2.1 Splitting integral intervals with Farey fractions

We set $r :=e^{-2\pi \varrho }$ with $\varrho =1/N^2$ for some $N>0$ . Let $h/k$ be a Farey fraction of order N, and let $\xi _{h,k} := [-\theta _{h,k}^{\prime },\theta _{h,k}^{\prime \prime }]$ with $\theta _{h,k}^{\prime },\theta _{h,k}^{\prime \prime }$ being the distance from $h/k$ to its neighboring mediants. If we set

$$\begin{align*}s:= r e^{2\pi i \theta} = e^{-2\pi \varrho} e^{2\pi i \theta},\end{align*}$$

and let $\theta := h/k + \varphi $ on each $\xi _{h,k}$ , then

$$\begin{align*}ds = 2\pi i e^{-2\pi \varrho} e^{2\pi i h/k} e^{2\pi i \varphi} d\varphi ,\end{align*}$$

so that

$$ \begin{align*} \alpha(n) = \sum_{\substack{0 \leq h<k \leq N \\ (h,k) = 1}} e^{-2\pi i nh/k} \int_{\xi_{h,k}} f(e^{2\pi i(h/k + i\varrho + \varphi)}) e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi. \end{align*} $$

Set $z:=k(\varrho -i\varphi )$ and $\tau :=(h+iz)/k$ to get

(2.1) $$ \begin{align} \begin{aligned} \alpha(n) &= \sum_{\substack{0 \leq h<k \leq N \\ (h,k) = 1}} e^{-2\pi i nh/k} \int_{\xi_{h,k}} f(e^{2\pi i(h+iz)/k}) e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ &= \sum_{\substack{0 \leq h<k \leq N \\ (h,k) = 1}} e^{-2\pi i nh/k} \int_{\xi_{h,k}} f(e^{2\pi i \tau}) e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi. \end{aligned} \end{align} $$

2.2 Dedekind’s eta-function

The Dedekind eta-function is defined by

$$\begin{align*}\eta(\tau):=q^{1/24}(q;q)_{\infty},\end{align*}$$

where $q=e^{2\pi i \tau }$ with $\mathrm {Im}(\tau )> 0.$ A well-known transformation formula for the Dedekind eta-function is as follows [Reference Apostol4, p. 52].

For $\gamma = \left (\begin {matrix} a & b \\ c & d \end {matrix}\right ) \in \varGamma :=\mathrm {SL}_2(\mathbb {Z})$ , $c>0$ , we have [Reference Apostol4, Theorem 3.4]

(2.2) $$ \begin{align} \eta(\gamma \tau) = e^{-\pi i s(d,c)} e^{\pi i (a+d)/(12c)} \sqrt{-i(c\tau + d)} \eta(\tau), \end{align} $$

where $s(d,c)$ is the Dedekind sum given by

$$ \begin{align*} s(d,c) := \sum_{n=0}^{c-1} \frac{n}{c} \left(\frac{dn}{c}-\left\lfloor \frac{dn}{c} \right\rfloor-\frac{1}{2} \right). \end{align*} $$

Set

$$ \begin{align*} F(e^{2\pi i \tau}) := \frac{1}{(q;q)_{\infty}} = \frac{e^{\pi i \tau/12}}{\eta(\tau)}. \end{align*} $$

It follows from (2.2) that

$$ \begin{align*} F(e^{2\pi i \tau}) = e^{\pi i (\tau - \gamma\tau)/12} e^{- \pi i s(d,c)} e^{\pi i (a + d)/(12c)} \sqrt{-i(c\tau + d)} F(e^{2\pi i \gamma\tau}). \end{align*} $$

For the interval $\xi _{h,k},$ if $\mathrm {gcd}(n,k)=1$ , then we choose an integer $h_n^{\prime }$ such that $nhh_n^{\prime } \equiv -1 ~ (\mathrm {mod} ~ k).$ Hence, there exists $b_n\in \mathbb {Z}$ such that $nhh_n^{\prime } - b_n k = -1.$ Set

$$ \begin{align*} \gamma_{(nh,k)} := \left(\begin{matrix} h_n^{\prime} & -b_n \\ k & -nh \end{matrix} \right) \in \varGamma. \end{align*} $$

Then,

$$ \begin{align*} \begin{aligned} \gamma_{(nh,k)} = \frac{h_n^{\prime} \frac{nh+inz}{k}-b_n}{k \frac{nh+inz}{k}-nh} = \frac{h_n^{\prime} nh - b_n k + in h_n^{\prime} z}{k i nz} = \frac{h_n^{\prime}}{k} + i\frac{1}{knz}, \end{aligned} \end{align*} $$

so that

(2.3) $$ \begin{align} \mathrm{Im}(\gamma_{(nh,k)}) = \mathrm{Re}\left(\frac{1}{knz}\right) = \frac{1}{kn} \mathrm{Re}\left(\frac{1}{z}\right), \end{align} $$

and

(2.4) $$ \begin{align} F(e^{2\pi i n \tau}) = e^{\frac{\pi}{12k}(\frac{1}{nz}-nz)}e^{\pi i s(nh,k)}\sqrt{nz} F(e^{2\pi i \gamma_{(nh,k)}(n\tau)}). \end{align} $$

2.3 Some bounds and an integral

It is obvious to observe that

$$ \begin{align*} \frac{1}{2kN} \leq \theta_{h,k}^{\prime}, \theta_{h,k}^{\prime \prime} \leq \frac{1}{kN}. \end{align*} $$

From this, we have

$$ \begin{align*} \frac{1}{kN} \leq |\xi_{h,k}| \leq \frac{2}{kN}, \end{align*} $$

and

$$ \begin{align*}|\varphi| \leq (kN)^{-1}.\end{align*} $$

Notice that $z=k(\varrho - i \varphi ).$ Then,

(2.5) $$ \begin{align} \mathrm{Re}\left(\frac{1}{z}\right) = \mathrm{Re}\left(\frac{1}{k(\varrho - i \varphi)}\right) = \frac{1}{k}\frac{\varrho}{\varrho^2 + \varphi^2} \geq \frac{1}{k}\frac{N^{-2}}{N^{-4} + k^{-2}N^{-2}} \geq \frac{k}{2}. \end{align} $$

Also, we have

(2.6) $$ \begin{align} |z|^{-1/2} = \frac{1}{k^{1/2}} \frac{1}{(\varrho^2 + \varphi^2)^{1/4}} \leq \frac{1}{(k^2\varrho^2)^{1/4}} = k^{-1/2} N. \end{align} $$

We now estimate the function $\mathrm {log}F(x)$ for $0 < x < 1.$ Since $0 < x <1$ , we have

$$ \begin{align*} mx^{m-1} \leq \frac{1-x^m}{1-x} = 1 + x + x^2 + \cdots + x^{m-1} \leq m. \end{align*} $$

From this, we deduce that

$$ \begin{align*} \frac{m(1-x)}{x} \leq \frac{1-x^m}{x^m} \leq \frac{m(1-x)}{x^m}, \end{align*} $$

so that

$$ \begin{align*} \frac{x^m}{m^2(1-x)} \leq \frac{x^m}{m(1-x^m)} \leq \frac{x}{m^2(1-x)}. \end{align*} $$

