Appendix
We provide a proof of (4.2). The proofs of (4.3)–(4.6), (5.6)–(5.10) and (5.12)–(5.16) are similar but lengthier.
We require the following results.
Lemma 9.1
Let
$$\begin{aligned} K=\frac{\chi (-q^5)^5}{\chi (-q)}. \end{aligned}$$
Then
$$\begin{aligned}&\displaystyle \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)} =\frac{4q}{K}, \end{aligned}$$
(9.1)
$$\begin{aligned}&\displaystyle \frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2 =K, \end{aligned}$$
(9.2)
$$\begin{aligned}&\displaystyle \frac{R(q)}{R(q^2)^3}+q^2\frac{R(q^2)^3}{R(q)} =K-2q+\frac{4q^2}{K}, \end{aligned}$$
(9.3)
$$\begin{aligned}&\displaystyle \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2) =K+2q+\frac{4q^2}{K}, \end{aligned}$$
(9.4)
$$\begin{aligned}&\displaystyle \frac{1}{R(q)^5}-q^2R(q)^5 =K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}, \end{aligned}$$
(9.5)
$$\begin{aligned}&\displaystyle \frac{R(q^2)}{R(q)^7}+q^2\frac{R(q)^7}{R(q^2)}=K+6q+\frac{20q^2}{K}+\frac{32q^3}{K^2}+\frac{64q^4}{K^3}, \end{aligned}$$
(9.6)
$$\begin{aligned}&\displaystyle \frac{1}{R(q)^{10}}+q^4R(q)^{10} =K^2+8qK+34q^2+\frac{96q^3}{K}\nonumber \\&\displaystyle \quad +\frac{192q^4}{K^2}+\frac{2546q^5}{K^3}+\frac{256q^6}{K^4}, \end{aligned}$$
(9.7)
$$\begin{aligned}&\displaystyle \frac{1}{R(q)^8R(q^2)}-q^4R(q)^8R(q^2) =K^2+6qK+20q^2+\frac{44q^3}{K}+\frac{64q^4}{K^2}+\frac{64q^5}{K^3}, \end{aligned}$$
(9.8)
$$\begin{aligned}&\displaystyle \frac{R(q^2)}{R(q)^{12}}-q^4\frac{R(q)^{12}}{R(q^2)} =K^2+10qK+52q^2+\frac{180q^3}{K}+\frac{448q^4}{K^2}\nonumber \\&\displaystyle \quad +\frac{832q^5}{K^3}+\frac{1024q^6}{K^4}+\frac{1024q^7}{K^5}, \end{aligned}$$
(9.9)
$$\begin{aligned}&\displaystyle K+q =\frac{E(q^2)^4E(q^5)^2}{E(q)^2E(q^{10})^4}, \end{aligned}$$
(9.10)
$$\begin{aligned}&\displaystyle 1-\frac{4q}{K} =\frac{E(q)^4E(q^{10})^2}{E(q^2)^2E(q^5)^4} \end{aligned}$$
(9.11)
and
$$\begin{aligned} \frac{q}{K-4q}=\zeta . \end{aligned}$$
(9.12)
Proofs of (9.1)–(9.4)
We see that (9.1) is (2.5), (9.2) is (2.7), (9.3) is (2.9) and (9.4) is (2.10). \(\square \)
Proof of (9.5)
$$\begin{aligned} \frac{1}{R(q)^5}-q^2R(q^5)&=\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \left( \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\right) \\&\qquad +\left( \frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2\right) \\&=\frac{4q}{K}\left( K+2q+\frac{4q^2}{K}\right) +K\\&=K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}. \end{aligned}$$
\(\square \)
Proof of (9.6)
$$\begin{aligned} \frac{R(q^2)}{R(q)^7}+q^2\frac{R(q)^7}{R(q^2)}&=\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \left( \frac{1}{R(q)^5}-q^2R(q)^5\right) \\&\quad +\left( \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\right) \\&=\frac{4q}{K}\left( K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}\right) +\left( K+2q+\frac{4q^2}{K}\right) \\&=K+6q+\frac{20q^2}{K}+\frac{32q^3}{K^2}+\frac{64q^4}{K^3}. \end{aligned}$$
\(\square \)
Proof of (9.7)
