2. Proof of Theorem 1.1

In this section, we always set $\zeta =e^{\pi \mathrm {i}/8}$. In order to prove Theorem 1.1, we first establish a lemma.

Lemma 2.1 We have

(2.1)\begin{equation} \prod_{k=0}^{\infty} \displaystyle\frac{1 }{(1-\sqrt{2}q^{4k+2}+q^{8k+4})} =\frac{E(q^{8})E(q^{32})^{3}}{E(q^{4})E(q^{16})^{2}E(q^{64}) }+\sqrt{2}q^{2}\frac{E(q^{8})E(q^{64})}{ E(q^{4})E(q^{16})}\end{equation}

and

(2.2)\begin{equation} \prod_{k=0}^{\infty} \displaystyle\frac{1 }{(1+\sqrt{2}q^{4k+2}+q^{8k+4})} =\frac{E(q^{8})E(q^{32})^{3}}{E(q^{4})E(q^{16})^{2}E(q^{64}) }-\sqrt{2}q^{2}\frac{E(q^{8})E(q^{64}) }{ E(q^{4})E(q^{16})}.\end{equation}

Proof. Noticing that $\zeta ^{2}=\frac {\sqrt {2}}{2}(1+\mathrm {i})$, we have

(2.3)\begin{align} \prod_{k=0}^{\infty} \displaystyle\frac{1 }{(1-\sqrt{2}q^{4k+2}+q^{8k+4})} & =\prod_{k=0}^{\infty} \frac{ (1+\sqrt{2}q^{4k+2}+q^{8k+4}) }{(1+q^{16k+8})} \nonumber\\ & =\frac{E(q^{8})E(q^{32})}{E(q^{4}) E(q^{16})^{2}}f(\zeta^{2}q^{2},q^{2}/\zeta^{2}), \end{align}

where Ramanujan's general theta function is given by

\[ f(a,b):=({-}a;ab)_\infty({-}b;ab)_\infty (ab;ab)_\infty. \]

It follows from Entry 30 (ii) and (iii) on page 46 of Berndt's book [Reference Berndt5] that

(2.4)\begin{equation} f(a,b)=f(a^{3}b,ab^{3})+af(b/a, a^{5}b^{3}). \end{equation}

Taking $a=\zeta ^{2}q^{2}$ and $b=q^{2}/\zeta ^{2}$ in (2.4) yields

(2.5)\begin{equation} f(\zeta^{2}q^{2},q^{2}/\zeta^{2}) =f(\zeta^{4}q^{8},q^{8}/\zeta^{4}) + \zeta^{2}q^{2} f(\zeta^{{-}4}, \zeta^{4}q^{16}). \end{equation}

By the fact that $\zeta ^{4}=\mathrm {i}$, we have

(2.6)\begin{equation} f(\zeta^{4}q^{8},q^{8}/\zeta^{4})=(-\mathrm{i}\,q^{8};q^{16})_\infty (\mathrm{i}\,q^{8};q^{16})_\infty E(q^{16})=\displaystyle\frac{E(q^{32})^{2} }{E(q^{64})} \end{equation}

and

(2.7)\begin{equation} f(\zeta^{{-}4}, \zeta^{4} q^{16}) =( \mathrm{i};q^{16})_\infty (-\mathrm{i}\,q^{16};q^{16})_\infty E(q^{16}) =(1-\mathrm{i}) \displaystyle\frac{E(q^{16})E(q^{64})}{E(q^{32})}. \end{equation}

Making use of (2.5)–(2.7) and the fact that $\zeta ^{2}=\frac {\sqrt {2}}{2}(1+\mathrm {i})$, we arrive at

(2.8)\begin{equation} f(\zeta^{2}q^{2},q^{2}/\zeta^{2})= \displaystyle\frac{E(q^{32})^{2}}{E(q^{64})} +\sqrt{2}q^{2}\frac{E(q^{16})E(q^{64})}{E(q^{32})}. \end{equation}

Now, (2.1) follows from (2.3) and (2.8). Also, replacing $q$ by $\mathrm {i}\,q$ in (2.1) leads to (2.2).

Now, we are ready to prove Theorem 1.1.

