Entry 127

Assume \(\tau=n\sqrt{-\color{blue}{3}}\) for some positive integer \(n\). Given \(_2F_1(a,b;c;z)\) where \(a+b=c=\frac12\) for the case \(a=\tfrac1{6}\). Let \(z_1 = (1-2w)^2\) where \(w\) is $$w=\frac{27}{27+\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12}}$$ Then \((z_1, z_2)\) are algebraic numbers in

$$_2F_1\left(\tfrac16,\tfrac13;\tfrac12;z_1\right) = z_2$$

Example: 

If \(n=2\) so \(\tau=2\sqrt{-3}\), then,

$$_2F_1\left(\frac16,\frac13;\frac12;\,\frac{25}{27}\right)=\frac{3\sqrt3}{4}$$ 

If \(n=4\) so \(\tau=4\sqrt{-3}\), then,

$$\qquad _2F_1\left(\frac16,\frac13;\frac12;\,\frac{45^2\,(\sqrt2+\sqrt3)^2}{(1+\sqrt6)^8}\right)=\frac58\big(1+\sqrt6\big)$$

Entry 126

Assume \(\tau=n\sqrt{-\color{blue}{2}}\) for some positive integer \(n\). Given \(_2F_1(a,b;c;z)\) where \(a+b=c=\frac12\) for the case \(a=\tfrac1{8}\). Let \(z_1 = (1-2w)^2\) where \(w\) is $$w=\frac{64}{64+\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24}}$$ Then \((z_1, z_2)\) are algebraic numbers in

$$_2F_1\left(\tfrac18,\tfrac38;\tfrac12;z_1\right) = z_2$$

Examples: 

If \(n=2\) so \(\tau=2\sqrt{-2}\), then,

$$_2F_1\left(\frac18,\frac38;\frac12;\frac{(70\sqrt{2})^2\,(1+\sqrt2)^2}{(2+3\sqrt2)^6}\right)=\frac34\left(\frac{2+3\sqrt2}2\right)^{1/2}$$

If \(n=3\) so \(\tau=3\sqrt{-2}\), then,

$$_2F_1\left(\frac18,\frac38;\frac12;\frac{2400}{2401}\right)=\tfrac23\cdot7^{1/2}$$

Entry 125

Assume \(\tau=n\sqrt{-\color{blue}{1}}\) for some positive integer \(n\). Given \(_2F_1(a,b;c;z)\) where \(a+b=c=\frac12\) for the case \(a=\tfrac1{12}\). Let \(j(\tau)\) be the j-function and \(z_1 = (1-2w)^2\) where \(w\) is a root of $$\frac{12^3}{4w(1-w)} =j(\tau)$$ Or more simply \(z_1=1-\frac{1728}{j(\tau)}\), then \((z_1, z_2)\) are algebraic numbers in

$$\,_2F_1\left(\tfrac1{12},\tfrac5{12};\tfrac12;z_1\right) = z_2$$

Examples: 

If \(n=2\) so \(\tau=2\sqrt{-1}\) and \(j(\tau)=66^3\), then,

$$_2F_1\left(\frac1{12},\frac5{12};\frac12;\frac{1323}{1331}\right)=\frac34\cdot11^{1/4}$$ 

If \(n=3\) so \(\tau=3\sqrt{-1}\), then,

$$\qquad _2F_1\left(\frac1{12},\frac5{12};\frac12;\frac{(14\sqrt6)^2\,(72+43\sqrt3)}{(21+20\sqrt3)^3}\right)=\frac23(21+20\sqrt3)^{1/4}$$

Entry 124

In Entry 123, the \(24\)th power of the golden ratio \(\phi\) and Gauss' constant \(G\) was discussed. We will use it again and connect it to the Dedekind eta function \(\eta(\tau)\) and Watson's triple integral $$\begin{align}I_1 &= \frac{1}{\pi^3}\int_0^\pi \int_0^\pi \int_0^\pi \frac{dx\, dy\, dz}{1-\cos x\cos y\cos z}\\ &=\frac{\Gamma^4(\frac{1}{4})}{4\pi^3}\\ &= 2\,G^2 = 4\,\eta(i)^4 = 1.393203\dots\end{align}$$ where $$I_1 =\frac{\,25\phi^6}{\sqrt{\phi^{24}-4}}\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \left(\frac{-\phi^{16}}{4(\phi^{24}-4)}\right)^{3n}=1.393203\dots$$ Notice that the \(24\)th power of the golden ratio is off by \(4\).

Entry 123

In the previous entry, we had the unusual hypergeometric function, call it \(\beta\) $$\beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G$$ with Gauss's constant \(G\). Ramanujan found the unusual equality $$\sqrt[8]{1+\sqrt{1-\left(\tfrac{-1+\sqrt{5}}{2}\right)^{24}}} = \tfrac{-1+\sqrt{5}}{2}\,\left(\tfrac{1+\sqrt[4]{5}}{\sqrt{2}}\right)$$ and one can notice the \(24\)th powers in both cases. After some manipulation, we find the similar $$\sqrt[8]{\left(\tfrac{1+\sqrt{5}}2\right)^{12}-\sqrt{\left(\tfrac{1+\sqrt{5}}2\right)^{24}-1}} = \sqrt{\tfrac{1+\sqrt{5}}2}\left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)\quad$$ therefore $$\beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\tfrac{1+\sqrt5}2\right) \left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)^2 G$$

Entry 122

Given Gauss' constant \(G\), the elliptic integral singular value \(K(k_r)\), $$\begin{align}G &=\frac{\sqrt2}{\pi} K(k_1) = \frac2{\pi}\int_0^{\pi/2}\frac{d\,\theta}{\sqrt{1+\sin^2\theta}}\\ &= \tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\tfrac12\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;-1\big)\\ &=(2\pi)^{-3/2}\,\Gamma^2\big(\tfrac14\big) = 0.834626\dots\end{align}$$ and modular lambda function \(\lambda(\tau)\) calculated by Mathematica as ModularLambda[tau]. The two hypergeometrics can be expressed as $$\tfrac1{\sqrt2}\,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(i)\big) = \,_2F_1\big(\tfrac12,\tfrac12;1;\lambda(1+i)\big) = G$$ while more complicated ones are $$\begin{align}\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/1\big)\Big) &= 2^{1/4}G\\ \,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/3\big)\Big) &= 6^{1/4}\left((1+\sqrt{3})^2-\sqrt{(2+\sqrt3)^4-1}\right)^{1/4}G\\ \quad\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) &= 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G\end{align}$$ and so on, with fundamental units \(U_3 = 2+\sqrt3\) and \(U_5= \frac{1+\sqrt5}2\), and where the \(\lambda(\tau)\) are actually radicals. Curiously, \(\lambda(i)\ = \lambda\big(\sqrt{3+4i}\big) = \frac12\).

Entry 121

This is the case \(a=\frac13\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$ \frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{4}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{4}\big)= \frac3{5^{5/6}}$$

$$ \frac{1}{48^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+\color{blue}{27}x^3}}=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\color{blue}{27}\big)=\frac{4}{7}$$ (To be continued.) 

Entry 120

This is the case \(a=\frac14\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{3}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}3\big) = \frac{2}{3^{3/4}}$$

$$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{80}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}{80}\big) = \frac35$$ (To be continued.) 

Entry 119

This is the case \(a=\frac16\) of $${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}$$ we have $$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{\tfrac{125}3}x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{\tfrac{125}{3}})=\frac{2}{3^{5/6}}$$

$$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{2^7\phi^9}\, x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{2^7\phi^9})=\frac{3}{5^{5/6}}\phi^{-1}$$ with golden ratio \(phi\). (To be continued.) 

Entry 118

If one has a palindromic quartic of form $$z^4-abz^3+(a^2+b^2-2)z^2-abz+1=0$$ then its roots can be factored as roots \((x,y)\) of quadratics $$x^2+ax+1=0\\ y^2+by+1= 0$$ $$z = xy = \left(\tfrac{-a+\sqrt{a^2-4}}2\right) \left(\tfrac{-b+\sqrt{b^2-4}}2\right)$$ hence are products of quadratic units. (To be continued.)

Entry 117

For fundamental discriminants \(d=4m\) with class number \(h(-d)=16\), there are exactly 60 \(m\) that are even. The largest is \(m = 3502 = 2\times17\times103\) hence has \(2^3 = 8\) divisors. But this set has no \(m\) with 32 divisors so it seems one can't express their modular lambda function \(\lambda(\sqrt{-m})\) with 16 quadratic units. (Unlike for \(h(-d)=8\) where we can express a few \(\lambda(\sqrt{-m})\) with 8 quadratic units.) However, using another function, we can have four quartic units. Given the nome \(q = e^{\pi i\tau}\), \(\tau=\sqrt{-n}\), and the Ramanujan G and g functions $$\begin{align}2^{1/4}G_n &= q^{-\frac{1}{24}}\prod_{k>0}(1+q^{2k-1}) = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)}\\ 2^{1/4}g_n &= q^{-\frac{1}{24}}\prod_{k>0}(1-q^{2k-1}) = \frac{\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\end{align}$$ discussed in Entry 116. Then $$\color{red}u  = (g_{3502})^4 = \small\left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^4 = \big(a+\sqrt{a^2-1}\big)^2 \big(b+\sqrt{b^2-1}\big)^2 \big(c+\sqrt{c^2-1}\big) \big(d+\sqrt{d^2-1}\big) \approx 1.43\times10^{13}$$ where \(\tau = \sqrt{-3502}\) and \((a,b,c,d)\) are $$\begin{align}a &= \tfrac{1}{2}(23+4\sqrt{34})\\ b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\ c &= (429+304\sqrt{2})\\ d &= \tfrac{1}{2}(627+442\sqrt{2})\end{align}$$A version of this was first found by Daniel Shanks in 1980 (Quartic Approximations for Pi) but this one is slightly different with smaller integers since I simplified the first two expressions as squares. These radicals imply a very close approximation to pi,$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2\color{red}u)^6+24\big)$$ which differs by just \(10^{-161}.\)

Entry 116

For fundamental discriminants \(d=4m\) with class number \(h(-d)=8\), there are exactly 29 \(m\) that are even. The three \(m = 210, 330, 462\) were discussed in Entry 114 since they have special properties. However, the four largest are \(m = 598, 658, 742, 862\). Given the nome \(q = e^{\pi i\tau}\), we can use the Ramanujan G and g functions $$\begin{align}2^{1/4}G_n &= q^{-\frac{1}{24}}\prod_{k>0}(1+q^{2k-1}) = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)}\\ 2^{1/4}g_n &= q^{-\frac{1}{24}}\prod_{k>0}(1-q^{2k-1}) = \frac{\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\end{align}$$ with \(\tau=\sqrt{-n}\) where Ramanujan uses \(G_n\) and \(g_n\) for odd \(n\) and even \(n\), respectively. For these four \(m = n\), then $$\begin{align}(g_{598})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \Big(6+\sqrt{26}+\sqrt{(6+\sqrt{26})^2-1}\Big)(1+\sqrt2)^2\Big(\tfrac{3+\sqrt{13}}2\Big)\\ (g_{658})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \, ?? \\ (g_{762})^2 &= \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2 = \Big(\tfrac{11+\sqrt{106}}2+\sqrt{\big(\tfrac{11+\sqrt{106}}2\big)^2-1}\Big)\,(1+\sqrt2)^2\,\Big(\tfrac{7+\sqrt{53}}2\Big)\\ (g_{862})^2 &=  \left(\frac{\;\eta\big(\tfrac{\tau}{2}\big)}{2^{1/4}\eta(\tau)}\right)^2  = \, ??\end{align}$$ I know the octics for the other two, though I don't know how to factor them into quartic units.

