Proof Substituting $\omega $ and $\tau $ by $\omega _1$ and $\tau _1$ in (2.6), respectively, we find that

(2.11) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =-2\sin\left(\pi\omega_1\right)\eta\left(\tau_1\right)\eta\left(2\tau_1\right) \vartheta^{-1}\left(\omega_1;\tau_1\right). \end{align} $$

By plugging (2.5) and (2.3) into (2.11), we infer that

(2.12) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =-2iz^{\frac{1}{2}}\sin\left(\pi\omega_1\right)\omega^3_{h,k} e^{\frac{\pi i}{4k}(h'-h)+\frac{\pi ka^2}{zc^2}} \eta\left(\tau_1\right)\eta\left(2\tau_1\right) \vartheta^{-1}\left(\frac{i\omega_1}{z};\tau_2\right). \end{align} $$

For k is even, since $(h,k)=1$ , we have $\left (h,\frac {k}{2}\right )=1$ . From (2.2) it follows that

(2.13) $$ \begin{align} \eta\left(\tau_1\right)\eta\left(2\tau_1\right) =\sqrt{\frac{i}{z}}\chi(h,h',k)\eta\left(\tau_2\right) \sqrt{\frac{i}{z}}\chi\left(h,h',\frac{k}{2}\right)\eta\left(2\tau_2\right). \end{align} $$

Inserting (2.3) into (2.13), we arrive at

$$ \begin{align*} \eta\left(\tau_1\right)\eta\left(2\tau_1\right) =z^{-1}\omega_{h,k}^{-1}\omega^{-1}_{h,\frac{k}{2}}e^{-\frac{\pi i}{4k}(h'-h)} \eta\left(\tau_2\right)\eta\left(2\tau_2\right). \end{align*} $$

Thus for k is even, (2.12) equals to

(2.14) $$ \begin{align} &\overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =-2iz^{-\frac{1}{2}}\sin\left(\pi\omega_1\right)\omega^2_{h,k}\omega^{-1}_{h,\frac{k}{2}} e^{\frac{\pi ka^2}{zc^2}} \frac{\eta\left(\tau_2\right)\eta\left(2\tau_2\right)} {\vartheta\left(\frac{i\omega_1}{z};\tau_2\right)}. \end{align} $$

We now assume that $c|k$ . It is obvious that $\frac {ak}{c}\in \mathbb {Z}$ . Rewriting $\eta \left (\tau _2\right )\eta \left (2\tau _2\right )$ by (2.6), we obtain

(2.15) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =iz^{-\frac{1}{2}}\frac{\sin\left(\pi \omega_1\right)}{\sin\left(\pi \omega_2\right)} \omega^2_{h,k}\omega^{-1}_{h,\frac{k}{2}}e^{\frac{\pi ka^2}{zc^2}} \overline{C}\left(e^{2\pi i\omega_2};e^{2\pi i\tau_2}\right) \frac{\vartheta\left(\omega_2;\tau_2\right)} {\vartheta\left(\frac{i\omega_1}{z};\tau_2\right)}. \end{align} $$

Recall the following identity in Rolon [Reference Zapata Rolon18]:

(2.16) $$ \begin{align} \frac{\vartheta\left(\omega_2;\tau_2\right)} {\vartheta\left(\frac{i\omega_1}{z};\tau_2\right)}=(-1)^{ak+1}e^{-\frac{\pi ia^2kh'}{c^2}-\frac{\pi a^2k}{zc^2}}. \end{align} $$

Then (2.15) becomes

$$ \begin{align*} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =(-1)^{ak+1}iz^{-\frac{1}{2}}\frac{\sin\left(\pi \omega_1\right)}{\sin\left(\pi \omega_2\right)} \omega^2_{h,k}\omega^{-1}_{h,\frac{k}{2}}e^{-\frac{\pi ia^2kh'}{c^2}} \overline{C}\left(e^{2\pi i\omega_2};e^{2\pi i\tau_2}\right), \end{align*} $$

which coincides with (2.7).

The case $c\nmid k$ presents more complexity, as typically $\frac {ak}{c}\not \in \mathbb {Z}$ . Given that l was determined as the solution to the congruence $l\equiv ak_1\pmod {c_1}$ by definition, it follows that $B=\frac {l-ak_1}{c_1}\in \mathbb {Z}$ . This allows for the adjustment of the theta function in the elliptic variable by a factor of B times the modular variable, as defined previously.

Recall the following identity in Rolon [Reference Zapata Rolon18]:

(2.17) $$ \begin{align} \vartheta\left(\frac{i\omega_1}{z};\tau_2\right) =(-1)^{ak+l}e^{\frac{\pi i(l-ak_1)^2\tau_2}{c_1^2}+2\pi i\left(\frac{l-ak_1}{c_1}\right)\cdot\frac{i\omega_1}{z}} \vartheta\left(\omega_3;\tau_2\right). \end{align} $$

Combining with (2.14), we get

(2.18) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) &=2(-1)^{ak+l+1}iz^{-\frac{1}{2}}\sin\left(\pi \omega_1\right) \omega^2_{h,k}\omega^{-1}_{h,\frac{k}{2}} \nonumber\\[6pt] &\qquad\times q_1^{-\frac{l^2}{2c_1}} e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2h'k_1}{cc_1}} \frac{\eta\left(\tau_2\right)\eta\left(2\tau_2\right)} {\vartheta\left(\omega_3;\tau_2\right)} \nonumber\\[6pt] &=(-1)^{ak+l}iz^{-\frac{1}{2}} \sin\left(\pi \omega_1\right)\omega^2_{h,k}\omega^{-1}_{h,\frac{k}{2}} \nonumber\\[6pt] &\qquad\times q_1^{-\frac{l^2}{2c_1}} e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2h'k_1}{cc_1}} \frac{\overline{C}\left(e^{2\pi i\omega_3};q_1\right)}{\sin\left(\pi \omega_3\right)}. \end{align} $$

We define $x=e^{2\pi i\omega _3}$ , and use the exponential representation of the sine to deduce

(2.19) $$ \begin{align} \frac{\overline{C}\left(e^{2\pi i\omega_3};q_1\right)}{\sin\left(\pi \omega_3\right)} &=\frac{2i(1-x)}{x^{\frac{1}{2}}-x^{-\frac{1}{2}}}\cdot\frac{(-q_1;q_1)_\infty}{(q_1;q_1)_\infty} \sum_{m=-\infty}^\infty\frac{(-1)^mq_1^{m+1\choose2}}{1-xq_1^m} \nonumber\\[5pt] &=-\frac{2ix^{\frac{1}{2}}(-q_1;q_1)_\infty}{(q_1;q_1)_\infty} \sum_{m=-\infty}^\infty\frac{(-1)^mq_1^{m+1\choose2}}{1-xq_1^m}\nonumber\\[5pt] &=-4\overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right). \end{align} $$

Thus, (2.9) can be given by inserting this expression into (2.18).

For k is odd, then we have $(2h,k)=1$ , by (2.2),

$$ \begin{align*} \eta\left(\tau_1\right)\eta\left(2\tau_1\right) =\sqrt{\frac{i}{z}}\chi(h,h',k)\eta\left(\tau_2\right) \sqrt{\frac{i}{2z}}\chi\left(2h,\frac{h'}2,k\right)\eta\left(\frac{\tau_2}{2}\right). \end{align*} $$

Using (2.3), we infer that

$$ \begin{align*} \eta\left(\tau_1\right)\eta\left(2\tau_1\right) =\frac{\sqrt2}{2}z^{-1}\omega_{h,k}^{-1}\omega^{-1}_{2h,k}e^{-\frac{\pi i}{4k}\left(\frac{h'}{2}-h\right)} \eta\left(\tau_2\right)\eta\left(\frac{\tau_2}{2}\right). \end{align*} $$

Thus for k is odd, (2.12) equals to

(2.20) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) =-\sqrt{2}iz^{-\frac{1}{2}}\sin\left(\pi\omega_1\right)\omega^2_{h,k}\omega^{-1}_{2h,k} e^{\frac{\pi ih'}{8k}+\frac{\pi ka^2}{zc^2}} \frac{\eta\left(\tau_2\right)\eta\left(\frac{\tau_2}{2}\right)} {\vartheta\left(\frac{i\omega_1}{z};\tau_2\right)}. \end{align} $$

Similarly, we consider the cases $c|k$ and $c\nmid k$ . For $c|k$ , we have $\frac {ak}{c}$ is an integer, and we obtain

(2.21) $$ \begin{align} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) & =i(2z)^{-\frac{1}{2}}\frac{\sin\left(\pi \omega_1\right)}{\sin\left(\pi \omega_2\right)}\omega^2_{h,k}\omega^{-1}_{2h,k} e^{\frac{\pi ih'}{8k}+\frac{\pi ka^2}{zc^2}} \frac{\eta\left(\frac{\tau_2}{2}\right)}{\eta\left(2\tau_2\right)} \nonumber \\ &\quad \cdot \frac{\vartheta\left(\omega_2;\tau_2\right)} {\vartheta\left(\frac{i\omega_1}{z};\tau_2\right)} \overline{C}\left(e^{2\pi i\omega_2};q_1\right). \end{align} $$

Utilizing (2.16), (2.21) can be rewritten as

$$ \begin{align*} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) &=(-1)^{ak+1}i(2z)^{-\frac{1}{2}}\frac{\sin\left(\pi \omega_1\right)}{\sin\left(\pi \omega_2\right)}\omega^2_{h,k}\omega^{-1}_{2h,k} e^{\frac{\pi ih'}{8k}-\frac{\pi ia^2k_1h'}{c}} \nonumber \\ &\quad \cdot\frac{\eta\left(\frac{\tau_2}{2}\right)} {\eta\left(2\tau_2\right)} \overline{C}\left(e^{2\pi i\omega_2};q_1\right), \end{align*} $$

which yields (2.8).

It remains to prove (2.10). Combining (2.17) and (2.19) with (2.20), we see that

$$ \begin{align*} \overline{C}\left(e^{2\pi i\omega_1};e^{2\pi i\tau_1}\right) &=\sqrt{2}(-1)^{ak+l+1}iz^{-\frac{1}{2}}\sin\left(\pi\omega_1\right) \omega^2_{h,k}\omega^{-1}_{2h,k}q_1^{-\frac{l^2}{2c_1}} e^{\frac{\pi ih'}{8k}+\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2h'k_1}{cc_1}} \nonumber\\[5pt] &\quad\cdot\frac{\eta\left(\tau_2\right)\eta\left(\frac{\tau_2}{2}\right)} {\vartheta\left(\omega_3;\tau_2\right)} \nonumber\\[5pt] &=(-1)^{ak+l}i(2z)^{-\frac{1}{2}} \sin\left(\pi\omega_1\right)\omega^2_{h,k}\omega^{-1}_{2h,k}q_1^{-\frac{l^2}{2c_1}} e^{\frac{\pi ih'}{8k}+\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2h'k_1}{cc_1}} \nonumber\\[5pt] &\quad\cdot \frac{\eta\left(\frac{\tau_2}{2}\right)}{\eta\left(2\tau_2\right)} \frac{\overline{C}\left(e^{2\pi i\omega_3};q_1\right)}{\sin\left(\pi\omega_3\right)} \nonumber\\[5pt] &=4(-1)^{ak+l+1}i(2z)^{-\frac{1}{2}} \sin\left(\pi \omega_1\right)\omega^2_{h,k}\omega^{-1}_{2h,k}q_1^{-\frac{l^2}{2c_1}} e^{\frac{\pi ih'}{8k}+\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2h'k_1}{cc_1}} \nonumber\\[5pt] &\quad\cdot \frac{\eta\left(\frac{\tau_2}{2}\right)}{\eta\left(2\tau_2\right)} \overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right). \end{align*} $$

This completes the proof.

