Entry 190

From Entry 184, the quintic \(y(y-5)^4 =  j_2(\tau)\) has a solvable Galois group. For example

$$\begin{align}y(y-5)^4 &= -(4\sqrt2)^4\\ y(y-5)^4 &=  -(12\sqrt2)^4\\ y(y-5)^4 &= 12^4\\ y(y-5)^4 &= 396^4\qquad\end{align}$$

But these can also be expressed as

$$\begin{align}x^4(x+5) &= (4\sqrt2)i\\ x^4(x+5) &=  (12\sqrt2)i \\ x^4(x-5) &= 12\\ x^4(x-5) &= 396\qquad\end{align}$$

In general, since there is a relationship between \(j_2(\alpha)\) and \(j_4(\beta)\) where the latter is expressible by Ramanujan's \(G_m\)-function, then,

Conjecture: The special Bring-Jerrard quintics below are solvable

$$\begin{align}x^5+5x &= \sqrt{2^3\left(\frac1{G_m^{12}}-G_m^{12}\right)}\\ x^5-5x &= \sqrt{2^3\left(\frac1{g_m^{12}}+g_m^{12}\right)}\end{align}$$

For example, \(G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}\,\) and \(g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\) yields

$$\begin{align}x^5+5x &= (4\sqrt2)i\\ x^5-5x &= 12\qquad\end{align}$$

respectively, where the latter is quite well-known as an example of a solvable quintic with small integer coefficients. And so on for all Ramanujan \(G\) and \(g\)-functions, with explicit examples in the previous entries.

Entry 189

Ramanujan's \(g_m\)-function has a consistent form for fundamental discriminants \(d=4m\) with class number \(2\) and \(4\), where \(m=2p\) for primes \(p\equiv 3\,\text{mod}\,4\). For example, for \(p=3,11\) and \(p=7,23,71\) then

$$\begin{align}g_{6}&= \sqrt[6]{1+\sqrt2}\\ g_{22}&= \sqrt{1+\sqrt2}\\ g_{14}&= \sqrt{\frac{-1+\sqrt2}4}+  \sqrt{\frac{4+(-1+\sqrt2)}4}\\ g_{46}&= \sqrt{\frac{1+\sqrt2}4}+  \sqrt{\frac{4+(1+\sqrt2)}4}\\ g_{142}&= \sqrt{\frac{(1+\sqrt2)^3}4}+  \sqrt{\frac{4+(1+\sqrt2)^3}4}\end{align}$$

It seems \(\sqrt{p}\) is not needed but only \(\sqrt{2}\), or the fundamental unit \(U_2 = 1+\sqrt{2}\) also known as the silver ratio. For the modular lambda function \(\lambda(\sqrt{-m})\) for these same \(m\), see also Entry 134 where

$$\frac1{\lambda(\sqrt{-14})} = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 ,\quad n=2(1+\sqrt2)$$

Entry 188

Ramanujan g-function \(g_m\) and \(\tau = \sqrt{-m}\) is,

$$2^{1/4}g_m = \frac{\eta\big(\frac12\tau\big)}{\eta(\tau)}$$ Ramanujan tabulated \(G_m\) and \(g_m\) for odd and even \(m\), respectively. For the latter and \(m\geq4\), we propose a slightly different formula,

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-m})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}g_m\big)^n}}$$

with modular lambda function \(\lambda(z)\). Values for \(g_m\) have been given in Entry 183. Using some of them, we conjecture

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-6})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}(1+\sqrt2)^{1/6}\big)^n}}$$

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-22})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}(1+\sqrt2)^{1/2}\big)^n}}$$

and so on.

Entry 187

For the fourth McKay-Thompson series for the Monster $$j_{4}(\tau) = \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}$$Since we have Ramanujan's \(G_m\)-function for \(\tau = \sqrt{-m}\) as

$$2^{1/4}G_m = \frac{\eta^2(\tau)}{\eta(\tfrac12\tau)\,\eta(2\tau)}$$

then \(j_4\big(\tfrac12\tau\big)\) is just the \(24\)th power of \(2^{1/4}G_m\). Hence we propose 

$$K(k_m) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/4}G_m\big)^{24n}}}$$

Examples. The values for \(G_5, G_{13}, G_{37}\) have already been given in Entry 181 and involve roots of quadratics, like \(G_5 = \phi^{1/4}\) with golden ratio \(\phi\). But for cubics, given the tribonacci constant \(T\), the plastic ratio \(P\) and the supergolden ratio \(\psi\), the real root of

$$\begin{align}T^3-T^2-T-1 &= 0\\ P^3-P-1 &=0\\ \psi^3-\psi^2-1 &=0 \end{align}$$

then \(G_{11}, G_{23}, G_{31}\) are

$$\begin{align}G_{11} &= 2^{-1/4}\left(\tfrac{T+1}T\right) \\ G_{23} &= 2^{1/4}P \\ \,G_{31} &=2^{1/4}\psi\end{align}$$

and we conjecture

$$\quad K(k_5) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/4}\phi^{1/4}\big)^{24n}}}$$

$$K(k_{11}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(\tfrac{T+1}T\big)^{24n}}}\quad$$

$$K(k_{23}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/2}P\big)^{24n}}}$$

$$K(k_{31}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/2}\psi\big)^{24n}}}$$

and so on.

Entry 186

Using the \(j_3(\tau)\) from the previous entry, then the following quintics have a solvable Galois group

$$\begin{align}\qquad y^3(y-5)^2 &= -3\cdot 4^3\\ y^3(y-5)^2 &= -5\cdot 12^3\\ y^3(y-5)^2 &= -7\cdot 36^3\\ y^3(y-5)^2 &= -\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3\end{align}$$

as well as

$$\begin{align}y^3(y-5)^2 &= -(2\sqrt3)^6\\ y^3(y-5)^2 &= -(4\sqrt3)^6\\ y^3(y-5)^2 &= -(10\sqrt3)^6\qquad \qquad\end{align}$$

as discussed in Entry 142.

Entry 185

For the third McKay-Thompson series of the Monster, or \(j_3(\tau)\) 

$$j_3(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2$$

I haven't yet found a general formula for the complete elliptic integral of the first kind \(K(k_m)\) except for the special case \(K(k_3)\). Note that,

$$\pi = \beta\big(\tfrac12,\tfrac12\big) = 2\int_0^1\frac1{(1-x^2)^{1/2}}dx = 3.14159 $$

$$\pi_3 = \beta\big(\tfrac13,\tfrac13\big) = 3\int_0^1\frac1{(1-x^3)^{1/3}}dx = 5.29992 $$

where \(\pi_3 = 2^{4/3}\,3^{1/4}K(k_3) = \frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)\), the real period of the Dixon elliptic functions, can be considered as a cubic analogue of \(\pi\). Let 

$$\begin{align}j_3\big(\tfrac{1+3\sqrt{-1/3}}2\big) &= -3\cdot 4^3\\ j_3\big(\tfrac{1+5\sqrt{-1/3}}2\big) &= -5\cdot 12^3\\ j_3\big(\tfrac{1+7\sqrt{-1/3}}2\big) &= -7\cdot 36^3\\ j_3\big(\tfrac{1+11\sqrt{-1/3}}2\big) &= -\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3\end{align}$$

Hence we propose

$$2^{1/3}K(k_{3}) = 3^{1/4}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-3\cdot 4^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{5^{2/3}}{3^{3/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-5\cdot 12^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{7^{5/6}}{3^{5/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-7\cdot 36^3}\big)^n}}$$

$$\frac{2^{1/3}K(k_{3})}{\sqrt{11}-\sqrt3\,} = \frac{11^{5/6}}{3^{3/4}}\,\frac{\pi}8 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3}\big)^n}}$$

Entry 184

Using the \(j_2(\tau)\) from the previous entries, then the following quintics have a solvable Galois group

$$\begin{align}y(y-5)^4 &= -(4\sqrt2)^4\\ y(y-5)^4 &=  -(12\sqrt2)^4\\ y(y-5)^4 &=  -(84\sqrt2)^4\quad\end{align}$$

as well as,

$$\begin{align}y(y-5)^4 &= (4\sqrt{3})^4\\ y(y-5)^4 &= 12^4\\ y(y-5)^4 &= (12\sqrt{11})^4\\ y(y-5)^4 &= 396^4\qquad\end{align}$$

and an example with class number \(4\)

$$\quad y(y-5)^4 = -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3$$

though for all radical \(j_2(\tau)\) as also discussed in Entry 141.

Entry 183

The more famous value of \(j_2(\tau)\) is

$$j_2\big(\tfrac12\sqrt{-58}\big) = 396^4$$

So for even \(m\) and since \(K(k) = \frac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12;1,k^2\big)\), we propose a slightly different formula,

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-m})\big) = \frac{(2^{1/4}g_m)^3}{\;\big(j_2(\tau)\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

with modular lambda function \(\lambda(z)\), Ramanujan g-function \(g_m\) for integer \(m\geq4\) and \(\tau = \frac12{\sqrt{-m}}\),

$$2^{1/4}g_m = \frac{\eta\big(\frac12\sqrt{-m}\big)}{\eta(\sqrt{-m})}$$

For Ramanujan's \(G_m\) and \(g_m\)-functions, he tabulated them for odd and even \(m\), respectively and found,

$$\begin{align}g_6 &=  (1+\sqrt2)^{1/6}\\ g_{10} &=  \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\\ g_{22} &=  (1+\sqrt2)^{1/2}\\ g_{58} & = \left(\tfrac{5+\sqrt{29}}2\right)^{1/2}\end{align}$$ Therefore we conjecture

$$\begin{align}\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-6})\big) &= \frac{(2^{1/4}g_{6})^3}{\;\sqrt{4\sqrt3}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(4\sqrt3)^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-10})\big) &= \frac{(2^{1/4}g_{10})^3}{\;\sqrt{12}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{12^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-22})\big) &= \frac{(2^{1/4}g_{22})^3}{\;\sqrt{12\sqrt{11}}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(12\sqrt{11})^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-58})\big) &= \frac{(2^{1/4}g_{58})^3}{\;\sqrt{396}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{396^{4n}}}\end{align}$$

Entry 182

Using class number \(4\), in Entry 68 we calculated \(j_{2}(\tau)\) at certain points and found,

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

$$\begin{align}G_{17} &=  \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &=  \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\\ G_{97} &=  \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\\ G_{193} &= \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$

These were used in Entry 162 to find formulas for \(K(k_m)\). But we can combine these to find new formulas using 

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^3}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

where \(\tau = \frac{1+\sqrt{-m}}2\). For example, let \(m = 73\)

$$K(k_{73}) = \frac{\pi}2  \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^3 \frac{2^{3/4}}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

where as given above

$$j_2(\tau) = j_2\Big(\tfrac{1+\sqrt{-73}}2\Big) = {-2^{9}}\cdot3^4\big(111+13\sqrt{73}\big)^3$$

and so on.

