# Exact formula for 1-lower run overpartitions

Lukas Mauth  Department of Mathematics and Computer Science
Division of Mathematics
University of Cologne
Weyertal 86-90
50931 Cologne
Germany<EMAIL_ADDRESS>
###### Abstract.

We are going to show an exact formula for lower $1$-run overpartitions. The
generating function is of mixed mock-modular type with an overall weight $0.$
We will apply an extended version of the classical Circle Method. The approach
requires bounding modified Kloosterman sums and Mordell integrals.

###### Key words and phrases:

Circle Method, $\eta$-function, partitions

###### 2020 Mathematics Subject Classification:

11B57, 11F03, 11F20, 11F30, 11F37, 11P82

##  1\. Introduction and statement of results

An overpartition of a non-negative integer $n$ is a partition of n in which
the final occurrence of a number may be overlined. We denote the number of
partitions of $n$ by $p(n)$ and the number of overpartitions of $n$ by
$\overline{p}(n).$ It is well known that their generating functions are

$\displaystyle P(q)$
$\displaystyle\coloneqq\sum_{n=0}^{\infty}p(n)q^{n}=\prod_{n=1}^{\infty}\frac{1}{1-q^{n}}=\frac{1}{(q;q)_{\infty}},$
$\displaystyle\overline{P}(q)$
$\displaystyle\coloneqq\sum_{n=0}^{\infty}\overline{p}(n)q^{n}=\prod_{n=1}^{\infty}\frac{1+q^{n}}{1-q^{n}}=\frac{(-q;q)_{\infty}}{(q;q)_{\infty}},$

where for $a\in\mathbb{C}$ and $n\in\mathbb{N}\cup\\{\infty\\}$ we define the
$q$-Pochhammer symbol
$(a)_{n}\coloneqq(a;q)_{n}\coloneqq\prod_{k=1}^{n-1}(1-aq^{k}).$ These
generating functions are essentially special cases of meromorphic modular
forms, so called eta-quotients. Classical questions are to determine
asymptotics or exact formulas for the Fourier coefficents of modular forms.
The Hardy$-$Ramanujan Tauberian Theorem [8] for eta-quotients gives the
following asymptotics as $n\rightarrow\infty$

$\displaystyle p(n)$
$\displaystyle\sim\frac{1}{4\sqrt{3}n}e^{\pi\sqrt{\frac{2\pi}{3}}},$
$\displaystyle\overline{p}(n)$ $\displaystyle\sim\frac{1}{8n}e^{\pi\sqrt{n}}.$

Later Rademacher perfected the Circle Method developed by Hardy and Ramanujan
and obtained an exact formula for $p(n).$ To state Rademacher’s result we
define the Kloosterman sums

$A_{k}(n):=\sum_{h\,\,({\rm mod}\,\,{k})^{*}}\omega_{h,k}e^{\frac{-2\pi
inh}{k}},$

where $*$ indicates that $h$ only runs over those residue classes that are
coprime to $k$ and $\omega_{h,k}$ is defined in Section $2.$ Let $I_{\kappa}$
denote the modified Bessel function of order $\kappa.$ Rademacher’s exact
formula for $p(n)$ then reads as follows

$p(n)=\frac{2\pi}{(24n-1)^{3/4}}\sum_{k=1}^{\infty}\frac{A_{k}(n)}{k}I_{\frac{3}{2}}\left(\frac{\pi\sqrt{24n-1}}{6k}\right).$

In [15] Zuckerman generalized the Circle Method to provide exact formulas for
the Fourier coefficients at any cusp for arbitrary modular forms of negative
weight for finite index subgroups of the modular group
$\operatorname{SL}_{2}(\mathbb{Z}).$ It is evident that modularity is the work
horse in those methods. However, one could ask if the Cirlce Method can be
used to find exact formulae for the Fourier coefficients of objects that still
have some automorphic structure. Natural candidates are mock theta functions.
Important examples are Ramanujan’s third order mock theta functions

$\displaystyle f(q)$ $\displaystyle\coloneqq\sum_{n\geq
0}\frac{q^{n^{2}}}{(-q;q)_{n}^{2}},$ $\displaystyle\varphi(q)$
$\displaystyle\coloneqq\sum_{n\geq 0}\frac{q^{n^{2}}}{(-q^{2};q^{2})_{n}}.$

Those are not modular forms, but can be understood as holomorphic parts of
harmonic Maas forms of half-integral weight. Bringmann and Ono have found
exact formulas for the coefficients of harmonic Maas forms of weight
$<\frac{1}{2}.$ This includes all the above examples but their technique does
not use the Cirlce Method and rather relies on Poincaré series. Another class
of automorphic objects which is not covered by the above cases is the space of
modular forms tensored with harmonic Maass forms. It is unlikely that a basis
of Poincaré series exists for this space, so the Circle Method is the method
of choice to study the coeffcients of these objects. In [5] Bringmann and
Mahlburg worked out a technique based on the Circle Method to obtain
asymptotics for the Fourier coeffcients and was later refined by Bridges and
Bringmann [3] to provide exact formulas of Rademacher type. They illustrated
the method for the function $p_{2}(n)$ which counts partitions without
sequences in its parts and it is the first example of an exact formula for the
Fourier coefficients of a mixed-mock modular form of weight $0.$ We follow
their developed method [3, 5] closely to study the Fourier coefficients of
lower $1$-run overpartitions, which are those overpartitons in which any
overlined part must occur within a run of exactly k consecutive overlined
parts that end below with a gap, where an overpartition is said to have a gap
at $m$ if there are no parts of size $m.$

######  Example.

The lower $1$-run overpartitions of size $4$ are

$\overline{4},\quad\overline{3}+1,\quad 3+\overline{1},\quad
2+\overline{2},\quad\overline{2}+1+1,\quad 2+1+\overline{1},\quad
1+1+1+\overline{1},$

together with the $5$ partitions of $4,$ thus $\overline{p_{1}}(n)=7+5=12.$

We will denote the lower $1$-run overpartitions of $n$ by
$\overline{p_{1}}(n).$ The generating function for $\overline{p_{1}}(n)$ was
worked by Bringmann, Holroyd, Mahlburg, and Vlasenko [4] (note that there is a
typo in their statement)

$G_{1}(q)\coloneqq\frac{(q^{4};q^{4})_{\infty}}{(q;q)_{\infty}(q^{2};q^{2})_{\infty}}\varphi(q).$

Using Ingham’s Tauberian Theorem they showed the following asymptotic

$\overline{p_{1}}(n)\sim\frac{\sqrt{5}}{4\sqrt{6}n}e^{\pi\sqrt{\frac{5}{6}n}},\quad(n\rightarrow\infty).$

The generating $G_{1}(q)$ is a mixed-mock modular form of overall weight $0$
and thus falls into the preivously mentioned framework developed by Bringmann
and Bridges [3]. We will follow their method closely to obtain the following
exact formula (Theorem 4.1)

######  Theorem.

We have, for $n\in\mathbb{N},$

$\displaystyle\overline{p_{1}}(n)$
$\displaystyle=\frac{\pi}{6\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=1\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{n+\nu}K_{k}^{[12]}(\nu,n)\mathcal{I}_{\frac{1}{24},k,\nu}(n)$
$\displaystyle+\frac{5\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{n+\nu}K_{k}^{[22]}(\nu,n)\mathcal{I}_{\frac{5}{12},k,\nu}(n).$

Here we have set for $b\in\mathbb{R},k\in\mathbb{N},$ and $\nu\in\mathbb{Z}$

$\mathcal{I}_{b,k,\nu}(n):=\int_{-1}^{1}\frac{\sqrt{1-x^{2}}I_{1}\left(\frac{2\pi}{k}\sqrt{2bn(1-x^{2})}\right)}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}dx.$