Proof Proof of Theorem 1.1

Employing (1.10), (1.11) and the fact that

(2.9)\begin{equation} \displaystyle\sum_{j=0}^{15} \zeta^{kj}= \begin{cases} 16, & \text{if }k\equiv 0 \pmod{16},\\ 0, & \text{if }k\not\equiv 0 \pmod{16}, \end{cases} \end{equation}

we have, for $0\le a\le 15$,

(2.10)\begin{align} \displaystyle\sum_{n=0}^{\infty} p(a,16;n)q^{n}& =\displaystyle\frac{1}{16} \sum_{j=0}^{15}\zeta^{{-}aj} \sum_{n=0}^{\infty} \sum_{r={-}\infty}^{\infty} p(r;n)\zeta^{jr} q^{n} \nonumber\\ & =\frac{1}{16}\frac{E(q^{2})^{2}}{E(q)E(q^{4})^{2} }\sum_{j=0}^{15}\zeta^{{-}aj} G(\zeta^{j},q), \end{align}

where

(2.11)\begin{equation} G(z,q)=\displaystyle\frac{1}{ (z^{2}q^{2};q^{4})_\infty(q^{2}/z^{2};q^{4})_\infty}. \end{equation}

It is easy to check that for $k,\, j\geq 0$,

(2.12)\begin{equation} (1-\zeta^{2j}q^{4k+2} )(1-q^{4k+2}/\zeta^{2j}) =1-(\zeta^{2j}+\zeta^{{-}2j})q^{4k+2} +q^{8k+4}. \end{equation}

In light of (2.11) and (2.12),

(2.13)\begin{equation} G(\zeta^{j},q)=\begin{cases} \displaystyle\displaystyle\frac{E(q^{4})^{2}}{E(q^{2})^{2}}, & \text{if }j\in\{0,8\},\\ \displaystyle\prod_{k=0}^{\infty} \frac{1}{(1-\sqrt{2}q^{4k+2}+q^{8k+4})}, & \text{if }j\in\{1,7,9,15\},\\ \displaystyle\frac{E(q^{4})E(q^{16})}{E(q^{8})^{2}}, & \text{if }j\in\{2,6,10,14\},\\ \displaystyle\prod_{k=0}^{\infty} \frac{1}{(1+\sqrt{2}q^{4k+2}+q^{8k+4})}, & \text{if }j\in\{3,5,11,13\},\\ \displaystyle\frac{E(q^{2})^{2}E(q^{8})^{2}}{E(q^{4})^{4}}, & \text{if }j\in\{4,12\}. \end{cases} \end{equation}

Using (2.1), (2.2), (2.10) and (2.13), we find that

(2.14)\begin{align} & \displaystyle\sum_{n=0}^{\infty} p(a,16;n)q^{n} = \displaystyle\frac{1}{16}\frac{E(q^{2})^{2}}{E(q)E(q^{4})^{2} } \Bigg\{(1+({-}1)^{a})\frac{E(q^{4})^{2} }{E(q^{2})^{2}}\nonumber\\ & \quad+(\zeta^{{-}a}+\zeta^{{-}7a}+\zeta^{{-}9a}+\zeta^{{-}15a})\left(\displaystyle\frac{E(q^{8})E(q^{32})^{3}}{E(q^{4})E(q^{16})^{2}E(q^{64}) }+\sqrt{2}q^{2}\frac{E(q^{8})E(q^{64})}{ E(q^{4})E(q^{16})}\right) \nonumber\\ & \quad +(\zeta^{{-}2a}+\zeta^{{-}6a}+\zeta^{{-}10a}+\zeta^{{-}14a})\frac{E(q^{4})E(q^{16})}{E(q^{8})^{2} }\nonumber\\ & \quad+(\zeta^{{-}3a}+\zeta^{{-}5a}+\zeta^{{-}11a}+\zeta^{{-}13a})\left(\frac{E(q^{8})E(q^{32})^{3}}{E(q^{4})E(q^{16})^{2}E(q^{64}) }-\sqrt{2}q^{2}\frac{E(q^{8})E(q^{64})}{ E(q^{4})E(q^{16})}\right) \nonumber\\ & \quad+(\mathrm{i}^{a}+(-\mathrm{i})^{a})\frac{E(q^{2})^{2}E(q^{8})^{2}}{E(q^{4})^{4} } \Bigg\}. \end{align}

Theorem 1.1 follows from (2.14) and the fact that $\zeta =\frac {\sqrt {2+\sqrt {2}}}{2}+ \frac {\sqrt {2-\sqrt {2}}}{2} \,\mathrm {i}$.