Entry 115

For fundamental discriminants \(d=4m\) with class number \(h(-d)=4\), there are exactly twelve \(m\) that are even. In Entry 113, the seven with 8 divisors were discussed. The remaining five are \(m = 14, 34, 46, 82, 142\) which are of form \(m=2p\) for prime \(p=7, 17, 23, 41, 71\). From experience, \(p\equiv 1\, (\text{mod}\, 4)\) are more well-behaved, hence for \(m=34,82\) $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-34})}} &= (1+\sqrt2)^2\sqrt{35+6\sqrt{34}} \left(\sqrt{\frac{5+\sqrt{17}}4}+\sqrt{\frac{1+\sqrt{17}}4}\right)^4 \\ \frac1{\sqrt{\lambda(\sqrt{-82})}} &\,=\, (1+\sqrt2)^4\, (9+\sqrt{82})\,\left(\sqrt{\frac{7+\sqrt{41}}2}+\sqrt{\frac{5+\sqrt{41}}2}\right)^4 \end{align}\quad$$ For \(m = 14, 46,142\), presumably they may be products of two quartic units $$ \frac1{\sqrt{\lambda(\sqrt{-m})}} = \big(a+\sqrt{a^2\pm1}\big)\big(b+\sqrt{b^2\pm1}\big)$$ where \((a,b)\) are roots of quadratics, but I haven't figured out the correct values yet.

Entry 114

This continues Entry 113. Recall the modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$calculated in Mathematica as ModularLambda[tau]. We now chose fundamental discriminants \(d=4m\) with class number \(h(-d)=8\) for even \(m\) with 16 divisors. There are only three, namely \(m = 210, 330, 462\), hence $$\begin{align}210 &= 2\times3\times5\times7\\ 330 &= 2\times3\times5\times11\\ 462 &= 2\times3\times7\times11\end{align}$$ Ramanujan found \(m=210\), though I'm not sure he did for the other two, so I went ahead and found them $$\small\begin{align}\frac1{\sqrt{\lambda(\sqrt{-210})}} &= (1+\sqrt{2})^2 (2+\sqrt{3}) (8+3\sqrt{7}) (3+\sqrt{10})^2 (4+\sqrt{15})^2 (6+\sqrt{35}) (\sqrt{6}+\sqrt{7})^2 (\sqrt{14}+\sqrt{15}) \\ \frac1{\sqrt{\lambda(\sqrt{-330})}} &= (1+\sqrt{2})^2(2+\sqrt{3})^3(3+\sqrt{10})^2(10+3\sqrt{11})(4+\sqrt{15})(65+8\sqrt{66}) (\sqrt{44}+\sqrt{45})^2(\sqrt{54}+\sqrt{55}) \\ \frac1{\sqrt{\lambda(\sqrt{-462})}} &= (2+\sqrt{3})^2(5+2\sqrt{6})^2 (8+3\sqrt{7})^2(10+3\sqrt{11})(15+4\sqrt{14})(76+5\sqrt{231})(7\sqrt{2}+3\sqrt{11})^2(\sqrt{21}+\sqrt{22})\end{align}$$ They are products of eight fundamental units \(U_n\). To find products of sixteen fundamental units \(U_n\), I checked class number \(h(-d)=16\) for even \(m\) with 32 divisors from the class number list. Surprisingly, there are none. So it seems the pattern of \(\lambda(\sqrt{-m})\) as a product of \(\color{blue}{2^k}\) quadratic units \(U_n\) where class number \(h(-4m)=\color{blue}{2^k}\) stops at \(8\). Note that $$\quad e^{\pi\sqrt{462}} = \frac{16}{\lambda\big(\sqrt{-462}\big)}-8.0000000000000000000000000000094\dots$$ or for 29 zeros.

Entry 113

This continues Entry 112. Recall the modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We chose fundamental discriminants \(d=4m\) with class number \(h(-d)=4\) for even \(m\) with eight divisors, hence only \(m = 30, 42, 78, 102\), and \(m =70, 130, 190\) which are evenly divisible by \(3\) and \(5\), respectively.  $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-30})}} &= (2+\sqrt3)(5+2\sqrt6)(4+\sqrt{15})\sqrt{11+2\sqrt{30}}\\  \frac1{\sqrt{\lambda(\sqrt{-42})}} &= (2+\sqrt3)^2(1+\sqrt2)^2(8+3\sqrt{7})\sqrt{13+2\sqrt{42}}\\ \frac1{\sqrt{\lambda(\sqrt{-78})}} &= (2+\sqrt3)^3(5+2\sqrt6)(25+4\sqrt{39})\sqrt{53+6\sqrt{78}}\\ \frac1{\sqrt{\lambda(\sqrt{-102})}} &=(2+\sqrt3)^2(5+2\sqrt6)^2(50+7\sqrt{51})\sqrt{101+10\sqrt{102}}\end{align}$$ as well as $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-70})}} &= (8+3\sqrt7)(15+4\sqrt{14})(6+\sqrt{35})\sqrt{251+30\sqrt{70}}\\  \frac1{\sqrt{\lambda(\sqrt{-130})}} &= (1+\sqrt2)^4\,(3+\sqrt{10})^2\,(5+\sqrt{26})\,(57+5\sqrt{130})\\ \frac1{\sqrt{\lambda(\sqrt{-190})}} &= (170+39\sqrt{19})(37+6\sqrt{38})(39+4\sqrt{95})\sqrt{52021+3774\sqrt{190}}\end{align}$$

For the last, note that \(U_{190}=52021+3774\sqrt{190} = (51\sqrt{10}+37\sqrt{19})^2\) and similarly for other \(U_n\) with composite \(n\) but I chose to retain the \(a+b\sqrt{n}\) form. Most of these do not appear in Mathworld's list. Arranged by class number, one can see some patterns like the \(m\) have eight divisors and \(\lambda(\tau)\) is a product of four \(U_n\). Next are \(m\) that have sixteen divisors and \(\lambda(\tau)\) is a product of eight \(U_n\).

Entry 112

Regarding the previous two entries, the more famous function with evaluations that involve fundamental units \(U_n\) is the modular lambda function \(\lambda(\tau)\). Define the McKay-Thompson series of class 4C for the Monster (A007248) $$j_{4C}(\tau) = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8+16 = \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8$$ compare the RHS to $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ which solves $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = -\tau\,i$$ We chose fundamental discriminants \(d=4m\) with class number \(h(-d)=2\) for even \(m = 6, 10, 22, 58\), hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)\sqrt{5+2\sqrt6}\\ \frac1{\sqrt{\lambda(\sqrt{-10})}} &= U_2^3\,U_{10} = (1+\sqrt2)^2(3+\sqrt{10})\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})\sqrt{197+42\sqrt{22}}\\ \frac1{\sqrt{\lambda(\sqrt{-58})}} &=\, U_2^6\,U_{58}\; =\; (1+\sqrt2)^6(99+13\sqrt{58})\end{align}$$ Note that these \(m\) have four divisors and \(\lambda(\tau)\) is a product of two \(U_n\). Also $$\begin{align}U_6 &= 5+2\sqrt6 = (\sqrt2+\sqrt3)^2\\ U_{22} &= 197+42\sqrt{22} = (7\sqrt2+3\sqrt{11})^2\end{align}$$ can be expressed as squares, which simplify things. For the next entry, the \(m\) have eight divisors and \(\lambda(\tau)\) is a product of four \(U_n\).