Then

(2.7) $$ \begin{align} \begin{aligned} \mathrm{log}F(x) &= \sum_{n=1}^{\infty} - \mathrm{log}(1-x^n) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{x^{nm}}{m} = \sum_{m=1}^{\infty} \frac{x^m}{m(1-x^m)}\\ &\leq \frac{x}{1-x} \sum_{m=1}^{\infty} \frac{1}{m^2} = \frac{\pi^2}{6} \frac{x}{1-x}. \end{aligned} \end{align} $$

Let $q_m(n)$ denote the number of m-colored partitions of n into an even number of distinct parts minus the number of m-colored partitions of n into an odd number of distinct parts. Then, by [Reference Andrews and Berndt2, eq.(6.1.3)]

$$ \begin{align*} \sum_{n=0}^{\infty} q_1(n) e^{2\pi i n \tau} =\frac{1}{F(e^{2\pi i \tau})}= (q;q)_{\infty}. \end{align*} $$

Define $\hat {a}_1(n)$ (resp. $\hat {a}_2(n)$ ) as the number of partitions of n into an even (resp. odd) number of parts with distinct elements. Then, we have $q_1(n) = \hat {a}_1(n) - \hat {a}_2(n).$ Since $0 \leq \hat {a}_1(n) \leq p(n)$ , $0 \leq \hat {a}_2(n) \leq p(n)$ , we have

(2.8) $$ \begin{align} |q_1(n)| = |\hat{a}_1(n) - \hat{a}_2(n)| \leq p(n). \end{align} $$

Then,

(2.9) $$ \begin{align} \begin{aligned} \left| \frac{1}{F(e^{2\pi i \tau})} \right| &\leq \sum_{n=1}^{\infty} |q_1(n)| |e^{2\pi i n \tau}| \leq \sum_{n=1}^{\infty} p(n) |e^{2\pi i n \tau}| \\ &=F(|e^{2\pi i \tau}|) = F(e^{-2\pi \mathrm{Im}(\tau)})\\ & \leq \mathrm{exp} \left(\frac{\pi^2 e^{-2\pi \mathrm{Im}(\tau)}}{6(1-e^{-2\pi \mathrm{Im}(\tau)})}\right), \end{aligned} \end{align} $$

where the last inequality follows from (2.7).

An inequality for $F(e^{2\pi i \tau })$ is given in the following lemma.

Lemma 2.1 Let $\tau _1,\tau _2,\cdots ,\tau _n \in \mathbb {C}$ and let $m_1,m_2,\cdots ,m_n \in \mathbb {Z}\backslash \{0\}.$ Then,

(2.10) $$ \begin{align} \left|\prod_{i=1}^{n} F(e^{2\pi i \tau_i})^{m_i} - 1\right| \leq \mathrm{exp}\left(\sum_{i=1}^{n} \frac{|m_i| \pi^2 e^{-2\pi \mathrm{Im}(\tau_i)}}{6(1-e^{-2\pi \mathrm{Im}(\tau_i)})} \right) - 1. \end{align} $$

Proof Let

$$ \begin{align*} F(e^{2\pi i \tau_i})^{m_i} =: \sum_{n=1}^{\infty} c_i(n) e^{2\pi i n \tau_i}, \quad F(e^{2\pi i \tau_i})^{|m_i|} =: \sum_{n=1}^{\infty} \tilde{c}_i(n) e^{2\pi i n \tau_i}. \end{align*} $$

If $m_i \geq 0$ , then $|c_i(n)| = \tilde {c}_i(n);$ if $m_i < 0$ , then, by (2.8),

$$ \begin{align*} |c_i(n)| \leq \sum_{\substack{1 \leq j \leq |m_i| \\ \sum n_j= n}} |q_1(n_j)| \leq \sum_{\substack{1 \leq j \leq |m_i| \\ \sum n_j = n}} p(n_j) = \tilde{c}_i(n). \end{align*} $$

Hence, by (2.7),

$$ \begin{align*} \begin{aligned} \left|\prod_{i=1}^{n} F(e^{2\pi i \tau_i})^{m_i} - 1\right| &\leq \left|\sum_{k_1=0}^{\infty}\cdots \sum_{k_n=0}^{\infty}c_1(k_1)\cdots c_n(k_n)e^{2\pi i k_1 \tau_1}\cdots e^{2\pi i k_n \tau_n} - 1\right|\\ & \leq \mathop{\sum_{k_1=0}^{\infty} \cdots\sum_{k_n=0}^{\infty}}\limits_{(k_1,\cdots,k_n)\neq (0,\cdots,0)} \tilde{c}_1(k_1)\cdots \tilde{c}_n(k_n)|e^{2\pi i k_1 \tau_1}|\cdots |e^{2\pi i k_n \tau_n}|\\&= \prod_{i=1}^{n} F(|e^{2\pi i \tau_i}|)^{|m_i|} - 1\\ &\leq \mathrm{exp}\left(\sum_{i=1}^{n} \frac{|m_i| \pi^2 e^{-2\pi \mathrm{Im}(\tau_i)}}{6(1-e^{-2\pi \mathrm{Im}(\tau_i)})} \right) - 1. \end{aligned} \end{align*} $$

This completes the proof.

To facilitate certain bounds in the proof of Theorem 1.1, we examine monotonicity of a function here.

Lemma 2.2 Let $u_1,u_2,\cdots ,u_n \in \mathbb {R}_{\geq 1}$ and let $m_1,m_2,\cdots ,m_n \in \mathbb {N}.$ Then,

$$ \begin{align*} W(x) := \frac{1}{x}\left(\mathrm{exp}\left(\sum_{i=1}^{n} \frac{m_i \pi^2 x^{u_i}}{6(1-x^{u_i})}\right) - 1\right) \end{align*} $$

is a nondecreasing function of x on $(0,1)$ .

Proof Taking derivative of $W(x)$ yields

$$ \begin{align*} \begin{aligned} W^{\prime}(x) = & -\frac{1}{x^2}\left(\mathrm{exp}\left(\sum_{i=1}^{n} \frac{m_i \pi^2 x^{u_i}}{6(1-x^{u_i})}\right) - 1\right) \\ & + \frac{1}{x}\mathrm{exp}\left(\sum_{i=1}^{n} \frac{m_i \pi^2 x^{u_i}}{6(1-x^{u_i})}\right) \sum_{i=1}^{n}\frac{\pi^2 m_i}{6} \left(\frac{u_ix^{u_i-1}}{(1-x^{u_i})^2}\right). \end{aligned} \end{align*} $$