3. Proof of Theorem 1.2

In this section, we prove Theorem 1.2. Throughout our proof, we always write $\omega =e^{\pi \mathrm {i} /12}$. We first prove the following lemma.

Lemma 3.1 We have

(3.1)\begin{equation} \prod_{k=0}^{\infty} \displaystyle\frac{1}{(1-\sqrt{3}q^{4k+2}+q^{8k+4})} =\frac{E(q^{8})E(q^{12})E(q^{16})}{E(q^{4})^{2}E(q^{24}) }+\sqrt{3}q^{2}\frac{E(q^{8})^{2}E(q^{12})E(q^{48})^{2} }{E(q^{4})^{2}E(q^{16})E(q^{24})^{2}} \end{equation}

and

(3.2)\begin{equation} \prod_{k=0}^{\infty} \displaystyle\frac{1}{(1+\sqrt{3}q^{4k+2}+q^{8k+4})} =\frac{E(q^{8})E(q^{12})E(q^{16})}{E(q^{4})^{2}E(q^{24}) }-\sqrt{3}q^{2}\frac{E(q^{8})^{2}E(q^{12})E(q^{48})^{2} }{E(q^{4})^{2}E(q^{16})E(q^{24})^{2}}. \end{equation}

Proof. Notice that $\omega ^{2}=\frac {\sqrt {3}}{2}+\frac {1}{2}\,\mathrm {i}$. Therefore,

(3.3)\begin{equation} \prod_{k=0}^{\infty} \displaystyle\frac{1}{(1-\sqrt{3}q^{4k+2}+q^{8k+4})} =\frac{E(q^{4})}{f(-\omega^{2}q^{2},-q^{2}/\omega^{2})} \end{equation}

where $f(a,\,b)$ is as defined in (2.4). It follows from the quintuple product identity [Reference Berndt5, (38.2), p. 80] that

(3.4)\begin{equation} \displaystyle\frac{E(q^{4})}{f(Bq^{2},q^{2}/B)} =\frac{1}{ f({-}B^{2},-q^{4}/B^{2})} (f(B^{3}q^{2},q^{10}/B^{3}) -B^{2}f(q^{2}/B^{3},B^{3}q^{10})). \end{equation}

Setting $B=-\omega ^{2}$ in (3.4), we deduce that

(3.5)\begin{equation} \displaystyle\frac{E(q^{4})}{f(-\omega^{2}q^{2},-q^{2}/\omega^{2})} =\frac{1}{ f(-\omega^{4},-q^{4}/\omega^{4})} (f(-\mathrm{i}\,q^{2},-q^{10}/\mathrm{i})-\omega^{4}f({-}q^{2}/\mathrm{i},-\mathrm{i}\,q^{10})). \end{equation}

By the fact that $\omega ^{4}=\frac {1}{2}+\frac {\sqrt {3}}{2}\,\mathrm {i}$,

\begin{align*} f(-\omega^{4},-q^{4}/\omega^{4})& =E(q^{4})(\omega^{4};q^{4})_\infty (q^{4}/\omega^{4};q^{4})_\infty\nonumber\\ & = (1-\omega^{4})E(q^{4}) \prod_{k=1}^{\infty}\left(1-\left(\omega^{4}+\displaystyle\frac{1}{\omega^{4}}\right) q^{4k}+q^{8k}\right)\nonumber\\ & = (1-\omega^{4})E(q^{4})\prod_{k=1}^{\infty}\frac{ 1+q^{12k} }{1+q^{4k}}\nonumber\\ & =(1-\omega^{4})\frac{E(q^{4})^{2} E(q^{24})}{E(q^{8})E(q^{12})}. \end{align*}

Therefore,

(3.6)\begin{equation} \displaystyle\frac{1}{f(-\omega^{4},-q^{4}/\omega^{4})} =\bigg(\frac{1}{2}+\frac{\sqrt{3}}{2}\,\mathrm{i}\bigg) \frac{E(q^{8})E(q^{12})}{E(q^{4})^{2} E(q^{24})}. \end{equation}

