove the following theorem.
\begin{theorem} \label{main2}
For all nonnegative integers $n$ and $\ell$, and positive integers $d$ and $t$,  
\[
p(n, \ell, d,t ) = p (n - T_{\ell,d, t} ) - p ( n - T_{\ell+1,d,t} ),
\]
where, for each nonnegative integer $k$,
\[
 T_{k,d,t}= \begin{cases}
          t + (t+d) + (t +2d) + \cdots + t + (k-1)d = \frac{dk^2  +( 2t - d ) k}{2} \text{, if $k \neq 0$,} \\
          0 \text{, if $k = 0$.}
         \end{cases}
\]
\end{theorem}


By summing even $\ell$'s, we see that, for a positive integer $a$ and an integer $b$ with $a+b>0$ and $a \equiv b \pmod{2}$, we find that
\begin{equation}\label{partial}
\frac{1}{(q;q)_{\infty}} \sum_{n=0}^{\infty} (-1)^n q^{(an^2 + bn)/2} =\sum_{n=0}^{\infty} p_{e} (n, a , (a+b)/2) q^n,
\end{equation}
where $(q;q)_{\infty} = \prod_{n=1}^{\infty} (1-q^n)$ and $p_{e} (n,d,t)$ is the number of partitions of $n$ having subpartitions of even length with gap $d$ and tail condition $t$. Here the assumption on $a$ and $b$ is for the positive integrality of $(an^2 + b n)/2$ for all positive integers $n$. From the representation of the partial theta function on the left side of \eqref{partial}, it is not clear at all the positivity of its $q$-expansion and what it counts. Since $n$ copies of $1$ is always counted by $p_{e} (n,a,(a+b)/2)$, the positivity of $q$-expansion is now clear from the combinatorial description. The case $a=b=1$ appears Andrews \cite{Andrews_Comp} as a generating function for the number of partitions of $n$ in which the first non-occurrence number as a part is odd, which is a conjugation of partition with subpartition of even length with gap $d$ as noted in \cite{J-Kim}.  When $a=1$ and $b=3$, we have
\[
\frac{1}{(q;q)_{\infty}} \sum_{n=0}^{\infty} (-1)^n q^{(n^2 + 3n)/2} = 1+q+q^2 + 2q^3 + 3q^4 + 5q^5 + \cdots.
\]
Among 7 partitions of 5, there are 5 partitions having subpartitions of even length with gap $1$ with tail condition $2$ as follows:
\[
4+1,\;\;3+2,\;\;2+2+1,\;\;2+1+1+1,\;\;1+1+1+1+1.
\]
 Moreover, by adopting the argument in \cite{J-Kim}, we can prove the following theorem.
\begin{theorem}\label{asym}
As $n$ tends to infinity, for positive integers $d$ and $t$, 
\[
p_{e} (n, d, t )  \sim \frac{1}{2} p(n).
\]
\end{theorem}

This is a generalization of \cite[Theorem 1]{J-Kim} and says that asymptotically half of the partitions of $n$ have subpartitions of even length. Much less obviously, there are inequalities between $p_{e} (n,a,(a+b)/2)$ and $p_{o} (n,a,(a+b)/2)$, where $p_{o} (n,a,(a+b)/2) = p(n) - p_{e} (n,a,(a+b)/2)$, i.e. the number of partitions of $n$ having subpartitions of odd length with gap $a$ and tail condition $(a+b)/2$.  These inequalities are unexpected since both $p_{e} (n,a,(a+b)/2)$ and $p_{o} (n,a,(a+b)/2)$ are asymptotically $p(n)/2$.