Therefore, it suffices to prove

(2.11) $$ \begin{align} \sum_{i=1}^{n}\frac{\pi^2 m_i}{6} \left(\frac{u_ix^{u_i}}{(1-x^{u_i})^2}\right) + \mathrm{exp}\left(-\sum_{i=1}^{n} \frac{\pi^2 m_i}{6}\frac{x^{u_i}}{1-x^{u_i}}\right) \geq 1 \end{align} $$

for $x \in (0,1)$ . Since

$$ \begin{align*} \frac{u_ix^{u_i}}{(1-x^{u_i})^2} - \frac{x^{u_i}}{1-x^{u_i}} = \frac{(u_i-1) x^{u_i} + x^{2u_i}}{(1-x^{u_i})^2}\geq 0 \end{align*} $$

for $u_i \geq 1$ , we have

(2.12) $$ \begin{align} \begin{aligned} & \sum_{i=1}^{n}\frac{\pi^2 m_i}{6} \left(\frac{u_ix^{u_i}}{(1-x^{u_i})^2}\right) + \mathrm{exp}\left(-\sum_{i=1}^{n} \frac{\pi^2 m_i}{6}\frac{x^{u_i}}{1-x^{u_i}}\right) \\ &\qquad \geq \sum_{i=1}^{n} \frac{\pi^2 m_i}{6}\frac{x^{u_i}}{1-x^{u_i}}+ \mathrm{exp}\left(-\sum_{i=1}^{n} \frac{\pi^2 m_i}{6}\frac{x^{u_i}}{1-x^{u_i}}\right). \end{aligned} \end{align} $$

Let

$$ \begin{align*} y := \sum_{i=1}^{n} \frac{\pi^2 m_i}{6}\frac{x^{u_i}}{1-x^{u_i}}. \end{align*} $$

Then, $y>0$ , and so $y + e^{-y} \geq 1.$ From this and (2.12), we easily obtain the inequality (2.11).

An estimate for a useful integral is given in the following form. Even though it contains the result in [Reference Chan9, Lemma 3.2] as a special case, its proof is similar to that of [Reference Chan9, Lemma 3.2], and we omit it here.

Lemma 2.3 Let $(h,k)=1$ , $b \in \mathbb {R}_{\geq 0}$ , and define

$$ \begin{align*} I := \int_{\xi_{h,k}} e^{\frac{\pi}{12k}(\frac{b}{z}-z)} z^{-\frac{1}{2}} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi. \end{align*} $$

Then,

$$ \begin{align*} I = \sqrt{\frac{2}{k(n-1/24)}} b^{1/4} \mathrm{cosh}\left(\frac{\pi}{k} \sqrt{\frac{2b}{3}\left(n-\frac{1}{24}\right)} \right) + E^{(b)}(I), \end{align*} $$

where

$$ \begin{align*} |E^{(b)}(I)| \leq \sqrt{2} \pi^{-1} e^{b \pi / 3 } \frac{e^{2\pi (n-1/24) \varrho} N^{1/2}}{n-1/24}. \end{align*} $$

The following inequalities are also important in the proof of Theorem 1.1.

$$ \begin{align*} \sum_{k=1}^{N} k^n \leq \begin{cases} N^{n+1}, & \mathrm{if} ~ n\geq0, \\ \frac{1}{n+1} N^{n+1}, & \mathrm{if} ~ -1 < n<0. \end{cases} \end{align*} $$

3.1 Transformation formulas for g and h

We apply (2.4) to transform g and h according to the values of $\mathrm {gcd}(k,27).$

Case 1: $\mathrm {gcd}(k,27)=1$ . By (2.4), we have

(3.5) $$ \begin{align} \begin{aligned} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (27\tau)})}{F(e^{2\pi i (9\tau)})^3}\ &= \frac{1}{3 \sqrt{3} \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})}{F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})^3} \\ &\quad\times e^{\pi i (s(h,k) + s(27h,k) - 3s(9h,k))} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} \end{aligned} \end{align} $$

and

(3.6) $$ \begin{align} \begin{aligned} &e^{2\pi i (2\tau)} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (9\tau)})}{F(e^{2\pi i (3\tau)})F(e^{2\pi i (27\tau)})^2}\\ &\quad= \frac{1}{9 \sqrt{3} \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})}{F(e^{2\pi i \gamma_{(3h,k)} (3\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})^2} \\ &\qquad\times e^{2\pi i (2h/k)} e^{\pi \frac{-48z}{12k}}e^{\pi i (s(h,k) + s(9h,k) - s(3h,k) - 2s(27h,k))} e^{\frac{\pi}{12k}(\frac{19}{27z} + 47z)} \\ &\quad= \frac{1}{9 \sqrt{3} \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})}{F(e^{2\pi i \gamma_{(3h,k)} (3\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})^2} \\ &\qquad \times e^{2\pi i (2h/k)} e^{\pi i (s(h,k) + s(9h,k) - s(3h,k)) - 2s(27h,k)} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)}. \end{aligned} \end{align} $$

Case 2: $\mathrm {gcd}(k,27)=3$ . Let $k = 3l_1.$ Then,

(3.7) $$ \begin{align} \begin{aligned} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (27\tau)})}{F(e^{2\pi i (9\tau)})^3} &= \frac{1}{\sqrt{3} \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,3l_1)} (\tau)})F(e^{2\pi i \gamma_{(9h,l_1)} (27\tau)})}{F(e^{2\pi i \gamma_{(3h,l_1)} (9\tau)})^3}\\ &\quad \times e^{\pi i (s(h,3l_1) + s(9h,l_1) - 3s(3h,l_1))} e^{\frac{\pi}{12k}(-\frac{5}{3z} - z)} \end{aligned} \end{align} $$

and

(3.8) $$ \begin{align} \begin{aligned} &e^{2\pi i (2\tau)} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (9\tau)})}{F(e^{2\pi i (3\tau)})F(e^{2\pi i (27\tau)})^2}\\ &\quad=\frac{1}{3 \sqrt{3} \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,3l_1)} (\tau)})F(e^{2\pi i \gamma_{(3h,l_1)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,l_1)} (3\tau)})F(e^{2\pi i \gamma_{(9h,l_1)} (27\tau)})^2}\\ & \qquad \times e^{2\pi i (2h/k)} e^{\pi i (s(h,3l_1) + s(3h,l_1) - s(h,l_1) - 2s(9h,l_1))} e^{\frac{\pi}{12k}(-\frac{5}{3z} - z)}. \end{aligned} \end{align} $$

Case 3: $\mathrm {gcd}(k,27)=9$ . Let $k = 9l_2.$ It follows that

$$ \begin{align*} \begin{aligned} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (27\tau)})}{F(e^{2\pi i (9\tau)})^3} &= \frac{\sqrt{3}}{\sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})}{F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})^3}\\ &\quad \times e^{\pi i (s(h,9l_2) + s(3h,l_2) - 3s(h,l_2))} e^{\frac{\pi}{12k}(-\frac{23}{z} - z)} \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} &e^{2\pi i (2\tau)} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (9\tau)})}{F(e^{2\pi i (3\tau)})F(e^{2\pi i (27\tau)})^2}\\ &\quad=\frac{1}{3 \sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,3l_2)} (3\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})^2}\\ &\qquad \times e^{2\pi i (2h/k)} e^{\pi i (s(h,9l_2) + s(h,l_2) - s(h,3l_1) - 2s(3h,l_2))} e^{\frac{\pi}{12k}(\frac{1}{z} - z)}. \end{aligned} \end{align*} $$