These integrals should be viewed as a natural result of mixing the Mordell
integrals and modular factors in the Circle Method. The Mordell integrals
appear in the transformation laws of mock theta functions [7] and Bessel
functions do appear in the exact formulas for eta-quotients [15]. In that
sense $\mathcal{I}_{b,k,\nu}$ is a natural result of combining Mordell
integrals and Bessel functions. As an immediate corollary, we will give
another proof of (1). Our strategy is as follows. In Section $2$ we will work
out all modular transformation laws needed for the Circle Method. It is
worthwhile to note that we will not work with $G_{1}(q)$ directly. Instead we
are going to twist the generating function by a root of unity and work instead
with

$\overline{G_{1}}(q)\coloneqq\sum_{n=0}^{\infty}(-1)^{n}\overline{p_{1}}(n)q^{n}=\frac{(q^{4};q^{4})_{\infty}}{(-q;q)_{\infty}(q^{2};q^{2})_{\infty}}\varphi(-q).$
(1.1)

The main reason for this choice is that the modular properties of (1.1) are
easier to work with, than for the other generating function, see Section $2.$
In Section $3$ we are going to give bounds for the Kloosterman sums and the
Mordell integrals that appear in the transformations laws for mock theta
functions. Finally in Section $4$ we are going to apply the Cirlce Method to
prove Theorem 4.1.

## Acknowledgements

The author wishes to thank Kathrin Bringmann for suggesting this problem and
helpful discussions. Furthermore, the author thanks Giulia Cesana for advice
on rewriting Kloosterman sums and aide to verify the main result numerically.
The author recieved funding from the European Research Council (ERC) under the
European Union’s Horizon 2020 research and innovation programme (grant
agreement No. 101001179).

##  2\. Modularity Properties

As stated in the introduction we define

$P(q):=(q;q)_{\infty}^{-1}.$

For coprime integers $h$ and $k,$ we find $h^{\prime}$ such that
$hh^{\prime}\equiv-1\,\,({\rm mod}\,\,{k}).$ We will assume that for even $k$
this congruence holds $\,\,({\rm mod}\,\,{4k})$ and for odd $k$ that
$8|h^{\prime}.$ Furthermore, we define by $z$ a complex variable that
satisfies $\text{Re}\left(z\right)>0$ and $q=e^{\frac{2\pi i}{k}(h+iz)},$ and
set $q_{1}:=e^{\frac{2\pi i}{k}(h^{\prime}+iz^{-1})}.$ The classical modular
transformation law is [1]

$P(q)=\omega_{h,k}z^{\frac{1}{2}}e^{\frac{\pi(z^{-1}-z)}{12k}}P(q_{1}),$ (2.1)

where $\omega_{h,k}$ is the $24k$-th root of unity defined by

$\omega_{h,k}\coloneqq\begin{cases}\left(\frac{-k}{h}\right)e^{-\pi
i\left(\frac{1}{4}(2-hk-h)+\frac{1}{12}\left(k-\frac{1}{k}\right)\left(2h-h^{\prime}+h^{2}h^{\prime}\right)\right)}&\text{
if }h\text{ is odd,}\\\ \left(\frac{-h}{k}\right)e^{-\pi
i\left(\frac{1}{4}(k-1)+\frac{1}{12}\left(k-\frac{1}{k}\right)\left(2h-h^{\prime}+h^{2}h^{\prime}\right)\right)}&\text{
if }k\text{ is odd,}\par\end{cases}$

and $\left(\frac{\cdot}{\cdot}\right)$ denotes the Kronecker symbol. From
(2.1) we obtain easily a transformation law for $P(q^{r}),$ where $r$ is a
positive integer. To state the transformation law in a convenient form, we
introduce some notation. First, we set $g_{r}:=\gcd(r,k),$ and define the
quantities $\rho_{r}:=\frac{r}{g_{r}}$ and $k_{r}:=\frac{k}{g_{r}}.$ Note that
$\gcd(\rho_{r},k_{r})=1.$ Furthermore, we define $q_{r}:=e^{\frac{2\pi
i\left(h_{r}^{\prime}+\frac{i}{\rho_{r}z}\right)}{k_{r}}}.$ It follows that up
to a root of unity $q_{r}=q_{1}^{\frac{g_{r}}{\rho_{r}}}.$ We will ignore this
technicality as it is not relevant for our estimates and just assume that
$q_{r}=q_{1}^{\frac{g_{r}}{\rho_{r}}}.$ Finally, we have the transformation
law

$P(q^{r})=\omega_{h\rho_{r},k_{r}}(\rho_{r}z)^{\frac{1}{2}}e^{\frac{\pi}{12k_{r}}\left(\frac{z^{-1}}{\rho_{r}}-\rho_{r}z\right)}P(q_{r}).$
(2.2)

We are now going to study the necessary transformation laws to sudy (1.1). We
exploit the following linear relation [13] between the two third-order mock
theta functions $\varphi(q)$ and $f(q):$

$2\varphi(-q)-f(q)=\frac{(q;q)_{\infty}^{2}}{(-q;q)_{\infty}(q^{2};q^{2})_{\infty}}.$

We can then rewrite the generating function (1.1) as

$\overline{G_{1}}(q)=\underbrace{\frac{(q^{4};q^{4})_{\infty}^{2}(q;q)_{\infty}}{2(q^{2};q^{2})_{\infty}^{4}}f(q)}_{\eqqcolon
g_{1}(q)}+\underbrace{\frac{(q^{4};q^{4})_{\infty}^{2}(q;q)_{\infty}^{4}}{2(q^{2};q^{2})_{\infty}^{6}}}_{\eqqcolon
g_{2}(q)}.$

We start with the transformation law for $g_{1}(q).$ Define
$\xi(q):=\frac{P(q^{2})^{4}}{P(q^{4})^{2}P(q)}.$ We will have to distinguish
three cases, depending on the different values of $\gcd(4,k).$ These
transformation all follow from iterated application of (2.1) and (2.2).

First, assume that $4|k.$ Then,

$\xi(q)=\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,k}\omega_{h,\frac{k}{4}}^{2}}z^{\frac{1}{2}}e^{-\frac{\pi}{12k}(z^{-1}-z)}\xi(q_{1}).$

Next, if $\gcd(4,k)=2,$ we have

$\xi(q)=\frac{\omega_{h,\frac{k}{2}}^{4}}{2\omega_{h,k}\omega_{2h,\frac{k}{2}}^{2}}z^{\frac{1}{2}}e^{\frac{5\pi}{12kz}+\frac{\pi
z}{12k}}\frac{P(q_{1}^{2})^{4}}{P(q_{1})P(q_{1})^{2}}.$

Finally, if $\gcd(4,k)=1,$

$\xi(q)=\frac{\omega_{2h,k}^{4}}{\omega_{h,k}\omega_{4h,k}^{2}}z^{\frac{1}{2}}e^{\frac{\pi}{24kz}+\frac{\pi
z}{12k}}\frac{P\left(q_{1}^{\frac{1}{2}}\right)^{4}}{P(q_{1})P\left(q_{1}^{\frac{1}{4}}\right)^{2}}.$

Now we turn to the third order mock theta function $f(q).$ Andrews [2] has
shown that $f(q)$ behaves essentially modular of level $2.$ If $k$ is even we
have

$\displaystyle f(q)$ $\displaystyle=(-1)^{\frac{k}{2}+1}e^{\pi
i\left(\frac{h^{\prime}}{2}-\frac{3h^{\prime}k}{4}\right)}\omega_{h,k}z^{-\frac{1}{2}}e^{\frac{\pi(z^{-1}-z)}{12k}}f(q_{1})$
$\displaystyle+\omega_{h,k}\frac{2}{k}z^{\frac{1}{2}}e^{-\frac{\pi
z}{12k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}+\nu)}{k}}I_{k,v}(z),$

where $I_{k,\nu}(z)$ is the Mordell-integral defined by

$I_{k,\nu}(z)\coloneqq\int_{-\infty}^{\infty}\frac{e^{-\frac{3\pi
zx^{2}}{k}}}{\cosh\left(\frac{\pi i\left(\nu-\frac{1}{6}\right)}{k}-\frac{\pi
zx}{k}\right)}dx.$