Proof The proof of equation 3.1 is essentially established in [Reference Andrews1], with only a slight modification required. Part two is a direct consequence of part one and the proof of Lemma 3.2 in [Reference Bringmann5].

Proof Substituting $z=\frac {k}{n}-ik\Phi $ into (3.3) gives

(3.6) $$ \begin{align} I_{k,t}=\frac{1}{ki}\int_{\frac{k}{n}-\frac{i}{N}}^{\frac{k}{n}+\frac{i}{N}}z^{-\frac{1}{2}} e^{\frac{2\pi }{k}\left(zn+\frac{t}{z}\right)}\mathrm{d}z. \end{align} $$

We now define $\Gamma $ as the circle passing through $\frac {k}{n}\pm \frac {i}{N}$ and tangent to the imaginary axis at $0$ . For $z=x+iy$ , $\Gamma $ is characterized by $x^2+y^2=\alpha x$ , where $\alpha =\frac {k}{n}+\frac {n}{N^2k}$ . Considering that $2>\alpha >\frac {1}{k}$ , $\mathrm {Re}(z)\leq \frac {k}{n}$ and $\mathrm {Re}\left (\frac {1}{z}\right )< k$ on the smaller arc, similar to the argument of Rolon [Reference Zapata Rolon18, Eq. (4.6)], we get that the integral over the smaller arc is:

$$ \begin{align*} I_{k,t}^{err}\leq \frac 2k e^{2\pi+2\pi t}\left(\frac 43+2^{\frac 54}\right)n^{-\frac 18}. \end{align*} $$

Furthermore, Cauchy’s theorem allows us to modify the path of integration in (3.6) to follow the larger arc of $\Gamma $ . Transforming the circle to a straight line by $s=\frac {2\pi t}{kz}$ gives:

$$\begin{align*}\frac{1}{ki}\int_{\Gamma}z^{-\frac{1}{2}} e^{\frac{2\pi }{k}\left(zn+\frac{t}{z}\right)}\mathrm{d}z=\frac{2\pi}{k}\left(\frac{2\pi t}{k}\right)^{\frac{1}{2}}\frac{1}{2\pi i} \int_{\gamma-i\infty}^{\gamma+i\infty} s^{-\frac{3}{2}}e^{s+\frac{\beta}{s}}\mathrm{d}s,\end{align*}$$

where $\gamma \in \mathbb {R}$ and $\beta =\frac {4\pi ^2 tn}{k^2}$ . Using the Hankel integral formula in [Reference Watson19], we compute

$$ \begin{align*} \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} t^{-\frac{3}{2}} \cdot e^{t+\frac{\alpha}{t}} \mathrm{d} t=\frac{1}{\sqrt{\pi \alpha}} \cdot \sinh (2 \sqrt{\alpha}). \end{align*} $$

We thus get (3.4). This completes the proof.

Proof By Cauchy’s theorem we have for $n>0$

$$\begin{align*}\overline{A}\left(\frac{a}{c};n\right)=\frac{1}{2\pi i}\int_{\mathcal{C}}\frac{\overline{C}\left(e^{\frac{2\pi ia}{c}};q\right)}{q^{n+1}}\mathrm{d}q,\end{align*}$$

where $\mathcal {C}$ denotes any path within the unit circle that encircles the origin in a counterclockwise direction. Opting for a circular path of radius $e^{-\frac {2\pi }{n}}$ and parameterizing it as $q=e^{-\frac {2\pi }{n}+2\pi it}$ for $0\leq t\leq 1$ yields

$$\begin{align*}\overline{A}\left(\frac{a}{c};n\right)=\int_0^1\overline{C}\left(e^{\frac{2\pi ia}{c}};e^{-\frac{2\pi}{n}+2\pi it}\right)e^{2\pi-2\pi int}\mathrm{d}t.\end{align*}$$

We define

$$\begin{align*}\vartheta^{\prime}_{h,k}=\frac{1}{k\left(\widetilde{k}_1+k\right)},~~~~~ \vartheta^{\prime\prime}_{h,k}=\frac{1}{k\left(\widetilde{k}_2+k\right)},\end{align*}$$

where $\frac {h_1}{\widetilde {k}_1}<\frac {h}{k}<\frac {h_2}{\widetilde {k}_2}$ are adjacent Farey fractions in the Farey sequence of order $N=\lfloor n^{1/2}\rfloor $ . We know that

$$\begin{align*}\frac{1}{k+\widetilde{k}_j}\leq\frac{1}{N+1}~~~~(j=1,2).\end{align*}$$

Now we decompose the path of integration along Farey arcs $-\vartheta ^{\prime }_{h,k}\leq \Phi \leq \vartheta ^{\prime \prime }_{h,k}$ , where $\Phi =t-\frac {h}{k}$ and $0\leq h<k\leq N$ with $(h,k)=1$ . From this decomposition of the path we can rewrite the integral along these arcs:

$$\begin{align*}\overline{A}\left(\frac{a}{c};n\right)=\sum_{h,k}e^{-\frac{2\pi ihn}{k}} \int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}}\overline{C}\left(e^{\frac{2\pi ia}{c}};e^{\frac{2\pi i}{k}(h+iz)}\right)e^{\frac{2\pi nz}{k}}\mathrm{d}\Phi,\end{align*}$$

where $z=\frac {k}{n}-k\Phi i$ . We incorporate our transformation formula into the integral and conclude the proof.

Proof We note that

$$ \begin{align*} \overline{C}\left(e^{\frac{2\pi iah'}{c}};q_1\right) &=1+\sum_{r=1}^\infty\sum_{m=-\infty}^\infty\overline{M}(m,r)e^{\frac{2\pi iah'm}{c}}q_1^r \\[5pt] &=1+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}\sum_{t=-\infty}^\infty\overline{M}(s+tc,r)e^{\frac{2\pi iah's}{c}}q_1^r. \end{align*} $$

Let

$$\begin{align*}\overline{a}(r,s)=\sum_{t=-\infty}^\infty\overline{M}(s+tc,r),\end{align*}$$

then we have

(3.9) $$ \begin{align} \overline{C}\left(e^{\frac{2\pi iah'}{c}};q_1\right) =1+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}\overline{a}(r,s)e^{\frac{2\pi ih'}{c}m_{r,s}}q_1^r, \end{align} $$

where $m_{r,s}$ takes values in $\mathbb {Z}$ . Consequently, $\Sigma _1$ can be expressed as follows:

$$ \begin{align*} \Sigma_1=\widehat{S}_1+\widehat{S}_2, \end{align*} $$

where

(3.10) $$ \begin{align} \widehat{S}_1&=i\sin\left(\frac{\pi a}{c}\right)\sum_{h}\sum_{c|k\atop 2|k} \frac{(-1)^{ak+1}}{\sin\left(\frac{\pi ah'}{c}\right)} \omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}} \int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}}z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}\mathrm{d}\Phi, \end{align} $$

(3.11) $$ \begin{align} \widehat{S}_2&=i\sin\left(\frac{\pi a}{c}\right)\sum_{h}\sum_{c|k\atop 2|k} \frac{(-1)^{ak+1}}{\sin\left(\frac{\pi ah'}{c}\right)} \omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}} \nonumber\\ &\quad\quad\quad \cdot\int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}}z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}} \sum_{r=1}^\infty\sum_{s\pmod{c}}\overline{a}(r,s)e^{\frac{2\pi ih'}{c}m_{r,s}}q_1^r\mathrm{d}\Phi, \end{align} $$

We first consider $\widehat {S}_1$ . Splitting the integration path into

(3.12) $$ \begin{align} \int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}}=\int_{-\frac{1}{kN}}^{\frac{1}{kN}} -\int_{-\frac{1}{kN}}^{-\frac{1}{k(\widetilde{k}_1+k)}} -\int_{\frac{1}{k(\widetilde{k}_2+k)}}^{\frac{1}{kN}}, \end{align} $$

and denoting the associated sums by $\widehat {S}_{11},\widehat {S}_{12}$ , and $\widehat {S}_{13}$ , then one can demonstrate that $\widehat {S}_{1i}$ $(i=1,2,3)$ contributes to the error term. Specifically, for $\widehat {S}_{11}$ , we find

$$ \begin{align*} \left|\widehat{S}_{11}\right|&\leq\sum_{c|k\atop 2|k} \int_{-\frac{1}{kN}}^{\frac{1}{kN}} \left|z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}\right|d\Phi \cdot\left|\frac{\sin\left(\frac{\pi a}{c}\right)}{(-1)^{ak+1}} \sum_{h\atop\ell\leq\widetilde{k}_1+k-1\leq N+k-1} \frac{\omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}}{\sin\left(\frac{\pi ah'}{c}\right)} e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}}\right|. \end{align*} $$

According to (3.2) and the facts that

(3.13) $$ \begin{align} \mathrm{Re}(z)=\frac{k}{n},\qquad |z|^{-\frac{1}{2}}\leq k^{-\frac{1}{2}}n^{\frac{1}{2}}, \end{align} $$

we deduce that

$$ \begin{align*} \left|\widehat{S}_{11}\right|&\leq C\sum_{c|k\atop 2|k}\gcd(24n+1,k)^{\frac{1}{2}}k^{\frac{1}{2}+\varepsilon} k^{-\frac{1}{2}}n^{\frac{1}{2}}e^{2\pi} \int_{-\frac{1}{kN}}^{\frac{1}{kN}}\mathrm{d}\Phi \\[3pt] &\leq C_1\sum_{k}\gcd(24n+1,k)^{\frac{1}{2}}k^{-1+\varepsilon} \leq C_2\sum_{k\leq N}k^{-1+\varepsilon}\sum_{d|k\atop d|24n+1}d^{\frac{1}{2}} \\[3pt] &\leq C_2\sum_{d|24n+1\atop d\leq N}d^{\frac{1}{2}}\sum_{k\leq N/d}(kd)^{-1+\varepsilon} \leq C_2\sum_{d|24n+1\atop d\leq N}d^{-\frac{1}{2}}\sum_{k\leq N/d}k^{-1}N^{\varepsilon} \\[3pt] &\leq C_3n^{\varepsilon}\sum_{d|24n+1\atop d\leq N}d^{-\frac{1}{2}} \leq C_4n^{\varepsilon}, \end{align*} $$

where $C_1, C_2, C_3$ and $C_4$ are constants.