Entry 111

Given \(q = e^{2\pi i \tau}\) and define the McKay-Thompson series of Class 10E for Monster (A138516) $$j_{10E}(\tau) =  \frac{\eta(2\tau)\,\eta^5(5\tau)}{\eta(\tau)\,\eta^5(10\tau)} + 1= \left(\frac{\eta^2(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta^2(10\tau)}\right)^2$$ On a hunch, I decided to test this since it seems similar to Class 6E of the previous entry. It turns out \(j_{10E}(\tau)\) is also product of fundamental units \(U_n\) for appropriate \(\tau\). (Here is a sample Wolfram calculation for \(U_5\).) Let \(d=20m\) with class number \(h(-d)=4\) for even \(m = 6,14,26,38\) and odd \(m=17\). $$\begin{align}j_{10E}\big(\tfrac{\sqrt{-6/5}}2\big) &= U_2\, U_ {10}\\ j_{10E}\big(\tfrac{\sqrt{-14/5}}2\big) &= U_2^3\, U_{10}\\ j_{10E}\big(\tfrac{\sqrt{-26/5}}2\big) &= U_{13}^3\, U_{65}\\ j_{10E}\big(\tfrac{\sqrt{-38/5}}2\big) &= U_2^4\, U_5^9 \\ j_{10E}\big(\tfrac{1+\sqrt{-17/5}}2\big) &= -U_5^6\, U_{17}\end{align}$$

Entry 110

Given \(q = e^{2\pi i \tau}\). Define the McKay-Thompson series of Class 6E for Monster (A128633) $$\begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align}$$ where \(\text{K}_6(q)\) is the cubic continued fraction. Just like the modular lambda function \(\lambda(\tau)\), it seems \(j_{6E}(\tau)\) is also a product of fundamental units \(U_n\) for appropriate \(\tau\). (Here is a sample Wolfram calculation for \(U_5\).) We choose \(d=12m\) with class number \(h(-d)=4\) for even \(m = 10, 14, 26, 34\) (also found in the previous entry)

$$\begin{align}j_{6E}\big(\tfrac{\sqrt{-10/3}}2\big) &= U_2^2\,U_5^6 \\ j_{6E}\big(\tfrac{\sqrt{-14/3}}2\big) &= U_6\,U_{14} \\ j_{6E}\big(\tfrac{\sqrt{-26/3}}2\big) &= U_2^4\,U_{26}^2 \\ j_{6E}\big(\tfrac{\sqrt{-34/3}}2\big) &= U_2^6\,U_{17}^2\end{align}$$ and odd \(m = 7, 11, 19, 31, 59\), 

$$\begin{align}j_{6E}\big(\tfrac{1+\sqrt{-7/3}}2\big) &= U_3\sqrt{U_{21}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-11/3}}2\big) &= U_6\sqrt{U_{33}} \\ j_{6E}\big(\tfrac{1+\sqrt{-19/3}}2\big) &= U_3^3\sqrt{U_{57}} \\ j_{6E}\big(\tfrac{1+\sqrt{-31/3}}2\big) &= U_3^3\sqrt{U_{93}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-59/3}}2\big) &= U_2^6\sqrt{U_{177}} \end{align}$$ Why this factors nicely I don't know. Also, some fundamental units \(U_n\) for composite \(n\) may actually be squares. For example, $$\begin{align}U_{21} &= \tfrac{5+\sqrt{21}}2=\Big(\tfrac{\sqrt{3}+\sqrt{7}}2\Big)^2\\ U_{33} &= 23+4\sqrt{33}=\big(2\sqrt{3}+\sqrt{11}\big)^2\end{align}$$ and so on, hence these \(\sqrt{U_n}\) simplifies a bit.

Entry 109

Let \(q = e^{2\pi i \tau}\) and Dedekind eta function \(\eta(\tau)\). Define the following $$\quad\text{K}_{12}(\tau) = \text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad $$ It is connected to the mod 6 version, or the cubic continued fraction $$\; \text{K}_{6}(\tau) = \text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\quad $$

by the quadratic relation 

$$\frac1{\text{K}_{12}(\tau)}+\text{K}_{12}(\tau) = \frac1{\text{K}_6(\tau)\,\text{K}_6(2\tau)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1$$ To find exact values of \(\text{K}_{12}(\tau)\), we use discriminants \(d = 12m\) with class number \(h(-d) = 4\) for even \(m=10,14,26,34\) to get the orderly $$\begin{align}\frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big) &= 1+\sqrt3\left(2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big) &= 1+\sqrt3\left(4+\sqrt{21}+\sqrt{1+(4+\sqrt{21})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big) &= 1+\sqrt3\left(15+4\sqrt{13}+\sqrt{1+(15+4\sqrt{13})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big) &= 1+\sqrt3\left(28+5\sqrt{34}+\sqrt{-1+(28+5\sqrt{34})^2}\right) \end{align}$$ Update: It turns out all the quartic roots can be factored into fundamentals units \(U_n\), for example $$2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}=(1+\sqrt2)\big(\tfrac{1+\sqrt5}2\big)^3 = U_2\,U_5^3$$ More on Entry 110.

Entry 108

Let \(q = e^{2\pi i \tau}\) and the Dedekind eta function \(\eta(\tau)\). Define the following $$\text{K}_{10}(\tau) = \text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})}$$ While no continued fraction is yet known for this, it is connected to the mod 5 version, $$R(\tau) =  \text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

or the Rogers-Ramanujan continued fraction \(R(\tau)\) by the simple relation

$$\text{K}_{10}(q) = R(q)\,R(q^2)\\ \text{K}_{10}(\tau) = R(\tau)\,R(2\tau)$$

It is known that$$\frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$ Thus a quadratic root $$R(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2$$ which allows for easily computation of \(\text{K}_{10}(\tau)\). For example, given the golden ratio \(\phi = \frac{1+\sqrt5}2\) and \(\sqrt{-1}=i\), $$\begin{align}R\big(\tfrac12 i\big) &= 0.511428\dots = \tfrac12(\sqrt{5\phi\,}-\phi^2)(+\sqrt[4]5\sqrt{\phi}+\phi^2)\\ R\big(i\big) &= 0.284079\dots = (\sqrt[4]5\sqrt{\phi}-\phi)\\ R\big(2i\big) &=  0.081002\dots =\tfrac12(\sqrt{5\phi\,}-\phi^2)(-\sqrt[4]5\sqrt{\phi}+\phi^2) \end{align}$$which implies the exact values$$\begin{align}\text{K}_{10}\big(\tfrac12 i\big) &= R\big(\tfrac12 i\big)\,R\big(i\big) = 0.145286\dots\\ \text{K}_{10}\big(i\big) &= R\big(i\big)\,R\big(2i\big) \,= 0.203011\dots \end{align}$$

Entry 107

Given \(q = e^{2\pi i \tau}\) and the two \(q\)-continued fractions $$\begin{align}\quad\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\quad\end{align}$$ $$\qquad \begin{align}\text{K}_8(q)\, &=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &=  \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\ddots}}}}\end{align}$$

then we can evaluate them using the modular lambda function \(\lambda(\tau)\) as $$\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}\quad$$ $$\quad\frac1{\text{K}_8(\tau)}-\text{K}_8(\tau)  = \frac{2}{\big(\lambda(4\tau)\big)^{1/4}}$$Mathematica can calculate \(\lambda(\tau)\) as ModularLambda[tau] and there is a list of exact values in Mathworld. We have previously given examples for \(\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}\). For \(\text{K}_8(\tau)\), then for \(d = 4m\) with class number \(h(-d)=2\) and \(m=10,58\), $$\begin{align}\frac1{\text{K}_8\big(\tfrac14\sqrt{-10}\big)}-\text{K}_8\big(\tfrac14\sqrt{-10}\big) &= 2(1+\sqrt2)\sqrt{3+\sqrt{10}} \;=\; 2\,U_2\sqrt{U_{10}}\\ \frac1{\text{K}_8\big(\tfrac14\sqrt{-58}\big)}-\text{K}_8\big(\tfrac14\sqrt{-58}\big) &= 2(1+\sqrt2)^3\sqrt{99+13\sqrt{58}} =2\,U_2^3\sqrt{U_{58}}\end{align}$$ where \(U_n\) are fundamental units.

Entry 106

Let \(q = e^{2\pi i \tau}\). In Entry 99, we gave the cubic continued fraction $$\begin{align}\text{K}_6(q)  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align}$$ Define the McKay-Thompson series of Class 6E for Monster (A128633)$$\begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align}$$ While it is possible to evaluate \(K_6(q)\), doing it for \(j_{6E}(\tau)\) seems more "easy" since it is a \(4\)th power. Let \(d=3m\) with class number \(h(-d)=2\) for \(m=17,41,89\). Then for $$\tau_1=\tfrac{1+\sqrt{-17/3}}2,\quad \tau_2=\tfrac{1+\sqrt{-41/3}}2,\quad \tau_3=\tfrac{1+\sqrt{-89/3}}2$$we get the evaluations$$\begin{align} -\sqrt{\frac{-3\;}{j_{6E}(\tau_1)}}+\sqrt{-\frac13\, j_{6E}(\tau_1)} &= 2^3+2(-2+2\sqrt{17})^{2/3}+2(2+2\sqrt{17})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_2)}}+\sqrt{-\frac13\, j_{6E}(\tau_2)} &= 4^3+4(-2+10\sqrt{41})^{2/3}+4(2+10\sqrt{41})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_3)}}+\sqrt{-\frac13\, j_{6E}(\tau_3)} &= 10^3+10(-2+106\sqrt{89})^{2/3}+10(2+106\sqrt{89})^{2/3}\end{align}$$ where the RHS are roots of cubics and all quadratic irrationals, for example \(72+8\sqrt{17} = (2+2\sqrt{17})^2\), are squares.

Entry 105

Let \(q = e^{2\pi i\tau}\). For convenience, define the Rogers-Ramanujan continued fraction in terms of \(\tau\), hence \(R(q) = R(\tau)\). Given the Dedekind eta function \(\eta(\tau)\), it is known that $$\frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$Thus a quadratic$$R(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2$$ For example, let \(\tau = \sqrt{-1}\), then the formula gives us Ramanujan's evaluation, $$R(\sqrt{-1}) = \sqrt[4]5\sqrt{\phi}-\phi = 0.284079\dots$$ with golden ratio \(\phi = \frac{1+\sqrt5}2\). The formula also works when \(\tau = \frac{1+\sqrt{-n}}2\) though \(R(\tau)\) is now negative. Since it contains \(q^{1/5}\), one has to be careful with \(5\)th roots. We give some nice new evaluations for discriminants \(d=5m\) with class number \(h(-d)=2\) for \(m=(3,7,23,47)\) also using the golden ratio \(\phi\)

$$\begin{align}\frac1{R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)-11 &= -(\sqrt5)^3\phi\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)-11 &= -(\sqrt5)^3\phi^3\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)-11 &= -(\sqrt5)^3\phi^9\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)-11 &= -(\sqrt5)^3\phi^{15}\\ \end{align}$$ Note the \(5\)th powers \(R^5(\tau)\) and how well-behaved their evaluations are.