Note that there is a typo regarding the term $e^{\frac{\pi(z^{-1}-z)}{12k}}$
as it is stated in the statement of Theorem $2.2$ in [2]. If $k$ is odd we
have

$\displaystyle f(q)$ $\displaystyle=2(-1)^{\frac{1}{2}(k-1)}e^{\frac{3\pi
ih^{\prime}}{4k}}\omega_{h,k}z^{-\frac{1}{2}}e^{-\frac{2\pi}{3kz}-\frac{\pi
z}{12k}}\omega\left(q_{1}^{\frac{1}{2}}\right)$
$\displaystyle+\frac{2}{k}\omega_{h,k}z^{\frac{1}{2}}e^{-\frac{\pi
z}{12k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}-\nu)}{k}}I_{k,\nu}(z),$

where $\omega(q)$ denoted the third order mock theta function

$\omega(q)\coloneqq\sum_{n\geq 0}\frac{q^{2n(n+1)}}{(q;q^{2})_{n+1}^{2}}.$

Finally, we are going to derive the transformation laws for $g_{2}(q)$
depending on $\gcd(4,k).$ If $4|k$ we have

$g_{2}(q)=\frac{\omega_{h,\frac{k}{2}}^{6}}{2\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}g_{2}(q_{1}).$

Next, if $\gcd(4,k)=2$ we have

$g_{2}(q)=\frac{\omega_{h,\frac{k}{2}}^{6}}{4\omega_{h,k}^{4}\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{\pi}{2kz}}\frac{P(q_{1}^{2})^{6}}{P(q_{1})^{6}}.$

Finally, if $\gcd(4,k)=1$ we have

$g_{2}(q)=\frac{\omega_{2h,k}^{6}}{4\omega_{h,k}^{4}\omega_{4h,k}^{2}}e^{-\frac{\pi}{8kz}}\frac{P\left(q_{1}^{\frac{1}{2}}\right)^{6}}{P(q_{1})^{4}P\left(q_{1}^{\frac{1}{4}}\right)^{2}}.$

##  3\. Bounds on Kloosterman sums and Mordell Integrals

In [12] Rademacher proved the following bound for Kloosterman sums. Recall
that $k_{1}$ is the denominator of the fraction proceeding $\frac{h}{k}$ in
the Farey sequence of order $N\in\mathbb{N}.$

######  Lemma 3.1.

We have for $k\in\mathbb{N},n,m,\ell\in\mathbb{Z},n\neq 0$ with
$N+1\leq\ell\leq N+k+1,$ for $\varepsilon>0,$

$\displaystyle K_{k}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}e^{-\frac{2\pi i}{k}(nh-
mh^{\prime})}=O_{\varepsilon}\left(k^{\frac{2}{3}+\varepsilon}\gcd(|n|,k)^{\frac{1}{3}}\right),$
$\displaystyle\mathbb{K}_{k,\ell}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\\\ N<k+k_{1}\leq\ell\end{subarray}}e^{-\frac{2\pi i}{k}(nh-
mh^{\prime})}=O_{\varepsilon}\left(k^{\frac{2}{3}+\varepsilon}\gcd(|n|,k)^{\frac{1}{3}}\right).$

In the case that $4|k$ define the following Kloosterman sums

$\displaystyle K_{k}^{[41]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})},$ $\displaystyle K_{k}^{[42]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}+\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle K_{k}^{[43]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

Furthermore, we define for $N+1\leq\ell\leq N+k-1,$

$\displaystyle\mathbb{K}_{k,\ell}^{[41]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})},$ $\displaystyle\mathbb{K}_{k,\ell}^{[42]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}+\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle\mathbb{K}_{k,\ell}^{[43]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

######  Lemma 3.2.

We have for $\varepsilon>0,$

$K_{k}^{[41]}(n,m),K_{k}^{[42]}(\nu,n,m),K_{k}^{[43]}(n,m),\mathbb{K}_{k,\ell}^{[41]}(n,m),\mathbb{K}_{k,\ell}^{[42]}(\nu,n,m),\mathbb{K}_{k,\ell}^{[43]}(n,m)\ll_{\varepsilon}n^{\frac{1}{3}}k^{\frac{2}{3}}.$

###### Proof.

We only give a complete proof for $K_{k}^{[41]}(n,m)$ as the idea is the same
for all other Kloosterman sums. The only hard part is to rewrite the
multiplier, so we just have to bound an ordinary Kloosterman sum, which we can
do with Lemma 3.1.

We have keeping in mind that $hh^{\prime}\equiv-1\,\,({\rm mod}\,\,{4k})$

$\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}=-e^{\frac{\pi
i}{8}(-h^{\prime}k+h^{2}h^{\prime}k-h(4+k))}=-e^{\frac{\pi
i}{8}h^{\prime}k+(4+2k)h}.$

Hence,

$\displaystyle K_{k}^{[41]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}$ $\displaystyle=-\sum_{\begin{subarray}{c}h\,\,({\rm
mod}\,\,{k})\\\ \gcd(h,k)=1\end{subarray}}e^{-\frac{2\pi
i}{k}\left(\left(n-\frac{2k^{2}+4k}{16}\right)h-\left(m+\frac{5k^{2}-4k}{16}\right)\right)}.$

Since $4|k$ we have

$\frac{2k^{2}+4k}{16}\in\mathbb{Z},\quad\frac{5k^{2}-4k}{16}\in\mathbb{Z}.$

Therefore, we can use Lemma 3.1 to obtain as desired

$\left|K_{k}^{[41]}(n,m)\right|=\left|K_{k}\left(n-\frac{2k^{2}+4k}{16},m+\frac{5k^{2}-4k}{16}\right)\right|\ll_{\varepsilon}k^{\frac{2}{3}+\varepsilon}\gcd\left(\left|n-\frac{2k^{2}+4k}{16}\right|,k\right)^{\frac{1}{3}}\ll_{\varepsilon}n^{\frac{1}{3}}k^{\frac{2}{3}+\varepsilon}.$

For the Kloosterman sum $K_{k}^{[43]}$ we distinguish two cases. If $8|k$ we
write

$\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}=e^{\frac{2\pi
i}{k}\left(-\frac{4k}{16}h-\frac{2k}{16}h^{\prime}\right)}.$

Thus, since $8|k$ we have

$\frac{4k}{16}\in\mathbb{Z},\quad\frac{2k}{16}\in\mathbb{Z}.$

In the case that $k\equiv 1\,\,({\rm mod}\,\,{4})$ we find using that
$h^{2}\equiv 1\,\,({\rm mod}\,\,{4})$ (since $k$ is even)

$\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}=ie^{\frac{2\pi
ik}{16}}e^{\frac{2\pi i}{k}\frac{4kh}{16}}.$

∎

In the case that $\gcd(4,k)=2$ define the following Kloosterman sums

$\displaystyle K_{k}^{[21]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})},$ $\displaystyle K_{k}^{[22]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}+\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle K_{k}^{[23]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

Furthermore, we need the incomplete Kloosterman sums for $N+1\leq\ell\leq
N+k-1,$ define

$\displaystyle\mathbb{K}_{k,\ell}^{[21]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})},$ $\displaystyle\mathbb{K}_{k,\ell}^{[22]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}+\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle\mathbb{K}_{k,\ell}^{[23]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{2h,\frac{k}{2}}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