\begin{theorem} \label{mainineq}
For integers $a$ and $b$ satisfying $a>0$, $a+b>0$, and $a \equiv b \pmod{2}$, we have  
\begin{align*}
p_{e}(n,a,(a+b)/2) > p_{o} (n,a,(a+b)/2), &\quad\text{ if $b>0$,}\\
p_{e}(n,a,(a+b)/2) < p_{o} (n,a,(a+b)/2), &\quad\text{ if $b<0$,}
\end{align*}
for large enough integers $n$. Moreover, for $b=0$ and even integers $a>2$, we have
\begin{align*}
p_{e} (n, a , a/2) > p_{o} (n, a, a/2),
\end{align*}
for all positive integers $n$ except that the equality holds when $n=2$ and $a=4$.
\end{theorem}

\begin{remark}
The case $a=2$ and $b=0$ was discussed in \cite[Theorem 2]{J-Kim}. In this case, the sign of $p_e (n,2,1) - p_o (n,2,1)$ is alternating. This difference is due to that the generating function is essentially modular in this case. The more precise statement in the second part is also due to that $1 - 2 \sum_{n=0}^{\infty} (-1)^n q^{(a n^2 + b n)/2}$ becomes a theta function in these cases. 
\end{remark}

\begin{remark}
The conditions on $a$ and $b$, i.e. $a>0$, $a+b>0$, and $a \equiv b \pmod{2}$, are needed just for having non-negative integer exponents  in the $q$-expansion.
\end{remark}


This paper is organized as follows. In Section 2, we prove the combinatorial results. By adopting the circle method and elementary $q$-series manipulation, we will prove Theorem \ref{mainineq} in Section 3.





\section{Proof of Combinatorial Results}

For a given partition $\lambda$, we always write it in the form $\la_1 \ge \la_2 \ge \cdots \ge \la_{m}$, and for convenience, we define $\la_{s} = 0$ for all integer $s > m$. It is well known \cite{Andrews_Partition} that
\[
\sum_{n=0}^{\infty} p(n) q^n = \frac{1}{(q;q)_{\infty}}.
\]
 Now we define $p(n,t,d)$ to be the number of partitions of $n$ having subpartitions of length $\ge m$ with gap $d$. Then, the following lemma immediately implies Theorem \ref{main}.

\begin{lemma}
For all nonnegative integers $m$, 
\begin{equation*}
p(n,m,d) = p ( n - S_{m,d} ).
\end{equation*}
\end{lemma}

\begin{proof}
It is enough to show that
\[
\sum_{n=0}^{\infty} p(n,m,d) q^n = \frac{q^{S_{m,d}}}{(q;q)_{\infty}}.
\]
By definition, it is clear that the above holds when $t=0$, and thus we may assume that $t \ge 1$. We first observe that $q^{S_{m,d}}$ generates the partition $\pi = (1+(t-1)d, 1+(t-2)d, \ldots, 1)$. Let $\la$ be a partition generated by $\frac{1}{(q)_{\infty}}$. We append each part of $\la$ to $\pi$ beginning with the largest part, and denote the resulting partition as $\mu$, i.e., $\mu_i = \pi_i + \la_i$ for all positive integers $i$. For example, when $\pi = (5,3,1)$ and $\la = (4,4,3,2,1)$, we obtain $\mu = (9,7,4,2,1)$. Since the gap between two consecutive parts of $\pi$ is larger than or equal to $d$ and $\mu_{m} > \mu_{m+1}$, we see that $\mu$ has a subpartition of length at least $t$, which completes the proof.
\end{proof}


By employing the same argument, we can easily see that $\frac{q^{T_{m,d,t}}}{(q;q)_{\infty}}$ is a generating function for the number of partitions of $n$ having subpartitions of length $\ge m$ with gap $d$ and tail condition $t$, which implies Theorem \ref{main2}.




Now we turn to the proof of Theorem \ref{asym}. To this end, we are going to employ Ingham's Tauberian theorem (\cite[Theorem 1]{Ing} and \cite[Theorem 5.3]{BM}). To apply the Tauberian theorem, we have to show that $p_{e} (n,d,t)$ is weakly increasing. To see this, suppose that $\la$ is a partition of $n$ with subpartition of even length. If there is no subpartition in $\la$, we add a part $1$ to the partition $\la$. Then, the resulting partition is a partition of $n+1$, and since the size of the first two parts remains the same, the length of the subpartition is $0$. If $\la$ contains the subpartition, we increase the largest part of $\la$ by $1$. Then, the resulting partition is a partition of $n+1$ and this operation does not affect the length of the subpartition. It is clear that this map is an injection, thus we observe that $p_{e} (n,d,t) \le p_{e} (n+1, d, t)$. Theorem \ref{asym} now immediately follows from \cite[Theorem 1]{BK} and Ingham's Tauberian theorem. 