Entry 104

Given \(q = e^{2\pi i \tau}\) and the \(q\)-continued fraction discussed in Entry 95. $$\begin{align}\quad\text{K}_4(q) &= \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\end{align}$$

and compare it to the modular lambda function $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta(\tfrac12\tau)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8\quad$$ therefore the two are related by $$\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}$$ Mathematica can calculate \(\lambda(\tau)\) as ModularLambda[tau] and there is list of exact values in Mathworld which also implies exact values for \(\text{K}_4(\tau)\). But we will give a nice consistent form when \(d = 4m\) has class number \(h(-d)=2\) for \(m=5,13,37\) as $$\begin{align}\lambda\big(\sqrt{-5}\big) &= \frac12-\sqrt{\left(\tfrac{{-1}+\sqrt{5}}2\right)^3}\\ \lambda\big(\sqrt{-13}\big) &= \frac12-3\sqrt{\left(\tfrac{{-3}+\sqrt{13}}2\right)^3}\\ \lambda\big(\sqrt{-37}\big) &= \frac12-21\sqrt{\left({-6}+\sqrt{37}\right)^3}\end{align}$$ as well as the complex values $$\begin{align}\frac1{\lambda\left(\frac{1+\sqrt{-5}}2\right)} &= \frac12-\sqrt{\left(\tfrac{{-1}-\sqrt{5}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-13}}2\right)} &= \frac12-3\sqrt{\left(\tfrac{{-3}-\sqrt{13}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-37}}2\right)} &= \frac12-21\sqrt{\left({-6}-\sqrt{37}\right)^3}\end{align}$$

Entry 103

Level 12. Given \(q = e^{2\pi i \tau}\),  the q-Pochhammer symbol \((a;q)_n\), and Ramanujan's functions \(f(a,b)\) and \(f(-q)\) discussed in Entry 92. We have the sum-product identities $$\begin{align}A(q) &= \frac{f(-q,-q^{11})}{f(-q^4)} = \sum_{n=0}^\infty \frac {q^{4n^2+4n}\,(q;q^2)_{2n+1}} {(q^4;q^4)_{2n+1}}  = \prod_{n=1}^\infty(1-q^{12n-1})(1-q^{12n-11})\,\alpha_{12}\\ B(q) &=\frac{f(-q^5,-q^{7})}{f(-q^4)} \,=\, \sum_{n=0}^\infty \frac {q^{4n^2}\,(q;q^2)_{2n}} {(q^4;q^4)_{2n}} \quad = \; \prod_{n=1}^\infty(1-q^{12n-5})(1-q^{12n-7})\,\alpha_{12}\end{align}$$

where \(\alpha_{12} = \dfrac{(1-q^{12n})}{(1-q^{4n})}\). 

Let \(a = q^{7/8}A(q)\) and \(b = q^{-1/8}B(q)\) then \((a,b)\) are radicals for appropriate \(\tau\). Define their ratio \(a/b\) $$\text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad\qquad$$ Surprisingly, this has a continued fraction (studied by Naika) based on a general form from Ramanujan for Level \(4m\), though its form is not as simple as the others. It is connected to the mod 6 version, $$\quad\text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})} = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$$ or the previously discussed cubic continued fraction \(\text{K}_{6}(q)\) by the quadratic relation 

$$j_{12I}=\frac1{\text{K}_{12}(q)}+\text{K}_{12}(q) = \frac1{\text{K}_6(q)\,\text{K}_6(q^2)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1$$

 where \(j_{12I}\) is the McKay-Thompson series of class 12I for the Monster (A187144).

Entry 102

Level 10. Given \(q = e^{2\pi i \tau}\),  the q-Pochhammer symbol \((a;q)_n\), and Ramanujan's functions \(f(a,b)\) and \(\varphi(q)\) discussed in Entry 92. We have the sum-product identities

$$\begin{align}A(q) &= \frac{f(-q,-q^9)}{\varphi(-q)} \;=\; \sum_{n=0}^\infty \frac {q^{n(n+3)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-3})(1-q^{10n-7})}\\ B(q) &= \frac{f(-q^3,-q^7)}{\varphi(-q)} = \sum_{n=0}^\infty \frac {q^{n(n+1)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-1})(1-q^{10n-9})}\end{align}$$

where \(\alpha_{10} = \dfrac{(1-q^{10n})^2}{(1-q^{n})(1-q^{5n})}\). 

Let \(a = q^{4/5}A(q)\) and \(b = q^{1/5}B(q)\) then \((a,b)\) are radicals for appropriate \(\tau\). Define their ratio \(a/b\), $$\text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})}$$ While no continued fraction is yet known for this, it is connected to the mod 5 version $$\text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

simply as, $$\text{K}_{10}(q) = \text{K}_5(q)\,\text{K}_5(q^2)$$

where \(\text{K}_5(q)=R(q)\) is just the Rogers-Ramanujan continued fraction.

Entry 101

Level 8. Recall the Level 4 continued fraction $$\begin{align}\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align}$$ Compare its similarity to the Level 8 version using the ratio of \((a,b)\) from Entry 100 $$\begin{align}\text{K}_8(q) &\,=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &= \cfrac{q^{1/2}}{1+\cfrac{q+q^2}{1+\cfrac{q^4}{1+\cfrac{q^3+q^6}{1+\cfrac{q^8}{1+\ddots}}}}} = \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\cfrac{q^8}{1+q^9+\ddots}}}}}\end{align}$$

In fact, they have the quadratic relation $$\quad\frac1{\text{K}_8(q)}-\text{K}_8(q) =\left(\frac{\sqrt2}{\text{K}_4(q^2)}\right)^2 = \left(\frac{\eta^3(4\tau)}{\eta(2\tau)\eta^2(8\tau)}\right)^2$$ while \(\text{K}_4(q)\) which is an eta quotient can also be expressed another way$$\frac1{\big(\text{K}_4(q)\big)^2}-\big(\text{K}_4(q)\big)^2 =\frac12\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^4\quad$$

Entry 100

Level 8. Given \(q = e^{2\pi i \tau}\),  the q-Pochhammer symbol, and Ramanujan's functions \(f(a,b)\) and \(\psi(q)\) discussed in Entry 92. We have the Gollnitz-Gordon sum-product identities,

$$\begin{align}A(q) &=\frac{f(-q,-q^7)}{\psi(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+2n}\,(-q;q^2)_n} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})}\\ B(q) &= \frac{f(-q^3,-q^5)}{\psi(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}\,(-q;q^2)_n} {(q^2;q^2)_n} \;\; = \;\; \prod_{n=1}^\infty\frac{1}{(1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})}\end{align}$$

Let \(a=q^{7/16}A(q)\) and \(b= q^{-1/16}\,B(q)\). Then \((a, b)\), for appropriate \(\tau\), are actually radicals.

Entry 99

Level 6. Using the eta quotients \((a,b)\) from Entry 98 their ratio \(\dfrac{a}{b}\) is $$\begin{align}\text{K}_6(q) &= q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ &=q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{2n-1})}{\; (1-q^{6n-3})^3} = q^{1/3}\prod_{n=1}^\infty \frac{(1+q^{3n})^3}{(1+q^{n})\;}\\  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align}$$

this is also known as the cubic continued fraction. For appropriate \(\tau\), then \((a, b, \text{K}_6)\) are radicals. The formula for the j-function using \(\text{K}_6(q)\) employs polynomial invariants of the tetrahedron and the integer \(12\) of \(b= q^{-1/12}B(q)\) in Entry 98 reflects the order \(12\) of the tetrahedral group. Note that the cube of the reciprocal of \(\text{K}_6(q)\) is the McKay-Thompson series of class 6E of the Monster (A105559)$$\left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4$$

Entry 98

Level 6. Given \(q = e^{2\pi i \tau}\), Ramanujan's theta function \(f(a,b)\), its one-parameter version \(f(-q) = f(-q,-q^2)\), and the q-Pochhammer symbol. We have the sum-product identities,

$$\begin{align}A(q) &= \frac{f(-q,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}(q,q^2)_n} {(-q,q)_{2n+1} (q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-3})(1-q^{6n-3})} = q^{-1/4}\frac{\eta(\tau)\eta^2(6\tau)}{\eta^2(2\tau)\eta(3\tau)}\\ B(q) &=  \frac{f(-q^3,-q^3)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}(q,q^2)_n} {(-q,q)_{2n} (q^2;q^2)_n} \;=\; \prod_{n=1}^\infty\frac{\alpha_6}{(1-q^{6n-1})(1-q^{6n-5})} = q^{1/12}\frac{\eta^2(3\tau)}{\eta(2\tau)\eta(6\tau)}\end{align}$$ where \(\alpha_6 = \dfrac{(1-q^{n})(1-q^{3n})}{(1-q^{2n})^2}\) 

Let \(a=q^{1/4}A(q)\) and \(b= q^{-1/12}\,B(q)\). Then \((a, b)\) for appropriate \(\tau\) are eta quotients like in Level 4 and are actually radicals.

Entry 97

Level 5 In Entry 96, we asserted that \(a=q^{11/60}H(q)\) and \(b= q^{-1/60}\,G(q)\) are radicals for appropriate \(\tau\). Proof: Given the j-function \(j(q)\) and let \(R(q)=r=a/b\)

$$\begin{align}a &= q^{11/60}H(q) = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{11/20}}{j(q)^{11/60}\,(r^{10}+11r^5-1)}\\ b &= q^{-1/60}G(q) =  \frac{j(q)^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}\end{align}$$ Getting their ratio \(a/b\), then it simplifies to

$$r = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3/5}}{j(q)^{1/5}\,(r^{10}+11r^5-1)}$$ Raising to the fifth power and moving \(j(q)\) to the LHS 

$$j(q)  = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3}}{r^5(r^{10}+11r^5-1)^5}$$ and we get the icosahedral equation for the j-function in terms of the Rogers-Ramanujan continued fraction \(R(q)=r\). Thus if \(j(q)\) and \(R(q)\) are radicals, then so are \((a, b)\) while the integer \(60\) of \(b= q^{-1/60}G(q)\) reflects the order \(60\) of the icosahedral group. 