######  Lemma 3.3.

We have for $\varepsilon>0,$

$K_{k}^{[21]}(n,m),K_{k}^{[22]}(\nu,n,m),K_{k}^{[23]}(n,m),\mathbb{K}_{k,\ell}^{[21]}(n,m),\mathbb{K}_{k,\ell}^{[22]}(\nu,n,m),\mathbb{K}_{k,\ell}^{[23]}(n,m)\ll_{\varepsilon}n^{\frac{1}{3}}k^{\frac{2}{3}}.$

###### Proof.

We will later show $K_{k}^{[21]}(n,m)=-K_{k}^{[23]}(n,m).$ Thus, is suffices
to consider $K_{k}^{[21]}(n,m).$ Since $h$ is odd we have $h^{2}\equiv
1\,\,({\rm mod}\,\,{4})$ and thus,

$\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}=ie^{-\frac{ki}{4}}e^{\frac{2\pi
i}{k}\left(\frac{-12+3k^{2}}{24}h^{\prime}\right)}.$

Now it suffices to note that since $2|k$ we have

$\frac{-12+3k^{2}}{24}\in\mathbb{Z}.$

∎

In the case that $\gcd(4,k)=1$ define the following Kloosterman sums

$\displaystyle K_{k}^{[11]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}e^{\frac{3\pi
ih^{\prime}}{4k}}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$ $\displaystyle
K_{k}^{[12]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}-\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle K_{k}^{[13]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{2h,k}^{6}}{\omega_{h,k}^{4}\omega_{4h,k}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

Furthermore, we need the incomplete Kloosterman sums for $N+1\leq\ell\leq
N+k-1,$ define

$\displaystyle\mathbb{K}_{k,\ell}^{[11]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}e^{\frac{3\pi
ih^{\prime}}{4k}}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle\mathbb{K}_{k,\ell}^{[12]}(\nu,n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}e^{\frac{\pi
ih^{\prime}}{k}(-3\nu^{2}-\nu)}e^{\frac{2\pi i}{k}(-nh+mh^{\prime})},$
$\displaystyle\mathbb{K}_{k,\ell}^{[13]}(n,m)$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\frac{\omega_{2h,k}^{6}}{\omega_{h,k}^{4}\omega_{4h,k}^{2}}e^{\frac{2\pi
i}{k}(-nh+mh^{\prime})}.$

######  Lemma 3.4.

We have for $\varepsilon>0,$

$K_{k}^{[11]}(n,m),K_{k}^{[12]}(\nu,n,m),K_{k}^{[13]}(n,m),\mathbb{K}_{k,\ell}^{[11]}(n,m),\mathbb{K}_{k,\ell}^{[12]}(\nu,n,m),\mathbb{K}_{k,\ell}^{[13]}(n,m)\ll_{\varepsilon}n^{\frac{1}{3}}k^{\frac{2}{3}}.$

###### Proof.

We start with $K_{k}^{[11]}(n,m).$ We will use the notation $[a]_{k}$ to
denote the inverse of $a\,\,({\rm mod}\,\,{k}).$ The multiplier evaluates to

$\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}=e^{\frac{2\pi
i}{k}\left(\frac{(k^{2}-1)(8h^{2}+1)h^{\prime}}{12}\right)}.$

In the case that $3|k$ we can write this as

$\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}=e^{\frac{2\pi
i}{k}12[12]_{k}(8h-h^{\prime})}.$

In this case

$K_{k}^{[11]}(n,m)=\sum_{\begin{subarray}{c}h\,\,({\rm mod}\,\,{k})\\\
\gcd(h,k)=1\end{subarray}}e^{\frac{2\pi
i}{k}\left(-\left(n-96[12]_{k}\right)h+\left(m-12[12]_{k}+\frac{3}{8}\right)h^{\prime}\right)}$

In the beginning we chose $h^{\prime}$ such that $8|h^{\prime}.$ To get rid of
this condition we perform the change of variables $h^{\prime}\mapsto
8h^{\prime}$ and $h\mapsto[8]_{k}h$ and find

$K_{k}^{[11]}(n,m)=K_{k}\left([8]_{k}n-96[96]_{k},8m-96[12]_{k}+3\right).$

In the case that $3\nmid k$ we can write

$3\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}=\frac{\left(k^{2}-1\right)(-8+8ak)h}{4}+\frac{\left(k^{2}-1\right)}{4}h^{\prime}.$

for some $a\in\mathbb{Z}.$ It is easy to see that the factors in front of $h$
and $h^{\prime}$ are both integers. Since $3|k$ we can extend the modulus of
the Kloosterman sum from $k$ to $3k$ and proceeding as before yields

$K_{k}^{[11]}(n,m)=\frac{1}{3}K_{3k}\left([8]_{k}n-[8]_{k}\left(k^{2}-1\right)(-8+8ak),8m+8\left(k^{2}-1\right)+3\right).$

For $K_{k}^{[13]}(n,m)$ we write

$\frac{\omega_{2h,k}^{6}}{\omega_{h,k}^{4}\omega_{4h,k}^{2}}=e^{\frac{\pi
i}{k}(h^{2}h^{\prime}(k-1))}$

Since $hh^{\prime}\equiv-1\,\,({\rm mod}\,\,{k})$ we find that this equals

$=e^{-2\pi i\frac{k^{2}-1}{2}}.$

The claim follows after recognizing that since $k$ is odd
$\frac{k^{2}-1}{2}\in\mathbb{Z}.$ ∎

We agree on the convention that for any Kloosterman sum above we set
$K_{k}(\nu,n)\coloneqq K_{k}(\nu,n,0).$

For $b\in\mathbb{R},k\in\mathbb{N}$ and $0<\nu\leq k$ define
$\mathcal{J}_{b,k,\nu}(z)\eqqcolon ze^{\frac{\pi b}{kz}}I_{k,\nu}(z)$ and
define the principal part truncation of $\mathcal{J}_{b,k,\nu}$ by

$\mathcal{J}_{b,k,\nu}^{*}(z)\coloneqq\sqrt{\frac{b}{3}}\int_{-1}^{1}\frac{e^{\frac{\pi
b}{kz}}}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}dx.$

We have the following estimate for the principal part integrals [5].

######  Lemma 3.5.

Let $b\in\mathbb{R},k\in\mathbb{N}$ and $0<\nu\leq k,$ then we have as
$z\rightarrow 0:$

  * (i)

If $b\leq 0,$ then we have

$|\mathcal{J}_{b,k,\nu}(z)|\ll\frac{1}{\left|\frac{\pi}{2}-\frac{\pi}{k}\left(\nu-\frac{1}{6}\right)\right|}.$

  * (ii)

If $b>0,$ then
$\mathcal{J}_{b,k,\nu}(z)=\mathcal{J}_{b,k,\nu}^{*}(z)+\mathcal{E}_{b,k,\nu},$
where

$|\mathcal{E}_{b,k,\nu}(z)|\ll\frac{1}{\left|\frac{\pi}{2}-\frac{\pi}{k}\left(\nu-\frac{1}{6}\right)\right|}.$

##  4\. Circle Method

In this section we are going to derive an exact formula for
$\overline{p_{1}}(n)$ using the Circle Method. For all $n\geq 1,$ Cauchy’s
Theorem yields

$(-1)^{n}\overline{p_{1}}(n)=\frac{1}{2\pi
i}\int_{C}\frac{\overline{G_{1}}(q)}{q^{n+1}dq},$

where we choose $C$ as the circle with radius $r=e^{-\frac{2\pi}{N^{2}}}$ with
$N\in\mathbb{N}$ which we take into the limit $N\rightarrow\infty.$
Futhermore, we parametrize the circle with $q=e^{-\frac{2\pi}{N^{2}}+2\pi it}$
for $0\leq t\leq 1.$ This yields

$(-1)^{n}\overline{p_{1}}(n)=\int_{0}^{1}\overline{G_{1}}\left(e^{-\frac{2\pi}{N^{2}}+2\pi
it}\right)\cdot e^{\frac{2\pi n}{N^{2}}-2\pi int}dt.$