Entry 181

For the second McKay-Thompson series of the Monster, or \(j_2(\tau)\) 

$$j_2(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2$$

we conjecture that the complete elliptic integral of the first kind \(K(k_m)\)

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^3}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

with Ramanujan G-function \(G_m\) for integer \(m\geq5\) and \(\tau = \frac{1+\sqrt{-m}}2\), 

$$2^{1/4}G_m = \frac{\eta^2(\sqrt{-m})}{\eta(\tfrac12\sqrt{-m})\,\eta(2\sqrt{-m})} = \zeta_{48}\frac{\eta(\tau)}{\eta(2\tau)}$$and \(48\)th root of unity \(\zeta_{48} = e^{2\pi i/48}\). 

Examples. For class number \(h(-4m) = 2\), we've already come across 

$$\begin{align}G_{5} &= \left(\frac{1+\sqrt{5}}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ \,G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$

hence,

$$K(k_{5}) = \frac{\pi}2  \frac{\big(1+\sqrt{5}\big)^{3/4}}{\big(4\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(4\sqrt2)^4}\big)^n}}$$

$$K(k_{13}) = \frac{\pi}2  \frac{\big(3+\sqrt{13}\big)^{3/4}}{\big(12\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(12\sqrt2)^4}\big)^n}}$$

$$K(k_{37}) = \frac{\pi}2  \frac{2^{3/4}\big(6+\sqrt{37}\big)^{3/4}}{\big(84\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(84\sqrt2)^4}\big)^n}}$$

and so on.

Entry 180

Using the \(j\)-functions from the previous entries, then the following quintics have a solvable galois group $$\begin{align}z^3(z^2+5z+40) &= -960^3\\ z^3(z^2+5z+40) &= -5280^3\\ z^3(z^2+5z+40) &= -640320^3\qquad\end{align}$$ and all other radical \(j(\tau)\). Examples from class number \(2\)$$\begin{align}z^3(z^2+5z+40) &= 2^3(25-13\sqrt{5})^3\\ z^3(z^2+5z+40) &= 30^3(31-9\sqrt{13})^3\\ z^3(z^2+5z+40) &= 60^3(2837-468\sqrt{37})^3\end{align}$$

a form related to the Brioschi quintic and discussed in Entry 140.

Entry 179

Continuing from Entry 178, we select some \(d=4m\) with class number \(h(-d) = 2\). It is known for \(m = 5,13,17\) 

$$\begin{align}G_{5} &= \left(\frac{1+\sqrt{5}}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ \,G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$

and calculate the j-function at the points,

$$\begin{align}j\big(\tfrac{1+\sqrt{-5}}2\big) &= 2^3(25-13\sqrt{5})^3\\ j\big(\tfrac{1+\sqrt{-13}}2\big) &= 30^3(31-9\sqrt{13})^3\\ j\big(\tfrac{1+\sqrt{-37}}2\big) &= 60^3(2837-468\sqrt{37})^3\end{align}$$

which are all negative values just like in the previous entry. The proposed relation has a bound \(m\geq7\), so the first value can't be used. But we conjecture 

$$K(k_{13}) = \frac{\pi}2  \frac{\big(3+\sqrt{13}\big)^{1/2}}{\big({-30^3}(31-9\sqrt{13})^3 \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(30^3(31-9\sqrt{13})^3\big)^n}}$$

$$K(k_{37}) = \frac{\pi}2  \frac{2^{3/4}\big(6+\sqrt{37}\big)^{3/4}}{\big({-60^3}(2837-468\sqrt{37} \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({60^3(2837-468\sqrt{37}}\big)^n}}$$

and so on for other \(j\big(\tfrac{1+\sqrt{-m}}2\big)\) with \(m\geq7\).

Entry 178

We go back to the first four McKay-Thompson series of the Monster. We conjecture that the complete elliptic integral of the first kind \(K(k_m)\) can be given by the first McKay-Thomson series \(j_{1A}(\tau) = j(\tau)\) or simply the j-function as,

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^2}{\big({-j(\tau)}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(j(\tau)\big)^n}}$$

with Ramanujan G-function \(G_m\) for integer \(m\geq7\) and \(\tau = \frac{1+\sqrt{-m}}2\), 

$$2^{1/4}G_m = \frac{\eta^2(\sqrt{-m})}{\eta(\tfrac12\sqrt{-m})\,\eta(2\sqrt{-m})} = \zeta_{48}\frac{\eta(\tau)}{\eta(2\tau)}$$and \(48\)th root of unity \(\zeta_{48} = e^{2\pi i/48}\). Let odd discriminant \(d=m\). Examples for class number \(h(-d) = 1\),

$$K(k_{7}) = \frac{\pi}2 \frac{2}{\big({15^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-15^3}\big)^n}}$$

while for higher Heegner numbers, \(x\) is a root of a simple cubic

$$K(k_{11}) = \frac{\pi}2 \left(\frac{1}{T}+1\right)^2\frac1{\big({32^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-32^3}\big)^n}}$$

with \(T\) the tribonacci constant, the real root of \(T^3-T^2-T-1=0\). For the largest

$$K(k_{163}) = \frac{\pi}2 \frac{x^2}{\big({640320^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-640320^3}\big)^n}}$$

where \(x\) is the real root of \(x^3-6x^2+4x-2 = 0\). 

P.S. The non-fundamental \(d=27\) also has class number \(1\) hence its j-function is an integer

$$K(k_{27}) = \frac{\pi}2 \frac{(2+2^{4/3}+2^{5/3})^{2/3}}{\big({{-3}\cdot160^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-3\cdot160^3}\big)^n}}$$

Entry 177

From Entry 176, we gave the fundamental unit \(U_d\) $$U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$ and from \(e^{\pi\sqrt{163}}\approx 640320^3+744\), observed that $$640320=80\,(\color{blue}{8005}-1)$$ This may be just coincidence, but not when \(U_{3d}\) for \(d = 7,11,19,43,67,163\). This was also observed by H. H. Chan. Given the fundamental units

$$\begin{align}U_{21} &=\left(\tfrac{\sqrt3+\sqrt{7}}2\right)^2\\ U_{33} &=\big(2\sqrt3+\sqrt{11}\big)^2\\ U_{57} &=\big(5\sqrt3+2\sqrt{19}\big)^2\\ U_{129} &=\big(53\sqrt3+14\sqrt{43}\big)^2\\ U_{201} &=\big(293\sqrt3+62\sqrt{67}\big)^2\\ U_{489} &=\big(35573\sqrt3+4826\sqrt{163}\big)^2 \end{align}$$

Define the function

$$F_d = 3\sqrt3\big(\sqrt{U_{3d}}-1/\sqrt{U_{3d}}\big)+6$$

then we get the rather familiar$$\begin{align}F_{7} &= 15\\ F_{11} &= 6(1+\sqrt{33})\\ F_{19} &= 96\\ F_{43} &= 960\\ F_{67} &= 5280\\ F_{163} &= 640320\end{align}$$

which (except for \(d=11\)) are the cube roots of the j-function (negated). 

P.S. I don't know why \(d=11\) does not obey the pattern, but it does yield the integer \(42\) if the positive sign of the second square root \(\pm 1/\sqrt{U_{3d}}\) is used, though the correct value should be \(32\).

Entry 176

Ramanujan found the exact value of the Ramanujan \(G\)-function $$G_{69} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}$$Note the fundamental units \(U_n\) $$\begin{align}U_{23} &= 24+5\sqrt{23} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^2\\ U_{69} &= \frac{25+3\sqrt{69}}2 = \left(\frac{3\sqrt3+\sqrt{23}}2\right)^2 \end{align}$$ and how he uses the squared version. As a second example

$$G_{77} = \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\quad$$

and fundamental units $$\begin{align}U_7 &= 8+3\sqrt7 =  \left(\frac{3+\sqrt{7}}{\sqrt2}\right)^2\\ U_{77} &= \frac{9+\sqrt{77}}2 = \left(\frac{\sqrt7+\sqrt{11}}2\right)^2 \end{align}$$

Ramanujan mostly uses the squared version of the \(U_n\) to get "simpler" expressions with smaller integers. For prime \(p \equiv 3\,\text{mod}\,4\), one can always do since 

$$x^2-py^2=-2\\ x^2-py^2=+2$$ are solvable by \(p \equiv 3\,\text{mod}\,8\,\) and \(p \equiv 7\,\text{mod}\,8\), respectively. Checking  \(U_{67}\) and \(U_{163}\), yields the reductions $$U_{67}= 48842 + 5967\sqrt{67} = \left(\frac{\color{blue}{221} + 27\sqrt{67}}{\sqrt2}\right)^2\\ U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$And from \(e^{\pi\sqrt{67}}\approx 5280^3+744\) and \(e^{\pi\sqrt{163}}\approx 640320^3+744\), we find the relations $$5280=24\,(\color{blue}{221}-1)\\ 640320=80\,(\color{blue}{8005}-1)$$ though it may be just coincidence.