Taking $a=-\mathrm {i}\,q^{2}$ and $b=-q^{10}/\mathrm {i}$ in (2.4) yields

(3.7)\begin{align} f(-\mathrm{i}\,q^{2},-q^{10}/\mathrm{i})& =f({-}q^{16},-q^{32}) -\mathrm{i}\,q^{2} f({-}q^{8},-q^{40})\nonumber\\ & =E(q^{16})-\mathrm{i}\,q^{2}\displaystyle\frac{E(q^{8})E(q^{48})^{2} }{E(q^{16})E(q^{24})}. \end{align}

On the other hand, if we put $a=- q^{2}/\mathrm {i}$ and $b=-\mathrm {i}\,q^{10}$ in (2.4), then

(3.8)\begin{align} f({-}q^{2}/\mathrm{i},-\mathrm{i}\,q^{10})& =f({-}q^{16},-q^{32}) -(q^{2}/\mathrm{i}) f({-}q^{8},-q^{40})\nonumber\\ & =E(q^{16})+\mathrm{i}\,q^{2}\displaystyle\frac{E(q^{8})E(q^{48})^{2} }{E(q^{16})E(q^{24})}. \end{align}

Finally, (3.1) follows from (3.3) and (3.5)–(3.8). Also, replacing $q$ by $\mathrm {i}\,q$ in (3.1) yields (3.2).

Now, we prove Theorem 1.2.

Proof of Theorem 1.2 Utilizing (1.10), (1.11) and the fact that

\[ \displaystyle\sum_{j=0}^{23}\omega^{kj}= \begin{cases} 24, & \text{if }k\equiv 0 \pmod{24},\\ 0, & \text{if }k\not\equiv 0 \pmod{24}, \end{cases} \]

we arrive at

(3.9)\begin{align} \displaystyle\sum_{n=0}^{\infty} p(a,24;n)q^{n}& =\displaystyle\frac{1}{24}\sum_{j=0}^{23}\omega^{{-}aj}\sum_{n=0}^{\infty}\sum_{r={-}\infty}^{\infty} p(r;n)\omega^{jr} q^{n} \nonumber\\ & =\frac{1}{24}\frac{E(q^{2})^{2}}{E(q)E(q^{4})^{2} }\sum_{j=0}^{23}\omega^{{-}aj} G(\omega^{j},q), \end{align}

where $G(z,\,q)$ is as defined in (2.11). In light of (2.11) and (2.12),

(3.10)\begin{equation} G(\omega^{j},q)= \begin{cases} \displaystyle \displaystyle\frac{E(q^{4})^{2}}{E(q^{2})^{2}} , & \text{if }j\in\{0,12\},\\ \displaystyle \prod_{k=0}^{\infty} \displaystyle\frac{1}{(1-\sqrt{3}q^{4k+2}+q^{8k+4})} , & \text{if }j\in\{1,11,13,23\},\\ \displaystyle \frac{E(q^{4})^{2}E(q^{6})E(q^{24})}{E(q^{2})E(q^{8})E(q^{12})^{2}} , & \text{if }j\in\{2,10,14,22\},\\ \displaystyle \frac{E(q^{4})E(q^{16})}{E(q^{8})^{2}} , & \text{if }j\in\{3,9,15,21\},\\ \displaystyle \frac{E(q^{2})E(q^{12})}{E(q^{4})E(q^{6})} , & \text{if }j\in\{4,8,16,20\},\\ \displaystyle \prod_{k=0}^{\infty} \frac{1}{(1+\sqrt{3}q^{4k+2}+q^{8k+4})} , & \text{if }j\in\{5,7,17,19\},\\ \displaystyle \frac{E(q^{2})^{2}E(q^{8})^{2}}{E(q^{4})^{4}}, & \text{if }j\in\{6,18\}. \end{cases} \end{equation}