We are now going to decompose the circle into standard Farey arcs. Throughout
we let $0\leq h<k\leq N$ with $\gcd(h,k)=1.$ We define

$\vartheta_{h,k}^{\prime}\coloneqq\frac{1}{k(k+k_{1})},\quad\vartheta_{h,k}^{\prime\prime}\coloneqq\frac{1}{k(k+k_{2})},$

where $\frac{h_{1}}{k_{1}}<\frac{h}{k}<\frac{h_{2}}{k_{2}}$ are neighbouring
Farey fractions in the Farey sequence of order $N.$ For notational convenience
write $z=k(N^{-2}-i\Phi)$ with $\Phi=t-\frac{h}{k},$ where we have that
$-\vartheta_{h,k}^{\prime}\leq\Phi\leq\vartheta_{h,k}^{\prime\prime}.$
Splitting the circle along the Farey arcs then yields

$\displaystyle(-1)^{n}\overline{p_{1}}(n)$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\end{subarray}}e^{-\frac{2\pi
ihn}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}\overline{G_{1}}\left(e^{\frac{2\pi
i}{k}(h+iz)}\right)\cdot e^{\frac{2\pi nz}{k}}d\Phi$ (4.1)
$\displaystyle=\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\end{subarray}}e^{-\frac{2\pi
ihn}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}\left[g_{1}\left(e^{\frac{2\pi
i}{k}(h+iz)}\right)+g_{2}\left(e^{\frac{2\pi i}{k}(h+iz)}\right)\right]\cdot
e^{\frac{2\pi nz}{k}}d\Phi.$

We will often use in subsequent calculations the well-known estimate

$\frac{1}{k+k_{j}}\leq\frac{1}{N+1},$ (4.2)

for $j=1,2,$ as well as the bound

$\text{Re}\left(z^{-1}\right)=\frac{N^{-2}}{kN^{-4}+k\Phi^{2}}\geq\frac{N^{2}}{k+N^{2}k^{-1}}\geq\frac{k}{2}.$
(4.3)

In the following we are going to split the sum in (4.1) according to the
different values of $\gcd(4,k)$ and we will estimate these contributions
individually. We write

$(-1)^{n}\overline{p_{1}}(n)=\sum_{4}+\sum_{2}+\sum_{1},$

where $\sum_{d}$ denoted the part of the sum (4.1) where $\gcd(4,k)=d.$ To
improve our estimates we will further decompose the integral long the Farey
arc as follows

$\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}=\int_{-\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}+\int_{-\frac{1}{k(k+k_{1})}}^{-\frac{1}{k(k+N)}}+\int_{\frac{1}{k(k+N)}}^{\frac{1}{k(k+k_{2})}}.$
(4.4)

Furthermore, we use the decompose the refined decomposition

$\int_{-\frac{1}{k(k+k_{1})}}^{-\frac{1}{k(k+N)}}=\sum_{\ell=k+k_{1}}^{k+N-1}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}.$

Thus, we obtain that

$\sum_{d}=\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\ \gcd(h,k)=1\\\
\gcd(4,k)=d\end{subarray}}\int_{-\frac{1}{k(k+k_{1})}}^{-\frac{1}{k(k+N)}}=\sum_{\begin{subarray}{c}1\leq
k\leq N\\\
\gcd(4,k)=d\end{subarray}}\sum_{\ell=N+1}^{k+N-1}\sum_{\begin{subarray}{c}0\leq
h<k\\\ \gcd(h,k)=1\\\
N<k+k_{1}\leq\ell\end{subarray}}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}.$
(4.5)

Before we start estimating we make three final definitions to simplify
notation. We define $a(n),b(n)$ and $r(n)$ by

$\sum_{n\geq 0}a(n)q^{n}=g_{1}(q),\quad\sum_{n\geq
0}b(n)q^{n}=g_{2}(q),\quad\sum_{n\geq 0}r(n)q^{n}=\xi(q).$

We start with $\sum_{4}.$ We have three sums, the first two coming from
$g_{1}(q)$ and the last one from $g_{2}(q).$

$\displaystyle S_{41}$
$\displaystyle\coloneqq\frac{1}{2}\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
4|k\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}(-1)^{\frac{k}{2}+1}e^{\pi
i\left(\frac{h^{\prime}}{2}-\frac{3h^{\prime}k}{4}\right)-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}}g_{1}(q_{1})d\Phi,$ $\displaystyle S_{42}$
$\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
4|k\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{k\omega_{h,\frac{k}{4}}^{2}}e^{-\frac{2\pi
inh}{k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}+\nu)}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}}e^{-\frac{\pi}{12kz}}\xi(q_{1})zI_{k,\nu}(z)d\Phi,$ $\displaystyle
S_{43}$ $\displaystyle\coloneqq\frac{1}{2}\sum_{\begin{subarray}{c}0\leq
h<k\leq N\\\ \gcd(h,k)=1\\\
4|k\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{h,\frac{k}{4}}^{2}}e^{-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}}g_{2}(q_{1})d\Phi.$

We start by estimating $S_{41}.$ We use the splittings (4.4) and (4.5) to
decompose $S_{41}$ into three sums $S_{41}^{[1]},S_{41}^{[2]}$ and
$S_{41}^{[3]}.$ We find

$S_{41}^{[1]}=\frac{1}{2}\sum_{\begin{subarray}{c}1\leq k\leq N\\\
4|k\end{subarray}}(-1)^{\frac{k}{2}+1}\sum_{\begin{subarray}{c}0\leq h<k\leq
N\\\
\gcd(h,k)=1\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{h,\frac{k}{4}}^{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)-\frac{2\pi
inh}{k}}\int_{-\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}e^{\frac{2\pi
nz}{k}}\sum_{m\geq 0}a(m)e^{\frac{2\pi
im}{k}\left(h^{\prime}+\frac{i}{z}\right)}d\Phi$

$=\sum_{\begin{subarray}{c}1\leq k\leq N\\\
4|k\end{subarray}}(-1)^{\frac{k}{2}+1}\sum_{m\geq
0}a(m)K_{k}^{[41]}(n,m)\int_{-\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}e^{\frac{2\pi}{k}(nz-\frac{m}{z})}d\Phi.$

Recall that $\text{Re}\left(z\right)=\frac{k}{N^{2}}$ and from (4.3) we infer
$\text{Re}\left(z^{-1}\right)\geq\frac{k}{2}.$ Together with Lemma 3.2 this
implies the bound

$S_{41}^{[1]}\ll\sum_{1\leq k\leq
N}\frac{n^{\frac{1}{3}}}{k^{\frac{1}{3}-\varepsilon}(k+N)}\rightarrow
0,\quad(N\rightarrow\infty).$

For $S_{41}^{[2]}$ we obtain with the same calculations

$S_{41}^{[2]}=\frac{1}{2}\sum_{1\leq k\leq N}\sum_{\begin{subarray}{c}1\leq
k\leq N\\\ 4|k\end{subarray}}(-1)^{\frac{k}{2}+1}\sum_{m\geq
0}a(m)\sum_{\ell=N+1}^{k+N-1}\mathbb{K}_{k,\ell}^{[41]}(n,m)\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}e^{\frac{2\pi}{k}\left(nz-\frac{m}{z}\right)}d\Phi.$

Bounding this as before with Lemma 3.2 shows $S_{41}^{[2]}\rightarrow 0$ as
$N\rightarrow\infty.$ The contribution $S_{41}^{[3]}$ is bounded in the exact
same way as $S_{41}^{[1]},$ using additionally (4.2) and thus, we see that
$S_{41}\rightarrow 0$ as $N\rightarrow\infty.$

Next, we are going to bound $S_{42}.$ We invoke the splittings (4.4) and (4.5)
to decompose $S_{42}$ into the three sums $S_{42}^{[1]},S_{42}^{[2]}$ and
$S_{42}^{[3]}.$ We have as before