Turning our attention to $\widehat {S}_{12}$ , the integral within this term can be reformulated as:

$$\begin{align*}\int_{-\frac{1}{kN}}^{-\frac{1}{k(\widetilde{k}_1+k)}} =\sum_{\ell=N}^{\widetilde{k}_1+k-1}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}.\end{align*}$$

Plugging into $\widehat {S}_{12}$ gives:

$$ \begin{align*} \widehat{S}_{12}=\sum_{c|k\atop 2|k} \sum_{\ell=N}^{\widetilde{k}_1+k-1}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}} z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}\mathrm{d}\Phi \frac{i\sin\left(\frac{\pi a}{c}\right)}{(-1)^{ak+1}}\sum_{h} \frac{\omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}}{\sin\left(\frac{\pi ah'}{c}\right)} e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}}. \end{align*} $$

Given the constraint $\widetilde {k}_1\leq N$ , it follows that $\ell \leq \widetilde {k}_1+k-1\leq N+k-1$ , thereby limiting the range of summation over h. Consequently, we can estimate $\widehat {S}_{12}$ by summing over more integrals:

$$ \begin{align*} \left|\widehat{S}_{12}\right|&\leq\sum_{c|k\atop 2|k} \sum_{\ell=N}^{N+k-1}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}} \left|z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}\right|\mathrm{d}\Phi \\[3pt] &\quad \cdot\left|\frac{\sin\left(\frac{\pi a}{c}\right)}{(-1)^{ak+1}} \sum_{h\atop\ell\leq\widetilde{k}_1+k-1\leq N+k-1} \frac{\omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}}{\sin\left(\frac{\pi ah'}{c}\right)} e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}}\right|. \end{align*} $$

Now by the theory of Farey fractions we have

$$\begin{align*}\widetilde{k}_1\equiv-h'\pmod{k},~~~\widetilde{k}_2\equiv h'\pmod{k},~~N-k<\widetilde{k}_i\leq N,\end{align*}$$

for $i=1,2$ . This means the restriction of $h'$ to one or two intervals in the range $0\leq h' < k$ . Similar to $\widehat {S}_{11}$ , using (3.2) and (3.13), we get $\widehat {S}_{12}= O(n^\varepsilon )$ . The shape of $\widehat {S}_{13}$ is the same as that of $\widehat {S}_{12}$ . Hence, representing the expression as

$$\begin{align*}\int_{\frac{1}{k(\widetilde{k}_2+k)}}^{\frac{1}{kN}}=\sum_{\ell=k+\widetilde{k}_2-1}^{N} \int_{\frac{1}{k(\ell+1)}}^{\frac{1}{k\ell}}\end{align*}$$

yields $\widehat {S}_{13}= O(n^\varepsilon )$ by the same argument.

Following this, our analysis shifts to $\widehat {S}_{2}$ . Analogously, the integral within $\widehat {S}_{2}$ can be expressed as

(3.14) $$ \begin{align} \int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}} =\int_{-\frac{1}{k(N+k)}}^{\frac{1}{k(N+k)}}+\int_{\frac{1}{k(N+k)}}^{\frac{1}{k(\widetilde{k}_2+k)}} +\int_{-\frac{1}{k(\widetilde{k}_1+k)}}^{-\frac{1}{k(N+k)}}, \end{align} $$

and we denote the associated sums by $\widehat {S}_{21},\widehat {S}_{22}$ and $\widehat {S}_{23}$ . Then taking the absolute value of this it is possible to bound the term. Before doing that we define

$$\begin{align*}\overline{a}(r)=\sum_{s\pmod c}\left|\overline{a}(r,s)\right|,\end{align*}$$

where $\overline {a}(r)$ are exactly $\overline {p}(n)$ except from some constant term ambiguity. Using (3.2) and (3.13) again, we have

$$ \begin{align*} \left|\widehat{S}_{21}\right| &\leq \sum_{r=1}^\infty\sum_{c|k\atop 2|k}\overline{a}(r) \left|(-1)^{ak+1}\sin\left(\frac{\pi a}{c}\right)\sum_{h}\frac{\omega_{h,k}^2\omega_{h,\frac{k}{2}}^{-1}}{\sin\left(\frac{\pi ah'}{c}\right)} e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}+\frac{2\pi ih'}{c}m_{r,s}}\right| \nonumber\\[5pt] &\quad \cdot k^{-\frac{1}{2}}n^{\frac{1}{2}}e^{2\pi-\pi r} \int_{-\frac{1}{k(N+k)}}^{\frac{1}{k(N+k)}}\mathrm{d}\Phi \nonumber\\[5pt] &=O(n^\varepsilon). \end{align*} $$

Since $\widehat {S}_{22}$ and $\widehat {S}_{23}$ are treated in exactly the same way as $\widehat {S}_{12}$ , by writting

$$ \begin{align*} \int_{\frac{1}{k(N+k)}}^{\frac{1}{k(\widetilde{k}_2+k)}}=\sum_{\ell=N+k-1}^{\widetilde{k}_2+k}\int_{\frac{1}{k(\ell+1)}} ^{\frac{1}{k\ell}},\qquad\qquad \int_{-\frac{1}{k(\widetilde{k}_1+k)}}^{-\frac{1}{k(N+k)}}=\sum_{\ell=k+\widetilde{k}_1}^{N+k-1}\int_{-\frac{1}{k\ell}}^{-\frac{1}{k(\ell+1)}}, \end{align*} $$

and repeating the steps of the estimation of $\widehat {S}_{12}$ , we derive that both $\widehat {S}_{22}$ and $\widehat {S}_{23}$ are of order $O(n^\varepsilon )$ . In conclusion, we have

(3.15) $$ \begin{align} \Sigma_1=O(n^\varepsilon). \end{align} $$

For the bounds of $\Sigma _1$ , using (3.13) and $\vartheta _{h,k}'+\vartheta" _{h,k}\leq \frac 2{k\sqrt {n}}$ , we have

$$ \begin{align*} |\Sigma_1|&\leq 2\bigg|\sin\left(\frac{\pi a}{c}\right)\bigg|e^{2\pi}\sum_{k\leq N\atop c|k}k^{-\frac 32}\sum_{h=1\atop (h,k)=1}^{k}\frac 1 {\left|\sin\left(\frac{\pi h}{c}\right)\right|} \max_z \left|\overline{C}\left(e^{\frac{2\pi ih}{c}};q_1\right)\right|. \end{align*} $$

Here we use a change of variables $ah'\rightarrow h$ . Applying (3.9) and $\mathrm {Re}\left (\frac 1z\right )>\frac k2$ , we have

$$ \begin{align*} \left|\overline{C}\left(e^{\frac{2\pi ih}{c}};q_1\right)\right| &\leq 1+\sum_{r=1}^\infty\overline{p}(r)e^{-\pi r}. \end{align*} $$

Recall the estimate in [Reference Ciolan10, Section 4.2]:

(3.16) $$ \begin{align} \sum_{\begin{subarray}{c} h=1\\ (h,k)=1 \end{subarray}}^k \frac 1{\left|\sin\left(\frac{\pi h}{c}\right)\right|} \leq \frac {2k\left(1+\log\left(\frac {c-1}2\right)\right)}{\pi(1-\pi^2/24)}. \end{align} $$

Using these inequalities above and setting $\epsilon _1=\sum _{r=1}^\infty \overline {p}(r)e^{-\pi r}$ , we can derive that

$$ \begin{align*} |\Sigma_1|&\leq 2\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}(1+\epsilon_1)\sum_{k\leq N\atop c|k}k^{-\frac 32}\sum_{h=1\atop (h,k)=1}^{k}\frac 1 {\left|\sin\left(\frac{\pi h}{c}\right)\right|} \notag\\[6pt] &\leq \frac{4\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}(1+\epsilon_1)\left(1+\log\left(\frac{c-1}2\right)\right)}{\pi \left(1-\pi^2/24\right)}\sum_{k\leq N\atop c|k}k^{-\frac 12} \notag\\[6pt] &\leq \frac{8\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}(1+\epsilon_1)\left(1+\log\left(\frac{c-1}2\right)\right)n^{\frac 14}}{c\pi \left(1-\pi^2/24\right)}. \end{align*} $$

This completes the proof.

Proof Note that

(3.18) $$ \begin{align} \overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right) &=\frac{i}{2}\frac{(-q_1;q_1)_\infty}{(q_1;q_1)_\infty} \nonumber \\ &\quad \times \left(\sum_{m=0}^\infty\frac{(-1)^me^{-\frac{\pi iah'}{c}}q_1^{{m+1\choose2}+\frac{l}{2c_1}}} {1-e^{-\frac{2\pi iah'}{c}q_1^{m+\frac{l}{c_1}}}}\right. \left.-\sum_{m=1}^\infty\frac{(-1)^me^{\frac{\pi iah'}{c}}q_1^{{m+1\choose2}-\frac{l}{2c_1}}} {1-e^{\frac{2\pi iah'}{c}q_1^{m-\frac{l}{c_1}}}}\right) \notag \\[5pt] &=\frac{i}{2}\frac{(-q_1;q_1)_\infty}{(q_1;q_1)_\infty} \left(\sum_{m=0}^\infty(-1)^m\sum_{r=0}^\infty e^{-\frac{\pi iah'}{c}-\frac{2\pi irah'}{c}} q_1^{{m+1\choose2}+\frac{l}{2c_1}+rm+\frac{rl}{c_1}}\right. \notag \\[5pt] &\quad -\left.\sum_{m=1}^\infty(-1)^m\sum_{r=0}^\infty e^{\frac{\pi iah'}{c}+\frac{2\pi irah'}{c}} q_1^{{m+1\choose2}-\frac{l}{2c_1}+rm-\frac{rl}{c_1}}\right). \end{align} $$

Then the product in $\Sigma _4$ can be written as

(3.19) $$ \begin{align} &e^{\frac{\pi ih'}{8k}+\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}} q_1^{-\frac{l^2}{2c_1^2}}\frac{\eta\left(\frac{h'+iz^{-1}}{2k}\right)} {\eta\left(\frac{2(h'+iz^{-1})}{k}\right)}\overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right) \nonumber\\[5pt] &\quad =\sum_{r=r^{\prime}_0}^\infty\sum_{s\pmod{c}}\overline{d}(r,s)e^{\frac{2\pi im^{\prime}_{r,s}h'}{k}}q_1^r. \end{align} $$

We aim to demonstrate that $m^{\prime }_{r,s}\in \mathbb {Z}$ and $r^{\prime }_0$ may be negative. The segments involving negative r contribute to the component. Considering $\tau _2=\frac {h'+iz^{-1}}{k}$ , $q_1=e^{2\pi i\tau _2}$ , we further elaborate (3.19) by employing

(3.20) $$ \begin{align} \frac{\eta\left(\frac{\tau_2}{2}\right)}{\eta(2\tau_2)} \frac{(-q_1;q_1)_\infty}{(q_1;q_1)_\infty} =q_1^{-\frac{1}{16}}\frac{(q_1^{\frac{1}{2}};q_1^{\frac{1}{2}})_\infty}{(q_1;q_1)^2_\infty} =q_1^{-\frac{1}{16}}\left(1+O\left(q_1^{\frac{1}{2}}\right)\right) \end{align} $$

inside of $\eta \left (\frac {h'+iz^{-1}}{2k}\right )\eta ^{-1}\left (\frac {2(h'+iz^{-1})}{k}\right ) \overline {C}\left (ah',\frac {lc}{c_1},c;q_1\right )$ . So, the main contribution of

$$ \begin{align*} e^{\frac{\pi ih'}{8k}+\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}} q_1^{-\frac{l^2}{2c_1^2}}\frac{\eta\left(\frac{h'+i(z)^{-1}}{2k}\right)} {\eta\left(\frac{2(h'+iz^{-1})}{k}\right)}\overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right) \end{align*} $$

comes from the following expression:

(3.21) $$ \begin{align} \pm\frac{i}{2}e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}+\frac{\pi ih'}{8k}} q_1^{-\frac{l^2}{2c_1^2}-\frac{1}{16}} (-1)^mq_1^{\frac{m}{2}(m+1)\pm\frac{l}{2c_1}+rm\pm\frac{rl}{c_1}} e^{\mp\frac{\pi iah'}{c}\mp\frac{2\pi iah'r}{c}}. \end{align} $$

From this is possible to split the expression into the roots of unity and to the part that depends on the variable z. The roots of unity look like:

$$\begin{align*}\exp\left(\frac{2\pi ih'}{k}\left(-\frac{a^2k_1k}{2cc_1}+\frac{lak}{cc_1} -\frac{l^2}{2c_1^2}+rm\pm\frac{rl}{c_1}\mp\frac{kra}{c} +\frac{m(m+1)}{2}\pm\frac{l}{2c_1}\mp\frac{ak}{2c}\right)\right).\end{align*}$$

Rewriting the expression in the second bracket, using the congruence condition $l\equiv ak_1\pmod {c_1}$ , $l^2+l$ is always even and rearranging the sum it is possible to show that the contribution of the roots of unity looks like $\exp \left (\frac {2\pi ih'm^{\prime }_{r,s}}{k}\right )$ where $m^{\prime }_{r,s}$ is a sequence in $\mathbb {Z}$ . The interesting part happens for $\exp \left (\frac {\pi }{kz}H_2\right )$ , where $H_2$ is defined in the following way:

$$\begin{align*}H_2=\frac{l^2}{c_1^2}+\frac{1}{8}-2rm\mp2r\frac{l}{c_1}-m(m+1)\mp\frac{l}{c_1}.\end{align*}$$

This part contributes to the circle method exactly if $H_2>0$ . A straightforward, but rather tedious computation shows that there are two contributions coming from each of the two terms of

$$ \begin{align*}\eta\left(\frac{h'+iz^{-1}}{2k}\right)\eta^{-1}\left(\frac{2(h'+iz^{-1})}{k}\right) \overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right).\end{align*} $$

The first one comes from the first sum, if $m=0$ , and this contributes with

$$ \begin{align*} \frac{i}{2}e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}-\frac{\pi iah'}{c}+\frac{\pi ih'}{8k}} q_1^{-\frac{l^2}{2c_1^2}-\frac{1}{16}+\frac{l}{2c_1}} \sum_{r\geq0\atop\delta^+_{a,c,k,r}>0}q_1^{\frac{rl}{c_1}} e^{-\frac{2\pi iah'r}{c}}, \end{align*} $$

where $\delta ^{+}_{a,c,k,r}=\frac {l^2}{2c_1^2}+\frac {1}{16}-\left (r+\frac {1}{2}\right )\frac {l}{c_1}$ . The second contribution comes from the second sum, if $m=1$ , and this contributes with

$$ \begin{align*} \frac{i}{2}e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}+\frac{\pi ih'}{8k}} q_1^{-\frac{l^2}{2c_1^2}-\frac{1}{16}} q_1^{1-\frac{l}{2c_1}+r-\frac{rl}{c_1}} e^{\frac{\pi iah'}{c}+\frac{2\pi iah'r}{c}}, \end{align*} $$

where $\delta ^{-}_{a,c,k,r}=\frac {l^2}{2c_1^2}-\frac {15}{16}-r\left (1-\frac {l}{c_1}\right )+\frac {l}{2c_1}$ . Thus, $\Sigma _4$ can be rewritten as

(3.22) $$ \begin{align} \Sigma_4&=\sqrt{2}\sin\left(\frac{\pi a}{c}\right)\sum_{c\nmid k,~2\nmid k\atop r\geq0,~i\in\{-,+\}} (-1)^{ak+l}\sum_{h}\omega_{h,k}^2\omega_{2h,k}^{-1} e^{\frac{2\pi i}{k}(-nh+m^i_{a,c,k,r}h')} \nonumber\\[5pt] &\quad \cdot\int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}}z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}+\frac{2\pi}{kz}\delta_{a,c,k,r}^i}\mathrm{d}\Phi. \end{align} $$

Now it is possible to rewrite the sum over k into the sum where the k’s have the same values for $c_1$ and l and thus the $\delta _{a,c,k,r}^i$ are constant in each class and the condition $\delta _{a,c,k,r}^i>0$ is independent of k in each class. Moreover it is clear that there are only finitely many solution to the inequality $\delta _{a,c,k,r}^i>0$ . This means it is possible to split the sum over r into positive and negative $\delta _{a,c,k,r}^i$ , where the part with positive $\delta _{a,c,k,r}^i$ contributes to the main term. Therefore, using (3.1) it is possible to evaluate $\Sigma _4$ by the same argument as for $\Sigma _1$ :

(3.23) $$ \begin{align} \Sigma_4&=\sqrt{2}\sin\left(\frac{\pi a}{c}\right)\sum_{r\geq0}\sum_{2\nmid k\atop c\nmid k}\sum_{\delta_{a,c,k,r}^i>0\atop i\in\{+,-\}}(-1)^{ak+l}\sum_{h\pmod{k}^*}\omega_{h,k}^2\omega_{2h,k}^{-1} e^{\frac{2\pi i}{k}\left(-nh+m_{a,c,k,r}^ih'\right)} \notag\\[5pt] &\quad\quad\quad \cdot\int_{-\frac{1}{kN}}^{\frac{1}{kN}} z^{-\frac{1}{2}}e^{\frac{2\pi zn}{k}+\frac{2\pi}{kz}\delta_{a,c,k,r}^i}\mathrm{d}\Phi+O(n^{\varepsilon}). \end{align} $$

Then substituting (3.4) into (3.23), we obtain (3.17).

Now we give a specific bound for the error coming from $\Sigma _4$ , which is composed of three parts. Firstly, we consider the error corresponding to the terms where $\delta _{a,c,k,r}^i$ is not positive, denoting it by $S_4^{err}$ . Define $\overline {M}(jh',l,c; q_1)$ to be the terms with positive exponents in the $q_1$ -expansion of

$$ \begin{align*} q_1^{-\frac{l^2}{2c_1^2}} \frac{\eta\left(\frac{\tau_2}2\right)}{\eta\left(2\tau_2\right)}\overline{C}\left(jh',\frac{lc}{c_1},c;q_1\right)\!, \end{align*} $$

where $\tau _2=\frac {h'+iz^{-1}}{k}$ . Then using the usual change of variable $ah'\rightarrow h$ , we find

$$ \begin{align*} |S_4^{err}|\leq 4\sqrt{2}\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}\sum_{h, k\atop c\nmid k}k^{-\frac 32} \max_z \left|\overline{M}(h,\frac{lc}{c_1},c; q_1)\right|. \end{align*} $$

By (3.18) and (3.20), we have

$$ \begin{align*} &\frac{\eta\left(\frac{\tau_2}2\right)}{\eta\left(2\tau_2\right)} \overline{C}\left(h,\frac{lc}{c_1},c;q_1\right) \\[5pt] &\quad=\frac{i}{2}q_1^{-\frac{1}{16}}\frac{(q_1^{\frac 12};q_1^{\frac 12})_\infty}{(q_1;q_1)_\infty^2} \left(\sum_{r\geq0} e^{-\frac{\pi ih}{c}-\frac{2\pi irh}{c}} q_1^{\frac{l}{2c_1}+\frac{rl}{c_1}}+\sum_{r\geq0}e^{\frac{\pi ih}{c}+\frac{2\pi irh}{c}} q_1^{-\frac{l}{2c_1}-\frac{rl}{c_1}+r+1}\right. \\[5pt] &\qquad +\left.\sum_{m\geq1}\frac{(-1)^m e^{-\frac{\pi ih}{c}}q_1^{\frac{m^2+m}{2}+\frac{l}{2c_1}}}{1-e^{-\frac{2\pi ih}{c}}q_1^{m+\frac l{c_1}}}-\sum_{m\geq2}\frac{(-1)^m e^{\frac{\pi ih}{c}}q_1^{\frac{m^2+m}{2}-\frac{l}{2c_1}}}{1-e^{\frac{2\pi ih}{c}}q_1^{m-\frac l{c_1}}}\right). \end{align*} $$

We bound the contributions of $\overline {M}$ term by term beginning with the first one. We find that

$$ \begin{align*} \left|\frac{(q_1^{1/2};q_1^{1/2})_\infty}{(q_1;q_1)_\infty^2}\right| =\left|\frac{(q_1^{1/2};q_1)_\infty}{(q_1;q_1)_\infty}\right|\leq \frac{2(-|q_1|;|q_1|)_\infty}{(|q_1|;|q_1|)_\infty} =\sum_{m\geq0}2\overline{p}(m)|q_1|^m. \end{align*} $$

Then we gain the following contribution to $\overline {M}$

$$ \begin{align*} & \left|q_1^{-\frac{1}{16}-\frac{l^2}{2c_1^2}+\frac{l}{2c_1}} \sum_{m=0}^{\infty}\overline{p}(m)|q_1|^m\sum_{r\geq r_0}q_1^{\frac{rl}{c_1}}\right| \\ & \quad\leq e^{\frac{\pi}{16}+\frac{\pi l^2}{2c_1^2}-\frac{\pi l}{2c_1}}\left( \sum_{r\geq r_0}e^{-\frac{\pi rl}{c_1}}+ \sum_{r\geq 0}e^{-\frac{\pi rl}{c_1}}\sum_{m\geq1}\overline{p}(m)e^{-\pi m}\right), \end{align*} $$

where $r_0:=\lceil \frac {l}{2c_1}+\frac {c_1}{16l}-\frac 12\rceil $ and $r\geq r_0$ is a solution of $\delta _{a,c,k,r}^+<0$ . Using

$$ \begin{align*} \sum_{r\geq r_0}e^{-\frac{\pi rl}{c_1}}=\frac{e^{-\frac{\pi r_0l}{c_1}}}{1-e^{-\frac{\pi l}{c_1}}}, \end{align*} $$

and the usual geometric series we can bound the term further by

$$ \begin{align*} \frac {e^{\frac{\pi}{16}+\frac{\pi l^2}{2c_1^2}-\frac{\pi l}{2c_1}-\frac{\pi r_0l}{c_1}}}{\left(1-e^{-\frac{\pi l}{c_1}}\right)}+\frac {e^{\frac{\pi}{16}+\frac{\pi l^2}{2c_1^2}-\frac{\pi l}{2c_1}}\epsilon_1}{\left(1-e^{-\frac{\pi l}{c_1}}\right)} =\frac{e^{\frac{\pi l}{c_1}\left(\frac{c_1}{16l}+\frac{l}{2c_1}-\frac 12-r_0\right)}\left(1+\epsilon_1e^{\frac{\pi r_0l}{c_1}}\right)}{\left(1-e^{-\frac{\pi l}{c_1}}\right)} \leq \frac {1+\epsilon_1e^{\pi\delta_0}}{\left(1-e^{-\frac{\pi }{c}}\right)}. \end{align*} $$