Entry 175

For discriminant \(d=4m\) with class number \(8\) and semiprime \(m\): 

If \(m =5p\), then distinguish between \(p \equiv 1\,\text{mod}\, 8\) and \(p \equiv 5\,\text{mod}\, 8\). 

If \(m =7p\), then distinguish between \(p \equiv 3\,\text{mod}\, 8\) and \(p \equiv 7\,\text{mod}\, 8\). 

The semiprime \(m =5p\) was in the previous entry. For \(m =7p\), there are only four and Ramanujan found the radicals below. 

For the 1st case, \(p = 11, 43\), thus \(m = 7p = 77,\, 301\)  

$$\begin{align}G_{77} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\\ G_{301} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{57\sqrt7+23\sqrt{43}}2\right)^{1/8} \left(\sqrt{\frac{42+7\sqrt{43}}4}+\sqrt{\frac{46+7\sqrt{43}}4}\right)^{1/2}\end{align}$$

For the 2nd case, \(p = 31, 79\), thus \(m = 7p = 217,\, 553\) 

$$\begin{align}G_{217} &= \left(\sqrt{x_1-\tfrac12}+\sqrt{x_1+\tfrac12}\right)^{1/2} \left(\sqrt{y_1-\tfrac12}+\sqrt{y_1+\tfrac12}\right)^{1/2}\\ G_{553} &= \left(\sqrt{x_2-\tfrac12}+\sqrt{x_2+\tfrac12}\right)^{1/2} \left(\sqrt{y_2-\tfrac12}+\sqrt{y_2+\tfrac12}\right)^{1/2} \quad\end{align}$$ where $$x_1=\frac{10+4\sqrt{7}}2,\quad y_1=\frac{14+5\sqrt{7}}4$$ $$x_2=\frac{142+16\sqrt{79}}2,\quad y_2=\frac{98+11\sqrt{79}}4$$

But it seems not noticed that a fundamental unit is imbedded in these radicals as

$$x_1+2y_1 = \frac32(8+3\sqrt{7}) = \frac32\left(\frac{3+\sqrt7}{\sqrt2}\right)^2 =\frac32\,U_7$$

$$x_2+2y_2 = \frac32(80+9\sqrt{79}) = \frac32\left(\frac{9+\sqrt{79}}{\sqrt2}\right)^2 = \frac32\,U_{79}$$

Entry 174

Continuing from the previous entry, for \(d=4m\) with class number \(8\), the semiprime \(m =5p\) with \(p \equiv 5\,\text{mod}\, 8\) is also well-behaved. And it involves the golden ratio. There are only four, namely \(p = 13, 29, 53, 101\), thus \(m = 5p = 65, 145, 265, 505\). Ramanujan found the radicals below and the \(G\)-function have a common form

$$G_{5p} = \phi^{k}\,U_{p}^{1/4}\,x_{p}^{1/2}$$

with powers of the golden ratio \(\phi\), fundamental unit \(U_n\), and \(x_{p}^2\) a root of a unit quartic

$$\begin{align}\frac{G_{65}}{\phi} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4} \left(\sqrt{\frac{1+\sqrt{65}}8}+\sqrt{\frac{9+\sqrt{65}}8}\right)^{1/2}\\ \frac{G_{145}}{\phi^3} &= \left(\frac{5+\sqrt{29}}2\right)^{1/4} \left(\sqrt{\frac{9+\sqrt{145}}8}+\sqrt{\frac{17+\sqrt{145}}8}\right)^{1/2}\\ \frac{G_{265}}{\phi^3} &= \left(\frac{7+\sqrt{53}}2\right)^{1/4} \left(\sqrt{\frac{81+5\sqrt{265}}8}+\sqrt{\frac{89+5\sqrt{265}}8}\right)^{1/2}\\ \frac{G_{505}}{\phi^7} &= \big(10+\sqrt{101}\big)^{1/4} \left(\sqrt{\frac{105+5\sqrt{505}}8}+\sqrt{\frac{113+5\sqrt{505}}8}\right)^{1/2}\end{align}$$

The case \(p \equiv 1\,\text{mod}\, 8\) or \(p = 41, 89\), thus \(m = 5p = 205, 445\) behaves slightly differently though

$$\begin{align}\frac{G_{205}}{\phi} &= \left(\frac{43+3\sqrt{205}}2\right)^{1/8} \left(\sqrt{\frac{-1+\sqrt{41}}8}+\sqrt{\frac{7+\sqrt{41}}8}\right)\\  \frac{G_{445}}{\phi^{3/2}} &= \,\left(\frac{21+\sqrt{445}}2\right)^{1/4}\, \left(\sqrt{\frac{5+\sqrt{89}}8}+\sqrt{\frac{13+\sqrt{89}}8}\right)\end{align}$$

How Ramanujan found these is a mystery.

Entry 173

There are many in the set \(d=4m\) with class number \(8\). When \(m\) is a prime or a semiprime (a product of two primes) like \(m =3p\), then it may be well-behaved. For this set, there are only three, namely \(p = 23, 47, 71\), thus \(m = 3p = 69, 141, 213\). Ramanujan found the radicals below and the \(G\)-function have a common form

$$G_{3p} = U_{p}^{1/24}\,U_{3p}^{1/16}\,x_{p}^{1/2}$$

with fundamental unit \(U_n\) and where \(x_{p}^2\) is a root of a unit quartic

$$\begin{align}G_{69} &= \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}\\ G_{141} &= \left(\frac{7+\sqrt{47}}{\sqrt2}\right)^{1/12}\; \big(4\sqrt3+\sqrt{47}\big)^{1/8}\; \left(\sqrt{\frac{14+9\sqrt3}4}+\sqrt{\frac{18+9\sqrt3}4}\right)^{1/2}\\ G_{213} &= \left(\frac{59+7\sqrt{71}}{\sqrt2}\right)^{1/12} \left(\frac{5\sqrt3+\sqrt{71}}2\right)^{1/8} \left(\sqrt{\frac{19+12\sqrt3}2}+\sqrt{\frac{21+12\sqrt3}2}\right)^{1/2} \end{align}$$ How Ramanujan found these is unknown as it is uncertain if he was aware of class field theory. Note also that without the factor \(3\), then \(d=23,47,71\) are the smallest \(d\) with class number \(h(-d) = 3,5,7\), respectively, while \(h(-3d) = 8\).

P.S. Checking \(h(-3d) = 16\), one finds the only primes are \(d=167,191,239,383,311\) which are the smallest \(d\) with class number \(h(-d) = 11,13,15,17,19\), respectively. Makes me wonder if their \(G_{3d}\) would be analogous.

Entry 172

There are many fundamental \(d=4m\) with class number \(8\), though only seven are prime namely \(p = 41, 113, 137, 313, 337, 457, 577\). Their Ramanujan \(G\)-functions are

$$\begin{align}G_{41} &= \left(\frac{x+\sqrt{x^2-4}}2\right)^{1/2} =\sqrt{\frac{x-2}4}+\sqrt{\frac{x+2}4} \\ G_{113} &=\left(\frac{y+\sqrt{y^2-4}}2\right)^{1/2} =\sqrt{\frac{y-2}4}+\sqrt{\frac{y+2}4} \end{align}$$ where $$x = \left(\frac{5+\sqrt{41}}4\right) \left(1+\sqrt{\frac{-5+\sqrt{41}}8}\right)\\ y = \left(\frac{9+\sqrt{113}}4\right) \left(1+\sqrt{\frac{-7+\sqrt{113}}2}\right)$$ and similarly for the other \(p\), though the quartic roots get more complicated.

Entry 171

Continuing from Entry 170, there is another way to express the Ramanujan \(G_n\)-function where \(d=4n\) for prime \(n\) has class number \(6\) by using fundamental units \(U_n\). Borrowing a trick from Ramanujan, he found

$$G_{169}=\frac13\Big(2+\sqrt{13}+\sqrt[3]{U_{13}(v+3\sqrt{3})\sqrt{13}}+\sqrt[3]{U_{13}(v-3\sqrt{3})\sqrt{13}}\Big)$$

where $$U_{13} = \frac{3+\sqrt{13}}2, \quad v = \frac{11+\sqrt{13}}2$$

Using a similar form, we propose that

$$\begin{align}G_{29} &=\frac1{3^{1/4}}\Big(\tfrac{9+\sqrt{29}}2+\sqrt[3]{U_{29}(x+24\sqrt{3})}+\sqrt[3]{U_{29}(x-24\sqrt{3})}\Big)^{1/4}\\ G_{53} &=\frac1{3^{1/4}}\Big(\tfrac{23+3\sqrt{53}}2+\sqrt[3]{U_{53}(y+120\sqrt{3})}+\sqrt[3]{U_{53}(y-120\sqrt{3})}\Big)^{1/4}\\ G_{61} &=\frac1{3^{1/4}}\Big(15+2\sqrt{61}+\sqrt[3]{U_{61}(z+72\sqrt{3})}+\sqrt[3]{U_{61}(z-72\sqrt{3})}\Big)^{1/4}\end{align}$$ where$$U_{29} = \frac{5+\sqrt{29}}2, \quad x = \frac{185+19\sqrt{29}}2$$ $$\quad U_{53} = \frac{7+\sqrt{53}}2, \quad y = \frac{1721+217\sqrt{53}}2$$ $$U_{61} = \frac{39+5\sqrt{61}}2, \quad z = \frac{601+93\sqrt{61}}2\;$$

and similarly for all seven prime \(p = 29, 53, 61, 109, 157, 277, 397\). Note they have form \(p \equiv 5\,\text{mod}\,8\). And all their fundamental units have form \(U_p=\frac{a+b\sqrt{p}}2\), hence have odd solutions to the Pell equation \(x^2-py^2=-4\).