$S_{42}^{[1]}=\sum_{\begin{subarray}{c}1\leq k\leq N\\\
4|k\end{subarray}}\frac{1}{k}\sum_{m\geq
0}r(m)(-1)^{\nu}K_{k}^{[42]}(\nu,n,m)\int_{-\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}e^{\frac{2\pi}{k}\left(nz-\frac{m}{z}\right)}e^{-\frac{\pi}{12kz}}zI_{k,\nu}(z)d\Phi.$

We use as before the bounds $\text{Re}\left(z\right)=\frac{k}{N^{2}}$ and
$\text{Re}\left(z^{-1}\right)\geq\frac{k}{2}$ as well as Lemma 3.2 and Lemma
3.5 to obtain as $N\rightarrow\infty,$

$\displaystyle S_{42}^{[1]}$ $\displaystyle\ll\sum_{\begin{subarray}{c}1\leq
k\leq N\\\ 4|k\end{subarray}}\frac{1}{k}e^{\frac{2\pi
n}{N^{2}}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}n^{\frac{1}{3}}k^{\frac{2}{3}+\varepsilon}\int_{-\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}\left|\mathcal{J}_{-\frac{1}{12},k,\nu}(z)\right|d\Phi$
$\displaystyle\ll n^{\frac{1}{3}}\sum_{1\leq k\leq
N}\frac{k^{-\frac{1}{3}+\varepsilon}}{k(k+N)}\sum_{\nu\,\,({\rm
mod}\,\,{k})}\frac{1}{\left|\frac{\pi}{2}-\frac{\pi}{k}\left(\nu-\frac{1}{6}\right)\right|}\ll\frac{n^{\frac{1}{3}}}{N}\sum_{k=1}^{N}k^{-\frac{1}{3}+\varepsilon}\log(k)$
$\displaystyle\ll
n^{\frac{1}{3}}N^{-\frac{1}{3}+\varepsilon}\log(N)\rightarrow 0.$

We bound $S_{42}^{[2]}$ in the same way by replacing the Kloosterman sum
$K_{k}^{[42]}$ with the sum of the Kloostermans sums
$\mathbb{K}_{k,\ell}^{[42]}(\nu,n,m).$ The sum $S_{42}^{[3]}$ is bounded
exactly like $S_{42}^{[1]}.$

Finally, the contribution $S_{43}\rightarrow 0$ as $N\rightarrow\infty$ by the
same argument as used for $S_{41},$ changing the Kloosterman sums
$K_{k}^{[41]}$ and $\mathbb{K}_{k,\ell}^{[41]}$ to the Kloosterman sums
$K_{k}^{[43]}$ and $\mathbb{K}_{k,\ell}^{[43]}.$ In total, we obtain that
$\sum_{4}\rightarrow 0,$ as $N\rightarrow\infty.$

We continue to estimate $\sum_{2}.$ Thus, suppose that $\gcd(4,k)=2.$ We
decompose $\sum_{2}$ into three sums

$\displaystyle S_{21}$
$\displaystyle\coloneqq\frac{1}{4}\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
\gcd(4,k)=2\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}(-1)^{\frac{k}{2}+1}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}+\frac{\pi}{2kz}}\frac{P(q_{1}^{2})^{4}}{P(q_{1})P(q_{1}^{4})^{2}}f(q_{1})d\Phi,$
$\displaystyle S_{22}$
$\displaystyle\coloneqq\frac{1}{2}\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
\gcd(4,k)=2\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{k\omega_{2h,\frac{k}{2}}^{2}}e^{-\frac{2\pi
inh}{k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}+\nu)}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}+\frac{5\pi}{12kz}}\frac{P(q_{1}^{2})^{4}}{P(q_{1})P(q_{1}^{4})^{2}}zI_{k,\nu}(z)d\Phi,$
$\displaystyle S_{23}$
$\displaystyle\coloneqq\frac{1}{4}\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
\gcd(4,k)=2\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{2h,\frac{k}{2}}^{2}}e^{-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}+\frac{\pi}{2kz}}\frac{P(q_{1}^{2})^{6}}{P(q_{1})^{6}}d\Phi.$

Eventhough, on a first glance $S_{21}$ and $S_{23}$ individually make a
significant contribution to the overall sum, they do in fact cancel each
other. This follows from the fact that they grow with the same factor of
$e^{\frac{\pi}{2kz}}$ and

$\frac{\omega_{h,\frac{k}{2}}^{4}}{\omega_{2h,\frac{k}{2}}^{2}}(-1)^{\frac{k}{2}+1}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}=-\frac{\omega_{h,\frac{k}{2}}^{6}}{\omega_{h,k}^{4}\omega_{2h,\frac{k}{2}}^{2}}.$
(4.6)

To see this, rewrite the equation as

$(-1)^{\frac{k}{2}}e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)}=\frac{\omega_{h,\frac{k}{2}}^{2}}{\omega_{h,k}^{4}}.$

Since $\gcd(4,k)=2$ we use the definition of $\omega_{h,k}$ and obtain that

$\frac{\omega_{h,\frac{k}{2}}^{2}}{\omega_{h,k}^{4}}=e^{\frac{\pi
i}{4}(4-h^{\prime}k+h^{2}h^{\prime}k-h(2+k))},$

which further simplifies, by using $hh^{\prime}\equiv-1\,\,({\rm
mod}\,\,{4k}),$ to

$=(-1)e^{-\frac{\pi i}{4}(h(2+k))}.$

Using again that $\gcd(4,k)=2$ and $hh^{\prime}\equiv-1\,\,({\rm
mod}\,\,{4k})$ we recognize $(-1)=(-1)^{\frac{k}{2}}$ and furthermore
$e^{-\frac{\pi i}{4}(h(2+k))}=e^{\frac{\pi
ih^{\prime}}{2}\left(1-\frac{3k}{2}\right)},$ thus proving (4.6).

We now continue to estimate $S_{22}.$ The non-principal part vanishes as for
$S_{42}$ in the limit $N\rightarrow\infty.$ We are left with

$\mathcal{S}_{22}\coloneqq\frac{1}{2}\sum_{\begin{subarray}{c}0\leq h<k\leq
N\\\ \gcd(h,k)=1\\\
\gcd(4,k)=2\end{subarray}}\frac{\omega_{h,\frac{k}{2}}^{4}}{k\omega_{2h,\frac{k}{2}}^{2}}e^{-\frac{2\pi
inh}{k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}+\nu)}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}}\mathcal{J}_{\frac{5}{12},k,\nu}(z)d\Phi.$

By Lemma 3.5 (ii) we can write
$\mathcal{J}_{\frac{5}{12},k,\nu}(z)=\mathcal{J}_{\frac{5}{12},k,\nu}^{*}(z)+\mathcal{E}_{\frac{5}{12},k,\nu},$
where the contribution from $\mathcal{E}_{\frac{5}{12},k,\nu}$ vanishes as
before when $N\rightarrow\infty.$ We decompose $\mathcal{S}_{22}$ into
$\mathcal{S}_{22}^{[1]},\mathcal{S}_{22}^{[2]}$ and $\mathcal{S}_{22}^{[3]}$
using the splittings (4.4) and (4.5).