\section{Proof of Theorem \ref{mainineq}}


For a positive integer $a$ and an integer $b$ with $a+b>0$ and $a \equiv b \pmod{2}$, define
\begin{equation*}
	S_{a,b}(q):=1+2 \sum_{n=1}^{\infty} (-1)^n q^{(an^2+bn)/2},
\end{equation*}
\begin{equation*}
	F_{a,b}(q)=\sum_{N=0}^{\infty} \alpha_{a,b}(N) q^{N}:=\frac{1}{(q;q)_{\infty}} S_{a,b}(q).
\end{equation*} 
Note that $\alpha_{a,b} (N) = p_{e} (N, a, (a+b)/2) - p_{o} (N,a,(a+b)/2)$. Therefore, to prove Theorem \ref{mainineq}, it suffices to see the sign of $\alpha_{a,b} (N)$. We are going to get an asymptotic formula for $\alpha_{a,b} (N)$ by similar argument of Bringmann and Mahlburg \cite{BM}. Main idea of the proof is that we can get an asymptotic formula by focusing on asymptotic behavior of $S_{a,b}(q)$ near $q=1$. 



Set $q=e^{2\pi i \tau}$ with $\tau=x+iy$. The following proposition describes an asymptotic behavior of $S_{a,b}(q)$ near $q=1$.
\begin{proposition}\label{S_near1}
	Assume $|x|\leq y$. As $y \rightarrow 0+$,
	\begin{equation*}
		S_{a,b}(q) = \frac{b}{4} (-2 \pi i \tau) + \mathcal{O}(y^2).
	\end{equation*}
\end{proposition}