Entry 96

Level 5. Given \(q = e^{2\pi i \tau}\), Ramanujan's theta function \(f(a,b)\), its one-parameter version \(f(-q) = f(-q,-q^2)\), and the q-Pochhammer symbol. We have the Rogers-Ramanujan sum-product identities,

$$\begin{align}H(q) &= \frac{f(-q,-q^4)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ G(q) &=  \frac{f(-q^2,-q^3)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n}  = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\end{align}$$ Their ratio is the Rogers-Ramanujan continued fraction

$$\begin{align}R(q) &= \frac{a}{b} = \frac{q^{11/60} H(q)}{q^{-1/60}\,G(q)} = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-2})}{(1-q^{5n-2})(1-q^{5n-3})}\\ &=\cfrac{q^{1/5}} {1+\cfrac{q} {1+\cfrac{q^2} {1+\cfrac{q^3} {1+\ddots}}}}\end{align}$$ Let \(a=q^{11/60}H(q)\) and \(b= q^{-1/60}\,G(q)\). Then \((a, b)\) for appropriate \(\tau\) are actually radicals. Proof in Entry 97.

Entry 95

Level 4. Using the eta quotients \((a,b)\) from Entry 94 we found that \(-a^8+b^8=c^8\) while their ratio is $$\begin{align}\text{K}_4(q) &= \frac{a}{b} = \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align}$$

For appropriate \(\tau\), then \((a, b, \text{K}_4)\) are radicals. The formula for the j-function using \(\text{K}_4(q)\) employs polynomial invariants of the octahedron and the integer \(24\) of \(b= q^{-1/24}B(q)\) in Entry 94 reflects the order \(24\) of the octahedral group. Note that the \(8\)th power of the reciprocal of \(\text{K}_4(q)\) without the \(\sqrt2\) is the McKay-Thompson series of class 4C of the Monster (A007248)$$\left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8-16=\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$

Entry 94

Level 4. Given \(q = e^{2\pi i \tau}\), Ramanujan's theta function \(f(a,b)\) and its one-parameter version as \(f(-q) = f(-q,-q^2)\). We have the sum-product identities,

$$\begin{align}A(q) &= \frac{f(-q,-q^3)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)}\\ B(q) &= \frac{f(-q^2,-q^2)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q^2;q^2)_n}  = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-1})(1-q^{4n-3})} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\end{align}$$

where \(\alpha_4 = \dfrac{(1-q^{2n})}{(1-q^{4n})}\). 

Let \(a = \sqrt2\,q^{1/12}A(q)\) and \(b= q^{-1/24}B(q)\), then \((a, b)\) are just eta quotients and for appropriate quadratic \(\tau\) are actually radicals. Furthermore, the difference of their \(8\)th powers is also an \(8\)th power $$-a^8+b^8 =-\left(\frac{\sqrt2\,\eta(4\tau)\,}{\;\eta(2\tau)}\right)^8+\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^8=\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8$$

Entry 93

Let \(q = e^{2\pi i \tau}\).  Ramanujan's theta function \(f(a,b)\) is given by $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2}\; b^{n(n-1)/2}$$ Then the following have nice \(q\)-continued fractions $$\begin{align}\text{K}_4(q) &= q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)} = q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ \text{K}_5(q) &= q^{1/5}\frac{f(-q,-q^4)}{f(-q^2,-q^3)} = q^{1/5} \prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\ \text{K}_6(q) &= q^{1/3}\frac{f(-q,-q^5)}{f(-q^3,-q^3)} = q^{1/3} \prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ \text{K}_8(q) &= q^{1/2}\frac{f(-q,-q^7)}{f(-q^3,-q^5)} = q^{1/2} \prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ \text{K}_{12}(q) &= q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)} = q^{1/1} \prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\end{align}$$ with the most famous being \(\text{K}_5(q)=R(q)\) since this is the Rogers-Ramanujan continued fraction. Each will be discussed in subsequent entries.

Entry 92

Given the q-Pochhammer symbol, $$\begin{align}(a;q)_n &= \prod_{k=0}^{n-1}(1-aq^k)\\ (a;q)_\infty &= \prod_{k=0}^{\infty}(1-aq^k)\end{align}$$

as well as the Ramanujan theta function,

$$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$

In his Notebooks, Ramanujan also defined four one-parameter versions he commonly used as,

$$\begin{align}\varphi(q) &= f(q,\, q)\\ f(-q) &= f(-q, -q^2)\\ \psi(q) &= f(q,\, q^3)\\ \chi(q) &=\frac{f(-q^2,-q^2)}{f(-q,-q^2)}\\ \end{align}$$

which we will also use in later entries. There are several simple relations between these four auxiliary functions, one of which I found using all four is the elegant Fermat curve of degree 8,

$$\big[f(-q)\,\chi(-q)\big]^8+\big[\sqrt2\,q^{1/8}\psi(-q)\big]^8 = \big[\varphi(-q^2)\big]^8$$

We normally assume \(q = e^{2\pi i \tau}\) unless otherwise specified. For simplicity, it will also be assumed \(\tau\) is an complex quadratic number so that certain functions later will also evaluate as radicals.

Entry 91

Here is a "bizarre" continued fraction from Ramanujan involving \(e\) and \(\pi\) $$\sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$ However, since \(\Gamma\big(\tfrac12\big) = \sqrt{\pi}\), we can express the LHS in terms of gamma functions and find a cubic counterpart using \(\Gamma\big(\tfrac13\big)\) 

$$\sum_{n=1}^\infty \frac{1}{2^n (\frac12)_n} = \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big) \sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}}$$

$$\sum_{n=1}^\infty \frac{1}{3^n (\frac13)_n} = \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} = 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}}$$

Ramanujan's identity then becomes a sum of two continued fractions with a cubic version,

$$\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} = 1+\cfrac{2/2}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cfrac{5/2}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

$$\Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}}= 1+\cfrac{2/3}{2+\cfrac{3/3}{3+\cfrac{4/3}{4+\cfrac{5/3}{5+\ddots}}}} \; \color{blue}+ \; \cfrac1{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{\color{red}5}{1+\ddots}}}}$$

where the 4th continued fraction (also by Ramanujan) is missing numerators \(P(n)=3n+1 = 4,7,10,13,\dots\) The four cfracs then have closed-forms as,

\begin{align}\Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} &= \Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \color{blue}+ \Big(\Gamma\big(\tfrac12,\tfrac12\big)\Big)\sqrt{\frac{e}{2}} \\ \Gamma\big(\tfrac13\big)\sqrt[3]{\frac{e}{9}} &= \Big(\Gamma\big(\tfrac13\big)-\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}} \color{blue}+ \Big(\Gamma\big(\tfrac13,\tfrac13\big)\Big)\sqrt[3]{\frac{e}{9}}\end{align}

Entry 90

The modular lambda function given by Mathematica as ModularLambda[tau] can solve $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1}$$ In the previous entry, we used the tribonacci constant for the case \(\tau=\sqrt{-11}\). More generally, to solve the higher Heegner numbers \(d = 11,19,43,67,163\) we can employ a slightly different method. For example, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)}  = \sqrt{163}$$ This also has an exact solution expressible in radicals \(x \approx 6.094\times10^{-17}\) as the smaller root of the quadratic $$16x^2-16x+1=y$$ and \(y\) is the real root of the cubic $$y^3+\frac{640320^3}{256}y-\frac{640320^3}{256}=0$$ The big number should be familiar and appears in Ramanujan's constant $$e^{\pi\sqrt{163}} = 640320^3 + 743.99999999999925\dots$$ For the other Heegner numbers \(d\), one just replaces the big number with the cube nearest to \(e^{\pi\sqrt{d}}-744\), like \(5280^3\) for \(\sqrt{67}\) and so on.

Entry 89

The modular lambda function \(\lambda(\tau)\)$$\lambda(\tau) = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8$$ given by Mathematica as ModularLambda[tau] can solve, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1}$$ For the special case \(\tau=\sqrt{-11}\) $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-u\big)}{_2F_1\big(\tfrac12,\tfrac12,1,u\big)}  = \sqrt{11}$$ we can can use the tribonacci constant \(T\), the real root of \(T^3-T^2-T-1=0\). The solution in terms of \(T\) is $$u=\lambda(\tau)=\frac14\left(2-\sqrt{\frac{2T+15}{2T+1}}\right) = \frac14\left(2-\sqrt{\frac{17S+2}{3S+2}}\right)= 0.00047741\dots$$ or in terms of \(S\) as the nice nested cube roots $$S = \frac1{T-1}=\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\sqrt[3]{\frac12+\dots}}}}$$ the cubic version of the golden ratio's $$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$

Entry 88

The modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8$$ discussed in the previous post solves, among other things, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1}$$ For example, let \(\tau=\sqrt{-2}\) so $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = \sqrt{2}$$

and the Mathematica command ModularLambda[tau] yields a real number equal to \(\lambda\big(\tau) = (1-\sqrt2)^2\). Other \(\tau=\sqrt{n}\) can be found in Mathworld's list. But we can use more general complex \(\tau\). For example, let \(\tau=\frac{1+\sqrt{-2}}2\) which is no longer in the list. So $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = {-\left(\tfrac{1+\sqrt{-2}}2\right)} \sqrt{-1}$$ and we find the complex number, $$\lambda(\tau) = 4\big({-1}+\sqrt2\big)^3\left(4+\sqrt{2(1-5\sqrt2)}\right) \approx 1.1370849 + 0.9905592 i$$ which is a root of quartic with two real roots and two complex roots. And so on for other complex quadratic irrationals \(\tau\).

Entry 87

Define \(d_k = \eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\). We have the nice $$\left(\frac{\sqrt2\,d_1\,d_4^2}{d_2^3}\right)^8+\left(\frac{d_1^2\,d_4}{d_2^3}\right)^8 = 1$$ Equivalently $$\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8=\left(\frac{d_1}{d_4}\right)^8+8$$ Focusing on the first term, note that the three similar Monster functions

$$j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8,\quad  j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad  j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2$$

being the McKay-Thompson series of class \(4C, 8E, 16B\), respectively, are necessary to the \(9\) dependencies found by Conway, Norton, and Atkins such that the moonshine functions span a linear space of \(172-9=163\) dimensions (discussed in previous entries). In fact, scaled and flipped over, it is an important function,  $$\lambda(\tau)=\left(\frac{\sqrt2\,d_{1/2}\,d_2^2}{d_1^3}\right)^8 = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8$$ known as the modular lambda function \(\lambda(\tau)\).