By (3.1), (3.2), (3.9) and (3.10),

(3.11)\begin{align} \displaystyle\sum_{n=0}^{\infty} p(a,24;n)q^{n} & = \displaystyle\frac{1}{24}\frac{E(q^{2})^{2}}{E(q)E(q^{4})^{2} } \Bigg\{(1+({-}1)^{a})\frac{E(q^{4})^{2} }{E(q^{2})^{2}}\nonumber\\ & \quad+(\omega^{{-}a}+\omega^{{-}11a}+\omega^{{-}13a}+\omega^{{-}23a})\nonumber\\ & \quad\times\left(\displaystyle\frac{E(q^{8})E(q^{12})E(q^{16})}{E(q^{4})^{2}E(q^{24}) }+\sqrt{3}q^{2}\frac{E(q^{8})^{2}E(q^{12})E(q^{48})^{2} }{E(q^{4})^{2}E(q^{16})E(q^{24})^{2}}\right) \nonumber\\ & \quad +(\omega^{{-}2a}+\omega^{{-}10a}+\omega^{{-}14a}+\omega^{{-}22a})\frac{E(q^{4})^{2}E(q^{6})E(q^{24}) }{E(q^{2})E(q^{8})E(q^{12})^{2} }\nonumber\\ & \quad +(\omega^{{-}3a}+\omega^{{-}9a}+\omega^{{-}15a}+\omega^{{-}21a})\frac{E(q^{4})E(q^{16}) }{E(q^{8})^{2} }\nonumber\\ & \quad +(\omega^{{-}4a}+\omega^{{-}8a}+\omega^{{-}16a}+\omega^{{-}20a})\frac{E(q^{2})E(q^{12}) }{E(q^{4})E(q^{6}) }\nonumber\\ & \quad+(\omega^{{-}5a}+\omega^{{-}7a}+\omega^{{-}17a}+\omega^{{-}19a})\nonumber\\ & \quad\times\left(\frac{E(q^{8})E(q^{12})E(q^{16})}{E(q^{4})^{2}E(q^{24}) }-\sqrt{3}q^{2}\frac{E(q^{8})^{2}E(q^{12})E(q^{48})^{2} }{E(q^{4})^{2}E(q^{16})E(q^{24})^{2}}\right) \nonumber\\ & \quad+(\mathrm{i}^{a}+(-\mathrm{i})^{a})\frac{E(q^{2})^{2}E(q^{8})^{2}}{E(q^{4})^{4} } \Bigg\}. \end{align}

Theorem 1.2 follows from (3.11) and the fact that $\omega = \frac {\sqrt {6}+\sqrt {2}}{4}+\frac {\sqrt {6} -\sqrt {2}}{4}\,\mathrm {i}$.

4. Proof of Theorem 1.3

In this section, we prove Theorem 1.3 using Theorems 1.1 and 1.2 along with a result due to Sussman [Reference Sussman11].

In [Reference Sussman11], applying the standard circle method due to Rademacher [Reference Rademacher8], Sussman obtained an exact formula for $g(n)$, the coefficients in

(4.1)\begin{equation} \displaystyle\sum_{n\ge 0} g(n) q^{n}:=\prod_{j=1}^{J} E(q^{m_j})^{\delta_j}, \end{equation}

where $\mathbf {m}=(m_1,\,\ldots,\,m_J)$ is a sequence of distinct positive integers and $\mathbf {d}=(\delta _1,\,\ldots,\,\delta _J)$ is a sequence of non-zero integers such that $\sum \nolimits _{j=1}^{J} \delta _{j}<0$.

To state Sussman's result, we first fix some notation. Let $k$ be a positive integer. We define

\begin{align*} \Sigma & :={-}\displaystyle\frac{1}{2}\displaystyle\sum_{j=1}^{J} \delta_{j}, \quad \Omega:=\sum_{j=1}^{J} \delta_j m_j,\\ \Delta(k)& :={-}\displaystyle\sum_{j=1}^{J} \displaystyle\frac{\delta_j \gcd^{2}(m_j,k)}{m_j},\quad \Pi(k):= \prod_{j=1}^{J} \left(\frac{m_j}{\gcd(m_j,k)}\right)^{-\frac{\delta_j}{2}}. \end{align*}

Further, for any integer $h$ such that $\gcd (h,\,k)=1$, we define

\[ \omega_{h,k}:=\exp\left(-\pi \mathrm{i}\displaystyle\sum_{j=1}^{J} \delta_j\cdot s\left(\displaystyle\frac{m_j h}{\gcd(m_j,k)},\frac{k}{\gcd(m_j, k)}\right)\right), \]