To prove this result, we need the following Zagier's result on asymptotic expansions for series (the first generalization of Proposition 3 in \cite{Zagier} with a correction on the sign),
\begin{lemma}\label{Z}
	Suppose that $h$ has the asymptotic expansion
	\[
		h(t) = \sum_{n=0}^{S} b_n t^n + \mathcal{O}\left( t^{S+1} \right)
	\]
	as $t \rightarrow 0+$ and that $h$ and all of its derivatives are of rapid decay at infinity, i.e. $\int_l^\infty |h^{(k)}(x)|\, dx$ converges for some $l>0$. Then, for $a>0$, as $t 	\rightarrow 0+$,
	\[
		\sum_{m=0}^{\infty} h((m+a)t)=\frac{1}{t} \int_{0}^{\infty} h(x) \, dx - \sum_{n=0}^{S} b_n \frac{B_{n+1}(a)}{n+1} t^n 	+\mathcal{O}\left( t^{S+1} \right),
	\]
where $B_n(x)$ is the $n$-th Bernoulli polynomial.
\end{lemma}
\begin{proof}[Proof of Proposition \ref{S_near1}]
	Let $f_{a,b}(\tau)=(S_{a,b}(q)-1)/2$, i.e.
	\[
		f_{a,b} ( \tau ) = \sum_{n=1}^{\infty} (-1)^n q^{(an^2 + bn)/2}.
	\]
	We can rewrite $f_{a,b}(\tau)$ as follows:
	\begin{equation*}
		f_{a,b}(\tau) = e^{-\frac{b^2}{4a}\pi i \tau} g_{a,b}(\tau),
	\end{equation*}
	where
	\begin{equation*}
		g_{a,b}(\tau)=\sum_{n=0}^{\infty} \left[ e^{4\pi a \left(n+1+\frac{b}{4a}\right)^2 i \tau} - e^{4\pi a \left(n+\frac{1}{2}+\frac{b}{4a}\right)^2 i \tau} \right].
	\end{equation*}
	We will find asymptotic formulas for the real and imaginary parts of $g_{a,b}(\tau)$.
	The real part of $g_{a,b}(\tau)$ can be written as
	\begin{equation*}
		\text{Re} \left(g_{a,b}(\tau)\right) = \sum_{n=0}^{\infty} \left[u_{\frac{x}{y}}\left( \left(n+1+\frac{b}{4a}\right)\sqrt{y} \right)-u_{\frac{x}{y}}\left( \left(n+\frac{1}{2}+\frac{b}{4a}\right)\sqrt{y} \right)\right],
	\end{equation*}
	where
	\begin{equation*}
		u_s(t)=e^{-4 \pi a t^2} \cos(4\pi a s t^2)=1 - 4 \pi a t^2  +\mathcal{O}(t^4) ~~\text{as}~~ t \rightarrow 0+.
	\end{equation*}
	By Lemma \ref{Z}, for $\dfrac{b}{4a}+\dfrac{1}{2}>0$ (this is the case as $a>0$, $a+b>0$),
	\begin{align*}
		\text{Re} \left(g_{a,b}(\tau)\right)
		=& \left[ \frac{I_u}{\sqrt{y}}-B_1\left(1+\frac{b}{4a}\right)-(- 4 \pi a)\frac{B_3\left(1+\frac{b}{4a}\right)}{3} y+\mathcal{O}_{\frac{x}{y}}(y^2) \right]\\
		&-\left[ \frac{I_u}{\sqrt{y}}-B_1\left(\frac{1}{2}+\frac{b}{4a}\right)-(- 4 \pi a)\frac{B_3\left(\frac{1}{2}+\frac{b}{4a}\right)}{3} y+ \mathcal{O}_{\frac{x}{y}}(y^2) \right]\\
		=&-\frac{1}{2}+\frac{b(2a+b)}{8} \pi y + \mathcal{O}_{\frac{x}{y}}(y^2) 
	\end{align*}
	with $|I_u|=\left| \int_{0}^{\infty} u_s(t) \, dt \right| < \infty$.
	The imaginary part can be treated similarly.
	\begin{align*}
		&\text{Im} \left(g_{a,b}(\tau)\right)\\
		&= \left[ \frac{I_v}{\sqrt{y}}-(4 \pi a)\frac{B_3\left(1+\frac{b}{4a}\right)}{3} x+\mathcal{O}_{\frac{x}{y}}(y^2) \right]
		-\left[ \frac{I_v}{\sqrt{y}}-(4 \pi a)\frac{B_3\left(\frac{1}{2}+\frac{b}{4a}\right)}{3} x + \mathcal{O}_{\frac{x}{y}}(y^2) \right]\\
		&= - \frac{b(2a+b)}{8} \pi x + \mathcal{O}_{\frac{x}{y}}(y^2) 
	\end{align*}
	where $v_s(t)=e^{-4 \pi a t^2} \sin(4\pi a s t^2)$ and $|I_v|=\left| \int_{0}^{\infty} v_s(t) \, dt \right| < \infty$.
	Together with the assumption $|x|\leq y$, we get
	\begin{equation*}
		g_{a,b}(\tau)=-\frac{1}{2}-\frac{b(2a+b)}{8a} (\pi i \tau) + \mathcal{O}(y^2).
	\end{equation*}

	Therefore, by considering the Taylor expansion of $e^{-\frac{b^2}{4a}\pi i \tau}$,
	\begin{align*}
		f_{a,b}(\tau)&= \left( 1-\frac{b^2}{4a}\pi i \tau + \mathcal{O}(y^2) \right) \left( -\frac{1}{2}-\frac{b(2a+b)}{8a} (\pi i \tau) + \mathcal{O}(y^2) \right) \\
		&=-\frac{1}{2}+\frac{b}{8} (-2 \pi i \tau)+\mathcal{O}(y^2),
	\end{align*}
	as $y\rightarrow 0+$ with $|x|\leq y$.
\end{proof}