For $0<l<c_1$ , one can easily verify that $\delta _{j,c,k,r}^{i}<\delta _{j,c,k,0}^{+}=\frac 12(\frac l{c_1}-\frac 12)^2-\frac 1{16}$ . Assuming $\frac {l}{c_1}<\frac {1}{2}$ , which we may do by the symmetry of the parabola in the argument $\frac l{c_1}$ , we see that $\delta _{j,c,k,0}^{+}$ arrives the maximum value $\delta =\frac 1{2{c_1}^2}-\frac 1{2c_1}+\frac 1{16}$ at $l=1$ . Since $\delta \leq \delta _0$ , we have $\delta _{j,c,k,r}^{i}\leq \delta _0$ . Therefore, we can bound the second term by the same way. In the third and fourth summand all the terms will contribute to the error as shown in Theorem 1.1. We obtain

$$ \begin{align*} & \left|\frac{i}{2}q_1^{-\frac{1}{16}-\frac{l^2}{2c_1^2}+\frac{l}{2c_1}}\frac{(q_1^{\frac 12};q_1^{\frac 12})_\infty}{(q_1;q_1)_\infty^2}\sum_{m=1}^\infty\frac{(-1)^m e^{-\frac{\pi ih}{c}}q_1^{\frac{m^2+m}{2}}}{1-e^{-\frac{2\pi ih}{c}}q_1^{m+\frac l{c_1}}}\right| \\ &\quad \leq e^{\frac{\pi}{16}+\frac{\pi l^2}{2c_1^2}-\frac{\pi l}{2c_1}}(1+\epsilon_1)\sum_{m=1}^\infty \frac{e^{-\frac{\pi m(m+1)}{2}}}{1-e^{-\pi m-\frac{\pi l}{c_1}}}\\ &\quad \leq e^{\pi \delta_0}(1+\epsilon_1)\epsilon_2. \end{align*} $$

Proceeding by repeating this step for the fourth sum in the expansion of $\frac {\eta \left (\frac {\tau _2}2\right )}{\eta \left (2\tau _2\right )}\overline {C}\left (h,\frac {lc}{c_1},c;q_1\right )$ , we can get a bound for $\overline {M}$

$$ \begin{align*} e^{\pi \delta_0+\pi}(1+\epsilon_1)\epsilon_3. \end{align*} $$

Based on the above estimates, we find that $\overline {M}$ can be bounded by the function $f(c)$ that just depends on c. Therefore the error can be bounded by

(3.24) $$ \begin{align} |S_4^{err}|\leq 4\sqrt{2}e^{2\pi}\sum_{h, k\atop c\nmid k}k^{-\frac 32}f(c)\leq 8\sqrt{2}e^{2\pi}n^{\frac 14}f(c). \end{align} $$

Here it used that the sum over h can be bounded trivially by k because the sum runs over the residue class modulo k and we estimated the sum over k by an integral expression.

As a next step we want to bound the second part error that comes from symmetrizing the integral, which is the associated sums of the last two integrals in (3.12), that is

$$ \begin{align*} \tilde{S}_{4}^{err}&:=-\sqrt{2}\sin\left(\frac{\pi a}{c}\right)\sum_{c\nmid k,~2\nmid k\atop r\geq0,~i\in\{-,+\}} (-1)^{ak+l}\sum_{h}\omega_{h,k}^2\omega_{2h,k}^{-1} e^{\frac{2\pi i}{k}(-nh+m^i_{a,c,k,r}h')} \nonumber\\[5pt] &\quad\quad\quad \cdot\left(\int_{-\frac{1}{kN}}^{-\frac{1}{k(\widetilde{k}_1+k)}}+\int_{\frac{1}{k(\widetilde{k}_2+k)}}^{\frac{1}{kN}}\right) z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}+\frac{2\pi}{kz}\delta_{a,c,k,r}^i}d\Phi. \end{align*} $$

Completely analogous to $\Sigma _1$ , it is possible to show

$$ \begin{align*} |\tilde{S}_{4}^{err}|&\leq4\sqrt{2}e^{2\pi}\sum_{\begin{subarray}{c} r, k\\ \delta_{a,c,k,r}^i>0\\ ~i\in\{+,-\} \end{subarray}} k^{-\frac 12}e^{2\pi\delta_{a,c,k,r}^i} \end{align*} $$

As we can bound the term with $i=-$ and $i=+$ in a same way, we restrict to the case $i=+$ . The sum over r gives

(3.25) $$ \begin{align} |\tilde{S}_{4}^{err}|&\leq 8\sqrt{2}e^{2\pi}\sum_{r, k\atop\delta_{j,c,k,r}^+>0}k^{-\frac 12}e^{2\pi\delta_{j,c,k,r}^+} \notag\\ \quad &=8\sqrt{2}e^{2\pi}\sum_k k^{-\frac 12}\sum_{r\leq r_0-1}e^{\frac {\pi l^2}{c_1^2}+\frac {\pi}{8}-\frac {2\pi rl}{c_1}-\frac {\pi l}{c_1}} \notag\\ \quad &=8\sqrt{2}e^{2\pi}\sum_k k^{-\frac 12}\frac{e^{\frac {\pi l^2}{c_1^2}+\frac {\pi}{8}-\frac {\pi l}{c_1}} \left(e^{-\frac{2\pi l r_0}{c_1}}-1\right)}{e^{-\frac{2\pi l}{c_1}}-1} \notag\\ \quad &\leq8\sqrt{2}e^{2\pi}\sum_k k^{-\frac 12}\frac {e^{2\pi \delta_0}}{1-e^{-\frac{2\pi}c}} \notag\\ \quad &\leq16\sqrt{2}e^{2\pi}\frac {e^{2\pi \delta_0}}{1-e^{-\frac{2\pi}c}}n^{\frac 14}. \end{align} $$

Here we have used that the summation over r is an error term if it starts with $r_0$ , so here we have to sum over all the r-terms where $r\leq r_0-1$ .

The last error we have to bound comes from the evaluation of the integral. There it is used that it is possible to change the path of integration if one accounts the integral over the smaller arc. This term has to be made explicit. Applying (3.5), and estimation of the sum over k, we bound the third part error by

(3.26) $$ \begin{align} 4\sqrt{2}\left(\frac 43+2^{\frac 54}\right)\frac {e^{2\pi \delta_0+2\pi}}{1-e^{-\frac{2\pi}c}}n^{\frac 38}. \end{align} $$

Summing up (3.24), (3.25), and (3.26), we obtain the bound of the error coming from $\Sigma _4$ . This completes the proof.

Proof Using (3.9), we obtain

$$ \begin{align*} \frac{\eta\left(\frac{\tau_2}{2}\right)} {\eta(2\tau_2)}\overline{C}\left(e^{\frac{2\pi iah'}{c}};q_1\right) &=\frac{q_1^{\frac{1}{48}}(q_1^{\frac{1}{2}};q_1^{\frac{1}{2}})_\infty} {q_1^{\frac{1}{12}}(q_1^2;q_1^2)_\infty} \left(1+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}\overline{a}(r,s)e^{\frac{2\pi ih'}{c}m_{r,s}}q_1^r\right) \\[5pt] &=q_1^{-\frac{1}{16}}(1+\widehat{T}_1), \end{align*} $$

where $\tau _2=\frac {h'+iz^{-1}}{k}$ , $q_1=e^{2\pi i\tau _2}$ and

(3.29) $$ \begin{align} \widehat{T}_1=\frac{(q_1^{\frac{1}{2}};q_1^{\frac{1}{2}})_\infty}{(q_1^2;q_1^2)_\infty} \left(1+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}\overline{a}(r,s)e^{\frac{2\pi ih'}{c}m_{r,s}}q_1^r\right)-1. \end{align} $$

Setting $\frac {(q_1^{1/2};q_1^{1/2})_\infty }{(q_1^2;q_1^2)_\infty }:=1+\sum _{n=1}^\infty b(n)q_1^{n/2}$ in (3.29), we obtain that

(3.30) $$ \begin{align} \widehat{T_1}&=\left(1+\sum_{n=1}^\infty b(n)q_1^{n/2}\right)\left(1+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}^\infty \overline{a}(r,s)e^{\frac{2\pi ih'}{c}m(r,s)}q_1^r\right)-1\nonumber \notag\\[5pt] &=\sum_{n=1}^\infty b(n)q_1^{n/2}+\sum_{r=1}^\infty\sum_{s~(\mathrm{mod}~{c})}^\infty \overline{a}(r,s)e^{\frac{2\pi ih'}{c}m(r,s)}q_1^r\nonumber \notag\\[5pt] &\mbox{}\quad+\left(\sum_{n=1}^\infty b(n)q_1^{n/2}\right)\!\cdot\left(\sum_{r=1}^\infty\sum_{s\pmod{c}}^\infty \overline{a}(r,s)e^{\frac{2\pi ih'}{c}m(r,s)}q_1^r\right)\!. \end{align} $$

Consequently, $\Sigma _2$ can be expressed as follows:

$$ \begin{align*} \Sigma_2=\widetilde{S}_1+\widetilde{S}_2, \end{align*} $$

where

(3.31) $$ \begin{align} \widetilde{S}_1&=\frac{\sqrt{2}}{2}i\sin\left(\frac{\pi a}{c}\right)\sum_{h}\sum_{c|k\atop 2\nmid k} \frac{(-1)^{ak+1}}{\sin\left(\frac{\pi ah'}{c}\right)} \omega_{h,k}^2\omega_{2h,k}^{-1}e^{-\frac{2\pi ihn}{k}+\frac{\pi ih'}{8k}-\frac{\pi ia^2k_1h'}{c}} \nonumber\\[5pt] &\quad\quad\quad \cdot\int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}} z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}q_1^{-\frac{1}{16}}\mathrm{d}\Phi, \nonumber\\[5pt] \widetilde{S}_2&=\frac{\sqrt{2}}{2}i\sin\left(\frac{\pi a}{c}\right)\sum_{h}\sum_{c|k\atop 2\nmid k} \frac{(-1)^{ak+1}}{\sin\left(\frac{\pi ah'}{c}\right)} \omega_{h,k}^2\omega_{2h,k}^{-1}e^{-\frac{2\pi ihn}{k}+\frac{\pi ih'}{8k}-\frac{\pi ia^2k_1h'}{c}} \nonumber\\[5pt] &\quad\quad\quad \cdot\int_{-\vartheta^{\prime}_{h,k}}^{\vartheta^{\prime\prime}_{h,k}} z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}}q_1^{-\frac{1}{16}}\widehat{T}_1\mathrm{d}\Phi. \end{align} $$

Using (3.2), it is possible to demonstrate that $\widetilde {S}_1$ contributes to the main term and $\widetilde {S}_2=O(n^\varepsilon )$ , that is

$$ \begin{align*} \Sigma_2&=\frac{\sqrt{2}}{2}i\sin\left(\frac{\pi a}{c}\right)\sum_{h}\sum_{c|k\atop 2\nmid k} \frac{(-1)^{ak+1}}{\sin\left(\frac{\pi ah'}{c}\right)} \omega_{h,k}^2\omega_{2h,k}^{-1}e^{-\frac{2\pi ihn}{k}-\frac{\pi ia^2k_1h'}{c}} \nonumber\\[5pt] &\quad\quad\quad \cdot \int_{-\frac{1}{kN}}^{\frac{1}{kN}} z^{-\frac{1}{2}}e^{\frac{2\pi nz}{k}+\frac{\pi}{8kz}}\mathrm{d}\Phi +O(n^\varepsilon). \end{align*} $$

Then by (3.4), we obtain (3.27).