Entry 170

There are only seven fundamental \(d=4p\) with class number \(6\), namely \(p = 29, 53, 61, 109, 157, 277, 397\). Hence their Ramanujan \(G\)-functions are

$$\begin{align}G_{29} &= \left(\frac{y_1^2+\sqrt{y_1^4+4}}2\right)^{1/4} \\ G_{53} &= \left(\frac{y_2^2+\sqrt{y_2^4+4}}2\right)^{1/4} \\ G_{61} &=  \left(\frac{y_3^2+\sqrt{y_3^4+36}}6\right)^{1/4} \end{align}$$ where$$y^3 - y^2 - 4y - 4 = 0\\ y^3 - 7y^2 + 13y - 11 = 0\\ y^3 - 6y^2 - 27y - 54 = 0$$ and similarly for the other \(p\). 

Entry 169

Recall Ramanujan's \(G\)-function$$2^{1/4}G_n=\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(\tau)}$$ where \(\tau=\sqrt{-n}\). There are only three fundamental \(d=4p\) with class number \(2\) for prime \(p\), namely \(p = 5,13,37\). Hence $$\begin{align}G_5 &= \left(\frac{1+\sqrt5}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$Going higher, there are only four fundamental \(d=4p\) with class number \(4\) for prime \(p\), namely \(p = 17,73,97,193\).

$$ \begin{align}G_{17} &= \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &= \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8} \\ G_{97} &= \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8} \\ G_{193} &=  \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$ all of which were already known to Ramanujan. But as was shown in Entry 161 and Entry 162, it turns out these also appear in the closed-form of the complete elliptic integral of the first kind \(K(k_n)\). The next entry will be for class number \(6\) where one had to extract \(4\)th roots again.

Entry 168

For class number \(5\), there are only four \(d\) of the kind \(d \equiv 7\,\text{mod}\,8\), namely \(d = 47, 79, 103, 127\). We illustrate only the first two and propose, $$\begin{align}K(k_{47}) &=\frac{\sqrt{2\pi}}{2\sqrt{47}}\,x^{2/3} \left(\prod_{m=1}^{47}\Big[\Gamma\big(\tfrac{m}{47}\big)\Big]^{\big(\tfrac{-47}{m}\big)}\right)^{\color{red}{1/10}}\\ K(k_{79}) &=\frac{\sqrt{2\pi}}{2\sqrt{79}}\,y^{2/3} \left(\prod_{m=1}^{79}\Big[\Gamma\big(\tfrac{m}{79}\big)\Big]^{\big(\tfrac{-79}{m}\big)}\right)^{\color{red}{1/10}}\end{align}$$ where \((x,y)\) are the real roots of the quintics solvable in radicals $$x^5 - 2x^4 - 10x^3 - 13x^2 - 6x - 1=0\\ y^5 - 11y^4 + 17y^3 - 2y^2 - 5y - 1=0$$ And similarly for \(d= 103,127\). But for the next discriminant or \(d=131\) which is of kind \(d \equiv 3\,\text{mod}\,8\), one now has to deal with an algebraic number of degree \(3\times5 = 15\). And it seems difficult to find the eta quotients responsible for \((x,y)\).

Entry 167

For odd class number \(h(-d)\), the kind that is \(d \equiv 7\,\text{mod}\,8\) seems more well-behaved than \(d \equiv 3\,\text{mod}\,8\). For class number \(3\), there are only two of the first kind: \(d = 23,31\). Hence, $$\begin{align}K(k_{23}) &=\frac{\sqrt{2\pi}}{2\sqrt{23}}\,x^{4/3} \left(\prod_{m=1}^{23}\Big[\Gamma\big(\tfrac{m}{23}\big)\Big]^{\big(\tfrac{-23}{m}\big)}\right)^{\color{red}{1/6}}\\ K(k_{31}) &=\frac{\sqrt{2\pi}}{2\sqrt{31}}\,y^{4/3} \left(\prod_{m=1}^{31}\Big[\Gamma\big(\tfrac{m}{31}\big)\Big]^{\big(\tfrac{-31}{m}\big)}\right)^{\color{red}{1/6}}\end{align}$$ where \((x,y)\) are the real roots of the cubics $$x^3-x-1=0\\ y^3-y^2-1=0$$ or the plastic ratio and supergolden ratio, respectively. But for \(d=59\) which is of the second kind, then the radical involved will be an algebraic number of degree \(3\times3 =9\).

Entry 166

The case \(d=8\) is special since it is even but has odd class number \(h(-d)=1\). Given the Kronecker symbol \(\big(\tfrac{-d}{m}\big)\), we propose a symmetrical closed-form 

$$\begin{align}K(k_8) &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\prod_{m=1}^{8}\Big[\Gamma\big(\tfrac{m}{8}\big)\Big]^{\big(\tfrac{-8}{m}\big)}\right)^{1/2}\\ &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\frac{\Gamma\big(\frac18\big)\,\Gamma\big(\frac38\big)}{\Gamma\big(\frac58\big)\,\Gamma\big(\frac78\big)}\right)^{1/2}\end{align}$$ with fundamental unit \(U_2=1+\sqrt2\) and form reminiscent of the odd \(d\) with class number \(1\).

Entry 165

The previous entries dealt with even class numbers. For odd class number \(h(-d)=1\), there are the nine Heegner numbers \(d=1,2,3,7,11,19,43,67,163\). Given the Kronecker symbol \(\big(\tfrac{-d}{m}\big)\), we propose 

Conjecture. Let \(d>3\) with class number \(h(-d)=1\). Then \(x\) below is an algebraic number $$\frac1{x_d^2}  = \frac1{K(k_d)}\frac{\sqrt{2\pi}}{\color{blue}4\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/2}}$$specifically$$x_d=2^{1/4}G_n=\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(\tau)}$$ with Ramanujan's \(G\)-function. This function \(x_d\) has been discussed in Entry 159. For \(d=7\), then \(x_7=\sqrt2\), but for the five \(d\geq11\), then \(x_d\) are the real roots of the following five simple cubics, 

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$ also discussed in Entry 160. 

Entry 164

To summarize, given fundamental discriminant \(d\), class number \(h(-d)=n\), complete elliptic integral of the first kind \(K(k_p)\), and Kronecker symbol \(\big(\tfrac{-d}{m}\big)\). 

Conjecture 1. Let even \(d=4p\) for prime \(p\equiv 1\,\text{mod}\,4\) with even class number \(h(-d)=n\) and $$x = \frac1{K(k_{\color{blue}{d/4}})}\frac{\sqrt{2\pi}}{2\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/(2n)}}$$ Conjecture 2. Let odd \(\,d=p\,\) for prime \(p\,\equiv\, 3\,\text{mod}\,4\,\) with odd class number \(h(-d)=n\) and $$y = \frac1{K(k_{d})}\frac{\sqrt{2\pi}}{2\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/(2n)}}$$ then \((x,y)\) are algebraic numbers as seen in entries \(160-163\) and \(165-168\). They seem to have a closed-form in term of eta quotients but I haven't found it yet.

Entry 163

There are only seven fundamental \(d = 4p\) with class number \(6\), namely \(p=29, 53, 61, 109, 157, 277, 397\).  To prevent clutter, only the first three will be stated. Given the Kronecker symbol \(\big(\frac{d}{m}\big)\), we propose

$$\begin{align}K(k_{29}) &= \frac{\sqrt{2\pi}}{2\sqrt{116}} \frac1{\;(x_1)^{1/3}} \left(\frac{y_1^2+\sqrt{y_1^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{116}\Big[\Gamma\big(\tfrac{m}{116}\big)\Big]^{\big(\tfrac{-116}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{53}) &= \frac{\sqrt{2\pi}}{2\sqrt{212}} \frac1{\;(x_2)^{1/3}} \left(\frac{y_2^2+\sqrt{y_2^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{212}\Big[\Gamma\big(\tfrac{m}{212}\big)\Big]^{\big(\tfrac{-212}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{61}) &= \frac{\sqrt{2\pi}}{2\sqrt{244}} \frac1{\;(x_3)^{1/3}} \left(\frac{y_3^2+\sqrt{y_3^4+36}}6\right)^{3/4} \left(\prod_{m=1}^{244}\Big[\Gamma\big(\tfrac{m}{244}\big)\Big]^{\big(\tfrac{-244}{m}\big)}\right)^{\color{red}{1/12}}\end{align}$$ where the \(x_n\) are the real roots of the following cubics $$x^3 - 5x^2 - 3x - 1 = 0\\ x^3 - 15x^2 - x - 1= 0\\ x^3 - 27x^2 - 5x - 1= 0$$ while the \(y_n\) are also the real roots of cubics $$y^3 - y^2 - 4y - 4 = 0\\ y^3 - 7y^2 + 13y - 11 = 0\\ y^3 - 6y^2 - 27y - 54 = 0$$ and similarly for the other \(p\).

Entry 162

There are only four fundamental \(d = 4p\) with class number \(4\), namely \(p=17,73,97,193\).  Given the Kronecker symbol \(\big(\frac{d}{m}\big)\), we propose

$$\begin{align}K(k_{17}) &= \frac{\sqrt{2\pi}}{2\sqrt{68}} \frac1{(U_{17})^{1/8}} \left(\sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8}\right)^{3} \left(\prod_{m=1}^{68}\Big[\Gamma\big(\tfrac{m}{68}\big)\Big]^{\big(\tfrac{-68}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{73}) &= \frac{\sqrt{2\pi}}{2\sqrt{292}}  \frac1{(U_{73})^{1/8}} \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^{3} \left(\prod_{m=1}^{292}\Big[\Gamma\big(\tfrac{m}{292}\big)\Big]^{\big(\tfrac{-292}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{97}) &= \frac{\sqrt{2\pi}}{2\sqrt{388}} \frac1{(U_{97})^{1/8}} \left(\sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\right)^{3}  \left(\prod_{m=1}^{388}\Big[\Gamma\big(\tfrac{m}{388}\big)\Big]^{\big(\tfrac{-388}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{193}) &= \frac{\sqrt{2\pi}}{2\sqrt{772}} \frac1{(U_{193})^{1/8}} \left(\sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8}\right)^{3} \left(\prod_{m=1}^{772}\Big[\Gamma\big(\tfrac{m}{772}\big)\Big]^{\big(\tfrac{-772}{m}\big)}\right)^{\color{red}{1/8}}\end{align}$$ where the \(U_n\) are fundamental units $$U_{17} = 4+\sqrt{17}$$ $$U_{73} = 1068+125\sqrt{73}$$ $$U_{97} = 5604 + 569\sqrt{97}$$ $$U_{193} = {1764132} + 126985\sqrt{193}$$ Going to class number \(6\), the red exponent will now be \(1/12\).