Next, we consider the behavior of $S_{a,b}(q)$ away from $q=1$.
\begin{proposition}\label{S_away1}
	For $y=\dfrac{1}{2 \sqrt{6N}}$ with $N>0$ and $y \le |x| \le \frac{1}{2}$, we have
	\begin{equation*}
		\left| S_{a,b}(q) \right| \ll N^{1/2}.
	\end{equation*}
\end{proposition}
\begin{proof}
	For $a>0$ and $a+b>0$, bounding each term in $S_{a,b}(q)$ gives
	\begin{equation*}
		\left| S_{a,b}(q) \right| \leq 1+2 \sum_{n=1}^{\infty} |q|^{n/2} \ll N^{1/2}.\qedhere
	\end{equation*}
\end{proof}

The following two corollaries describe the behavior of the generating function $F_{a,b}(q)$ near $q=1$ and away from $q=1$, respectively.

\begin{corollary}\label{F_near1}
	Assume $y=\dfrac{1}{2 \sqrt{6N}}$ and $|x|\leq y$. As $N \rightarrow \infty$,
	\begin{equation*}
		F_{a,b}(q) = \left(\frac{b}{4}\right)\frac{e^{\frac{\pi i}{12 \tau}}}{\sqrt{2 \pi}}(-2 \pi i \tau)^{3/2} + \mathcal{O}\left(N^{-5/4} e^{\pi \sqrt{\frac{N}{6}}}\right).
	\end{equation*}
\end{corollary}
\begin{proof}
	From the asymptotic expansion (3.8) in \cite{BM}
	\begin{equation*}
		\frac{1}{(q;q)_{\infty}}=\sqrt{-i \tau} e^{\frac{\pi i}{12 \tau}} \left(1+\frac{2\pi i \tau}{24} + \mathcal{O}(N^{-1})\right),
	\end{equation*}
	we derive
	\begin{align*}
		\frac{1}{(q;q)_{\infty}}S_{a,b}(q)&= \sqrt{-i \tau} e^{\frac{\pi i}{12 \tau}} \left(1+\frac{2\pi i \tau}{24} + \mathcal{O}(N^{-1})\right) \left(\frac{b}{4} (-2 \pi i \tau) + \mathcal{O}(N^{-1})) \right)\\
		&= \sqrt{-i \tau} e^{\frac{\pi i}{12 \tau}} \left( \frac{b}{4} (-2 \pi i \tau) +\mathcal{O}(N^{-1})\right)
	\end{align*}
	by combining with Proposition \ref{S_near1}.
\end{proof}


\begin{corollary}\label{F_away1}
	If $y=\dfrac{1}{2 \sqrt{6N}}$ with $N>0$ and $y\leq |x| \leq \frac{1}{2}$,
	\begin{equation*}
		\left|F_{a,b}(q)\right| \ll e^{ \pi \sqrt{\frac{N}{6}} - \frac{\sqrt{6N} }{5\pi} }.
	\end{equation*}
\end{corollary}
\begin{proof}
	Proposition \ref{S_away1} together with the bound from \cite[Lemma 3.5]{BD}
	\begin{equation*}
		\frac{1}{|(q;q)_{\infty}|} \ll \sqrt{y} \exp \left[ \frac{1}{y} \left( \frac{\pi}{12} - \frac{1}{2\pi} \left( 1- \frac{1}{\sqrt{2}} \right)\right)\right]
	\end{equation*}
	gives the corollary.
\end{proof}

Now, we use the Circle Method with the results on $F_{a,b}(q)$ to see the sign pattern of $\alpha_{a,b} (N)$. By Cauchy's Theorem, we find 
\begin{equation*}
	\alpha_{a,b}(N)=\frac{1}{2 \pi i} \int_{\mathcal{C}} \frac{F_{a,b}(q)}{q^{N+1}} \, dq =\int_{-1/2}^{1/2} F_{a,b}\left( e^{-\frac{\pi}{\sqrt{6N}}+2\pi i x} \right)e^{\pi \sqrt{\frac{N}{6}}-2\pi i N x} \, dx,
\end{equation*}
where $\mathcal{C}=\{|q|=e^{-\frac{\pi}{\sqrt{6N}}}\}$.
We separate this integral into two integrals,
\begin{align*}
I'&=\int_{|x|\leq \frac{1}{2\sqrt{6N}}} F_{a,b}\left( e^{-\frac{\pi}{\sqrt{6N}}+2\pi i x} \right)e^{\pi \sqrt{\frac{N}{6}}-2\pi i N x} \, dx\\
\intertext{and}
I''&=\int_{\frac{1}{2\sqrt{6N}} \leq |x|\leq \frac{1}{2}} F_{a,b}\left( e^{-\frac{\pi}{\sqrt{6N}}+2\pi i x} \right)e^{\pi \sqrt{\frac{N}{6}}-2\pi i N x} \, dx.
\end{align*}
The first integral $I'$ gives the main contribution to the coefficient $\alpha_{a,b} (N)$.