Case 4: $\mathrm {gcd}(k,27)=27$ . Let $k = 27l_3.$ We have

(3.9) $$ \begin{align} \begin{aligned} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (27\tau)})}{F(e^{2\pi i (9\tau)})^3} = & \frac{1}{\sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,27l_3)} (\tau)})F(e^{2\pi i \gamma_{(h,l_3)} (27\tau)})}{F(e^{2\pi i \gamma_{(h,3l_3)} (9\tau)})^3} \\ & \times e^{\pi i (s(h,27l_3) + s(h,l_3) - 3s(h,3l_3))} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} \end{aligned} \end{align} $$

and

$$ \begin{align*} \begin{aligned} &e^{2\pi i (2\tau)} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (9\tau)})}{F(e^{2\pi i (3\tau)})F(e^{2\pi i (27\tau)})^2}\\ &\quad=\frac{1}{\sqrt{z}} \frac{F(e^{2\pi i \gamma_{(h,27l_3)} (\tau)})F(e^{2\pi i \gamma_{(h,3l_3)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,9l_3)} (3\tau)})F(e^{2\pi i \gamma_{(h,l_3)} (27\tau)})^2} \\ &\qquad\times e^{2\pi i (2h/k)} e^{\pi i (s(h,27l_3) + s(h,3l_3) - s(h,9l_3) - 2s(h,l_3))} e^{\frac{\pi}{12k}(-\frac{47}{z} -z)}. \end{aligned} \end{align*} $$

From the above transformation formulas for g and h, we see that the integrands of the integrals in $S_2(A), S_2(B),S_3(A)$ , and $S_4(B)$ have factors of the forms $e^{\frac {\delta }{z}}$ with $\delta <0.$ For these integrals, we use (2.3), (2.5), (2.6), and (2.9) to give bounds. However, since the integrands of the integrals in $S_1(A), S_1(B), S_3(B)$ , and $S_4(A)$ contain factors of the forms $e^{\frac {\delta }{z}}$ with $\delta>0,$ we split these integrals into two parts and then apply Lemma 2.2, (2.3), (2.5), (2.6), and (2.10) to estimate the second parts. For the first parts, we employ Lemma 2.3 to tackle the integrals. It should be mentioned that our main term $P(n)$ (in Subsection 3.3) originates from $S_3(B).$

3.2 Bounding $S_2(A)$ , $S_2(B)$ , $S_3(A)$ , and $S_4(B)$

For $S_2(A)$ , we apply (3.7) to deduce that

$$ \begin{align*} \begin{aligned} A = & \frac{1}{\sqrt{3}} z^{-\frac{1}{2}} e^{\pi i (s(h,3l_1) + s(9h,l_1) - 3s(3h,l_1))} e^{\frac{\pi}{12k} \left(-\frac{5}{3z}-z\right)} \\ & \times \frac{F(e^{2\pi i \gamma_{(h,3l_1)} (\tau)})F(e^{2\pi i \gamma_{(9h,l_1)} (27\tau)})}{F(e^{2\pi i \gamma_{(3h,l_1)} (9\tau)})^3} e^{2 \pi n \varrho} e^{-2 \pi i n \varphi} \end{aligned} \end{align*} $$

with $k=3l_1$ . Using (2.3), (2.5), (2.6), and (2.9), we find that

$$ \begin{align*} \begin{aligned} |A| \leq & \frac{1}{\sqrt{3}} F(e^{-2\pi(\frac{1}{k})\mathrm{Re}(\frac{1}{z})}) F(e^{-2\pi(\frac{3}{k})\mathrm{Re}(\frac{1}{9z})}) F(e^{-2\pi(\frac{3}{k})\mathrm{Re}(\frac{1}{3z})})^3 \\ & \times e^{-\frac{5\pi}{36k}\mathrm{Re}(\frac{1}{z})} e^{-\frac{\pi}{12k}\mathrm{Re}(z)} |z|^{-\frac{1}{2}} e^{2\pi n \varrho} \\ \leq & \frac{1}{\sqrt{3}} \mathrm{exp}\left(\frac{2\pi^2 e^{-\pi}}{3(1-e^{-\pi})} + \frac{\pi^2 e^{-\pi/3}}{6(1-e^{-\pi/3})}\right) \times e^{-\frac{5\pi}{72}} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N \\ \leq & \frac{1}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N. \end{aligned} \end{align*} $$

So we have

$$ \begin{align*} \begin{aligned} |S_2(A)| \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 3}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \int_{\xi_{h,k}} \frac{1}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N d\varphi \\ \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 3}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \frac{1}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} \frac{2}{k^{\frac{3}{2}}} \\ \leq & \frac{2}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} \sum_{1 \leq k \leq N} k^{-\frac{1}{2}} \\ \leq & \frac{4}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}. \end{aligned} \end{align*} $$

For $S_2(B)$ , we apply (3.8) to give

$$ \begin{align*} \begin{aligned} B = & \frac{1}{3 \sqrt{3}} z^{-\frac{1}{2}} e^{2\pi i (2h/k)} e^{\pi i (s(h,3l_1) + s(3h,l_1) - s(h,l_1) - 2s(9h,l_1))} e^{\frac{\pi}{12k}(-\frac{5}{3z} - z)} \\ & \times \frac{F(e^{2\pi i \gamma_{(h,3l_1)} (\tau)})F(e^{2\pi i \gamma_{(3h,l_1)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,l_1)} (3\tau)})F(e^{2\pi i \gamma_{(9h,l_1)} (27\tau)})^2} e^{2 \pi n \varrho} e^{-2 \pi i n \varphi} \end{aligned} \end{align*} $$

with $k=3l_1$ . Using (2.3), (2.5), (2.6), and (2.9), we find that

$$ \begin{align*} \begin{aligned} |B| \leq & \frac{1}{3\sqrt{3}} F(e^{-2\pi(\frac{1}{k})\mathrm{Re}(\frac{1}{z})}) F(e^{-2\pi(\frac{3}{k})\mathrm{Re}(\frac{1}{3z})}) F(e^{-2\pi(\frac{3}{k})\mathrm{Re}(\frac{1}{z})}) F(e^{-2\pi(\frac{3}{k})\mathrm{Re}(\frac{1}{9z})})^2 \\ & \times e^{-\frac{5\pi}{36k}\mathrm{Re}(\frac{1}{z})} e^{-\frac{\pi}{12k}\mathrm{Re}(z)} |z|^{-\frac{1}{2}} e^{2\pi n \varrho} \\ \leq & \frac{1}{3\sqrt{3}} \mathrm{exp}\left(\frac{\pi^2 e^{-\pi}}{3(1-e^{-\pi})} + \frac{\pi^2 e^{-3\pi}}{6(1-e^{-3\pi})} + \frac{\pi^2 e^{-\pi/3}}{3(1-e^{-\pi/3})}\right) e^{-\frac{5\pi}{72}} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N \\ \leq & \frac{1}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} |S_2(B)| \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 3}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \int_{\xi_{h,k}} \frac{1}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} k^{-\frac{1}{2}} N d\varphi \\ \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 3}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \frac{1}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} \frac{2}{k^{\frac{3}{2}}} \\ \leq & \frac{2}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} \sum_{1 \leq k \leq N} k^{-\frac{1}{2}} \\ \leq & \frac{4}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}. \end{aligned} \end{align*} $$