We next estimate the error coming from $\widetilde {S}_1$ , which consists of two parts: the error arises from integrating over the remaining parts of the interval and the error coming from integrating along a smaller arc. Using (3.5), (3.16) and $k\leq \sqrt {n}$ , we can bound the error coming from integrating along a smaller arc:

(3.32) $$ \begin{align} \widetilde{S}_{11}^{err}\leq\frac{2\sqrt{2}e^{2\pi+\frac{\pi}{8}}\left(1+\log\left(\frac{c-1}2\right)\right) \left(\frac 43+2^{\frac 54}\right)n^{\frac 38}}{\pi(1-\pi^2/24)}. \end{align} $$

Utilizing (3.16) again, the error arises from integrating over the remaining parts of the interval can be bounded by

(3.33) $$ \begin{align} \widetilde{S}_{1err}\leq\frac{8\sqrt{2}e^{2\pi+\frac{\pi}{8}}\left(1+\log\left(\frac{c-1}2\right)\right)n^{\frac 14}}{c\pi(1-\pi^2/24)}. \end{align} $$

We now give a bound for $\widetilde {S}_2$ . Applying (3.31), we deduce that

$$ \begin{align*} \left|\widetilde{S}_2\right|\leq \sqrt{2}\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}\sum_{1\leq k\leq \sqrt{n}\atop c|k}k^{-\frac 32} \sum_{h=1\atop(h,k)=1}^{k} \frac{1}{\left|\sin\left(\frac{\pi h}{c}\right)\right|} \max_z \left|e^{\frac{\pi}{8kz}}\widehat{T}_1\right|\!. \end{align*} $$

Here we also use the change of variables $ah'\rightarrow h$ . Utilizing the triangle inequality and $|q_1|<1$ , we get that

$$ \begin{align*} \left|\frac{(q_1^{1/2};q_1^{1/2})_\infty}{(q_1^2;q_1^2)_\infty}\right|=\left|\frac{(q_1^{1/2};q_1)_\infty}{(-q_1;q_1)_\infty}\right|\leq \frac{2(-|q_1|;|q_1|)_\infty}{(|q_1|;|q_1|)_\infty}. \end{align*} $$

Combining this with the identity

$$\begin{align*}\frac{(-q_1;q_1)_\infty}{(q_1;q_1)_\infty}=\sum_{r=0}^{\infty} \overline{p}(r)q_1^r\end{align*}$$

and (3.30), we can infer that

$$ \begin{align*} \left|\widehat{T_1}\right| \leq 3\sum_{r=1}^\infty\overline{p}(r)e^{-\pi r}+2\left(\sum_{r=1}^\infty\overline{p}(r)e^{-\pi r}\right)^2=3\epsilon_1+2\epsilon_1^2. \end{align*} $$

Applying (3.16), we have

(3.34) $$ \begin{align} \left|\widetilde{S}_2\right|\leq \frac{4\sqrt{2}\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi+\frac{\pi}8}(3\epsilon_1+2\epsilon_1^2)\left(1+\log\left(\frac{c-1}2\right)\right)n^{\frac 14}}{c\pi \left(1-\pi^2/24\right)}. \end{align} $$

Summing up (3.32), (3.33), and (3.34) gives (3.28). This completes the proof.

Proof By (3.18), it is easy to see that

(3.35) $$ \begin{align} &e^{\frac{2\pi ialh'}{cc_1}-\frac{\pi ia^2k_1h'}{cc_1}} q_1^{-\frac{l^2}{2c_1^2}}\overline{C}\left(ah',\frac{lc}{c_1},c;q_1\right) =\sum_{r=r_0}^\infty\sum_{s\pmod{c}}\overline{b}(r,s)e^{\frac{2\pi im_{r,s}h'}{k}}q_1^r, \end{align} $$

where $m_{r,s}\in \mathbb {Z}$ . By the same argument as for $\Sigma _4$ , we can find that the there is no part contributing to the main term, resulting in

$$ \begin{align*} \Sigma_3=O(n^\varepsilon). \end{align*} $$

Now we consider the bound of $\Sigma _3$ . Similar to the proof for $\Sigma _1$ , we have

$$ \begin{align*} |\Sigma_3|\leq 8\left|\sin\left(\frac{\pi a}{c}\right)\right|e^{2\pi}\sum_{h, k\atop c\nmid k}k^{-\frac 32}\max_z \left|q_1^{-\frac {l^2}{2c_1^2}}\overline{C}\left(h,\frac{lc}{c_1},c;q_1\right)\right|. \end{align*} $$

Combining (3.18), we have

$$ \begin{align*} \left|\overline{C}\left(h,\frac{lc}{c_1},c;q_1\right)\right| &\leq\frac{1}{2}\sum_{n=0}^\infty\overline{p}(n)e^{-\pi n} \left(\sum_{m=0}^\infty e^{-\pi m(m+1)/2-\frac{\pi l}{2c_1}}\left|\frac{e^{-\frac{\pi ih}{c}}} {1-e^{-\frac{2\pi ih}{c}q_1^{m+\frac{l}{c_1}}}}\right| \right. \\[5pt] &\quad \left.+\sum_{m=1}^\infty e^{-\pi m(m+1)/2+\frac{\pi l}{2c_1}}\left|\frac{e^{\frac{\pi ih}{c}}} {1-e^{\frac{2\pi ih}{c}q_1^{m-\frac{l}{c_1}}}}\right|\right). \end{align*} $$

Since

$$ \begin{align*} &\left|\frac{e^{-\frac{\pi ih}{c}}}{1-e^{-\frac{2\pi ih}{c}q_1^{m+\frac{l}{c_1}}}}\right|\leq \frac {1}{1-e^{-\pi m-\frac{\pi }{c}}}, \qquad\left|\frac{e^{\frac{\pi ih}{c}}}{1-e^{\frac{2\pi ih}{c}q_1^{m-\frac{l}{c_1}}}}\right|\leq \frac {1}{1-e^{-\pi m+\pi\left(1-\frac{1}{c}\right)}}, \end{align*} $$

we have

$$ \begin{align*} |\Sigma_3| \leq 4 e^{\frac{5}{2}\pi}(1+\epsilon_1)(\epsilon_2+e^{\frac{\pi}{2}}\epsilon_3)\sum_{1\leq k\leq \sqrt{n}\atop c\nmid k}k^{-\frac 12} \leq 8 e^{\frac{5}{2}\pi}(1+\epsilon_1)(\epsilon_2+e^{\frac{\pi}{2}}\epsilon_3)n^{\frac 14}. \end{align*} $$

This completes the proof.

Proof of Theorem 1.1

In light of Lemmas 3.3–3.7, we derive the asymptotic formula for $\overline {A}\left (\frac {a}{c};n\right )$ , including an initial estimation of the error term. Given that this initial error bound is overly broad and lacks precision, we proceed to provide a more detailed and specific error bound.

We first give the estimates of constants $\epsilon _1$ , $\epsilon _2$ and $\epsilon _3$ in (3.8). Using the fact that for $n\geq 1$ ,

$$ \begin{align*} \overline p(n)<e^{\pi\sqrt{n}}, \end{align*} $$

we have

$$ \begin{align*} \epsilon_1\leq\int_1^\infty e^{\pi\sqrt{r}}\cdot e^{-\pi r}dr<0.5488. \end{align*} $$

It can be easily verify by derivation that,

(3.36) $$ \begin{align} \frac{1}{1-e^{-\frac{k\pi}{c}}}\leq k\pi c,~~k=1,2. \end{align} $$

Hence, we also can compute that

$$ \begin{align*}\epsilon_2<\frac{1}{1-e^{-\frac{\pi}{c}}}+\int_1^\infty\frac{e^{-\pi r(r+1)/2}}{1-e^{-\pi r}}\mathrm{d}r<\pi c+0.0085.\end{align*} $$

Similarly, $\epsilon _3<\pi c+0.0001$ . Moveover, we have the following inequality [Reference Zhang and Zhong20, Eq. (3.13)]

(3.37) $$ \begin{align} \frac{1+\log(\frac{c-1}{2})}{\pi\left(1-\frac{\pi^2}{24}\right)}<0.2704c. \end{align} $$

In view of Lemma 3.4 and (3.37), we have

(3.38) $$ \begin{align} |\Sigma_1| &\leq 8e^{2\pi}(1+\epsilon_1)\cdot 0.2704n^{\frac 14}<1794.0921n^{\frac 14}. \end{align} $$

Similarly, by Lemma 3.7, we have

(3.39) $$ \begin{align} |\Sigma_3| &\leq (582623.3677c+286.6509)n^{\frac 14}. \end{align} $$

By Lemma 3.6 and (3.37), we obtain the bound of the error arising from $\Sigma _2$ :

(3.40) $$ \begin{align} 5153.9942n^{\frac 14}+2251.2819cn^{\frac 38}. \end{align} $$

Recall that

$$\begin{align*}\delta_0=\frac 1{2c^2}-\frac 1{2c}+\frac 1{16}<\frac 1{16}.\end{align*}$$

Hence, in view of Lemma 3.5 and (3.36), we can bound the error arising from $\Sigma _4$ by

(3.41) $$ \begin{align} (123.4854+1042254.902c)n^{\frac 14}+104624.4148cn^{\frac 38}. \end{align} $$

Summing up (3.38), (3.39), (3.40), and (3.41) yields

$$ \begin{align*} \overline{R}_c(n) &\leq(1624878.2697c+7358.2224)n^{\frac 14}+106875.6967cn^{\frac 38} \\ &\leq(1700009.4994c + 7358.2224) n^{\frac 38}. \end{align*} $$

This completes the proof.

Proof of Corollary 1.2

By the orthogonality of roots of unity, we obtain

(4.1) $$ \begin{align} \sum_{n=0}^\infty\overline{M}(a,c,n)q^n=\frac{1}{c}\sum_{n=0}^\infty \overline{p}(n)q^n +\frac{1}{c}\sum_{j=1}^{c-1}\zeta_c^{-aj}\overline{C}\left(\zeta_c^j;q\right)\!. \end{align} $$

Recall the following asymptotic formula for $\overline {p}(n)$ in [Reference Zuckerman21]:

(4.2) $$ \begin{align} \overline{p}(n)=\frac{1}{2\pi}\sum_{k=1\atop 2\nmid k}^\infty\sqrt{k}\sum_{h=0\atop (h,k)=1}^k \frac{\omega_{h,k}^2}{\omega_{2h,k}}e^{-\frac{2\pi inh}{k}}\frac{\mathrm{d}}{\mathrm{d}n} \left(\frac{\sinh\frac{\pi\sqrt{n}}{k}}{\sqrt{n}}\right). \end{align} $$

By substituting Theorem 1.1 and (4.2) into (4.1), the proof is concluded.