Recall that $z=k(N^{-2}-i\Phi).$ We have for $b>0$

$\int_{\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}e^{\frac{2\pi
nz}{k}}\mathcal{J}_{b,k,\nu}^{*}(z)d\Phi=\int_{\frac{1}{k(k+N)}}^{\frac{1}{k(k+N)}}e^{\frac{2\pi
nz}{k}}\sqrt{\frac{b}{3}}\int_{-1}^{1}\frac{e^{\frac{\pi
b}{kz}(1-x^{2})}}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}dxd\Phi$

$=\frac{\sqrt{b}}{ik\sqrt{3}}\int_{-1}^{1}\frac{1}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}\int_{\frac{k}{N^{2}}-\frac{i}{k+N}}^{\frac{k}{N^{2}}+\frac{i}{k+N}}e^{\frac{\pi
b}{kz}(1-x^{2})+\frac{2\pi nz}{k}}dzdx.$

After the change of variables $w=\frac{z}{k}$ the last expression equals

$=\frac{\sqrt{b}}{i\sqrt{3}}\int_{-1}^{1}\frac{1}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}\int_{\frac{1}{N^{2}}-\frac{i}{k(k+N)}}^{\frac{1}{N^{2}}+\frac{i}{k(k+N)}}e^{2\pi
nw+\frac{2\pi}{k^{2}w}\frac{b}{2}(1-x^{2})}dwdx.$

Thus, we find that

$\displaystyle\mathcal{S}_{22}^{[1]}$
$\displaystyle=\frac{\sqrt{5}\pi}{6}\sum_{\begin{subarray}{c}1\leq k\leq N\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}K_{k}^{[22]}(\nu,n)\int_{-1}^{1}\frac{L_{k}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}dx$
$\displaystyle+\frac{\sqrt{5}\pi}{6}\sum_{\begin{subarray}{c}1\leq k\leq N\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}K_{k}^{[22]}(\nu,n)\int_{-1}^{1}\frac{1}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}$
$\displaystyle\times\left(\mathcal{E}_{k}^{[1]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)+\mathcal{E}_{k}^{[2]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)+\mathcal{E}_{k}^{[3]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)\right)dx,$

where we denote by the $R$ the rectangle with edges
$\pm\frac{1}{N^{2}}\pm\frac{i}{k(k+N)}$ around $0$ with counterclockwise
orientation and set

$\displaystyle L_{k}(n,y)$ $\displaystyle:=\frac{1}{2\pi i}\int_{R}e^{2\pi
nw+\frac{2\pi y}{k^{2}w}}dw,\quad\mathcal{E}_{k}^{[1]}(n,y):=\frac{1}{2\pi
i}\int_{\frac{1}{N^{2}}+\frac{i}{k(k+N)}}^{-\frac{1}{N^{2}}+\frac{i}{k(k+N)}}e^{2\pi
nw+\frac{2\pi y}{k^{2}w}}dw,$ $\displaystyle\mathcal{E}_{k}^{[2]}(n,y)$
$\displaystyle:=\frac{1}{2\pi
i}\int_{-\frac{1}{N^{2}}+\frac{i}{k(k+N)}}^{-\frac{1}{N^{2}}-\frac{i}{k(k+N)}}e^{2\pi
nw+\frac{2\pi y}{k^{2}w}}dw,\quad\mathcal{E}_{k}^{[3]}(n,y):=\frac{1}{2\pi
i}\int_{-\frac{1}{N^{2}}-\frac{i}{k(k+N)}}^{\frac{1}{N^{2}}-\frac{i}{k(k+N)}}e^{2\pi
nw+\frac{2\pi y}{k^{2}w}}dw.$

We first bound $\mathcal{E}_{k}^{[1]}$ and $\mathcal{E}_{k}^{[2]}.$ On the
horizontal edges of $R$ we have, following [12]

$w=u\pm\frac{i}{k(k+N)},\quad-\frac{1}{N^{2}}\leq
u\leq\frac{1}{N^{2}},\quad\text{Re}\left(w\right)=u\leq\frac{1}{N^{2}},\quad\text{Re}\left(\frac{1}{w}\right)\leq
4k^{2}.$

Hence,

$\left|\mathcal{E}_{k}^{[1]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)\right|,\left|\mathcal{E}_{k}^{[1]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)\right|\leq\frac{1}{N^{2}\pi}e^{\frac{5\pi}{6}\left(1-x^{2}\right)+\frac{2\pi
n}{N^{2}}}.$

Following again [12] we have on the vertical part of $R$

$w=-\frac{1}{N^{2}}+iv,\quad-\frac{1}{k(k+N)}\leq
v\leq\frac{1}{k(k+N)},\quad\text{Re}\left(w\right)<0,\quad\text{Re}\left(\frac{1}{w}\right)<0.$

Thus,

$\left|\mathcal{E}_{k}^{[2]}\left(n,\frac{5}{24}\left(1-x^{2}\right)\right)\right|\leq\frac{1}{\pi
kN}.$

These estimates together with Lemma 3.3 imply
$\mathcal{E}_{k}^{[1]},\mathcal{E}_{k}^{[2]}$ and $\mathcal{E}_{k}^{[3]}$
contribute at most

$\ll e^{\frac{2\pi
n}{N^{2}}}\sum_{k=1}^{N}\frac{1}{k}\sum_{\nu=1}^{k}k^{\frac{2}{3}+\varepsilon}n^{\frac{1}{3}}\int_{-1}^{1}\frac{1}{\left|\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)\right|}\frac{1}{kN}e^{\frac{5\pi}{6}\left(1-x^{2}\right)}dx.$

For $\alpha\geq 0$ and $0<\beta<\pi$ we have the bound

$|\cosh(\alpha+i\beta)|\geq\left|\sin\left(\frac{\pi}{2}-\beta\right)\right|\gg\left|\frac{\pi}{2}-\beta\right|.$

Since $\nu-\frac{1}{6}>0$ we can bound the above as $N\rightarrow\infty$ by

$\displaystyle\ll\frac{n^{\frac{1}{3}}}{N}\sum_{k=1}^{N}k^{-\frac{4}{3}+\varepsilon}\sum_{\nu=1}^{k}\frac{1}{\left|\frac{\pi}{2}-\frac{\pi}{k}\left(\nu-\frac{1}{6}\right)\right|}\int_{-1}^{1}e^{\frac{5\pi}{6}\left(1-x^{2}\right)}dx\ll\frac{n^{\frac{1}{3}}}{N}\sum_{k=1}^{N}k^{-\frac{1}{3}+\varepsilon}\log(k)$
$\displaystyle\ll
n^{\frac{1}{3}}N^{-\frac{1}{3}+\varepsilon}\log(N)\rightarrow 0.$

We now show that

$L_{k}(n,y)=\frac{1}{k}\sqrt{\frac{y}{n}}I_{1}\left(\frac{4\pi\sqrt{ny}}{k}\right).$

By the Residue Theorem we have

$L_{k}(n,y)=\text{Res}_{w=0}\;e^{2\pi nw+\frac{2\pi y}{k^{2}w}}.$

Using the series expansion of the exponential function we find that

$\displaystyle\text{Res}_{w=0}\;e^{2\pi nw+\frac{2\pi y}{k^{2}w}}$
$\displaystyle=\sum_{m\geq
0}\frac{1}{m!(m+1)!}(2\pi)^{2m+1}\frac{n^{m}y^{m+1}}{k^{2m+2}}$
$\displaystyle=\frac{1}{k}\sqrt{\frac{y}{n}}\sum_{m\geq
0}\frac{1}{m(m+1)}(2\pi)^{2m+1}\frac{n^{m+\frac{1}{2}}y^{m+\frac{1}{2}}}{k^{2m+1}}.$

We recognize the last sum as the series representation of the Bessel function
[14]

$I_{\ell}(z)\coloneqq\sum_{m\geq
0}\frac{1}{m!\Gamma(m+\ell+1)}\left(\frac{x}{2}\right)^{2m+\ell},$

which proves the claim. Therefore, as $N\rightarrow\infty$ we find

$\mathcal{S}_{22}^{[1]}=\frac{5\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq
1\\\ \gcd(4,k)=2\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}K_{k}^{[22]}(\nu,n)\mathcal{I}_{\frac{5}{12},k,\nu}(n).$

We continue to bound $\mathcal{S}_{22}^{[2]}.$ The contribution of
$\mathcal{S}_{22}^{[3]}$ can be bounded in the exact same way. We find