Similarly, for $S_3(A)$ and $S_4(B)$ , we get

$$ \begin{align*} |S_3(A)| \leq 4\sqrt{3} e^{-2.047} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}} \end{align*} $$

and

$$ \begin{align*} |S_4(B)| \leq 4e^{-6.077} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}. \end{align*} $$

3.3 Tackling $S_1(A)$ , $S_1(B)$ , $S_3(B)$ , and $S_4(A)$

For $S_1(A)$ , set

$$ \begin{align*}\omega^{(1)}_{(h,k)} := e^{\pi i (s(h,k) + s(27h,k) - 3s(9h,k))}.\end{align*} $$

We apply (3.5) to deduce

$$ \begin{align*} S_1(A) &= \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \int_{\xi_{h,k}} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (27\tau)})}{F(e^{2\pi i (9\tau)})^3} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ &= \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} \\ & \quad\times \frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})}{F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})^3} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ \notag &= \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ \notag &\quad + \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} \\ \notag & \quad\times \left(\frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})}{F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})^3} - 1 \right) e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ \notag & =: T_{1} + T_{2}. \notag \end{align*} $$

By (2.9) and (2.10), we have

$$ \begin{align*} \begin{aligned} \left| e^{\frac{\pi}{12k}\left(\frac{19}{27z}\right)} \left(\frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})}{F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})^3} - 1 \right) \right| \leq e^{\frac{19\pi}{27\times12k}\mathrm{Re}\left(\frac{1}{z}\right)} \left( e^{f(k,z)}- 1\right), \end{aligned} \end{align*} $$

where

$$ \begin{align*}f(k,z)=\frac{\pi^2 e^{-2\pi \frac{1}{k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{1}{k} \mathrm{Re}(\frac{1}{z})})} + \frac{\pi^2 e^{-2\pi \frac{1}{27k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{1}{27k} \mathrm{Re}(\frac{1}{z})})} + \frac{3\pi^2 e^{-2\pi \frac{1}{9k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{1}{9k} \mathrm{Re}(\frac{1}{z})})} .\end{align*} $$

Let

$$ \begin{align*} x := e^{-\frac{19\pi}{27\times12k}\mathrm{Re}\left(\frac{1}{z}\right)}. \end{align*} $$

Then,

$$ \begin{align*} \begin{aligned} & \left| e^{\frac{\pi}{12k}\left(\frac{19}{27z}\right)} \left(\frac{F(e^{2\pi i \gamma_{(h,k)} (\tau)})F(e^{2\pi i \gamma_{(27h,k)} (27\tau)})}{F(e^{2\pi i \gamma_{(9h,k)} (9\tau)})^3} - 1 \right) \right| \\ &\quad \leq \frac{1}{x} \left(\mathrm{exp}\left(\frac{\pi^2 x^{\frac{648}{19}}}{6(1-x^{\frac{648}{19}})}+\frac{\pi^2 x^{\frac{24}{19}}}{6(1-x^{\frac{24}{19}})}+\frac{3\pi^2 x^{\frac{72}{19}}}{6(1-x^{\frac{72}{19}})}\right) - 1\right)=: W_1(x). \end{aligned} \end{align*} $$

By Lemma 2.2 and (2.5), we get

$$ \begin{align*} \begin{aligned} W_1(x) & \leq W_1(e^{-\frac{19\pi}{648}}) = e^{\frac{19\pi}{648}} \left(\mathrm{exp}\left(\frac{\pi^2 e^{-\pi}}{6(1-e^{-\pi})}+\frac{\pi^2 e^{-\pi/27}}{6(1-e^{-\pi/27})}+\frac{3\pi^2 e^{-\pi/9}}{6(1-e^{-\pi/9})}\right) - 1\right) \\ & \leq 9.819 \times 10^{10}. \end{aligned} \end{align*} $$

Then, by (2.6),

$$ \begin{align*} \begin{aligned} |T_{2}| \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \int_{\xi_{h,k}} 1.890 \times 10^{10} e^{-\frac{\pi}{12k}\mathrm{Re}(z)} |z|^{-1/2} e^{-2\pi i n \varrho} d\varphi \\ \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} 1.890 \times 10^{10} e^{2\pi (n - 1/24) \varrho} \frac{2}{k^{3/2}} \\ \leq & \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} 1.890 \times 10^{10} e^{2\pi (n - 1/24) \varrho} \frac{2}{k^{1/2}} \\ \leq & 7.560 \times 10^{10} e^{2\pi (n - 1/24) \varrho} N^{1/2}. \end{aligned} \end{align*} $$

Similarly, we use (3.6) to derive that

$$ \begin{align*} S_1(B)=R_{1} + R_{2}, \end{align*} $$

where

$$ \begin{align*} \begin{aligned} R_{1} := \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \omega^{(2)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{9 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi, \end{aligned} \end{align*} $$

and

$$ \begin{align*} |R_{2}| \leq 1.435 \times 10^{13} e^{2\pi i (n - 1/24) \varrho} N^{1/2}. \end{align*} $$

Here, $\omega ^{(2)}_{(h,k)} := e^{\pi i (s(h,k) + s(9h,k) - s(3h,k) - 2s(27h,k))}.$

Using the software Mathematica, we find thatFootnote ²

(3.10) $$ \begin{align} 4s(9h,k) - s(3h,k) -3s(27h,k) + \frac{4h}{k} \equiv 0 ~~ (\mathrm{mod} ~ 2), \end{align} $$

where $0\leq h < k \leq 17, (k,3)=1$ , and $(h,k)=1.$ From this, we have

$$ \begin{align*} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} = \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \omega^{(2)}_{(h,k)} \end{align*} $$

with $k \leq 17$ and $\mathrm {gcd}(k,27)=1.$ Then,

$$ \begin{align*} \begin{aligned} |T_{1} - 3R_{1}| &= \bigg|\sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ &\quad - \big. \sum_{\substack{1 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \omega^{(2)}_{(h,k)}\\&\quad \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \bigg| \\& \leq \bigg|\sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(1)}_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi\bigg| \\ &\quad + \bigg|\sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \omega^{(2)}_{(h,k)}\\&\quad \int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi\bigg|. \end{aligned} \end{align*} $$

By Lemma 2.3, we have

$$ \begin{align*} &\sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \left|\int_{\xi_{h,k}} \frac{1}{3 \sqrt{3} \sqrt{z}} e^{\frac{\pi}{12k}(\frac{19}{27z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi\right| \\ \notag &\leq \sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \sqrt{\frac{2}{k(n-1/24)}} \left(\frac{19}{27}\right)^{1/4} \mathrm{cosh}\left(\frac{\pi}{k} \sqrt{\frac{19}{27}\times\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) \\ &\quad+ \sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \sqrt{2} \pi^{-1} e^{19\pi/81} \frac{e^{2\pi (n-1/24) \varrho} N^{1/2}}{n-1/24} \\\notag &\leq \sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} \sqrt{\frac{2k}{n-1/24}} \mathrm{cosh}\left(\frac{\pi}{18} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) \\ \notag &\; + \sum_{\substack{19 \leq k \leq N \\ (k,27) = 1}} 1.2828 \frac{e^{2\pi (n-1/24) \varrho} k N^{1/2}}{n-1/24} \\ \notag & \leq N^{3/2} \sqrt{\frac{2}{n-1/24}} \mathrm{cosh}\left(\frac{\pi}{18} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) + 1.2828 \frac{e^{2\pi (n-1/24) \varrho} N^{5/2}}{n-1/24}. \notag \end{align*} $$