\begin{proposition}
	As $N \rightarrow  \infty$,
	\begin{equation*}
		I'= b \frac{\pi^2}{24 \sqrt[4]{24}}N^{-5/4}I_{-5/2}\left(\pi\sqrt{\frac{2N}{3}}\right)+\mathcal{O}\left(N^{-7/4}e^{\pi \sqrt{\frac{2N}{3}}}\right),
	\end{equation*}
	where $I_{s}(z)$ is the modified Bessel function of the first kind.
\end{proposition}
\begin{proof}
	By Corollary \ref{F_near1} with the change of variable $u=2\sqrt{6N}x$, we deduce
	\begin{align*}
		I'&= \frac{1}{2\sqrt{6N}}\int_{-1}^1 F_{a,b} \left( e^{\frac{\pi}{\sqrt{6N}}(-1+iu)} \right) e^{\pi \sqrt{\frac{N}{6}}(1-iu)} \, du\\
		&= \frac{1}{2\sqrt{6N}}\int_{-1}^1 \left[\frac{b}{4 \sqrt{2\pi}} \left( \frac{\pi(1-iu)}{\sqrt{6N}} \right)^{3/2} e^{\pi\sqrt{\frac{N}{6}}\left(\frac{1}{1-iu}+(1-iu)\right)} +\mathcal{O}\left(N^{-5/4}e^{2\pi\sqrt{\frac{N}{6}}}\right)\right] \, du\\
		&= \frac{b}{4 \sqrt{2\pi}} \left(\frac{\pi}{\sqrt{6N}}\right)^{5/2} P_{3/2} + \mathcal{O}\left(N^{-7/4}e^{\pi\sqrt{\frac{2N}{3}}}\right),
	\end{align*}
	where 
	\begin{equation*}
		P_s := \frac{1}{2 \pi i} \int_{1-i}^{1+i} v^s e^{\pi\sqrt{\frac{N}{6}} \left(v+\frac{1}{v}\right)} \, dv.
	\end{equation*}
	Lemma 4.2 in \cite{BM} shows that
	\begin{equation*}
		P_s = I_{-s-1}\left(\pi \sqrt{\frac{2N}{3}}\right) + \mathcal{O}\left(e^{\frac{3\pi}{2}\sqrt{\frac{N}{6}}}\right),
	\end{equation*}
	which completes the proof.
\end{proof}

The next proposition shows that the second integral $I''$ is smaller than the error term in the main asymptotic formula of $I'$.
\begin{proposition}
	As $N \rightarrow \infty$,
	\begin{equation*}
		|I''| \ll e^{ 2\pi \sqrt{\frac{N}{6}} - \frac{\sqrt{6N} }{5\pi} }
	\end{equation*}
\end{proposition}
\begin{proof}
	By Corollary \ref{F_away1},
	\begin{equation*}
		|I''| \leq \int_{\frac{1}{2\sqrt{6N}} \leq |x|\leq \frac{1}{2}} \left|F_{a,b}\left( e^{-\pi \sqrt{\frac{N}{6}}+2\pi i x} \right)e^{\pi \sqrt{\frac{N}{6}}-2\pi i N x}\right| \, dx \ll e^{\pi \sqrt{\frac{N}{6}}} e^{ \pi \sqrt{\frac{N}{6}} - \frac{\sqrt{6N} }{5 \pi} }.\qedhere
	\end{equation*}
\end{proof}