Proof With the fact that $|\exp (\pi i x)|=1$ for $x\in \mathbb {R}$ , letting $jh'\rightarrow h$ in the summation of sine, we have

$$ \begin{align*} \left|\sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c'|k,~2\nmid k \end{subarray}} \frac{\overline{B}_{j',c',k}(-n,0)}{\sqrt{k}}\sinh\left(\frac{\pi\sqrt{n}}{k}\right)\right| \leq \sinh\left(\frac{\pi\sqrt{n}}{3}\right) \sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c|k,~2\nmid k \end{subarray}} \frac 1{\sqrt{k}} \sum_{\begin{subarray}{c} h=1\\ (h,k)=1 \end{subarray}}^k \frac 1{\left|\sin\left(\frac{\pi h}{c}\right)\right|}. \end{align*} $$

Using the estimate (3.16), we obtain that

(4.5) $$ \begin{align} \left|\sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c'|k,~2\nmid k \end{subarray}} \frac{\overline{B}_{j,c,k}(-n,0)}{\sqrt{k}}\sinh\left(\frac{\pi\sqrt{n}}{k}\right)\right| &\leq \sinh\left(\frac{\pi\sqrt{n}}{3}\right) \frac {2\left(1+\log\left(\frac {c-1}2\right)\right)}{\pi(1-\pi^2/24)} \sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c|k,~2\nmid k \end{subarray}} k^{\frac 12} \notag\\[5pt] &\leq \frac {4n^{3/4}\left(1+\log\left(\frac {c-1}2\right)\right)}{3c\pi(1-\pi^2/24)} \sinh\left(\frac {\pi \sqrt{n}}3\right). \end{align} $$

Here it is used that following estimation of the sum:

$$ \begin{align*} \sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c|k,~2\nmid k \end{subarray}} k^{\frac 12} \leq c^{\frac 12}\sum_{1\leq j\leq \lfloor\frac Nc\rfloor}j^{\frac 12}\leq c^{\frac 12}\int_1^{\lfloor\frac Nc\rfloor} x^{\frac 12}dx\leq \frac 2{3c}n^{\frac 34}. \end{align*} $$

By (4.5) and (3.37), we have

$$ \begin{align*} |\overline G_1| \leq\frac{2n^{\frac 14}\left(1+\log(\frac{c-1}{2})\right)}{3\pi c\left(1-\frac{\pi^2}{24}\right)}\cdot e^{\frac{\pi\sqrt{n}}{c}} \leq 0.1803 n^{\frac 14}\cdot e^{\frac{\pi\sqrt{n}}{3}}. \end{align*} $$

We next turn to $\overline G_2$ . Recall that $\delta _{j,c,k,r}^i\leq \delta _0$ . Combining $|\overline {D}_{j,c,k}(-n,m_{j,c,k,r}^i)|\leq k$ , we have

(4.6) $$ \begin{align} |\overline G_2|\leq n^{-\frac 12} e^{4\pi\sqrt{n\delta_0}}\sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c'\nmid k,~2\nmid k\\ \delta_{j',c',k,r}^i>0\\ r\geq0,~i\in\{+,-\} \end{subarray}} \sqrt{k}. \end{align} $$

Noting that the number of r satisfying $\delta _{j,c,k,r}^i>0$ is less than $\frac {c}{16}$ , we have

(4.7) $$ \begin{align} \sum_{\substack{ r\geq 0,~i\in\{+,-\}\\ \delta_{j',c',k,r}^i>0 }}1\leq \frac c{8}. \end{align} $$

Moreover, one can see that

(4.8) $$ \begin{align} \sum_{1\leq k\leq \sqrt{n} \atop c'\nmid k,~2\nmid k}\sqrt{k}\leq \int_1^{\sqrt{n}}x^{\frac 12}dx\leq \frac 23 n^{\frac 34}. \end{align} $$

Substituting (4.7) and (4.8) into (4.6), we can obtain (4.4). This completes the proof.

Proof of Theorem 1.3

From [Reference Zhang and Zhong20, Theorem 6], we get the following lower bound for $\overline p(n)$ , that is, for $n\geq 1$ ,

(4.9) $$ \begin{align} \overline p(n)> \frac{1}{8n}\left(1-\frac 1{\sqrt{n}}\right)e^{\pi\sqrt{n}}. \end{align} $$

Combining with (1.2), (4.3), (4.9), and Lemma 4.1, for odd c and positive integer n, we have

$$ \begin{align*} \left|\overline R(a,c,n)\right| &\leq\left(0.1803 n^{\frac 14}e^{\frac{\pi\sqrt{n}}{3}}+\frac{c}{12} n^{\frac 14} e^{4\pi\sqrt{n\delta_0}} +(1700009.4994c + 7358.2224) n^{\frac 38}\right) \\ &\quad \cdot\frac{8n e^{-\pi\sqrt{n}}}{1-\frac{1}{\sqrt{n}}}. \end{align*} $$

Since for $n\geq 2$ ,

$$\begin{align*}\frac{1}{1-\frac{1}{\sqrt{n}}}\leq\frac{1}{1-\frac{1}{\sqrt{2}}}\leq3.415,\end{align*}$$

which yields that

$$ \begin{align*} \left|\overline R(a,c,n)\right| &\leq 4.93 n^{\frac 54} e^{-\frac{2}{3}\pi\sqrt{n}}+2.28 cn^{\frac 54} e^{(4\sqrt{\delta_0}-1)\pi\sqrt{n}} \\ &\quad + 10^5 (464.443c + 2.011)n^{\frac{11}{8}}e^{-\pi\sqrt{n}}. \end{align*} $$

We take the sum of all three coefficients, the highest order exponential, and the highest power of n from the three terms and put them together in one term. Note that $4\sqrt {\delta _0}>\frac 13$ for $c\geq 9$ . Therefore, we get the bounds of $\overline R(a,c,n)$ , which completes the proof.

Proof of Theorem 1.4

By (4.1) we have

(5.1) $$ \begin{align} \sum_{n=0}^\infty\left(\overline{M}(a,c,n)-\overline{M}(b,c,n)\right)q^n =\frac{2}{c}\sum_{j=1}^{\frac{c-1}{2}}\rho_j(a,b,c)\overline{C}(\zeta_c^j;q), \end{align} $$

where

$$\begin{align*}\rho_j(a,b,c)=\cos\left(\frac{2\pi aj}{c}\right) -\cos\left(\frac{2\pi bj}{c}\right).\end{align*}$$

We use Theorem 1.1, with a change of variables that does not effect the theorem. If $c\nmid k$ , we redefine $0<l<c$ to be the unique solution to the congruence $l\equiv jk$ (mod c). Inserting Theorem 1.1 into (5.1), we obtain

$$ \begin{align*} \overline{M}(a,c,n)-\overline{M}(b,c,n) =\sum_{j=1}^{\frac{c-1}{2}}\left(\overline{S}_j(a,b,c;n)+\sum_{i\in\{+,-\}} \overline{T}_j^i(a,b,c;n)+O(n^\varepsilon)\right), \end{align*} $$

where

$$ \begin{align*} \overline{S}_j(a,b,c;n)&=\rho_j(a,b,c)\frac{2i}{c\sqrt{n}} \sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c|k,~2\nmid k \end{subarray}} \frac{\overline{B}_{j,c,k}(-n,0)}{\sqrt{k}}\sinh\left(\frac{\pi\sqrt{n}}{k}\right), \nonumber\\[5pt] \overline{T}_j^i(a,b,c;n)&=\rho_j(a,b,c)\frac{4}{c\sqrt{n}}\sin\left(\frac{\pi j}{c}\right) \sum_{\begin{subarray}{c} 1\leq k\leq \sqrt{n}\\ c\nmid k,~2\nmid k\\ \delta_{j,c,k,r}^i>0\\ r\geq0 \end{subarray}} \frac{\overline{D}_{j,c,k}(-n,m_{j,c,k,r}^i)}{\sqrt{k}} \\ &\quad\times\sinh\left(\frac{4\pi}{k}\sqrt{n\delta_{j,c,k,r}^i}\right). \end{align*} $$

Then, we need to find the main term in $\overline {M}(a,c,n)-\overline {M}(b,c,n)$ .

The main term:

Firstly, we detect the main contribution coming from sinh. In $\overline {S}_j$ , the main term occurs for $k=c$ , giving $\sinh \left (\frac {\pi \sqrt {n}}{c}\right )$ as the main hyperbolic sine argument, whereas in $\overline {T}_j^i$ it is given by $k=1$ . Recall that $\delta _{j,c,k,r}^{i}<\delta _{j,c,k,0}^{+}$ and $\delta _{j,c,k,0}^{+}$ arrives the maximum value $\delta _0=\frac 1{2{c}^2}-\frac 1{2c}+\frac 1{16}$ at $l=1$ . Moreover, $l\equiv jk$ (mod c) together with $k=l=1$ gives that $j=l=1$ . These imply that the largest argument occurs for $k=1, r=0$ and $j=1$ . For $k=1, r=0,$ and any j, we easily see that $\overline {D}_{j,c,1}(-n,m_{j,c,1,0}^+)=\overline {D}_{j,c,1}(-n,0)=1$ . Now, since

$$ \begin{align*} 4\sqrt{\delta_0}>\frac 1c \end{align*} $$

for $c\geq 9$ , the main term is

$$ \begin{align*} \overline{T}_1^+=\rho_1(a,b,c)\frac{4}{c\sqrt{n}}\sinh\left(4\pi \sqrt{n\delta_0}\right), \end{align*} $$

and this is positive for all $0\leq a<b\leq \frac {c-1}2$ . Therefore, for sufficiently large n, we obtain $\overline M(a,c,n)>\overline M(b,c,n)$ . The following step entails providing a clear definition of “sufficiently large” by establishing bounds for the remaining terms.

Bounding contribution of $\overline {S}_j$ :

Since $|\rho _j(a,b,c)|<2$ , by (4.5), we have

$$ \begin{align*} |\overline{S}_j(a,b,c;n)|\leq \frac {16n^{1/4}\left(1+\log\left(\frac {c-1}2\right)\right)}{3c^2\pi(1-\pi^2/24)} \sinh\left(\frac {\pi \sqrt{n}}c\right). \end{align*} $$

Next we want to bound the $\overline {T}_j^i$ for $j\geq 2$ and $\overline {T}_1^-$ .

Bounding the contribution of $\overline {T}_j^i$ for $j\geq 2$ :

For $k\geq 2$ , we trivially bound

$$ \begin{align*} |\overline{D}_{j,c,k}(-n,m_{j,c,k,r}^+)|\leq k. \end{align*} $$

Since $\sinh (x)=(e^x-e^{-x})/2$ , we have

$$ \begin{align*} \sinh\left(\frac{4\pi}{k}\sqrt{n\delta_{j,c,k,r}^i}\right)\leq \frac {e^{2\pi \sqrt{n\delta_0}}}2. \end{align*} $$

Next, recall that the number of r satisfying $\delta _{j,c,k,r}^i>0$ is less than $\frac {c}{16}$ . Thus we can bound $\overline {T}_j^i$ for $k\geq 2$ by the following expression:

$$ \begin{align*} \frac {1}{6} n^{1/4} e^{2\pi \sqrt{n\delta_0}}. \end{align*} $$

Since $\delta _{j,c,1,0}^i<\delta _0$ for $j\geq 2$ , we bound the $k=1$ contribution by the argument

$$ \begin{align*} \frac {1}{4\sqrt{n}}e^{4\pi \sqrt{n\delta_{2,c,1,0}^i}}. \end{align*} $$

Before coming to the error terms of the circle method we have to bound the contribution of $\overline {T}_1^-$ .