Entry 86

As discussed in the previous entry, the eta quotients \(\left(\frac{\eta(\tau)}{\eta(k\tau)}\right)\) are useful for j-function formulas. The easy levels \(k\) are when \(m=24/(k-1)\) is an integer. For square \(k=(4,9,25)\) yields \(m=(8,3,1)\) which can be found in the exponents below. $$j(\tau) =\frac{(x^2-48)^3}{x^2-64}, \quad\text{with}\quad x=\left(\frac {\sqrt4\,\eta(4\tau)}{\eta(\tau)}\right)^8+8$$

$$j(\tau) =\frac{x^3(x^3-24)^3}{x^3-27}, \quad\text{with}\quad x=\left(\frac {\sqrt9\,\eta(9\tau)}{\eta(\tau)}\right)^3+3$$

$$j(\tau) = \frac{-(x^{20}+12 x^{15}+14 x^{10}-12 x^5+1)^3}{x^{25} (x^{10}+11 x^5-1)} , \quad\text{with}\;\; x^{-1}-x=\left(\frac {\sqrt{25}\,\eta(25\tau)}{\eta(\tau)}\right)^1+1$$

Alternatively,

$$j(\tau) =\frac{(x^2+192)^3}{(x^2-64)^2}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$

$$j(\tau) =\frac{x^3(x^3+216)^3}{(x^3-27)^3}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$

$$j(\tau) = \frac{-(r^{20}-228 r^{15}+494 r^{10}+228 r^5+1)^3}{r^5 (r^{10}+11 r^5-1)^5} , \quad\text{with}\;\; r^{-1}-r=\left(\frac {\eta(\tau/5)}{\eta(5\tau)}\right)^1+1$$ After some manipulation, these eta quotients are related to q-continued fractions with octahedral, tetrahedral, and icosahedral symmetries, with \(r=R(q)\) being the well-known Rogers-Ramanujan continued fraction.

Entry 85

The eta quotients \(\left(\frac{\eta(\tau)}{\eta(k\tau)}\right)\) discussed previously are useful for j-function formulas. The easy levels \(k\) are when \(m=24/(k-1)\) is an integer. For prime \(k=(2,3,5,7,13)\) yields \(m=(24,12,6,4,2)\) which are the exponents of \(x\) below. $$j(\tau) =\frac{(x-16)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt2\,\eta(2\tau)}{\;\eta(\tau)}\right)^{24}$$

$$\; j(\tau) =\frac{(x+3)^3(x+27)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt3\,\eta(3\tau)}{\;\eta(\tau)}\right)^{12}$$

$$j(\tau) \,=\,\frac{(x^2+10x+5)^3}x, \;\quad\text{with}\;\; x=\left(\frac {\sqrt5\,\eta(5\tau)}{\;\eta(\tau)}\right)^6$$

$$j(\tau) =\frac{(x^2+5x+1)^3(x^2+13x+49)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4$$

$$j(\tau) =\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x, \;\text{with}\;\; x=\left(\frac {\sqrt{13}\,\eta(13\tau)}{\;\eta(\tau)}\right)^2$$

Entry 84

Summarizing, we find tentative patterns for $$\text{I. Levels}\; 4n+1 = (5,13)\\ \text{II. Levels}\; 4n+2 = (6,10)\\ \text{III. Levels}\; 4n+3 = (3,7)\,$$The definitions for these functions are in previous entries, but it may be good to have a few selected values together to get an overall picture. Note that \(U_n\) are fundamental units, important to Pell equations. 

I. Levels \(4n+3 = (3,7)\)

$$\quad \begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -(\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -(2\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &= -(4\sqrt3)^6 \\j_{3B}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -3^3U_{5}^2 =-3^3\left(\tfrac{1+\sqrt{5}}2\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -3^3U_{17}^2 =-3^3\left(4+\sqrt{17}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &=-3^3U_{41}^2 =-3^3\left(32+5\sqrt{41}\right)^2 \end{align}$$ and one can observe its similarity to Level 7 $$\begin{align}j_{7A}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -(\sqrt7)^2\\ j_{7A}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -(3\sqrt7)^2 \\ j_{7A}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -(39\sqrt7)^2\\ j_{7B}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\,U_{5}^{2} = -7\left(\tfrac{1+\sqrt5}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\,U_{13}^{2} = -7\left(\tfrac{3+\sqrt{13}}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\,U_{61}^{2} = -7\left(\tfrac{39+5\sqrt{61}}2\right)^{2}\end{align}$$

II. Levels \(4n+1 = (5,13)\)

$$\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &= -(6\sqrt{23})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &= -(18\sqrt{47})^2\\ j_{5B}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &=  -(\sqrt5)^3\left(\tfrac{1+\sqrt{5}}2\right)^9\\ j_{5B}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &=  -(\sqrt5)^3\left(\tfrac{1+\sqrt{5}}2\right)^{15}\end{align}$$ where we see the golden ratio above and the bronze ratio below $$\begin{align}j_{13A}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -(\sqrt7)^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -(2\sqrt{31})^2\\ j_{13B}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = {-\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)}\\ j_{13B}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)^3 \end{align}$$

III. Levels \(4n+2 = (6,10)\)

$$\quad \begin{align}j_{6A}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -20^2\\  j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -1060^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -U_{11}^2 = -\big(10+3\sqrt{11}\big)^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -U_{59}^2 = -\big(530+69\sqrt{59}\big)^2 \end{align}$$ compared to $$\begin{align}j_{10A}\big(\tfrac12\sqrt{-6/5}\big) &= 6^2\\ j_{10A}\big(\tfrac12\sqrt{-38/5}\big) &= 76^2\\ j_{10D}\big(\tfrac12\sqrt{-6/5}\big) &=\, U_{10}^{2}\, = \left(3+\sqrt{10}\right)^{2}\\ j_{10D}\big(\tfrac12\sqrt{-38/5}\big) &=\, U_{5}^{18}\, = \left(\tfrac{1+\sqrt5}2\right)^{18} \end{align}$$

Entry 83

Level 13. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 13B $$j_{13B}(\tau) = \left(\frac{d_1}{d_{13}}\right)^2$$

Examples. Let \(d=13m\) with class number \(h(-d)=2\) for \(m=7, 31\) and we find a phenomenon similar to Level 5 $$\begin{align}j_{13B}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)\\ j_{13B}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -\sqrt{13}\left(\tfrac{3+\sqrt{13}}2\right)^3 \end{align}$$

But where Class 5B involves the fundamental unit \(U_5\) or the golden ratio, Class 13B involves \(U_{13} = \tfrac{3+\sqrt{13}}2\) also known as the bronze ratio.

Entry 82

Level 13. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 13A for the Monster. $$j_{13A}(\tau) = \left(\frac{d_1}{d_{13}}\right)^2+13\left(\frac{d_{13}}{d_1}\right)^2+6$$

Examples. We select \(d=13m\) for class number \(h(-d)=2\) with \(m=7, 31\) as well as class number \(h(-d)=6\) with \(m=19, 151\) and find a well-behaved pattern just like for level 5 $$\begin{align}j_{13A}\Big(\tfrac{1+\sqrt{-7/13}}{2}\Big) & = -(\sqrt7)^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-31/13}}{2}\Big) & = -(2\sqrt{31})^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-19/13}}{2}\Big) & = -(x\sqrt{19})^2\\ j_{13A}\Big(\tfrac{1+\sqrt{-151/13}}{2}\Big) & = -(y\sqrt{151})^2\end{align}$$

where \((x,y)\) are roots of cubics $$19x^3 - 38x^2 + 21x - 9 = 0\\ 151y^3 - 2567y^2 - 512y - 44 = 0$$ and so on for class number \(h(-d)=10\), etc.

Entry 81

Level 10. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 10D for the Monster (A132130). $$j_{10D}(\tau) = \left(\frac{d_2\,d_5}{d_1\,d_{10}}\right)^6$$

Examples. We select \(d=20m\) with class number \(h(-d)=4\) and find odd \(m=17\) as well as $$m=6, 14, 26, 38$$ such that the following are special quadratic irrationals $$\begin{align}j_{10D}\Big(\tfrac{1+\sqrt{-17/5}}{2}\Big) &= -U_{5}^{12} = -\left(\tfrac{1+\sqrt5}2\right)^{12}\\ j_{10D}\big(\tfrac12\sqrt{-6/5}\big) &=\, U_{10}^{2}\, = \left(3+\sqrt{10}\right)^{2}\\ j_{10D}\big(\tfrac12\sqrt{-14/5}\big) &=\, U_{2}^{6}\, = \left(1+\sqrt2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-26/5}\big) &=\, U_{13}^{6}\, = \left(\tfrac{3+\sqrt{13}}2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-38/5}\big) &=\, U_{5}^{18}\, = \left(\tfrac{1+\sqrt5}2\right)^{18} \end{align}$$

as they are fundamental units \(U_n\). Since \(U_5=\phi\) is the golden ratio, then the first and last implies the integers $$\left(\sqrt{-\phi^{12}}-1/\sqrt{-\phi^{12}}\right)^2 = -18^2\\ \left(\sqrt{\phi^{18}}-1/\sqrt{\phi^{18}}\right)^2 = 76^2$$ and similarly for the others as discussed in the previous entry.

Entry 80

Level 10. Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and the McKay-Thompson series of class 10A for the Monster. 

$$j_{10A}(\tau) = \left(\left(\frac{d_2\,d_5}{d_1\,d_{10}}\right)^3-\left(\frac{d_1\,d_{10}}{d_2\,d_5}\right)^3\right)^2$$ Examples. We select \(d=20m\) with class number \(h(-d)=4\) and find odd \(m=17\) as well as $$m=6, 14, 26, 38$$ such that the following are well-behaved integers $$\begin{align}j_{10A}\Big(\tfrac{1+\sqrt{-17/5}}{2}\Big) &= -18^2\\ j_{10A}\big(\tfrac12\sqrt{-6/5}\big) &= 6^2\\ j_{10A}\big(\tfrac12\sqrt{-14/5}\big) &= 14^2\\ j_{10A}\big(\tfrac12\sqrt{-26/5}\big) &= 36^2\\ j_{10A}\big(\tfrac12\sqrt{-38/5}\big) &= 76^2 \end{align}$$ Note that the last is responsible for the prime-generating polynomial $$F(n)=10n^2+19$$ which has discriminant \(d = -5\times 38 = -190\) and is prime for \(19\) consecutive values \(n=0 - 18\).