In summary, we have obtained the following asymptotic formula for the coefficient $\alpha_{a,b} (N)$.
\begin{corollary} \label{asymcor}
	For an positive integer $a$ and an integer $b$ with $a+b>0$ and $a \equiv b \pmod{2}$, as $N \rightarrow \infty$,
	\begin{equation*}
		\alpha_{a,b}(N)=b \frac{\pi^2}{24 \sqrt[4]{24}}N^{-5/4}I_{-5/2}\left(\pi\sqrt{\frac{2N}{3}}\right)+\mathcal{O}\left(N^{-7/4}e^{\pi \sqrt{\frac{2N}{3}}}\right).
	\end{equation*}
\end{corollary}

Now, we are ready to prove Theorem \ref{mainineq}. Since
\[
I_{s} (z) \sim \frac{e^z}{\sqrt{2\pi z}},
\]
the first part of Theorem \ref{mainineq} follows immediately from Corollary \ref{asymcor}. 

 
For the second part of Theorem \ref{mainineq}, note that
\[
\sum_{n=0}^{\infty} \alpha_{2k, 0} (n) q^n = \frac{ (q^k ; q^{2k})_{\infty}^{2} (q^{2k} ; q^{2k} )_{\infty}}{ (q;q)_{\infty} },
\]
where we set $a=2k>2$ and have applied Jacobi's triple product identity \cite[Theorem 2.8]{Andrews_Partition}. From this, we deduce that
\begin{align}
\frac{ (q^k ; q^{2k})_{\infty}^{2} (q^{2k} ; q^{2k} )_{\infty}}{ (q;q)_{\infty} } &= \frac{ (q^k ; q^{2k})_{\infty}^{2} (q^{2k} ; q^{2k} )_{\infty}^{3}}{ (q)_{\infty} (q^{2k} ; q^{2k} )_{\infty}^{2}} \nonumber \\
&=\frac{ (q^k ; q^{k})_{\infty}^{2} }{ (q;q)_{\infty} (q^{2k} ; q^{2k} )_{\infty}} \nonumber \\
&=\frac{ (q^k ; q^{k})_{\infty}^{k} }{ (q;q)_{\infty}(q^k ; q^{k})_{\infty}^{k-2} (q^{2k} ; q^{2k} )_{\infty}}. \label{b0form}
\end{align}
Since $\frac{(q^k ;q^k)_{\infty}^{k}}{(q;q)_{\infty}}$ is a generating function for the number of $k$-core partition, it is clear that $q$-expansion of the above infinite product is nonnegative. By the result of A. Granvile and K. Ono \cite{GO}, we know that there is a $k$-core partition of $n$ if $k \ge 4$. Thus, when $k \ge 4$, the $q$-expansion of \eqref{b0form} has positive coefficients. When $k=3$, since the partitions $1$ and $1+1$ are $3$-core partitions and since $\frac{1}{(q^3 ;q^3)_{\infty}} = \sum_{n=0}^{\infty} p(n) q^{3n}$, the coefficients of $q$-expansion of \eqref{b0form} are always positive. Finally, when $k=2$, we see that \eqref{b0form} is
\[
\frac{ (q^2 ; q^{2})_{\infty}^{2} }{ (q;q)_{\infty} (q^{4} ; q^{4} )_{\infty}}.
\]
Note that the partitions $1$, $1+2$, and $1+2+3$ are $2$-core partitions and $\frac{1}{(q^4 ; q^4 )_{\infty}} = \sum_{n=0}^{\infty} p(n) q^{4n}$. Therefore, the $n$-th coefficient of $q$-expansion has positive coefficient provided $n \equiv 0,1,3,6$. Hence, the $q$-expansion of \eqref{b0form} has positive coefficients except $n=2$. This completes the proof of the second part of Theorem \ref{mainineq}.



\subsection*{Acknowledgments}
The authors would like to thank Bruce Berndt, Jehanne Dousse, and the referee for their careful reading and comments on an earlier version of the paper. 


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%               References              %
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