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

\[ s(d,c):=\displaystyle\sum_{n \bmod{c}} \left(\left(\displaystyle\frac{dn}{c}\right)\right)\left(\left(\frac{n}{c}\right)\right) \]

with

\[ ((x)):= \begin{cases} x-\lfloor x\rfloor -1/2 & \text{if }x\not\in \mathbb{Z},\\ 0 & \text{if }x\in \mathbb{Z}. \end{cases} \]

Let $L=\operatorname {lcm}(m_1,\,\ldots,\,m_J)$. We divide the set $\{1,\,2,\,\ldots,\,L\}$ into two disjoint subsets:

\begin{align*} \mathcal {L}_{{>}0}& :=\{1\le \ell \le L :\Delta(\ell)>0\},\\ \mathcal {L}_{\le 0}& :=\{1\le \ell \le L :\Delta(\ell)\le 0\}. \end{align*}

Theorem 4.1 Sussman

If $\Sigma > 0$ and the inequality

(4.2)\begin{equation} \min_{1\le j\le J}\left(\displaystyle\frac{\gcd^{2}(m_j,\ell)}{m_j}\right)\ge \frac{\Delta(\ell)}{24} \end{equation}

holds for all $1\le \ell \le L,$ then for positive integers $n>-\Omega /24,$

(4.3)\begin{align} g(n)& =2\pi \displaystyle\sum_{\ell\in \mathcal {L}_{{>}0}} \Pi(\ell) \left(\displaystyle\frac{24n+\Omega}{\Delta(\ell)}\right)^{-\frac{\Sigma+1}{2}}\nonumber\\ & \quad\ \times \sum_{\substack{k\ge 1\\k \equiv \ell \bmod{L}}} \frac{1}{k} I_{\Sigma+1}\left(\frac{\pi }{6k}\sqrt{\Delta(\ell)(24n+\Omega)}\right) \sum_{\substack{0\le h< k\\ \gcd(h,k)=1}} e^{-\frac{2\pi \mathrm{i} nh}{k}}\omega_{h,k}, \end{align}

where $I_s(x)$ is the modified Bessel function of the first kind.

Remark We also frequently make use of the asymptotic expansion of $I_s(x)$ (see [Reference Abramowitz and Stegun1, p. 377, (9.7.1)]): for fixed $s$, when $|\arg x|<\frac {\pi }{2}$,

(4.4)\begin{equation} I_{s}(x)\sim \displaystyle\frac{e^{x}}{\sqrt{2\pi x}}\left(1-\frac{4s^{2}-1}{8x}+\frac{(4s^{2}-1)(4s^{2}-9)}{2!(8x)^{2}}-\cdots\right). \end{equation}

Let us define, for $i=1,\,\ldots,\,5$,

\[ \displaystyle\sum_{n= 0}^{\infty}s_i(n)q^{n}=S_i(q), \]

where $S_i(q)$'s are as defined in Theorem 1.1. First, we know from a famous result due to Hardy and Ramanujan [Reference Hardy and Ramanujan7] that, as $n\to \infty$,

(4.5)\begin{equation} s_1(n)=p(n)\sim \displaystyle\frac{1}{4\cdot 3^{1/2}\cdot n}\exp\left(2\pi\sqrt{\frac{n}{6}}\right). \end{equation}

Next, we show that, as $n\to \infty$,

(4.6)\begin{align} s_2(n)& \sim \displaystyle\frac{43^{1/2}}{2^{5}\cdot 3^{1/2}\cdot n}\exp\left(\frac{\pi}{4}\sqrt{\frac{43n}{6}}\right), \end{align}

(4.7)\begin{align} s_3(n)& \sim \displaystyle\frac{7^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\exp\left(\frac{\pi}{2}\sqrt{\frac{7n}{6}}\right), \end{align}

(4.8)\begin{align} s_4(n)& \sim \frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}({-}1)^{n}\cos\left(\frac{n\pi}{2}+\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right), \end{align}

(4.9)\begin{align} s_5(n)& \sim \frac{43^{1/2}}{2^{11/2}\cdot 3^{1/2}\cdot n}\exp\left(\frac{\pi}{4}\sqrt{\frac{43n}{6}}\right). \end{align}

We only prove (4.6) and (4.8) as instances. The rest can be shown analogously by Sussman's result (4.3).