Entry 161

Entry 160 dealt with discriminants \(d\) with class number \(1\). Going higher, there are only three fundamental \(d = 4p\) with class number \(2\), namely \(p=5,13,17\).  Given the Kronecker symbol \(\big(\frac{d}{m}\big)\), we propose

$$\begin{align}K(k_5) &= \frac{\sqrt{2\pi}}{2\sqrt{20}}\left(\frac{1+\sqrt5}2\right)^{3/4}\left(\prod_{m=1}^{20}\Big[\Gamma\big(\tfrac{m}{20}\big)\Big]^{\big(\tfrac{-20}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{13}) &= \frac{\sqrt{2\pi}}{2\sqrt{52}}\left(\frac{3+\sqrt{13}}2\right)^{3/4}\left(\prod_{m=1}^{52}\Big[\Gamma\big(\tfrac{m}{52}\big)\Big]^{\big(\tfrac{-52}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{37}) &= \frac{\sqrt{2\pi}}{2\sqrt{148}}\left(6+\sqrt{37}\right)^{3/4}\left(\prod_{m=1}^{148}\Big[\Gamma\big(\tfrac{m}{148}\big)\Big]^{\big(\tfrac{-148}{m}\big)}\right)^{\color{red}{1/4}}\end{align}$$Going to class number \(4\), the red exponent will now be \(1/8\).

Entry 160

We continue with closed-forms for the Dedekind eta function \(\eta^2(\sqrt{-d})\) for \(d=11,19,43,67,163\). As discussed in Entry 159, the closed-form for the complete elliptic integral of the first kind \(K(k_d)\) then necessarily follows. 

$$\eta^2(\sqrt{-11}) = \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}$$ $$\eta^2(\sqrt{-19}) = \frac1{x_{19}^2}\frac{\Gamma\big(\tfrac1{19}\big)\,\Gamma\big(\tfrac4{19}\big)\,\Gamma\big(\tfrac5{19}\big)\,\Gamma\big(\tfrac6{19}\big)\dots\Gamma\big(\tfrac{17}{19}\big)}{19^{1/4}\,(2\pi)^5}$$

$$\eta^2(\sqrt{-43}) = \frac1{x_{43}^2}\frac{\Gamma\big(\tfrac1{43}\big)\,\Gamma\big(\tfrac4{43}\big)\,\Gamma\big(\tfrac6{43}\big)\,\Gamma\big(\tfrac9{43}\big)\dots\Gamma\big(\tfrac{41}{43}\big)}{43^{1/4}\,(2\pi)^{11}}$$

$$\eta^2(\sqrt{-67}) = \frac1{x_{67}^2}\frac{\Gamma\big(\tfrac1{67}\big)\,\Gamma\big(\tfrac4{67}\big)\,\Gamma\big(\tfrac6{67}\big)\,\Gamma\big(\tfrac9{67}\big)\dots\Gamma\big(\tfrac{65}{67}\big)}{67^{1/4}\,(2\pi)^{17}}$$

$$\eta^2(\sqrt{-163}) = \frac1{x_{163}^2}\frac{\Gamma\big(\tfrac1{163}\big)\,\Gamma\big(\tfrac4{163}\big)\,\Gamma\big(\tfrac6{163}\big)\,\Gamma\big(\tfrac9{163}\big)\dots\Gamma\big(\tfrac{161}{163}\big)}{163^{1/4}\,(2\pi)^{41}}$$

where the \(x_d\) are the real roots of the following simple cubics, respectively

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$

The numerators, for example, of \(\Gamma\big(\frac{n}{19}\big)\) can be found using the Kronecker symbol and this Wolfram command which yields the \(\frac{19-1}2 = 9\) numerators as \(n = 1, 4, 5, 6, 7, 9, 11, 16, 17\).

Entry 159

Given the McKay-Thompson series of Class 4A for the Monster (A097340)

$$j_{4A}(\tau)=\left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)}\right)^{24} = \frac1q+24 + 276q + 2048q^2 +\dots$$ We take the \(6\)th root, scale \(\tau \to \frac{\tau}2\), and use the version, $$(x_d)^4=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4=\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}{\eta^2(\tau)}=\frac{\frac2{\pi}\times K(k_d)}{\eta^2(\tau)}$$ where \(K(k_d)\) is a complete elliptic integral of the first kind. Compared also to Ramanujan's G-function, $$2^{1/4}G_n = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)} = q^{-\frac{1}{24}}\prod_{n>0}(1+q^{2n-1})\qquad\qquad $$ where Ramanujan used odd \(n\) for \(G_n\) and calculated a lot of \(n\). Mathworld also has a partial list of closed-forms for \(K(k_d)\). Knowing \(x_d\) or \(G_n\) immediately leads to a closed-form for \(\eta^2(\tau)\) as well. Examples, 

$$\begin{align}\frac{2}{\pi}\times K(k_7) &= x_7^2\,\frac{\Gamma\big(\tfrac17\big)\,\Gamma\big(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\\ \eta^2(\sqrt{-7}) &= \frac1{x_7^2}\frac{\Gamma\big(\tfrac17\big)\,\Gamma(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\end{align}$$

where \(x_7 =\sqrt2\). And

$$\begin{align}\frac{2}{\pi}\times K(k_{11}) &= x_{11}^2\,\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\\ \eta^2(\sqrt{-11}) &= \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\end{align}$$

where \(x_{11} =\dfrac1T+1\) and \(T\) is the tribonacci constant, the real root of \(x^3-x^2-x-1=0\). For \(d=11,19,43,67,163\), then \(x_d\) is just the real root of a cubic.

Entry 158

In the previous entry, given the Jacobi theta functions \(\vartheta_n(0,q)\) with nome \(q = e^{\pi i\tau}\), modular lambda function \(\lambda(\tau)\), and complete elliptic integral of the first kind \(K(k) = \tfrac{\pi}2\, {_2F_1}\big(\tfrac12,\tfrac12,1,k^2) \) we proposed three identities, one as,

$$\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 \overset{\color{red}?}=\lambda$$ where \(\lambda = \lambda(\tau)\). Using the known definitions of \(\vartheta_n(0,q)\) and \(\lambda(\tau)\), the proposed identity implies 

$$\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}{\eta^2(\tau)}=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4$$

For appropriate complex quadratics such as \(\tau=\sqrt{-n}\), then the RHS is a radical. But if the numerator of the LHS has a closed-form (implied in this list), then this also gives a closed-form for the denominator \(\eta^2(\tau)\) such as

$$\begin{align}\eta^2(\sqrt{-1}) &=\frac{\Gamma^2(\tfrac14)}{(2\pi)^{3/2}}\frac1{2^{1/2}}\\ \eta^2(\sqrt{-2}) &=\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)}{(2\pi)^{3/2}}\frac1{2^{5/4}}\\ \eta^2(\sqrt{-3}) &=\frac{\Gamma^3(\tfrac13)}{(2\pi)^{2}}\frac{3^{1/4}}{2^{2/3}}\end{align}$$

and so on. Since an eta quotient is involved, then \(\eta(\sqrt{-d})\) for \(d\) with class number \(1\) will behave quite orderly and will be discussed in the next entry.

Entry 157

Given the Jacobi theta functions \(\vartheta_n(0,q)\) which traditionally uses the nome \(q = e^{\pi i\tau}\). Define the modular lambda function \(\lambda(\tau)\),

$$\lambda(\tau) = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$

Then we propose the three identities below and for appropriate \(\tau\) such as \(\tau = \sqrt{-d}\) and \(\lambda = \lambda(\tau)\) that the ratios below are radicals,

$$\begin{align}\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=\lambda\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1\end{align}$$

Adding the first two implies the third. Hence, after removing the common denominator

$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$

which is known to be true. As eta quotients in the same order above, 

$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4$$

Also, the equalities $$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} = \frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}$$which has a cubic version given in Entry 156.

Entry 156

Given the square of the nome, so \(q = e^{2\pi i\tau}\) and the Borwein cubic theta functions \(a(q),b(q),c(q)\). Define,

$$\beta = \left(\frac{3}{\left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3=\left(\frac{c(q)}{a(q)}\right)^3$$

Then we conjecture,

$$\begin{align}\left(\frac{c(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=\beta\\ \left(\frac{b(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1-\beta\\ \left(\frac{a(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1\end{align}$$

Adding the first two implies the third,

$$\big(c(q)\big)^3+\big(b(q)\big)^3=\big(a(q)\big)^3$$

which is a known relationship of the Borwein cubic theta functions. As eta quotients,

$$\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3+\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$

Like in the previous entry, the RHS is also a sum,

$$a(q) = \sum_{m,n\,=-\infty}^\infty q^{m^2+mn+n^2} = \frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}$$

Entry 155

It turns out we can use the previous entries to parameterize the equation $$x^n+y^n = z^n$$ for \(n=2,3,4\). Given the square of the nome \(q = e^{2\pi i\tau}\) and assume the functions \(A(q), B(q), C(q)\). Define

$$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$

Then we conjecture,

$$\begin{align}\left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1\end{align}$$

where \(C(q), B(q), A(q)\) are defined by the relation and eta quotients below,

$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$

$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$ Note also that $$\begin{align}A(q) &= 1+24\sum_{n=1}^\infty\frac{n q^n}{1+q^n}\\ &=\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_2(0,q)\big)^4\\ &=\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\end{align}$$ where, in this particular instance, we let the Jacobi theta functions \(\vartheta_n(0,q)\) use \(q = e^{2\pi i\tau}\). 