Entry 79

Level 7. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 7B for the Monster $$j_{7B}(\tau) = \left(\frac{d_1}{d_7}\right)^4$$ Examples. We select \(d=7m\) with class number \(h(-d)=2\) and find \(m=5,13,61\) such that the following are special quadratic irrationals $$\begin{align}j_{7B}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\,U_{5}^{2} = -7\left(\tfrac{1+\sqrt5}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\,U_{13}^{2} = -7\left(\tfrac{3+\sqrt{13}}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\,U_{61}^{2} = -7\left(\tfrac{39+5\sqrt{61}}2\right)^{2}\end{align}$$

as they involve fundamental units \(U_n\). These are analogous to the examples in Level 3B which have the form \(-3^3\,U_n^2\). The last implies the integer $$\left(\sqrt{-7\,U_{61}}+7/\sqrt{-7\,U_{61}}\right)^2 = -7\times39^2=-22^3+1$$ and solutions to the curve \(x^3-1 = 7y^2\) as discussed in the previous entry.

Entry 78

Level 7. This level is special since we have to consider the curve \(x^3-1 = 7y^2\). Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and the McKay-Thompson series of class 7A for the Monster.  $$j_{7A}(\tau) = \left(\left(\frac{d_1}{d_7}\right)^2+7\left(\frac{d_7}{d_1}\right)^2\right)^2$$ Examples. We select \(d=7m\) with class number \(h(-d)=2\) and find \(m=5,13,61\) such that the following are well-behaved integers $$\begin{align}j_{7A}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\times1^2 = -2^3+1\\ j_{7A}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\times3^2 = -4^3+1\\ j_{7A}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\times39^2 = -22^3+1 \end{align}$$ A WolframAlpha search for positive integer solutions to \(x^3-1 = 7y^2\) reveals only these three. However, if we allow \((x,y)\) to be higher algebraic integers with the same odd degree and a solvable Galois group, then it seems there are infinitely many. For example, we select \(d=7m\) with class number \(h(-d)=6\) and find \(m=101\) and others so $$j_{7A}\Big(\tfrac{1+\sqrt{-101/7}}{2}\Big) =-x^3+1=-7y^2$$ where \((x,y)\) are the real roots of cubics $$x^3 - 46x^2 - 380x - 800 = 0\\ y^3 - 145y^2 - 357y - 1235=0$$ For class number \(h(-d)=10\), we find \(m=17\) and others so $$j_{7A}\Big(\tfrac{1+\sqrt{-17/7}}{2}\Big) =-x^3+1=-7y^2$$ where \((x,y)\) are the real roots of solvable quintics $$x^5 - 5x^4 + 4x^3 - 15x^2 - 23x - 11 = 0\\ y^5 - 5y^4 + 5y^3 - 7y^2 - 1=0$$ and so on.

Entry 77

Level 6. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 6B for the Monster.

$$j_{6B}(\tau) = \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12}$$

Example. We select \(d=12m\) with class number \(h(-d)=4\) and find  

$$m=10, 14, 26, 34\\ m=7, 11, 19, 31, 59$$ such that the following are special quadratic irrationals $$\begin{align}j_{6B}\big(\tfrac12\sqrt{-10/3}\big) &= U_5^{12}=\big(\tfrac{1+\sqrt5}2\big)^{12}\\ j_{6B}\big(\tfrac12\sqrt{-14/3}\big) &= U_{14}^2 =\big(15+4\sqrt{14}\big)^2 \\ j_{6B}\big(\tfrac12\sqrt{-26/3}\big) &= U_{26}^4 =\big(5+\sqrt{26}\big)^4\\ j_{6B}\big(\tfrac12\sqrt{-34/3}\big) &= U_2^{12}=\big(1+\sqrt{2}\big)^{12}\end{align}$$

$$\quad \begin{align}j_{6B}\Big(\tfrac{1+\sqrt{-7/3}}{2}\Big) &= -U_{21}^{3}= -\big(\tfrac{5+\sqrt{21}}2\big)^{3}\\ j_{6B}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -U_{11}^2 = -\big(10+3\sqrt{11}\big)^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-19/3}}{2}\Big) &= -U_{3}^6 = -\big(2+\sqrt{3}\big)^6\\ j_{6B}\Big(\tfrac{1+\sqrt{-31/3}}{2}\Big) &= -U_{93}^{3}= -\big(\tfrac{29+3\sqrt{93}}2\big)^{3}\\ j_{6B}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -U_{59}^2 = -\big(530+69\sqrt{59}\big)^2 \end{align}$$

since they are fundamental units \(U_n\). Note the integer $$\left(\sqrt{-\big(530+69\sqrt{59}\big)^2}-1/\sqrt{-\big(530+69\sqrt{59}\big)^2}\right)^2 = -1060^2$$ and similarly for the others as discussed in the previous entry.

Entry 76

Level 6. Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and the McKay-Thompson series of class 6A for the Monster. 

$$j_{6A}(\tau) = \left(\left(\frac{d_2\,d_3}{d_1\,d_6}\right)^6-\left(\frac{d_1\,d_6}{d_2\,d_3}\right)^6\right)^2$$ Examples. We select \(d=12m\) with class number \(h(-d)=4\) and find  

$$m=10, 14, 26, 34\\ m=7, 11, 19, 31, 59$$ such that the following are well-behaved integers $$\begin{align}j_{6A}\big(\tfrac12\sqrt{-10/3}\big) &= (8\sqrt5)^2\\ j_{6A}\big(\tfrac12\sqrt{-14/3}\big) &= (8\sqrt{14})^2\\ j_{6A}\big(\tfrac12\sqrt{-26/3}\big) &= (20\sqrt{26})^2\\ j_{6A}\big(\tfrac12\sqrt{-34/3}\big) &= (140\sqrt2)^2 \end{align}$$ as well as $$\begin{align}j_{6A}\Big(\tfrac{1+\sqrt{-7/3}}{2}\Big) &= -(4\sqrt7)^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -20^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-19/3}}{2}\Big) &= -52^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-31/3}}{2}\Big) &= -(28\sqrt{31})^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -1060^2\end{align}$$ Note that the prime-generating polynomials $$F(n)=n^2-n+41\\ F(n)=6n^2-6n+31$$ where the latter is prime for \(30\) consecutive values \(n=0 - 29\). Solving \(F(n)=0\) yields \(n=\frac{1+\sqrt{-163}}{2}\) and \(n=\frac{1+\sqrt{-59/3}}{2}\), respectively hence $$j_{1A}\Big(\tfrac{1+\sqrt{-163}}2\Big) = -640320^3\\ j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}2\Big) = -1060^2\quad$$

Entry 75

V. Level 5. The McKay-Thompson series of class 5B for the Monster.

$$j_{5B}(\tau) =\left(\frac{\eta(\tau)}{\eta{(5\tau)}}\right)^{6}$$

Examples. Surprisingly, the following "cubes" are powers of the golden ratio \(\phi=\frac{1+\sqrt5}2\)  $$\begin{align}j_{5B}\Big(\tfrac{1+\sqrt{-7/5}}{2}\Big) &= -(\sqrt5)^3\,\phi^3\\ j_{5B}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &=  -(\sqrt5)^3\,\phi^9\\ j_{5B}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &=  -(\sqrt5)^3\,\phi^{15}\end{align}$$ These discriminants \(d=5m\) have class number \(h(-d)=2\).

Entry 74

In the previous entries, Levels \(1, 2, 3, 4\) were discussed. Level \(5\) is intimately connected to the famous Rogers-Ramanujan continued fraction \(R(q)\). Let \(q = e^{2\pi i \tau}\) then 

$$x=\frac1{R(q)}-R(q) =  \frac{\eta(\tau/5)}{\eta(5\tau)}+1\\ \quad y=\frac1{R^5(q)}-R^5(q) = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11$$ Eliminating \(R(q)\) between the two and we get the relationship between \((x,y)\) as the solvable DeMoivre quintic,

$$x^5+5x^3+5x=y$$ which is solvable in radicals for any \(y\). As usual, define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and $$j_{5A}(\tau) = \left(\frac{d_1}{d_5}\right)^6+5^3\left(\frac{d_5}{d_1}\right)^6+22$$ Examples

$$\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-7/5}}{2}\Big) &= -(2\sqrt7)^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &= -(6\sqrt{23})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &= -(18\sqrt{47})^2\\ \end{align}$$ Compare to a similar phenomenon for Level 5. These have discriminant \(d=5p\) for prime \(p=7,23,47\) and have class number 2. For class number 6,

$$\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-103/5}}{2}\Big) &= -(x\sqrt{103})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-167/5}}{2}\Big) &= -(y\sqrt{167})^2\\ \end{align}$$

where \((x,y)\) are the real roots of the cubics

$$103 x^3 + 12566 x^2 - 12316 x + 15272 = 0\\ 167 y^3 + 113226 y^2 + 6372 y + 216= 0$$ and so on.