First, we show (4.6). In (4.1), let us put

\[ \mathbf{m}=(1, 2, 4, 8, 16, 32, 64),\quad \mathbf{d}=({-}1, 2, -3, 1, -2, 3, -1). \]

Thus, we have $\Sigma =\frac {1}{2}$ and $\Omega =-1$. Also, $L=64$. We compute that

\begin{align*} \mathcal {L}_{{>}0}& =\{1,3,4,5,7,8,9,11,12,13,15,16,17,19,20,21,23,24,\\ \quad & 25,27,28,29,31,33,35,36,37,39,40,41,43,44,45,\\ \quad & 47,48,49,51,52,53,55,56,57,59,60,61,63,64\}. \end{align*}

Next, we verify that assumption (4.2) is satisfied. Then, it can be computed that when $k=1$, the $I$-Bessel term has the largest order, which is

\[ I_{3/2}\left(\frac{\sqrt{43}\pi}{48}\sqrt{24n-1}\right). \]

Further, when $k=1$, we have

\[ \sum_{\substack{0\le h< k\\ \gcd(h,k)=1}} e^{-\frac{2\pi \mathrm{i} nh}{k}}\omega_{h,k}=1. \]

It follows from (4.3), with (4.4) utilized, that

\[ s_2(n)\sim \frac{43^{1/2}}{2^{5}\cdot 3^{1/2}\cdot n}\exp\left(\frac{\pi}{4}\sqrt{\frac{43n}{6}}\right). \]

For (4.8), we put

\[ \mathbf{m}=(1, 2, 4, 8),\quad \mathbf{d}=({-}1, 4, -6, 2) \]

in (4.1). Thus, $\Sigma =\frac {1}{2}$ and $\Omega =-1$. Further, $L=8$. We compute that

\[ \mathcal {L}_{{>}0}=\{1,3,4,5,7,8\}. \]

Next, we verify that assumption (4.2) is satisfied. Then, it can be computed that when $k=4$, the $I$-Bessel term has the largest order, which is

\[ I_{3/2}\left(\frac{\sqrt{13}\pi}{24}\sqrt{24n-1}\right). \]

Further, when $k=4$, we have

\[ \sum_{\substack{0\le h< k\\ \gcd(h,k)=1}} e^{-\frac{2\pi \mathrm{i} nh}{k}}\omega_{h,k}=2({-}1)^{n} \cos\left(\frac{n\pi}{2}+\frac{\pi}{8}\right). \]

It follows from (4.3), with (4.4) utilized, that

\[ s_4(n)\sim \frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}({-}1)^{n}\cos\left(\frac{n\pi}{2}+\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right). \]

Notice that, for the exponential terms in (4.5)–(4.9), we have, numerically,

(4.10)\begin{align} & 2\pi\sqrt{\frac{1}{6}}=2.56\cdots,\quad \frac{\pi}{4}\sqrt{\frac{43}{6}}=2.10\cdots,\ \frac{\pi}{2}\sqrt{\frac{7}{6}}=1.69\cdots,\nonumber\\ & \frac{\pi}{2}\sqrt{\frac{13}{6}}=2.31\cdots,\quad \frac{\pi}{4}\sqrt{\frac{43}{6}}=2.10\cdots. \end{align}

Recall that, for any integer $i$ with $1\le i\le 4$, we have $p(2i,\,16,\,n)=p(16-2i,\,16,\,n)$. We conclude from the numerical calculations in (4.10) that

\[ p(2i,16;n)\sim \displaystyle\frac{s_1(n)}{8}=\frac{p(n)}{8} \]

as $n\to \infty$ for any integer $i$ with $0\le i\le 7$, and therefore (1.24) follows when $m=4$.

We also deduce from the numerical calculations in (4.10) that, for $0\le i<4$,

\[ p(4i,16; n)-p(4i+2,16; n)\sim \displaystyle\frac{s_4(n)}{4} \]

as $n\to \infty$. We know from (4.8) that

\[ s_4(n)\sim\begin{cases} \displaystyle\displaystyle\frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\cos\left(\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right) & \text{if }n\equiv 0 \pmod{4},\\ \displaystyle\frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\sin\left(\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right) & \text{if }n\equiv 1 \pmod{4},\\ \displaystyle-\frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\cos\left(\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right) & \text{if }n\equiv 2 \pmod{4},\\ \displaystyle-\frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\sin\left(\frac{\pi}{8}\right)\exp\left(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\right) & \text{if }n\equiv 3 \pmod{4}. \end{cases} \]

Hence, (1.26) and (1.27) hold when $m=4$. Finally, since

\[ \frac{\cos(\pi/8)}{\sin(\pi/8)}=1+\sqrt{2}, \]

we see that (1.25) is true when $m=4$.

