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 what is not is 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 $d=4n$ has class number $6$ using fundamental units $U_n$. Borrowing a trick from Ramanujan, he found

$$G_{169}=\frac13\Big(2+\sqrt{13}+\sqrt[3]{U_{13}(v+3\sqrt{39})}+\sqrt[3]{U_{13}(v-3\sqrt{39})}\Big)$$

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

Using this form, we propose that

$$\begin{align}U_{29} &=\frac1{3^{1/4}}\Big(\tfrac{9+\sqrt{29}}2+\sqrt[3]{U_{29}(x+24\sqrt{3})}+\sqrt[3]{U_{29}(x-24\sqrt{3})}\Big)^{1/4}\\ U_{53} &=\frac1{3^{1/4}}\Big(\tfrac{23+3\sqrt{53}}2+\sqrt[3]{U_{53}(y+120\sqrt{3})}+\sqrt[3]{U_{53}(y-120\sqrt{3})}\Big)^{1/4}\\ U_{61} &=\frac1{3^{1/4}}\Big(15+2\sqrt{61}+\sqrt[3]{U_{61}(z+72\sqrt{3})}+\sqrt[3]{U_{61}(z-72\sqrt{3})}\Big)^{1/4}\end{align}$$ 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 $d=4p$ with class number $6$. All their seven 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 

$$\begin{align}\beta(n_1) &= U_{7}\,\sqrt{U_{14}}=\big(8+3\sqrt{2}\big) \big(2\sqrt2+\sqrt{7}\big) \\ \beta(n_2)  &= U_{23}\,\sqrt{U_{46}}=\big(25+5\sqrt{23}\big) \big(78\sqrt2+23\sqrt{23}\big) \\ \beta(n_3)  &= U_{71}\,\sqrt{U_{142}}=\big(3480+413\sqrt{71}\big) \big(1710\sqrt2+287\sqrt{71}\big) \\ \end{align}$$ 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,

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

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 $$\frac1{\big(\text{K}_4(q)\big)^2}-\big(\text{K}_4(q)\big)^2 =\frac12\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^4\quad$$