Bounding the contribution of $\overline {T}_1^-$ :

By the same analysis, it is possible to bound this term by a similar expression as the ones before. By bounding the sinh by the exponential function, using $|\rho _j(a,b,c)|<2$ and showing $\delta _{1,c,k,r}^-$ is smaller than $\delta _{2,c,1,0}^+$ and so choosing the right argument in the exponential function, $\overline {T}_1^-$ can be bounded by

$$ \begin{align*} \frac {1}{4\sqrt{n}}e^{4\pi \sqrt{n\delta_{2,c,1,0}^+}}. \end{align*} $$

The explicit constant:

Denoting the four different error terms in Lemmas 3.4–3.7, $\overline {S}_j$ , $\overline {T}_j^i$ for $j\geq 2$ and $\overline {T}_1^-$ by $\Sigma _{err,m}$ we can conclude that

$$ \begin{align*} \overline{N}_{a,b,c}=\min\left\{n\in\mathbb{N} \left|\overline T_1^+-\sum_m\Sigma_{err,m}(c,n)>0\right.\right\}. \end{align*} $$

This finishes the proof.

Proof For c being an odd integer less than $9$ , these inequalities can be readily verified using mathematical software. Specifically, for cases where $c < 7$ and $k = c$ , this setting results in the largest argument for the hyperbolic sine function in $\overline {S}j$ . Consequently, it is sufficient to determine the sign of $\sum _j \rho _j(a, b, c) \overline {B}{j, c, c}(-n, 0)$ to identify which inequality the crank differences satisfy. For the case where $c=7$ , the arguments of the hyperbolic sines in $\overline {S}_j$ and $\overline {T}_1^+$ may align and potentially cancel each other out. Therefore, we compute the sum of $\sum _j i\rho _j(a, b, c) \overline {B}_{j, c, c}(-n, 0)/\sqrt {7}$ and $ 2\rho _1(a, b, c)\sin {\frac {\pi }{7}}$ to determine which inequality is met by the crank differences.

Proof of Theorem 1.5

In light of (1.3), to prove $\overline M(a,c,n)>0$ , it is sufficient to prove

$$ \begin{align*} \left|\overline R(a,c,n)\right|<\frac 1c \end{align*} $$

if $c<9$ and $n\geq 196$ , and

(5.2) $$ \begin{align} \left|\overline R(a,c,n)\right|<\frac 1{2c} \end{align} $$

if $c\geq 9$ and $n>\overline {Q}_c$ , where $\overline {Q}_c$ is defined in (1.4). According to the bound of $\overline R(a,c,n)$ in Theorem 1.3, we just need to show for $c<9$

(5.3) $$ \begin{align} 10^5(464.45c + 2.02)n^{\frac{11}{8}}e^{-\frac{2}{3}\pi\sqrt{n}}<\frac 1c \end{align} $$

and for $c\geq 9$

(5.4) $$ \begin{align} 10^5(464.45c + 2.02)n^{\frac{11}{8}}e^{-\left(1-\sqrt{1-\frac 8{c}+\frac 8{c^2}}\right)\pi\sqrt{n}}<\frac 1{2c}. \end{align} $$

In order to verify (5.3), it suffices to prove that

$$ \begin{align*} 10^5(464.45\times9 + 2.02)n^{\frac{11}{8}}e^{-\frac{2}{3}\pi\sqrt{n}}<\frac 1{9}, \end{align*} $$

which holds for $n\geq 196$ . We next deal with the case of $c\geq 9$ . To prove (5.4), it is equivalent to prove that

(5.5) $$ \begin{align} \left(1-\sqrt{1-\frac 8{c}+\frac 8{c^2}}\right)\pi\sqrt{n}-\frac{11}{8}\log n>\log\left(2\times10^5\times(464.45c^2 + 2.02c)\right). \end{align} $$

Since for $x\geq 104$ ,

$$ \begin{align*} \log x<\frac{16}{11}x^{\frac 14}, \end{align*} $$

and for $c\geq 9$ ,

$$ \begin{align*} \left(1-\sqrt{1-\frac 8{c}+\frac 8{c^2}}\right)>\frac 4c, \end{align*} $$

we have

(5.6) $$ \begin{align} \left(1-\sqrt{1-\frac 8{c}+\frac 8{c^2}}\right)\pi\sqrt{n}-\frac{11}{8}\log n>\frac{4\pi\sqrt{n}}c-2n^{\frac 14}. \end{align} $$

In view of (5.5) and (5.6), it is sufficient to verify

$$ \begin{align*} \frac{4\pi\sqrt{n}}c-2n^{\frac 14}>\log\left(2\times10^5\times(464.45c^2 + 2.02c)\right), \end{align*} $$

which holds for $n\geq \overline Q_c$ . This completes the proof.

Proof of Theorem 1.6

We first consider the case $c<9$ . By Theorem 1.3, we can give a lower and an upper bound for $\overline M(a,c,n)$ , that is,

$$ \begin{align*} \overline L(c,n)< \overline M(a,c,n)<\overline U(c,n), \end{align*} $$

where

$$ \begin{align*} \overline L(c,n):=\overline p(n)\left(\frac 1c-10^5(464.45c+2.02)n^{\frac{11}{8}}e^{-\frac{2\pi}{3}\sqrt{n}}\right), \\ \overline U(c,n):=\overline p(n)\left(\frac 1c+10^5(464.45c+2.02)n^{\frac{11}{8}}e^{-\frac{2\pi}{3}\sqrt{n}}\right). \end{align*} $$

For $3\leq c<9$ , we have

$$ \begin{align*} \overline L(c,n)>\overline p(n)\left(\frac 19-10^5(464.45\times 9+2.02)n^{\frac{11}{8}}e^{-\frac{2\pi}{3}\sqrt{n}}\right), \end{align*} $$

Noting that $n^{\frac {11}{8}}e^{-\frac {2\pi }{3}\sqrt {n}}$ is decreasing for $n\geq 196$ , it follows that

$$ \begin{align*} \overline L(c,n)>0.0017\overline p(n). \end{align*} $$

Similarly, we get

$$ \begin{align*} \overline U(c,n)<\overline p(n)\left(\frac 13+10^5(464.45\times 9+2.02)n^{\frac{11}{8}}e^{-\frac{2\pi}{3}\sqrt{n}}\right)<0.4428\overline p(n). \end{align*} $$

Hence, when $3\leq c<9$ and $n\geq 196$ , we have

(5.7) $$ \begin{align} 0.0017~\overline{p}(n)< \overline M(a,c,n)<0.4428~\overline{p}(n). \end{align} $$

From [Reference Zhang and Zhong20, Theorem 6], we know that for $n\geq 1$ ,

(5.8) $$ \begin{align} \frac{1}{8n}\left(1-\frac 1{\sqrt{n}}\right)e^{\pi\sqrt{n}}<\overline p(n)< \frac{1}{8n}\left(1+\frac 1{\sqrt{n}}\right)e^{\pi\sqrt{n}}. \end{align} $$

For $c<9$ , let $n_1=\lambda n_2$ with $\lambda \geq 1$ . Then by (5.7) and (5.8), we have

$$ \begin{align*} \overline M(a,c,n_1)\overline M(a,c,n_2)>\frac{0.0017^2}{64\lambda n_2^2}\cdot\left(1-\frac{1}{\sqrt{\lambda n_2}}\right)\left(1-\frac{1}{\sqrt{n_2}}\right)\cdot e^{\pi\sqrt{\lambda n_2}+\pi\sqrt{n_2}} \end{align*} $$

and

$$ \begin{align*} \overline M(a,c,n_1+n_2)<\frac{0.4428}{8(\lambda n_2+n_2)}\left(1+\frac 1{\sqrt{\lambda n_2+n_2}}\right)e^{\pi\sqrt{\lambda n_2+n_2}} \end{align*} $$

for $n\geq 196$ . Hence it suffices to show that

$$ \begin{align*} T_{n_2}(\lambda)>\log(V_{n_2}(\lambda))+\log(S_{n_2}(\lambda)), \end{align*} $$

where

$$ \begin{align*} T_{n_2}(\lambda)&:=\pi\sqrt{\lambda n_2}+\pi\sqrt{n_2}-\pi\sqrt{\lambda n_2+n_2},\\ V_{n_2}(\lambda)&:=\frac{0.4428}{0.0017^2}\cdot\frac{8\lambda n_2}{\lambda+1},\\ S_{n_2}(\lambda)&:=\frac{1+\frac 1{\sqrt{\lambda n_2+n_2}}}{\left(1-\frac{1}{\sqrt{\lambda n_2}}\right)\left(1-\frac{1}{\sqrt{n_2}}\right)}. \end{align*} $$

On the one hand, it can be shown that $T_{n_2}(\lambda )$ is increasing and $S_{n_2}(\lambda )$ is decreasing for $\lambda \geq 1$ . On the other hand,

$$ \begin{align*} V_{n_2}(\lambda)<\frac{0.4428\cdot8 n_2}{0.0017^2}. \end{align*} $$

So it is sufficient to verify that

$$ \begin{align*} T_{n_2}(1)&=2\pi\sqrt{n_2}-\pi\sqrt{2n_2}>\log\left(\frac{0.4428\cdot8 n_2}{0.0017^2}\right)+\log\left(\frac{1+\frac 1{\sqrt{2n_2}}}{\left(1-\frac{1}{\sqrt{n_2}}\right)^2}\right), \end{align*} $$

which holds for $n_2\geq 107$ .

Finally, we deal with the case $c\geq 9$ . In this case, we bound $ \overline M(a,c,n)$ by a different way. Applying (1.3) and (5.2), we obtain that

$$ \begin{align*} \frac{1}{2c}\overline p(n)<\overline M(a,c,n)< \frac{3}{2c}\overline p(n) \end{align*} $$

for $n\geq \overline Q_c$ . By the same argument of the case $c<9$ , it suffices to show that

$$ \begin{align*} T_{n_2}(\lambda)>\log(\overline{V}_{n_2}(\lambda))+\log(S_{n_2}(\lambda)), \end{align*} $$

where

$$ \begin{align*} \overline{V}_{n_2}(\lambda):=6c\cdot\frac{8\lambda n_2}{\lambda+1}. \end{align*} $$

It is obvious that

$$ \begin{align*} \overline{V}_{n_2}(\lambda)<48cn_2. \end{align*} $$

So it is equivalent to verify that

(5.9) $$ \begin{align} T_{n_2}(1)=2\pi\sqrt{n_2}-\pi\sqrt{2n_2}>\log\left(48cn_2\right)+\log\left(\frac{1+\frac 1{\sqrt{2n_2}}}{\left(1-\frac{1}{\sqrt{n_2}}\right)^2}\right). \end{align} $$

On the one hand, when $n_2\geq 2$ , we get

(5.10) $$ \begin{align} \log\left(48cn_2\right)+\log\left(\frac{1+\frac 1{\sqrt{2n_2}}}{\left(1-\frac{1}{\sqrt{n_2}}\right)^2}\right)<\log\left(48cn_2\right)+\log\left(\frac{1+\frac 1{\sqrt{4}}}{\left(1-\frac{1}{\sqrt{2}}\right)^2}\right)<\log\left(840cn_2\right). \end{align} $$

On the other hand, when $n_2>840c$ , we have

(5.11) $$ \begin{align} T_{n_2}(1)=2\pi\sqrt{n_2}-\pi\sqrt{2n_2}>2\log(n_2)>\log(n_2)+\log(840c)=\log\left(840cn_2\right). \end{align} $$

Therefore, in light of (5.9), (5.10), and (5.11), we know (1.5) holds for $c\geq 9$ and $n_1,n_2\geq \overline Q_c$ .