Next, we prove Theorem 1.3 when $m=6$. Let us define, for $i=1,\,\ldots,\,7$,

\[ \displaystyle\sum_{n= 0}^{\infty} f_i(n)q^{n}=F_i(q), \]

where $F_i(q)$'s are as defined in Theorem 1.2. Applying Sussman's result (4.3), we have, as $n\to \infty$,

(4.11)\begin{align} f_1(n)& =p(n)\sim \displaystyle\frac{1}{4\cdot 3^{1/2}\cdot n}\exp\left(2\pi\sqrt{\frac{n}{6}}\right), \end{align}

(4.12)\begin{align} f_2(n)& \sim \displaystyle\frac{37^{1/2}}{2^{4}\cdot 3\cdot n}\exp\left(\frac{\pi}{6}\sqrt{\frac{37n}{2}}\right), \end{align}

(4.13)\begin{align} f_3(n)& \sim \frac{7^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}\exp\left(\frac{\pi}{2}\sqrt{\frac{7n}{6}}\right), \end{align}

(4.14)\begin{align} f_4(n)& \sim \frac{7^{1/2}}{2^{2}\cdot 3\cdot n}\exp\left(\frac{\pi}{3}\sqrt{\frac{7n}{2}}\right), \end{align}

(4.15)\begin{align} f_5(n)& \sim \frac{19^{1/2}}{2^{2}\cdot 3\cdot n}({-}1)^{n}\cos\left(\frac{n\pi}{2}+\frac{\pi}{8}\right)\exp\left(\frac{\pi}{6}\sqrt{\frac{19n}{2}}\right), \end{align}

(4.16)\begin{align} f_6(n)& \sim \frac{13^{1/2}}{2^{3}\cdot 3^{1/2}\cdot n}({-}1)^{n}\cos\left(\frac{n\pi}{2}+\frac{\pi}{8}\right)\exp\bigg(\frac{\pi}{2}\sqrt{\frac{13n}{6}}\bigg), \end{align}

(4.17)\begin{align} f_7(n)& \sim \frac{37^{1/2}}{2^{4}\cdot 3^{3/2}\cdot n}\exp\left(\frac{\pi}{6}\sqrt{\frac{37n}{2}}\right). \end{align}

Moreover, we notice that, for the exponential terms in (4.11)–(4.17), we have, numerically,

(4.18)\begin{align} & 2\pi\sqrt{\frac{1}{6}}=2.56\cdots,\quad \frac{\pi}{6}\sqrt{\frac{37}{2}}=2.25\cdots,\ \frac{\pi}{2}\sqrt{\frac{7}{6}}=1.69\cdots,\nonumber\\ & \frac{\pi}{3}\sqrt{\frac{7}{2}}=1.95\cdots,\quad \frac{\pi}{6}\sqrt{\frac{19}{2}}=1.61\cdots,\ \frac{\pi}{2}\sqrt{\frac{13}{6}}=2.31\cdots,\nonumber\\ & \frac{\pi}{6}\sqrt{\frac{37}{2}}=2.25\cdots. \end{align}

Recall that, for any integer $i$ with $1\le i\le 6$, we have $p(2i,\,24;n) =p(24-2i,\,24;n)$. We conclude from the numerical calculations in (4.18) that

\[ p(2i,24;n)\sim \displaystyle\frac{f_1(n)}{12}=\frac{p(n)}{12} \]

as $n\to \infty$ for any integer $i$ with $0\le i\le 11$, and therefore (1.24) follows when $m=6$. We also have, for $0\le i<6$,

\[ p(4i,24; n)-p(4i+2,24; n)\sim \displaystyle\frac{f_6(n)}{6} \]

as $n\to \infty$. Hence, in (1.25)–(1.27), the case of $m=6$ follows by arguments akin to those for the case of $m=4$. Therefore, the proof of Theorem 1.3 is completed.