Entry 73

IV. Level 4. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\). Then the McKay-Thompson series of class 4C for the Monster (A007248) $$j_{4C}(\tau) = \left(\frac{d_1}{d_4}\right)^8$$ Examples. $$\begin{align}j_{4C}(\tfrac12\sqrt{-3}) &= 2^4\,U_3^2 = 2^4(2+\sqrt3)^2\\ j_{4C}(\tfrac12\sqrt{-7}) &= 2^4\,U_7^2 = 2^4(8+3\sqrt7)^2\end{align}$$ where \(U_n\) are fundamental units. Since $$\left(\frac{d_1}{d_4}\right)^8+16 = \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8$$ where the RHS is related to the modular lambda function, then we can use \( \left(\frac{d_1}{d_4}\right)^8\) for the case \(\tau = \tfrac12\sqrt{-n}\) to solve $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)} = \sqrt{n}$$ where \(x=\dfrac{16}{\left(\frac{d_1}{d_4}\right)^8+16}\). For example, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)} = \sqrt{3}$$ has solution $$x=\frac{16}{16(2+\sqrt3)^2+16}$$

Entry 72

 IV. Level 4. The McKay-Thompson series of class 4A for the Monster (A097340)

$$\begin{align}j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+4^2\left(\frac{d_4}{d_1}\right)^4\right)^2\\ &=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\end{align}$$ The second form shows they can be \(12\)th powers. Examples: 

$$j_{4A}\big(\tfrac12\sqrt{-7}\big)=2^{12}$$

which has class number 1 (but non-fundamental \(d\)). For class number 2,

$$\begin{align}j_{4A}\Big(\tfrac{1+\sqrt{-6}}{2}\Big) &= -2^6\left(1+\sqrt2\right)^{4}\\ j_{4A}\Big(\tfrac{1+\sqrt{-10}}{2}\Big) &= -2^6\left(\frac{1+\sqrt5}2\right)^{12}\\ j_{4A}\Big(\tfrac{1+\sqrt{-22}}{2}\Big) &= -2^6\left(1+\sqrt2\right)^{12}\\ j_{4A}\Big(\tfrac{1+\sqrt{-58}}{2}\Big) &= -2^6\left(\frac{5+\sqrt{29}}2\right)^{12}\end{align}$$

Entry 71

III. Level 3. The McKay-Thompson series of class 3B for the Monster.

$$j_{3B}(\tau) =\left(\frac{d_1}{d_3}\right)^{12}$$ Examples: $$\begin{align}j_{3B}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -3^3U_{5}^2 =-3^3\left(\tfrac{1+\sqrt{5}}2\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -3^3U_{17}^2 =-3^3\left(4+\sqrt{17}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &=-3^3U_{41}^2 =-3^3\left(32+5\sqrt{41}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-89/3}}{2}\Big) &= -3^3U_{89}^2=-3^3\left(500+53\sqrt{89}\right)^2\end{align}$$ These \(d=3m\) have class number \(h(-d)=2\). The quadratic irrationals have already appeared in Level 1 and are fundamental units \(U_n\), solutions to Pell equations \(x^2-ny^2=\pm1\). See also class 7B.

Entry 70

 III. Level 3. The McKay-Thompson series of class 3A for the Monster (A030197)

$$\begin{align}j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\ &=  \left(\left(\frac{d_1}{d_3}\right)^2+9\Big(\frac{d_9^3}{d_1\,d_3^2}\Big)\right)^6\end{align}$$ The second form shows they may be \(6\)th powers. Examples: 

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -(\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -(2\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &= -(4\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-89/3}}{2}\Big) &= -(10\sqrt3)^6\end{align}$$ Compare to a similar phenomenon for Level 2. These have discriminant \(d=3p\) for prime \(p=5,17,41,89\) and have class number 2. For class number 6,

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-29/3}}{2}\Big) &= -(x\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-113/3}}{2}\Big) &= -(y\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-137/3}}{2}\Big) &= -(z\sqrt3)^6\end{align}$$ where \((x,y,z)\) are the real roots of the cubics

$$x^3 - x^2 - 4x - 5 = 0\\ y^3 - 14y^2 - 4y - 16 =0\\ z^3 - 22z^2 + 44z - 32 = 0$$ and so on. For class number 10, $$j_{3A}\Big(\tfrac{1+\sqrt{-53/3}}{2}\Big) = -(u\sqrt3)^6$$ where \(u\) is the real roof of the solvable quintic $$u^5 - 8u^4 + 19u^3 - 26u^2 + 16u - 11 = 0$$ as well as for other \(d\).

Entry 69

II. Level 2. The McKay-Thompson series of class 2B for the Monster.

$$j_{2B}(\tau) =\left(\frac{d_1}{d_2}\right)^{24}$$ Examples: 

$$\begin{align}j_{2B}\big(\tfrac12\sqrt{-10}\big) & =\, 2^6\,U_{5}^{12} = 2^6\left(\tfrac{1+\sqrt{5}}2\right)^{12}\\ j_{2B}\big(\tfrac12\sqrt{-58}\big) & = \,2^6\,U_{29}^{12} = 2^6\left(\tfrac{5+\sqrt{29}}2\right)^{12}\\  j_{2B}\Big(\tfrac{1+\sqrt{-5}}2\Big) & = -2^6\,U_{5}^6 = -2^6\left(\tfrac{1+\sqrt{5}}2\right)^6\\ j_{2B}\Big(\tfrac{1+\sqrt{-13}}2\Big) & = -2^6\,U_{13}^6 = -2^6\left(\tfrac{3+\sqrt{13}}2\right)^6\\ j_{2B}\Big(\tfrac{1+\sqrt{-37}}2\Big) & = -2^6\,U_{37}^6 = -2^6\big(6+\sqrt{37}\big)^6\end{align}$$ which have class number \(h(-d)=2\).  The quadratic irrationals \(U_n\) are fundamental units, solutions to Pell equations \(x^2-ny^2=-1\) as discussed in Level 1.

Entry 68

II. Level 2. The McKay-Thompson series of class 2A for the Monster

$$\begin{align}j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\ &= \left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4\end{align}$$ The second form shows they may be \(4\)th powers. Examples: 

$$j_{2A}\big(\tfrac12\sqrt{-10}\big)=12^4\\ \; j_{2A}\big(\tfrac12\sqrt{-58}\big)=396^4\\  j_{2A}\Big(\tfrac{1+\sqrt{-5}}2\Big) = -\big(4\sqrt2\big)^4\\ j_{2A}\Big(\tfrac{1+\sqrt{-13}}2\Big) = -\big(12\sqrt2\big)^4\\ j_{2A}\Big(\tfrac{1+\sqrt{-37}}2\Big) = -\big(84\sqrt2\big)^4$$ which have class number \(h(-d)=2\). For class number \(h(-d)=4\)

$$\begin{align}j_{2A}\Big(\tfrac{1+\sqrt{-17}}2\Big) &= -2^{11}\big(4+\sqrt{17}\big)^2 \big({-1}+\sqrt{17}\big)\\ j_{2A}\Big(\tfrac{1+\sqrt{-73}}2\Big) &= -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3\\ j_{2A}\Big(\tfrac{1+\sqrt{-97}}2\Big) &= -2^{11}\cdot3^4\big(59+6\sqrt{97}\big)^4\big({-9}+\sqrt{97}\big)\\ j_{2A}\Big(\tfrac{1+\sqrt{-193}}2\Big) &= -2^{11}\cdot3^4\big(208+15\sqrt{193}\big)^4\big(903+65\sqrt{193}\big)\end{align}$$ All the quadratic irrationals with odd powers are odd fundamental solutions to Pell equations \(x^2-dy^2=-16\). For example, the initial solution to \(x^2-193y^2=-16\) is \((x,y)=(903,\,65)\) which appears above.

Entry 67

Let \(j=j(\tau)\) be the j-function. Define $$j_{1B}(\tau)=432\frac{\sqrt{j}+\sqrt{j-1728}}{\sqrt{j}-\sqrt{j-1728}}=\frac1q-120+10260q-901120q^2+\dots$$ which is (A299954). Then the following values $$\begin{align}j_{1B}\big(\tfrac{1+\sqrt{-19}}2\big) &= -432\,U_{19a}=-432\left(\sqrt{96^3/12^3}+\sqrt{96^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) &= -432\,U_{43b}=-432\left(\sqrt{960^3/12^3}+\sqrt{960^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-67}}2\big) &= -432\,U_{67c}=-432\left(\sqrt{5280^3/12^3}+\sqrt{5280^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-163}}2\big) &= -432\,U_{163c}=-432\left(\sqrt{640320^3/12^3}+\sqrt{640320^3/12^3+1}\right)^2\end{align}$$ are fundamental units, solutions to Pell equations \(x^2-ny^2=\pm1\) and where \((a,b,c,d) = (6, 15, 330, 10005)\). The smaller Heegner numbers like \(d=11\) don't yield fundamental units. And the second one \(U_{43b}=U_{645}\) involves a cube (Wolfram computation), $$j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) = -432\,U_{645}=-432\left(\frac{127+5\sqrt{645}}2\right)^3$$ which seems surprising. Note that $$432\big(\sqrt{-U_{645}}+1/\sqrt{-U_{645}}\big)^2=-960^3$$

Entry 66

Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\).

The McKay-Thompson series of class 1A for the Monster (A007240), disregarding the constant term \(744\), is the well-known j-function \(j(\tau)\). Given the three Weber modular functions \(\mathfrak{f}_n\) 

$$\begin{align}j_{1A}(\tau) &=\left(\frac{\mathfrak{f}(\tau)^{16}+\mathfrak{f}_1(\tau)^{16}+\mathfrak{f}_2(\tau)^{16}}{2}\right)^3\\ &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3\end{align}$$ They tend to be cubes, but not always. Examples: 

$$\begin{align}j_{1A}\big(\tfrac{1+\sqrt{-67}}{2}\big) &= -12^3\big(21^2-1\big)^3 \,=\, -5280^3\\ j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big) &= -12^3\big(231^2-1\big)^3=-640320^3\end{align}$$ while the two highest \(d\) with class number \(h(-d)=2\) are

$$\; j_{1A}\big(\tfrac{1+\sqrt{-403}}{2}\big)=-12^3\big((5301 + 1470\sqrt{13})^2-1\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-427}}{2}\big)=-12^3\big((7215 + 924\sqrt{61})^2-1\big)^3$$ and so on, though if \(d\) is a multiple of \(3\) then it is almost a cube $$\begin{align}j_{1A}\big(\tfrac{1+\sqrt{-15}}2\big) &=-3^3\,U_5^2\,\big(5+4\sqrt{5}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-51}}2\big) &=-48^3\,U_{17}^2\,\big(5+\sqrt{17}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-123}}2\big) &=-480^3\,U_{41}^2\,\big(8+\sqrt{41}\big)^3\\ j_{1A}\big(\tfrac{1+\sqrt{-267}}2\big) &=-240^3\,U_{89}^2\,\big(625+53\sqrt{89}\big)^3\end{align}$$ where \(U_n\) are fundamental units, $$\begin{align}U_5 &= \tfrac{1+\sqrt{5}}2\\ U_{17} &= 4+\sqrt{17}\\ U_{41} &= 32+5\sqrt{41}\\ U_{89} &= 500+53\sqrt{89}\end{align}$$ or fundamental solutions to the Pell equation \(x^2-ny^2 = -1\) with the first as the golden ratio \(\phi\ =U_5\). These will appear later in Class 3B.