$$\begin{aligned} \frac{1}{R(q)^{10}}+q^4R(q)^{10}&=\left( \frac{1}{R(q)^5}-q^2R(q)^5\right) ^2+2q^2\\&=\left( K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}\right) ^2+2q^2\\&=K^2+8qK+34q^2+\frac{96q^3}{K}+\frac{192q^4}{K^2}+\frac{256q^5}{K^3}+\frac{256q^7}{K^4}. \end{aligned}$$
\(\square \)
Proof of (9.8)
$$\begin{aligned}&\frac{1}{R(q)^8R(q^2)}-q^4R(q)^8R(q^2)\\&\quad =\left( \frac{1}{R(q)^5}-q^2R(q)^5\right) \left( \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\right) \\&\qquad -q^2\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \\&\quad =\left( K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}\right) \left( K+2q+\frac{4q^2}{K}\right) -q^2\left( \frac{4q}{K}\right) \\&\quad =K^2+6qK+20q^2+\frac{44q^3}{K}+\frac{64q^4}{K^2}+\frac{64q^5}{K^3}. \end{aligned}$$
\(\square \)
Proof of (9.9)
$$\begin{aligned}&\frac{R(q^2)}{R(q)^{12}}-q^4\frac{R(q)^{12}}{R(q^2)}\\&\quad =\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \left( \frac{1}{R(q)^{10}}+q^4R(q)^{10}\right) \\&\qquad +\left( \frac{1}{R(q)^8R(q^2)}-q^4R(q)^8R(q^2)\right) \\&\quad =\frac{4q}{K}\left( K^2+8qK+34q^2+\frac{96q^3}{K}+\frac{192q^4}{K^2}+\frac{256q^5}{K^3}+\frac{256q^5}{K^3}\right) \\&\qquad +\left( K^2+6qK+20q^2+\frac{44q^3}{K}+\frac{64q^4}{K^2}+\frac{64q^5}{K^3}\right) \\&\quad =K^2+10qK+52q^2+\frac{180q^3}{K}+\frac{448q^4}{K^2}+\frac{832q^5}{K^3}+\frac{1024q^6}{K^4}+\frac{1024q^7}{K^5}. \end{aligned}$$
\(\square \)
Proofs of (9.10) and (9.11)
We see that (9.10) is (2.11) and (9.11) is (2.12).
\(\square \)
Proof of (9.12)
$$\begin{aligned} K-4q=K\left( 1-\frac{4q}{K}\right) =\frac{E(q^2)E(q^5)^5}{E(q)E(q^{10})^5}\cdot \frac{E(q)^4E(q^{10})^2}{E(q^2)^2E(q^5)^4}=\frac{E(q)^3E(q^5)}{E(q^2)E(q^{10})^3}=\frac{q}{\zeta }, \end{aligned}$$
from which the result follows. \(\square \)
Proof of (4.2)
We start by noting that (4.2) is equivalent to
$$\begin{aligned} U(\zeta )=41\zeta +860\zeta ^2+6800\zeta ^3+24000\zeta ^4+32000\zeta ^5. \end{aligned}$$
We have
$$\begin{aligned}&U(\zeta )=U\left( q\frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)}\right) \\&\quad =\frac{E(q^2)^3}{E(q)}\,U\left( q\frac{E(q^2)}{E(q)^3}\right) \\&\quad =\frac{E(q^2)^3}{E(q)}\,U\left( qE(q^{50})\left( \frac{1}{R(q^{10})}-q^2-q^4R(q^{10})\right) \right. \\&\qquad \left. \times \left( \frac{E(q^{25})^5}{E(q^5)^6}\right) ^3\left( \frac{1}{R(q^5)^4}+\frac{q}{R(q^5)^3}+\frac{2q^2}{R(q^5)^2}+\frac{3q^3}{R(q^5)}+5q^4\right. \right. \\&\qquad \left. \left. -3q^5R(q^5)+2q^6R(q^5)^2-q^7R(q^5)^3+q^8R(q^5)^4\right) ^3\right) \\&\quad =\frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\left( 51q\left( \frac{1}{R(q)^8R(q^2)}-q^4R(q)^8R(q^2)\right) \right. \\&\qquad \left. -9q\left( \frac{1}{R(q)^{10}}+q^4R(q)^{10}\right) -q\left( \frac{R(q^2)}{R(q)^{12}}-q^4\frac{R(q)^{12}}{R(q^2)}\right) \right. \\&\qquad \left. +153q^2\left( \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\right) -177q^2\left( \frac{1}{R(q)^5}-q^2R(q)^5\right) \right. \\&\qquad \left. -78q^2\left( \frac{R(q^2)}{R(q)^7}+q^2\frac{R(q)^7}{R(q^2)}\right) -219q^3\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) -71q^3\right) \\&\quad =\frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\\&\qquad \times \left( 51q\left( K^2+6qK+20q^2+\frac{44q^3}{K}+\frac{64q^4}{K^2}+\frac{64q^5}{K^3}\right) \right. \\&\qquad \left. -9q\left( K^2+8qK+34q^2+\frac{96q^3}{K}+\frac{192q^4}{K^2}+\frac{256q^5}{K^3}+\frac{256q^6}{K^4}\right) \right. \\&\qquad \left. -q\left( K^2+10qK+52q^2+\frac{180q^3}{K}+\frac{448q^4}{K^2}+\frac{832q^5}{K^3}+\frac{1024q^6}{K^4}+\frac{1024q^7}{K^5}\right) \right. \\&\qquad \left. +153q^2\left( K+2q+\frac{4q^2}{K}\right) -177q^2\left( K+4q+\frac{8q^2}{K}+\frac{16q^3}{K^2}\right) \right. \\&\qquad \left. -78q^2\left( K+6q+\frac{20q^2}{K}+\frac{32q^3}{K^2}+\frac{64q^4}{K^3}\right) -219q^3\left( \frac{4q}{K}\right) -71q^3\right) \\&\quad =\frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\\&\qquad \times \frac{(K+q)^2(K-4q)}{K^5}(41qK^4+204q^2K^3+416q^3K^2+384q^4K+256q^5)\\&\quad =\frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\cdot \frac{(K+q)^2(K-4q)}{K^5}\\&\qquad \times (41q(K-4q)^4+860q^2(K-4q)^3+6800q^3(K-4q)^2\\&\qquad +24000q^4(K-4q)+32000q^5)\\&\quad =\frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\cdot \frac{(K+q)^2(K-4q)^6}{K^5}\\&\qquad \times \left( \frac{41q}{K-4q}+\frac{860q^2}{(K-4q)^2}+\frac{6800q^3}{(K-4q)^3}+\frac{24000q^4}{(K-4q)^4}+\frac{32000q^5}{(K-4q)^5}\right) \\&\quad =\left( \frac{E(q^2)^3E(q^5)^{15}E(q^{10})}{E(q)^{19}}\right) \left( \frac{E(q^2)^4E(q^5)^2}{E(q)^2E(q^{10})^4}\right) ^2 \left( \frac{E(q)^3E(q^5)}{E(q^2)E(q^{10})^3}\right) ^6 \\&\qquad \times \left( \frac{E(q)E(q^{10})^5}{E(q^2)E(q^5)^5}\right) ^5 (41\zeta +860\zeta ^2+6800\zeta ^3+24000\zeta ^4+32000\zeta ^5)\\&\quad =41\zeta +860\zeta ^2+6800\zeta ^3+24000\zeta ^4+32000\zeta ^5. \end{aligned}$$
\(\square \)
In proceeding in the same manner with proofs of (4.3)–(4.6), (5.6)–(5.10) and (5.12)–(5.16), we encounter terms of the form
$$\begin{aligned} P(\alpha ,\beta )=\frac{1}{R(q)^{\alpha +2\beta }R(q^2)^{2\alpha -\beta }}+(-1)^{\alpha +\beta }q^{2\alpha }R(q)^{\alpha +2\beta }R(q^2)^{2\alpha -\beta } \end{aligned}$$
with \(\alpha \ge 0\).
These terms can be expressed in terms of \(q,\ K\) and \(K^{-1}\) by making use of the recurrence relations
$$\begin{aligned} P(\alpha ,\beta +1)&=\frac{4q}{K}P(\alpha ,\beta )+P(\alpha ,\beta -1), \end{aligned}$$
(9.13)
$$\begin{aligned} P(\alpha +2,0)&=KP(\alpha +1,0)+q^2P(\alpha ,0) \end{aligned}$$
(9.14)
and
$$\begin{aligned} P(\alpha +2,-1)&=\left( K-2q+\frac{4q^2}{K}\right) P(\alpha +1,0)-q^2P(\alpha ,1), \end{aligned}$$
(9.15)
together with the initial values
$$\begin{aligned} P(0,0)&=2, \end{aligned}$$
(9.16)
$$\begin{aligned} P(0,1)&=\frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}=\frac{4q}{K}, \end{aligned}$$
(9.17)
$$\begin{aligned} P(1,0)&=\frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2=K \end{aligned}$$
(9.18)
and
$$\begin{aligned} P(1,-1)&=\frac{R(q)}{R(q^2)^3}+q^2\frac{R(q^2)^3}{R(q)}=K-2q+\frac{4q^2}{K}. \end{aligned}$$
(9.19)
We see that (9.17) is (9.1), (9.18) is (9.2) and (9.19) is (9.3).
Proofs of (9.13)–(9.15) were given by Chern and Tang [2].