So we have

$$ \begin{align*} |T_{1} - 3R_{1}| \leq 2N^{3/2} \sqrt{\frac{2}{(n-1/24)}} \mathrm{cosh}\left(\frac{\pi}{18} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) + 2.5656 \frac{e^{2\pi (n-1/24) \varrho} N^{5/2}}{n-1/24}. \end{align*} $$

Setting $N:=\sqrt {2\pi (n-1/24)}$ , we obtain

$$ \begin{align*} \begin{aligned} |T_{1} - 3R_{1}| &\leq 2 (2\pi)^{3/4} \sqrt{2} (n-1/24)^{(1/4)} (e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}+1)\\ &+ 2.5656 e (2\pi)^{5/4} (n-1/24)^{(1/4)} \\ &\leq 11.225 (n-1/24)^{(1/4)} e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 80.601 (n-1/24)^{(1/4)}. \end{aligned} \end{align*} $$

For $S_3(B)$ , set $\Omega _{(h,k)} := e^{\pi i (s(h,9l_2) + s(h,l_2) - s(h,3l_1) - 2s(3h,l_2))}$ . We apply (3.1) to conclude that

$$ \begin{align*} \begin{aligned}\! S_3(B)& = \!\sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \!\!e^{-2\pi i nh/k}\! \int_{\xi_{h,k}} \!e^{2\pi i (2\tau)} \frac{F(e^{2\pi i \tau})F(e^{2\pi i (9\tau)})}{F(e^{2\pi i (3\tau)})F(e^{2\pi i (27\tau)})^2} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ & = \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{z}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} \\ & \quad \times \frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,3l_2)} (3\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})^2} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ &= \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{z}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ & \quad + \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \int_{\xi_{h,k}} \frac{1}{3 \sqrt{z}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} \\ &\quad\times \left(\frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,3l_2)} (3\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})^2} - 1 \right) e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ & =:P_{1} + P_{2} \end{aligned} \end{align*} $$

with $k=9l_2$ .

For $P_2$ , by (2.10), we have

$$ \begin{align*} \begin{aligned} \left| e^{\frac{\pi}{12k}(\frac{1}{z})} \left(\frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,3l_2)} (3\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})^2} - 1 \right) \right| \leq e^{\frac{\pi}{12k}\mathrm{Re}\left(\frac{1}{z}\right)} \left( e^{g(k,z)}- 1\right), \end{aligned} \end{align*} $$

where

$$ \begin{align*}g(k,z)=\frac{\pi^2 e^{-2\pi \frac{1}{k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{1}{k} \mathrm{Re}(\frac{1}{z})})} + \frac{\pi^2 e^{-2\pi \frac{9}{k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{9}{k} \mathrm{Re}(\frac{1}{z})})} + \frac{3\pi^2 e^{-2\pi \frac{3}{k} \mathrm{Re}(\frac{1}{z})}}{6(1-e^{-2\pi \frac{3}{k} \mathrm{Re}(\frac{1}{z})})}.\end{align*} $$

Let

$$ \begin{align*} x := e^{-\frac{\pi}{12k}\mathrm{Re}(\frac{1}{z})}. \end{align*} $$

Then,

$$ \begin{align*} \begin{aligned} & \left| e^{\frac{\pi}{12k}(\frac{1}{z})} \left(\frac{F(e^{2\pi i \gamma_{(h,9l_2)} (\tau)})F(e^{2\pi i \gamma_{(h,l_2)} (9\tau)})}{F(e^{2\pi i \gamma_{(h,3l_2)} (3\tau)})F(e^{2\pi i \gamma_{(3h,l_2)} (27\tau)})^2} - 1 \right) \right| \\ & \quad\leq \frac{1}{x} \left(\mathrm{exp}\left(\frac{\pi^2 x^{24}}{6(1-x^{24})}+\frac{\pi^2 x^{216}}{6(1-x^{216})}+\frac{3\pi^2 x^{72}}{6(1-x^{72})}\right) - 1\right) =: W_3(x). \end{aligned} \end{align*} $$

By Lemma 2.2 and (2.5), we have

$$ \begin{align*} \begin{aligned} W_3(x) &\leq W_3(e^{-\frac{\pi}{24}}) = e^{\frac{\pi}{24}} \left(\mathrm{exp}\left(\frac{\pi^2 e^{-\pi}}{6(1-e^{-\pi})}+\frac{\pi^2 e^{-9\pi}}{6(1-e^{-9\pi})} + \frac{\pi^2 e^{-3\pi}}{2(1-e^{-3\pi})}\right) - 1\right) \\ & \leq 0.089. \end{aligned} \end{align*} $$

Hence,

$$ \begin{align*} \begin{aligned} |P_2| &\leq 0.0297 \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \int_{\xi_{h,k}} |z|^{-\frac{1}{2}} e^{\frac{-\pi}{12k}\mathrm{Re}(z)} e^{2\pi n \varrho} d\varphi \\ &\leq 0.0297 \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \int_{\xi_{h,k}} k^{-\frac{1}{2}} N e^{2\pi (n - 1/24) \varrho} d\varphi \\ &\leq 0.0594 \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} k^{-\frac{3}{2}} e^{2\pi (n - 1/24) \varrho} \\ &\leq 0.0594 \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} k^{-\frac{1}{2}} e^{2\pi (n - 1/24) \varrho} \\&\leq 0.119 e^{2\pi (n - 1/24) \varrho} N^{\frac{1}{2}}.\\ \end{aligned} \end{align*} $$

For $P_1,$ applying Lemma 2.3 with $b=1$ , we establish

$$ \begin{align*} P_1 &= \frac{1}{3} \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \int_{\xi_{h,k}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} z^{-\frac{1}{2}} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \notag \\ &=\frac{1}{3} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)} \sqrt{\frac{2}{9(n-1/24)}} \mathrm{cosh}\left(\frac{\pi}{9} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) \notag \\ & \quad+\frac{1}{3} \sum_{\substack{18 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \sqrt{\frac{2}{k(n-1/24)}} \mathrm{cosh}\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right)\\ &\quad + \frac{1}{3} \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} E^{(1)}(I) \\ &= \frac{1}{3} \bigg(\sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg) \sqrt{\frac{2}{9(n-1/24)}} e^{\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} \notag \\ &\quad + \frac{1}{3} \bigg(\sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg) \sqrt{\frac{2}{9(n-1/24)}} e^{-\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} \notag \\ & \quad+ \frac{1}{3} \sum_{\substack{18 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} \sqrt{\frac{2}{k(n-1/24)}} \mathrm{cosh}\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right) \notag \\ &\quad+ \frac{1}{3} \sum_{\substack{1 \leq k \leq N \\ (k,27) = 9}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)} E^{(1)}(I) \notag \\ & =:P(n) + Q \notag \end{align*} $$