Entry 154

As an overview of the previous four entries, Ramanujan's theory of elliptic functions to alternative bases uses the hypergeometric function \(_2F_1(a,b;c;z)\) with \(a+b=c=1\) for the cases \(a=\frac16, \frac14, \frac13, \frac12\). This can be related to the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145 for \(n = 1,2,3,4\). Consider the equations, 

$$\begin{align}\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)} &=-\tau\sqrt{-1}\\ \frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)} &=-\tau\sqrt{-2}\\ \frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)} &=-\tau\sqrt{-3}\\ \frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)} &=-\tau\sqrt{-4}\end{align}$$ Then the \(\alpha_n\) can be solved and expressed in terms of the \(j_n(\tau)\) as discussed in the said entries.

Entry 153

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145. Define,

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

Let \(\alpha_4 = \alpha_4(\tau)\). Then we conjecture that,

$$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}=-\tau\sqrt{-4}$$ as well as

$$\begin{align}j_{4}(\tau) &=  \frac{16}{\alpha_4\,(1-\alpha_4)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}\\ &= \left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}\end{align}$$

Example. Let \(\tau =\sqrt{-3}\). Then \(\alpha_4=(\sqrt3-\sqrt2)^4(\sqrt2-1)^4\) solves $$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}=\sqrt4\times\sqrt3$$ Alternatively and more familiar

$$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\lambda(\sqrt{-r})\big)}{_2F_1\big(\frac12,\frac12,1,\,\lambda(\sqrt{-r})\big)}=\sqrt{r}$$ where \(\lambda(\tau)\) is the modular lambda function.

Entry 152

We relate Ramanujan's theory of elliptic functions to alternative bases to the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145. Define,

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

Let \(\alpha_3 = \alpha_3(\tau)\). Then we conjecture that,

$$\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}=-\tau\sqrt{-3}$$ as well as

$$\begin{align}j_{3}(\tau) &=  \frac{27}{\alpha_3\,(1-\alpha_3)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2\\ &= \left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}\end{align}$$

Example. Let \(\tau =\sqrt{-3}\). Then \(\alpha_3=\frac1{250}(187-171\cdot2^{1/3}+18\cdot2^{2/3})\) solves $$\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}=\sqrt3\times\sqrt3$$

Entry 151

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145. Define,

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2 $$

Let \(\alpha_2 = \alpha_2(\tau)\). Then we conjecture that,

$$\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}=-\tau\sqrt{-2}$$ as well as

$$\begin{align}j_{2}(\tau) &=  \frac{64}{\alpha_2\,(1-\alpha_2)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2\\ &= \left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}\end{align}$$

Example. Let \(\tau =\sqrt{-3}\). Then \(\alpha_2=\frac1{3(2+\sqrt3)^2(13+4\sqrt3)}\) solves $$\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}=\sqrt2\times\sqrt3$$

Entry 150

Ramanujan's theory of elliptic functions to alternative bases considers the hypergeometric function \(_2F_1(a,b;c;z)\) with \(a+b=c=1\) for the cases \(a=\frac16, \frac14, \frac13, \frac12\). We can relate this to the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145 for \(n = 1,2,3,4\). Define,

$$\alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1}(\tau)}}\right) $$

where \(j = j_1(\tau)\) is the j-function. Let \(\alpha_1 = \alpha_1(\tau)\). Then we conjecture that,

$$\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}=-\tau\sqrt{-1}$$ as well as

$$\begin{align}j_{1}(\tau)  &=\frac{432}{\alpha_1\,(1-\alpha_1)}\\ &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3\\  &=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\end{align}$$

Example. Let \(\tau =\sqrt{-3}\). Then \(\alpha_1=\frac1{5\sqrt5}\left(\frac{-1+\sqrt5}2\right)^5\) solves $$\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}=\sqrt3$$ since \(-\tau\sqrt{-1}=-\sqrt3\, i\times i=\sqrt3\).

Entry 149

This is the octic overview. Using the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145, if \(\tau\) are complex quadratics such that \(j_n(\tau)\) is a radical, then the following octics have a solvable Galois group, hence solvable in radicals $$ \begin{align}\qquad{j_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\ {j_2}^2 &=\frac{j_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\ {j_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\ {j_4}^2 &=\frac{7 j_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x \end{align}$$

It would be good if the one for \(j_2\) can be simplified. They have discriminants (with another factor of \(D_2\) suppressed),

$$\begin{align}D_1 &= -7^7(j_1-1728)^4\,{j_1}^4\\ D_2 &= -7^7(j_2-256)^4\,{j_2}^6 \\ D_3 &= -7^7(j_3-108)^3\,{j_3}^5\\ D_4 &= -7^7(j_4-64)^4\,{j_4}^{12}\quad \end{align}$$

Examples. Let \(\tau = \tfrac{1+\sqrt{-41/3}}2\) so \(j_3(\tau) = -(4\sqrt3)^6\). Let \(\tau = \tfrac{1+\sqrt{-89/3}}2\) so \(j_3(\tau) = -(10\sqrt3)^6\). Then $$\frac{(x + 1)^6(x^2 + x + 7)}x = -(4\sqrt3)^6\; \\ \frac{(x + 1)^6(x^2 + x + 7)}x =  -(10\sqrt3)^6$$ are octics both solvable in radicals. And also $$e^{\pi\sqrt{41/3}} =  (4\sqrt3)^6+41.993\dots\quad\\ e^{\pi\sqrt{89/3}} =  (10\sqrt3)^6+41.99997\dots$$ There are infinitely many \(\tau\) but special ones such that \(j_n(\tau)\) are integers can be found in Entry 145.

Entry 148

This is the 7th-deg overview though only results by Klein for the j-function \(j = j_1(\tau)\) are known. In Klein's "On the Order-Seven Transformations of Elliptic Functions", he gave two elegant resolvents of degrees 7 and 8 in pages 306 and 313. Translated to more understandable notation, we have,

$$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$$

$$y^8+14y^6+63y^4+70y^2-7 = y\sqrt{j-1728}$$

If \(\tau\) are complex quadratics such that \(j= j_1(\tau)\) is a radical, then the two resolvents have a solvable Galois group, hence solvable in radicals

Example. Let \(\tau = \tfrac{1+\sqrt{-163}}2\), then \(j = -640320^3\) and $$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3  = -640320^3$$ is solvable in radicals. There are infinitely many such \(\tau\) and some can be found in Entry 145. Note also that $$\frac{(y^4 + 14y^3 + 63y^2 + 70y - 7)^2}y + 1728 = \frac{(y^2 + 5y + 1)^3 (y^2 + 13y + 49)}y$$ where the octic on the RHS will appear in the next entry.

Entry 147

This is the sextic overview. Using the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145, if \(\tau\) are complex quadratics such that \(j_n(\tau)\) is a radical, then the following sextics have a solvable Galois group, hence solvable in radicals $$ \begin{align}\qquad j_1 &=\frac{(x^2 + 10x + 5)^3}x\\ \color{red}{j_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\ \color{red}{{j_3}^2} &= \frac{j_3\,(2 x^6 + 29x^5 + 85x^4 + 50x^3) + 5^4 x^6}{(2x - 1)}\\ j_4 &=\frac{(x + 1)^5 (x + 5)}x \end{align}$$

with the ones in red by Joachim König. (It would be nice if the one for \(j_3\) can be simplified.) They have discriminants,

$$\begin{align}D_1 &= 5^5\,(j_1-1728)^2\,{j_1}^4\\ D_2 &= 5^5\,(j_2-256)^3\,{j_2}^3 \\ D_3 &= 5^5(j_3-108)^3\,{j_3}^{11}\\ D_4 &= 5^5\,(j_4-64)^2\,{j_4}^4\qquad\end{align}$$

Examples. Let \(\tau = \tfrac{1+\sqrt{-163}}2\) so \(j_1(\tau) = -640320^3\). Let \(\tau = \tfrac{\sqrt{-58}}2\) so \(j_2(\tau) = 396^4\). Then $$\frac{(x^2 + 10x + 5)^3}x = -640320^3\\ \frac{(x+1)^4(x^2+6x+25)}x=396^4$$ are sextics both solvable in radicals. There are infinitely many \(\tau\) but special ones such that \(j_n(\tau)\) are integers can be found in Entry 145.

Entry 146

This is the quintic overview of Entries 140-144. Using the McKay-Thompson series \(j_n = j_n(\tau)\) for the Monster defined in Entry 145, if \(\tau\) are complex quadratics such that \(j_n(\tau)\) is a radical, then the following simple quintics have a solvable Galois group hence solvable in radicals $$\begin{align}x^5 + 5x^4 + 40x^3 &= j_1\\ x(x - 5)^4 &= j_2\\ x^3(x - 5)^2 &= j_3\\ x^5+5x &= \sqrt{\frac{64}{\sqrt{j_4}}-\sqrt{j_4}}\\ x^5 + 5x^3 - 10x^2 &= j_6\end{align}$$ Example. Let \(\tau = \tfrac{1+\sqrt{-163}}2\) so \(j_1(\tau) = -640320^3\). Let \(\tau = \tfrac{\sqrt{-58}}2\) so \(j_2(\tau) = 396^4\). Then $$x^5 + 5x^4 + 40x^3 = -640320^3\\ x(x - 5)^4 = 396^4$$ are quintics both solvable in radicals. Of course, it is well-known that $$\qquad e^{\pi\sqrt{163}} = 640320^3+743.99999999999925\dots\\ e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$There are infinitely many \(\tau\) but special ones such that \(j_n(\tau)\) are integers can be found in Entry 145.