$\displaystyle\mathcal{S}_{22}^{[2]}$
$\displaystyle=\frac{1}{2}\sum_{\begin{subarray}{c}1\leq k\leq N\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}\sum_{\ell=N+1}^{k+N-1}\mathbb{K}_{k,\ell}^{22}(v,n)\sqrt{\frac{5}{12\cdot
3}}$ $\displaystyle\times\int_{-1}^{1}\frac{1}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}e^{2\pi
nw+\frac{5\pi}{12k^{2}w}\left(1-x^{2}\right)}d\Phi dx.$

Following [12] we bound

$\text{Re}\left(2\pi
nw+\frac{5\pi}{12k^{2}w}\left(1-x^{2}\right)\right)\leq\frac{2\pi
n}{N^{2}}+\frac{5\pi}{3}\left(1-x^{2}\right).$

Thus, using Lemma 3.3 we find as $N\rightarrow\infty,$

$\displaystyle\mathcal{S}_{22}^{[2]}$ $\displaystyle\ll e^{\frac{2\pi
n}{N^{2}}}\sum_{k=1}^{N}\sum_{k=1}^{\nu}\sum_{\ell=N+1}^{k+N-1}k^{\frac{2}{3}+\varepsilon}n^{\frac{1}{3}}\int_{-1}^{1}\frac{e^{\frac{5\pi}{3}\left(1-x^{2}\right)}}{\cosh\left(\frac{\pi
i}{k}\left(\nu-\frac{1}{6}\right)-\frac{\pi}{k}\sqrt{\frac{b}{3}}x\right)}dx\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}d\Phi$
$\displaystyle\ll
n^{\frac{1}{3}}\sum_{k=1}^{N}k^{-\frac{1}{3}+\varepsilon}k\log(k)\frac{1}{k(N+k)}\ll
n^{\frac{1}{3}}N^{-\frac{1}{3}+\varepsilon}\log(N)\rightarrow 0.$

Therefore, in total as $N\rightarrow\infty$ we find that

$\Sigma_{2}=\frac{5\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}K_{k}^{[22]}(\nu,n)\mathcal{I}_{\frac{5}{12},k,\nu}(n).$

Finally, we have to evaluate $\sum_{1}.$ Thus, we assume that $\gcd(4,k)=1.$
We have the following three sums

$\displaystyle S_{11}$ $\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq
h<k\leq N\\\ \gcd(h,k)=1\\\
\gcd(4,k)=1\end{subarray}}\frac{\omega_{2h,k}^{4}}{\omega_{4h,k}^{2}}(-1)^{\frac{k-1}{2}}e^{\frac{3\pi
ih^{\prime}}{4k}-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}-\frac{5\pi}{8kz}}\frac{P\left(q_{1}^{\frac{1}{2}}\right)^{4}}{P(q_{1})P\left(q^{\frac{1}{4}}\right)^{2}}\cdot\omega\left(q^{\frac{1}{2}}\right)d\Phi,$
$\displaystyle S_{12}$ $\displaystyle\coloneqq\sum_{\begin{subarray}{c}0\leq
h<k\leq N\\\ \gcd(h,k)=1\\\
\gcd(4,k)=1\end{subarray}}\frac{\omega_{2h,k}^{4}}{k\omega_{4h,k}^{2}}e^{-\frac{2\pi
nh}{k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}-\nu)}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}+\frac{\pi}{24kz}}\frac{P\left(q_{1}^{\frac{1}{2}}\right)^{4}}{P(q_{1})P\left(q^{\frac{1}{4}}\right)^{2}}zI_{k,\nu}(z)d\Phi,$
$\displaystyle S_{13}$
$\displaystyle\coloneqq\frac{1}{4}\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\
\gcd(h,k)=1\\\
\gcd(4,k)=1\end{subarray}}\frac{\omega_{2h,k}^{6}}{\omega_{h,k}^{4}\omega_{4h,k}^{2}}e^{-\frac{2\pi
inh}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}-\frac{\pi}{8kz}}\frac{P\left(q_{1}^{\frac{1}{2}}\right)^{6}}{P(q_{1})^{4}P\left(q_{1}^{\frac{1}{4}}\right)^{2}}d\Phi.$

Similar to $S_{41}$ and $S_{43}$ we can show that as $N\rightarrow\infty$ we
have $S_{11}\rightarrow 0$ and $S_{12}\rightarrow 0.$ Furthermore for $S_{22}$
the non-principal part vanishes as $N\rightarrow\infty$ and we are left with

$\mathcal{S}_{12}=\sum_{\begin{subarray}{c}0\leq h<k\leq N\\\ \gcd(h,k)=1\\\
\gcd(4,k)=1\end{subarray}}\frac{\omega_{2h,k}^{4}}{k\omega_{4h,k}^{2}}e^{-\frac{2\pi
nh}{k}}\sum_{\nu\,\,({\rm mod}\,\,{k})}(-1)^{\nu}e^{\frac{\pi
ih^{\prime}(-3\nu^{2}-\nu)}{k}}\int_{-\vartheta_{h,k}^{\prime}}^{\vartheta_{h,k}^{\prime\prime}}e^{\frac{2\pi
nz}{k}}\mathcal{J}_{\frac{1}{24},k,\nu}d\Phi.$

Similar to $\mathcal{S}_{22},$ using Lemma 3.4, one can show that as
$N\rightarrow\infty$

$\Sigma_{1}=\frac{\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=1\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{\nu}K_{k}^{[12]}(\nu,n)\mathcal{I}_{\frac{1}{24},k,\nu}(n).$

Combining our results we have the following

######  Theorem 4.1.

We have for $n\geq 1,$

$\displaystyle\overline{p_{1}}(n)$
$\displaystyle=\frac{\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=1\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{n+\nu}K_{k}^{[12]}(\nu,n)\mathcal{I}_{\frac{1}{24},k,\nu}(n)$
$\displaystyle+\frac{5\pi}{12\sqrt{6n}}\sum_{\begin{subarray}{c}k\geq 1\\\
\gcd(4,k)=2\end{subarray}}\frac{1}{k^{2}}\sum_{\nu\,\,({\rm
mod}\,\,{k})}(-1)^{n+\nu}K_{k}^{[22]}(\nu,n)\mathcal{I}_{\frac{5}{12},k,\nu}(n).$

It is easy to see that the summand $k=2$ dominates the rest of the sum and
thus we get as a consequence the following asymptotics as $n\rightarrow\infty$

$\overline{p_{1}}(n)\sim\frac{5\pi}{48\sqrt{6n}}\sum_{\nu=0,1}(-1)^{n+\nu}K_{2}^{[22]}(\nu,n)\mathcal{I}_{\frac{5}{12},2,\nu}(n).$

We can simplify things quite a bit, by recognizing that

$K_{2}^{[22]}(0,n)=(-1)^{n},\quad K_{2}^{[22]}(1,\nu)=(-1)^{n+1}.$

Hence, we see that this removes the twist by the root of unity introtuced at
the beginning and we find

$\overline{p_{1}}(n)\sim\frac{5\pi}{48\sqrt{6n}}\left(\mathcal{I}_{\frac{5}{12},2,0}(n)+\mathcal{I}_{\frac{5}{12},2,1}(n)\right),\quad(n\rightarrow\infty).$

In [9] the author used in detail the saddle point method to find a strong
asymptotic formula for $\mathcal{I}_{\frac{1}{18},1,0}(n).$ The same method
can be applied here to find asymptotic formulas for
$\mathcal{I}_{\frac{5}{12},2,0}(n)$ and $\mathcal{I}_{\frac{5}{12},2,1}(n).$
We leave out the details. The result is the following

######  Corollary 4.2.

We have for $n\rightarrow\infty$

$\overline{p_{1}}(n)\sim\frac{\sqrt{5}}{4\sqrt{6}n}e^{\pi\sqrt{\frac{5}{6}n}}.$

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