with

$$ \begin{align*} P(n) = \bigg(\frac{1}{3} \sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg) \sqrt{\frac{2}{9(n-1/24)}} e^{\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} |Q| \leq & \sum_{\substack{18 \leq k \leq N \\ (k,27) = 9}} \bigg(\frac{1}{3} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n-2)h/k} \Omega_{(h,k)}\bigg) \sqrt{\frac{2}{k(n-1/24)}} \\ & \times \left(e^{\frac{\pi}{k} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + e^{-\frac{\pi}{k} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}\right) + \frac{2\sqrt{2}}{3} + 11.563 (n-1/24)^{(1/4)}\\ \leq & \sum^{\lfloor\sqrt{2\pi(n-1/24)}/9\rfloor}_{2 \leq l_2 \leq N} \frac{\sqrt{k}}{3} \sqrt{\frac{2}{n-1/24}} \left(e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 1\right)\\ & + \frac{2\sqrt{2}}{3} + 11.563 (n-1/24)^{(1/4)} \\ \leq & \left(\frac{\sqrt{2\pi(n-1/24)}}{9}\right)^{\frac{3}{2}} \sqrt{\frac{2}{n-1/24}} \left(e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 1\right) \\ & + \frac{2\sqrt{2}}{3} + 11.563 (n-1/24)^{(1/4)} \\ \leq & 0.208(n-1/24)^{(1/4)} e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 11.771(n-1/24)^{(1/4)} + 0.943. \end{aligned} \end{align*} $$

Similarly, applying (3.9), we deduce that

$$ \begin{align*} S_4(A)= U_{1} + U_{2}, \end{align*} $$

with

$$ \begin{align*} U_{1} = \sum_{\substack{1 \leq k \leq N \\ (k,27) = 27}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i nh/k} \omega^{(3)}_{(h,k)} \int_{\xi_{h,k}} z^{-\frac{1}{2}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi, \end{align*} $$

and

$$ \begin{align*} |U_{2}| \leq 0.3536 e^{2\pi (n - 1/24) \varrho} N^{1/2}. \end{align*} $$

Here, $\omega ^{(3)}_{(h,k)} := e^{\pi i (s(h,27l_3) + s(h,l_3) - 3s(h,3l_3))}.$

For $U_1$ , we have

$$ \begin{align*} \begin{aligned} |U_1| &= \sum_{\substack{1 \leq k \leq N \\ (k,27) = 27}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n)h/k} \omega^{(3)}_{(h,k)} \int_{\xi_{h,k}} z^{-\frac{1}{2}} e^{\frac{\pi}{12k}(\frac{1}{z} - z)} e^{2\pi n \varrho} e^{-2\pi i n \varphi} d\varphi \\ &\leq \bigg|\sum_{\substack{1 \leq k \leq N \\ (k,27) = 27}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n)h/k} \omega^{(3)}_{(h,k)} \sqrt{\frac{2}{k(n-1/24)}} \mathrm{cosh}\left(\frac{\pi}{k} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)} \right)\bigg| \\ &\quad + \bigg|\sum_{\substack{1 \leq k \leq N \\ (k,27) = 27}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} e^{-2\pi i (n)h/k} \omega^{(3)}_{(h,k)} E^{(1)}(I)\bigg| \\ &\leq \sum_{\substack{1 \leq k \leq N \\ (k,27) = 27}} \sum_{\substack{0 \leq h < k \\ (h,k) = 1}} \sqrt{\frac{2}{k(n-1/24)}} (e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}+1) + 34.698 (n-1/24)^{(1/4)} \\ &\leq \sum^{\lfloor\sqrt{2\pi (n-1/24)}/27 \rfloor}_{l_3=1} \sqrt{\frac{54l_3}{n-1/24}} (e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}+1) + 34.698 (n-1/24)^{(1/4)} \\ &\leq \left(\frac{\sqrt{2\pi (n-\frac{1}{24})}}{27}\right)^{\frac{3}{2}} \sqrt{\frac{54}{n-1/24}} (e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}+1) + 34.698 (n-1/24)^{(1/4)} \\& \leq 0.208(n-1/24)^{(1/4)}e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 34.897(n-1/24)^{(1/4)}. \end{aligned} \end{align*} $$

3.4 Signs of the Fourier coefficients of $g(q)-3h(q)$

It follows from (3.1) and (3.2) that

$$ \begin{align*}g(q)-3h(q)=\sum_{n=1}^{\infty} (a(n) - 3b(n))q^n.\end{align*} $$

From (3.3), (3.4) and the results in Subsections 3.2 and 3.3, we derive that

$$ \begin{align*} \begin{aligned} a(n) - 3b(n)& = S_1(A) + S_2(A) + S_3(A) + S_4(A) - 3(S_1(B) + S_2(B) + S_3(B) + S_4(B)) \\ & = T_1 - 3R_1 + T_2 - 3R_2 + S_2(A) + S_3(A) - 3S_2(B) \\ &\quad - 3P(n) -3P_2 -3Q + U_1 + U_2 - 3S_4(B) \\ &=: -3P(n) + E(n) \end{aligned} \end{align*} $$

with

$$ \begin{align*} -3P(n) = -\bigg(\sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg) \sqrt{\frac{2}{9(n-1/24)}} e^{\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}. \end{align*} $$

From Subsctions 3.2 and 3.3, we know that

$$ \begin{align*} \begin{cases} |T_1 - 3R_1| & \leq \quad 11.225 (n-1/24)^{(1/4)} e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 80.601 (n-1/24)^{(1/4)}, \\ |T_2| & \leq \quad 7.560 \times 10^{10} e^{2\pi (n - 1/24) \varrho} N^{\frac{1}{2}} = 3.254 \times 10^{11}(n-1/24)^{(1/4)}\\ |R_2| & \leq \quad 1.435 \times 10^{13} e^{2\pi i (n - 1/24) \varrho} N^{\frac{1}{2}} = 6.176 \times 10^{13}(n-1/24)^{(1/4)}, \\ |S_2(A)| & \leq \quad \frac{4}{\sqrt{3}} e^{0.969} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}} = 26.193(n-1/24)^{(1/4)}, \\ |S_3(A)| & \leq \quad 4\sqrt{3} e^{-2.047} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}= 3.850(n-1/24)^{(1/4)}, \\ |S_2(B)| & \leq \quad \frac{4}{3\sqrt{3}} e^{1.710} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}=18.318(n-1/24)^{(1/4)}, \\ |P_2| & \leq \quad 0.119 e^{2\pi (n - 1/24) \varrho} N^{\frac{1}{2}}=0.513(n-1/24)^{(1/4)}, \\ |Q| & \leq \quad 0.208(n-1/24)^{(1/4)} e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 11.771(n-1/24)^{(1/4)} + 0.943, \\ |U_1| & \leq \quad 0.208(n-1/24)^{(1/4)}e^{(\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})})} + 34.897(n-1/24)^{(1/4)}, \\ |U_2| & \leq \quad 0.3536 e^{2\pi (n - 1/24) \varrho} N^{\frac{1}{2}} = 1.522(n-1/24)^{(1/4)}, \\ |S_4(B)| & \leq \quad 4e^{-6.077} e^{2\pi(n-\frac{1}{24})\varrho} N^{\frac{1}{2}}=0.040(n-1/24)^{(1/4)}, \end{cases} \end{align*} $$