Entry 145

Summarizing the McKay-Thompson series of the Monster discussed in Entries 140-144, 

$$\begin{align}\quad j_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\ \quad j_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\ \quad j_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ \quad j_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}\\ j_{6} &=\left( \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^3 + 2^3\left(\frac{\eta(2\tau)\,\eta(6\tau)}{\eta(\tau)\,\eta(3\tau)}\right)^3\right)^2  \end{align}$$

where \(j_1(\tau)\) is the j-function. Let \(\tau\) be complex quadratics \(\tau = \tfrac12\sqrt{-d}\) or \(\tau =\frac12+ \sqrt{-d}\) such that the \(j_n(\tau)\) are radicals. For the following special \(\tau\), then \(j_n(\tau)\) are integers 

$$j_1(\tau)\; \text{where}\; \tau=\sqrt{-d}\;\text{for}\; d = 1, 2, 3, 4, 7,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =1, 3, 7, 11, 19, 127, 43, 67, 163.$$

$$j_2(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d}}2\;\text{for}\; d = 4, 6, 10, 18, 22, 58,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =5, 7, 9, 13, 25, 37.$$

$$j_3(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d/3}}2\;\text{for}\; d = 4,8,16,20,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d/3}}2\,\text{for}\; d =5, 9, 17, 25, 41, 49, 89.$$

$$j_4(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d}}2\;\text{for}\; d = 3, 7,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =1, 2, 4.$$

$$j_6(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d/3}}2\;\text{for}\; d = 10, 14, 26, 34,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d/3}}2\,\text{for}\; d = 7, 11, 19, 31, 59.$$

Entry 144

Define the McKay-Thompson series of Class 6A for the Monster $$j_6 = j_{6}(\tau) =\left( \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^3 + 2^3\left(\frac{\eta(2\tau)\,\eta(6\tau)}{\eta(\tau)\,\eta(3\tau)}\right)^3\right)^2$$ and the quintic $$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0$$ where \(\small\alpha = -\dfrac1{j_6-32}.\) Alternatively $$(y^2+15)^2(y-5) = 32\big(j_6-32\big) $$ $$z^5 + 5z^3 - 10z^2 = j_6 $$

Conjecture: "If \(\tau\) is a complex quadratic such that \(j_6=j_{6}(\tau)\) is an algebraic number with \(j_6\neq 32\), then the quintics above have a solvable Galois group."

Example: Let \(j_6\big(\tfrac{1\sqrt{-59/3}}2\big)=-1060^2\), then $$(y^2+15)^2(y-5) = 32\big({-1060^2}-32\big)$$ $$z^5 + 5z^3 - 10z^2=-1060^2 $$ are solvable in radicals.

Entry 143

Define the McKay-Thompson series of Class 4A for the Monster $$j_4 = j_{4}(\tau) = \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}$$ and the Bring-Jerrard quintic $$y^5+5y=\left(\frac{64}{\sqrt{j_4}}-\sqrt{j_4}\right)^{1/2}$$

Conjecture: "If \(\tau\) is a complex quadratic such that \(j_4=j_{4}(\tau)\) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let \(j_4\big(\tfrac12\sqrt{-7}\big)=2^{12}\), then $$y^5+5y=\sqrt{-63}$$ is solvable in radicals.

Entry 142

Define the McKay-Thompson series of Class 3A for the Monster $$j_3 = j_{3}(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2$$ and the Euler-Jerrard quintic $$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0$$ Alternatively $$y^3(y-5)^2 =j_3$$

Conjecture: "If \(\tau\) is a complex quadratic such that \(j_3=j_{3}(\tau)\) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let \(j_3\big(\tfrac{1+\sqrt{-89/3}}2\big)=-300^3\), then $$y^3(y-5)^2 = -300^3$$ is solvable in radicals.

Entry 141

Define the McKay-Thompson series of Class 2A for the Monster $$j_2 = j_{2}(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2$$ and the Bring-Jerrard quintic $$x^5-5\alpha x -\alpha=0$$ Alternatively $$y(y-5)^4 =j_2$$

Conjecture: "If \(\tau\) is a complex quadratic such that \(j_2=j_{2}(\tau)\) is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let \(j_2\big(\tfrac12\sqrt{-10}\big)=12^4\), then $$\quad y(y-5)^4=12^4\\ z^5-5z-12=0$$ are solvable in radicals.

Entry 140

Given the Dedekind eta function \(\eta(\tau)\) and define the McKay-Thompson series of Class 1A for the Monster, or better known as the j-function \(j\) $$j=j_1(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3$$ and the Brioschi quintic $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ where \(\small\alpha = -\dfrac1{j-1728}\). Alternatively $$(y^2+20)^2(y-5) = j-1728$$ $$z^5 + 5z^4 + 40z^3=j$$Conjecture: "If \(\tau\) is a complex quadratic such that \(j\) is an algebraic number \(j\neq1728\), then the quintics above have a solvable Galois group." 

Example. Let \(j\big(\tfrac{1+\sqrt{-163}}2\big) = -640320^3\) and \(\alpha = \tfrac1{640320^3+1728}\), then $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ $$(y^2+20)^2(y-5) = -640320^3-1728$$ $$z^5 + 5z^4 + 40z^3=-640320^3$$ are quintics solvable in radicals. (In fact, they factor into a quadratic and a cubic.)

Entry 139

The general quintic can be reduced to the following one-parameter forms

$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$

$$x^5-5\alpha x -\alpha = 0\tag2$$

$$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$

$$\; x^5+5x+\left(\frac1{\sqrt{\alpha}}-64\sqrt{\alpha}\right)^{1/2} = 0\tag4$$

$$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0\tag5$$ with the last found by yours truly. They have neat discriminants

$$\begin{align}D_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\ D_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\ D_3 &= 5^5\,(1-108\alpha)\,\alpha^2\\ D_4 &= 5^5\,(1+64\alpha)^2\,\alpha^{-1}\\ D_5 &= 5^5\,(1-36\alpha)(1-32\alpha)\,\alpha^8\end{align}$$ The integers \((1728, 256, 108, 64)\) appear in Ramanujan's theory of elliptic functions to alternative bases and we will connect these quintics to the McKay-Thompson series of class 1A, 2A, 3A, 4A, 6A for the Monster in subsequent entries. 

Entry 138

This summarizes the last several entries. Let fundamental discriminant \(d = 4m\) with class number \(h(-d)=2^k\) and even \(m = 2p\) for prime \(p\). Previously, \(p \equiv 1\,\text{mod}\,4\) and \(p \equiv 3\,\text{mod}\,4\) were distinguished, but we can have a more unified approach. Given the modular lambda function \(\lambda(\tau)\) and define the three simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2\\ \gamma(n) &= \Big(n+\sqrt{n^2+1}\Big)^2 \end{align}$$ If \(p \equiv 3\,\text{mod}\,4\), then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n),\quad \text{where}\, n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ If \(p \equiv 1\,\text{mod}\,4\), then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\gamma(n),\quad \text{where}\, n = \frac{\lambda+1}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with \(\lambda=\lambda(\tau)\) for simplicity and \(\alpha(n),\,\beta(n),\,\gamma(n)\) are units. Examples. Let \(m = 2p\) with class number \(4\). 

For \(p = 7,23\) $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n)\,\beta(n),\quad n=2(1+\sqrt2)\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n)\,\beta(n),\quad n=2(13+9\sqrt2) \end{align}$$ For \(p=17,41\) $$\begin{align}\frac1{\lambda(\sqrt{-34})} &= \alpha(n)\,\gamma(n),\quad n=3(4+\sqrt{17})\\ \frac1{\lambda(\sqrt{-82})} &= \alpha(n)\,\gamma(n),\quad n=3(51+8\sqrt{41})\end{align}$$ One can observe that \(n\) is an algebraic number of degree half that of the class number. For \(m = 2p\) with class number \(8\), examples for \(p \equiv 3\,\text{mod}\,4\) were given in Entry 136. For \(p \equiv 1\,\text{mod}\,4\) like \(p=89\), then $$n^4 - 8886 n^3 + 648 n^2 - 10314 n + 5751=0$$ and so on for other \(p\).

Entry 137

This continues Entry 136. Recall the function $$\beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ and the examples of \(n\) which were quartic roots. It turns out these \(n\) have additional properties which yield fundamental units \(U_k\) though I don't know why.

For \(p=31\), let \(n_{\color{red}\pm} = 2(1+\sqrt2)^2\Big(1+3\sqrt2\color{red}{\pm}2\sqrt{1+4\sqrt2}\Big)\) or the two real roots of the quartic. Then $$\frac{\beta(n_{+})}{\beta(n_{-})} = U_{31}\sqrt{U_{62}} = (1520+273\sqrt{31})\,(4\sqrt2+\sqrt{31})$$ For \(p=47\), let \(n_{\color{red}\pm} = 2(1+\sqrt2)^3\Big(9\color{red}{\pm}2\sqrt{9+8\sqrt2}\Big)\) or again the two real roots. Then $$\frac{\beta(n_{+})}{\beta(n_{-})} = U_{47}\sqrt{U_{94}} = (48+7\sqrt{47})\,(732\sqrt2+151\sqrt{47})$$ and so on for \(p = 31, 47, 79, 191, 239, 431\).