so that

$$ \begin{align*} \begin{aligned} |E(n)| &\leq |T_1 - 3R_1| + |T_2| + 3|R_2| + |S_2(A)| + |S_3(A)| + 3|S_2(B)| \\ &\quad + 3|P_2| + 3|Q| + |U_1| + |U_2| + 3|S_4(B)| \\ &\leq 12.057(n-1/24)^{(1/4)}e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 3.772 \\ &\quad + (3.254 \times 10^{11} + 3\times6.176 \times 10^{13} + 26.193 + 3.850 + 3\times 18.318 + \\ &\qquad 3\times 0.513 + 3\times 12.969 + 12.969 + 1.522 + 3\times 0.040)\bigg(n-\frac{1}{24}\bigg)^{\frac{1}{4}} \\ &\leq 12.057(n-1/24)^{(1/4)}e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 1.857\times 10^{14}\bigg(n-\frac{1}{24}\bigg)^{\frac{1}{4}}. \end{aligned} \end{align*} $$

It is observed that the signs of the main term $P(n)$ are determined by

$$ \begin{align*} \sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)} \end{align*} $$

with $\Omega _{(h,9)}:= e^{\pi i (s(h,9l_2) + s(h,l_2) - s(h,3l_1) - 2s(3h,l_2))}$ . After calculations, we have

$$ \begin{align*} \begin{cases} \Omega_{(1,9)} = e^{\frac{25}{54}\pi i},\quad \Omega_{(2,9)} = e^{\frac{11}{54}\pi i}, \\ \Omega_{(4,9)} = e^{-\frac{11}{54}\pi i},\quad \Omega_{(5,9)} = e^{\frac{11}{54}\pi i}, \\ \Omega_{(7,9)} = e^{-\frac{11}{54}\pi i},\quad \Omega_{(8,9)} = e^{-\frac{25}{54}\pi i}. \end{cases} \end{align*} $$

Therefore,

$$ \begin{align*} \sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)} = \begin{cases} 2\mathrm{cos}(23\pi/54) + 4\mathrm{cos}(49\pi/54) & \mathrm{if} ~n \equiv 0 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(35\pi/54) + 4\mathrm{cos}(37\pi/54) & \mathrm{if} ~n \equiv 1 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(25\pi/54) + 4\mathrm{cos}(11\pi/54) & \mathrm{if} ~n \equiv 2 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(49\pi/54) + 4\mathrm{cos}(13\pi/54) & \mathrm{if} ~n \equiv 3 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(37\pi/54) + 4\mathrm{cos}(\pi/54) & \mathrm{if} ~n \equiv 4 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(11\pi/54) + 4\mathrm{cos}(47\pi/54) & \mathrm{if} ~n \equiv 5 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(13\pi/54) + 4\mathrm{cos}(23\pi/54) & \mathrm{if} ~n \equiv 6 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(\pi/54) + 4\mathrm{cos}(35\pi/54) & \mathrm{if} ~n \equiv 7 ~(\mathrm{mod} ~ 9) \\ 2\mathrm{cos}(47\pi/54) + 4\mathrm{cos}(25\pi/54) & \mathrm{if} ~n \equiv 8 ~(\mathrm{mod} ~ 9). \\ \end{cases} \end{align*} $$

It is easily computed that

$$ \begin{align*} \bigg|\sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg|\geq 0.20142 \end{align*} $$

for $n\in \mathbb {N}.$ When $n \geq 22472$ ,

$$ \begin{align*} \begin{aligned} |-3P(n)| &= \bigg|\bigg(\sum_{\substack{0 \leq h < 9 \\ (h,9) = 1}} e^{-2\pi i (n-2)h/9} \Omega_{(h,9)}\bigg) \sqrt{\frac{2}{9(n-1/24)}} e^{\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}}\bigg| \\ & \geq 0.20142 \sqrt{\frac{2}{9(n-1/24)}} e^{\frac{\pi}{9} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} \\ &> 12.057(n-1/24)^{(1/4)}e^{\frac{\pi}{18} \sqrt{\frac{2}{3}(n-\frac{1}{24})}} + 1.857\times 10^{14}\bigg(n-\frac{1}{24}\bigg)^{\frac{1}{4}}\\ & \geq |E(n)|. \end{aligned} \end{align*} $$

From this, we obtain that

(3.11) $$ \begin{align} \begin{cases} a(n) - 3b(n)> 0 & \mathrm{if} ~ n \equiv 0,1,5,8 ~ (\mathrm{mod}~9) \\ a(n) - 3b(n) < 0 & \mathrm{if} ~ n \equiv 2,3,4,6,7 ~ (\mathrm{mod}~9) \end{cases} \end{align} $$

holds for $n\geq 22472$ . Calculating the first $22471$ coefficients of $g(q)-3h(q)$ by using the software Mathematica, we find that (3.11) is also true for $467 \leq n \leq 22471.$

3.5 Proof of Theorem 1.1

It follows from the identities in [Reference Aygin and Chan6, eqs.(2.10) and (2.11)] that

$$ \begin{align*} \sum_{n=1}^{\infty} M(0,1,1;9;n) q^n - \frac{1}{3(q;q)_{\infty}} = \frac{2}{3}(g(q)-3h(q)), \end{align*} $$

and

$$ \begin{align*} \sum_{n=1}^{\infty} M(2,3,4;9;n) q^n - \frac{1}{3(q;q)_{\infty}} = -\frac{1}{3}(g(q)-3h(q)), \end{align*} $$

so that

$$ \begin{align*} \begin{aligned} M(0,1,1;9;n) - \frac{p(n)}{3} &= \frac{2}{3}(a(n) - 3b(n) ), \\ M(2,3,4;9;n) - \frac{p(n)}{3} &= -\frac{1}{3}(a(n) - 3b(n) ). \end{aligned} \end{align*} $$

It is known from Subsection 3.4 that (3.11) holds for $n \geq 467.$ Hence, when $n \geq 467,$

$$ \begin{align*} \begin{cases} M(0,1,1;9;n)> \frac{p(n)}{3}, & \mathrm{if} ~ n \equiv 0,1,5,8 ~ (\mathrm{mod}~9), \\[1ex] M(0,1,1;9;n) < \frac{p(n)}{3}, & \mathrm{if} ~ n \equiv 2,3,4,6,7 ~ (\mathrm{mod}~9), \\[1ex] M(2,3,4;9;n) < \frac{p(n)}{3}, & \mathrm{if} ~ n \equiv 0,1,5,8 ~ (\mathrm{mod}~9), \\[1ex] M(2,3,4;9;n)> \frac{p(n)}{3}, & \mathrm{if} ~ n \equiv 2,3,4,6,7 ~ (\mathrm{mod}~9). \\[1ex] \end{cases} \end{align*} $$

This finishes the proof of Theorem 1.1.