Entry 136

Given fundamental discriminants \(d = 4m\) with class number \(h(-d)=8\). Then there are exactly ten \(m = 2p\) for prime \(p\), namely \(p \equiv 1\,\text{mod}\,4 = 89, 113, 233, 281\) and \(p \equiv 3\,\text{mod}\,4 = 31, 47, 79, 191, 239, 431\). The modular lambda function \(\lambda(\sqrt{-2p})\) for both is a root of a deg-\(16\) equation, but the latter is easier to factor into two deg-\(8\) equations. Define the two simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$ It seems for the \(p \equiv 3\,\text{mod}\,4\), then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n)$$ where \(\alpha(n),\beta(n)\) are octic units and \(n\) is just a quartic root given by $$n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with \(\lambda=\lambda(\tau)\) for simplicity. Examples,

Let \(p=31\) and \(n= 2(1+\sqrt2)^2\Big(1+3\sqrt2+2\sqrt{1+4\sqrt2}\Big)\) then $$\frac1{\lambda(\sqrt{-62})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ Let \(p=47\) and \( n= 2(1+\sqrt2)^3\Big(9+2\sqrt{9+8\sqrt2}\Big)\) then $$\frac1{\lambda(\sqrt{-94})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ and so on for \(p = 31, 47, 79, 191, 239, 431\). P.S. Note that solutions to the Pell-like equation \(x^2-2y^2 = -p\) appear in the nested radicals \(\sqrt{x+y\sqrt2\,}\) above, namely$$1^2-2\cdot4^2=-31\\ 9^2-2\cdot8^2=-47$$

Entry 135

Mathworld has a list of the modular lambda function \(\lambda(\tau)\) with the particular case \(\tau=\sqrt{-14}\) as the rather complicated $$ \sqrt{\lambda(\sqrt{-14})}=-11-8\sqrt2-2(2+\sqrt2)\sqrt{5+4\sqrt2}\qquad \\\ \qquad+\sqrt{11+8\sqrt2}\,\Big(2+2\sqrt2+\sqrt2\sqrt{5+4\sqrt2}\Big)\qquad $$ which is approximately \(0.011208.\) It can be calculated in Mathematica or WolframAlpha as ModularLambda[tau]. However, we can simplify and factor that into two quartic units as $$\begin{align}\frac1{\lambda(\sqrt{-14})} & =\frac1{2^8}(8+3\sqrt7)\,(\sqrt7+\sqrt8)\,\Big(2^{1/4}+\sqrt{4+\sqrt2}\Big)^8\\ &=7960.423255\dots\end{align}$$ It is then just a matter of getting the reciprocal and square roots. One can do so similarly for discriminants \(d=4m\) with class number \(h(-d)=4\) and even \(m=14,62,142\) as described in Entry 134.

Entry 134

Given fundamental discriminants \(d = 4m\) with class number \(h(-d)=4\), there are exactly five \(m = 2p\) for prime \(p\), namely \(p=7,17,23,41,71\). The cases \(p = 17, 41\) were in Entry 115 while \(p = 7,23,71\) which are \(p \equiv 3\,\text{mod}\,4\) will be tackled here. Define the two simple functions$$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$and let$$\begin{align}n_1 &=2+2\sqrt2 \\ n_2 &= 26+18\sqrt2 \\ n_3 &=1450+1026\sqrt2\end{align}$$Then the first quartic unit can be an eighth power 

$$\begin{align}\alpha(n_1) &=\frac1{2^8}\Big(\sqrt{0+\sqrt2}+\sqrt{4+\sqrt2}\Big)^8\\ \alpha(n_2)  &=\frac1{2^8}\Big(\sqrt{4+3\sqrt2}+\sqrt{8+3\sqrt2}\Big)^8\\ \alpha(n_3)  &=\frac1{2^8}\Big(\sqrt{36+27\sqrt2}+\sqrt{40+27\sqrt2}\Big)^8 \end{align}$$ while the second is a product of quadratic units 

$$\begin{align}\beta(n_1) &= U_{7}\,\sqrt{U_{14}}=\big(8+3\sqrt{7}\big) \big(2\sqrt2+\sqrt{7}\big) \\ \beta(n_2)  &= U_{23}\,\sqrt{U_{46}}=\big(25+5\sqrt{23}\big) \big(78\sqrt2+23\sqrt{23}\big) \\ \beta(n_3)  &= U_{71}\,\sqrt{U_{142}}=\big(3480+413\sqrt{71}\big) \big(1710\sqrt2+287\sqrt{71}\big) \\ \end{align}$$ which gives the radical expressions of  $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n_1)\,\beta(n_1),\quad n_1=2+2\sqrt2\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n_2)\,\beta(n_2),\quad n_2=26+18\sqrt2 \\ \frac1{\lambda(\sqrt{-142})} &= \alpha(n_3)\,\beta(n_3),\quad n_3=1450+1026\sqrt2 \end{align}$$ And since \(\beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2\), then they are also nested radicals that use only \(\sqrt2\). 

Entry 133

Given fundamental discriminants \(d=4m\) with class number \(h(-d)=2\), there are exactly four even \(m = 6, 10, 22, 58\). The most well-known is \(m=58\) because of the near-integer $$\quad e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ and the appearance of \(396^4\) in the denominator of Ramanujan's famous \(1/\pi\) formula. These \(m=2p\) for prime \(p\) have other interesting properties. Recall the modular lambda function \(\lambda(\tau)\) also discussed in Entry 112 $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We focus on \(m=2p\) for prime \(p = 3\,\text{mod}\,4\) hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)(\sqrt2+\sqrt3)\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})(7\sqrt2+3\sqrt{11}) \end{align}$$ with fundamental units \(U_n\). However, special \(d\) with class number \(h(-d)=2^k\) surprisingly can be expressed by nested radicals using only the square root of 2. So,$$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= (1+\sqrt2)^2+\sqrt{1+ (1+\sqrt2)^4}\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &=  (1+\sqrt2)^6+\sqrt{1+ (1+\sqrt2)^{12}} \end{align}\quad$$ Similar behavior can also be observed for \(2p\) for \(p=7,23,71\) which now have class number 4.

Entry 132

Given \(_2F_1(a,b;c;z)\) and Dedekind eta function \(\eta(\tau)\) where \(\tau = \frac{1+n\sqrt{-3}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac56}\) $$\begin{align}&\,_2F_1\big(\tfrac12,\tfrac13;\tfrac56;(1-2\delta_1)^2\big),\quad\;\delta_1 = \delta_2\\ &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\sqrt{\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}}\\ &\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1\,=\,\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. Note that \( _2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big)\) so the first form is superfluous. Examples: Let \(\tau = \frac{1+5\sqrt{-3}}2\), $$_2F_1\Big(\frac13,\frac12;\frac56;\frac45\Big)=\;\frac35\sqrt5$$ $$\quad _2F_1\Big(\frac16,\frac23;\frac56;\frac{80}{81}\Big)=\frac35\,(9\sqrt5)^{1/3}$$

Entry 131

Given \(_2F_1(a,b;c;z)\) and Dedekind eta function \(\eta(\tau)\) where \(\tau = \frac{1+n\sqrt{-1}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac34}\) $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\sqrt{-\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}}\\ &\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\ &\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1\,=\, -\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}\\ &\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. Examples: Let \(\tau = \frac{1+5\sqrt{-1}}2\), 

$$_2F_1\Big(\frac14,\frac12;\frac34;\frac{80}{81}\Big)=\frac95$$

$$\quad _2F_1\Big(\frac18,\frac58;\frac34;\frac{25920}{25921}\Big)=\frac35\,161^{1/4}$$

Entry 130

Given \(_2F_1(a,b;c;z)\) and j-function \(j = j(\tau)\) where \(\tau = \frac{1+n\sqrt{-3}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac23}\) $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;(1-2\beta_1)^2\big),\qquad \color{red}{\beta_1 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;(1-2\beta_2)^2\big),\qquad \frac{1}{\beta_2}-1=\sqrt{\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}}\\ &\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;(1-2\beta_3)^2\big),\qquad \color{red}{\beta_3 =\,?} \\ &\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;(1-2\beta_4)^2\big),\qquad \frac{1}{\beta_4}-1=\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}\\\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. Examples: Let \(\tau = \frac{1+3\sqrt{-3}}2\), $$_2F_1\Big(\frac16,\frac12;\frac23;\frac{125}{128}\Big) =\frac43\times2^{1/6}$$ $$_2F_1\Big(\frac1{12},\frac7{12};\frac23;\frac{64000}{64009}\Big) =\frac23\times253^{1/6}$$ Let \(\tau = \frac{1+5\sqrt{-3}}2\), $$_2F_1\left(\frac16,\frac12;\frac23;\Big(\frac45\Big)^2\Big(\frac{15-\sqrt5}{11}\Big)^3\right) =\frac35(5+4\sqrt5)^{1/6}$$

Entry 129

Given \(_2F_1(a,b;c;z)\) where \(a+b=c\). The previous four (Entries 125-128) discuss closed-forms and can be neatly summarized as type \(a+b=c=\color{blue}{\tfrac12}\)  $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\qquad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\qquad \tau_1=n\sqrt{-4}\\ &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\qquad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\qquad \tau_2=n\sqrt{-3}\\ &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\qquad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\qquad \tau_3=n\sqrt{-2}\\ &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad\; \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\qquad\tau_4=n\sqrt{-1}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. However, for the parameters \(c=\frac12, \frac23,\frac34. \frac56\), it seems only the first is easy. For type \(a+b=c=\color{blue}{\tfrac23}\), it's more challenging and will be discussed in the next entry.

Entry 128

Assume \(\tau=n\sqrt{-\color{blue}{4}}\) for some positive integer \(n\). Given \(_2F_1(a,b;c;z)\) where \(a+b=c=\frac12\) for the case \(a=\tfrac1{4}\). Let \(z_1 = (1-2w)^2\) where \(w\) is $$w=\frac{16}{16+\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8}$$ Then \((z_1, z_2)\) are algebraic numbers in

$$_2F_1\left(\tfrac14,\tfrac14;\tfrac12;z_1\right) = z_2$$

Examples: 

If \(n=2\) so \(\tau=2\sqrt{-4}\), then,

$$_2F_1\left(\frac14,\frac14;\frac12;\,\frac{9\,(3+\sqrt2)^2}{(1+\sqrt2)^6}\right)=\frac38\big(2+\sqrt2\big)$$ 

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

$$\quad _2F_1\left(\frac14,\frac14;\frac12;\,\frac{8\sqrt3}{(2+\sqrt3)^2}\right)=\frac23\big(3+2\sqrt3\big)^{1/2}$$