** Question**: Any other interesting consequences of $d=163$ having class number $h(-d)=1$ aside from the list below?

Let $\tau = \tfrac{1+\sqrt{-163}}{2}$. We have (see notes at end of list),

$$e^{\pi\sqrt{163}}\approx 640320^3+743.99999999999925\dots\tag{1}$$

$$B(n) = 4n^2+163\tag{2}$$

$$F(n) = n^2+n+41\tag{3a}$$

$$P(n) = 9n^2-163\cdot3n+163\cdot41\tag{3b}$$

$$1^2+40^2=42^2-163\tag{4}$$

$$u^3-6u^2+4u \approx 1.999999999999999999999999999999977\dots\tag{5}$$

$$5y^6 - 640320y^5 - 10y^3 + 1 =0\tag{6}$$

$$163(12^3+640320^3) = 12^2\cdot \color{blue}{545140134}^2\tag{7}$$

$$12\sum_{n=0}^\infty (-1)^n \frac{(6n)!}{n!^3(3n)!} \frac{\color{blue}{545140134}\,n+13591409}{(640320^3)^{n+1/2}} = \frac{1}{\pi}\tag{8}$$

$$12\sum_{n=0}^\infty (-1)^n\, G_1 \frac{\color{blue}{545140134}\,(n+m)+13591409}{(640320^3)^{n+m+1/2}} = \frac{1}{2^6}\ln\left(\frac{3^{21}\cdot5^{13}\cdot29^5}{2^{38}\cdot23^{11}}\right)\tag{9}$$

$$\frac{E_{4}(\tau)}{\left(E_2(\tau)-\frac{6}{\pi\sqrt{163}}\right)^2}=\frac{5\cdot23\cdot29\cdot163}{2^2\cdot3\cdot181^2}\tag{10a}$$

$$\frac{E_{6}(\tau)}{\left(E_2(\tau)-\frac{6}{\pi\sqrt{163}}\right)^3}=\frac{7\cdot11\cdot19\cdot127\cdot163^2}{2^9\cdot181^3}\tag{10b}$$

$$640320 = 2^6\cdot 3\cdot 5\cdot 23\cdot 29\tag{11a}$$

$$\color{blue}{545140134}/163 = 2\cdot 3^2\cdot 7\cdot 11\cdot 19\cdot 127\tag{11b}$$

$$x^3-6x^2+4x-2=0,\;\;\text{where}\; x = e^{\pi i/24}\frac{\eta(\tau)}{\eta(2\tau)}=5.31863\dots\tag{12}$$

$$x = 2+2\sqrt[3]{a+2\sqrt[3]{a+2\sqrt[3]{a+2\sqrt[3]{a+\cdots}}}} = 5.31863\dots\;\text{where}\;a=\tfrac{5}{4}\tag{13}$$

$$\frac{x^{24}-256}{x^{16}} = 640320\tag{14}$$

$$K(k_{163}) = \frac{\pi}{2} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{x^{24n}} }=1.57079\dots\tag{15a}$$

$$K(k_{163}) = \frac{\pi}{2} \frac{x^2 }{640320^{1/4}} \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{n!^3(3n)!} \frac{1}{(-640320^3)^{n}} }\tag{15b}$$

$$K(k_{163}) = \frac{\pi}{2} \frac{x^2 }{640320^{1/4}}\,_2F_1(\tfrac{1}{6},\tfrac{5}{6},1,\alpha) \tag{16a}$$

$$\frac{\,_2F_1(\tfrac{1}{6},\tfrac{5}{6},1,1-\alpha)}{\,_2F_1(\tfrac{1}{6},\tfrac{5}{6},1,\alpha)}=-\frac{1+\sqrt{-163}}{2}\,i \tag{16b}$$

$$\alpha = \frac{1}{2}\Big(1-\sqrt{1+1728/640320^3}\Big)\tag{17}$$

$$K(k_{163}) = \frac{\pi}{2} \frac{x^2 }{163^{1/4}(2\pi)^{41}}\Gamma(\tfrac{1}{163})\,\Gamma(\tfrac{4}{163})\,\Gamma(\tfrac{6}{163})\dots\Gamma(\tfrac{161}{163}) \tag{18}$$

$$\frac{(e^{\pi\sqrt{163}})^{1/24}}{x} = 1+\cfrac{q}{1-q+\cfrac{q^3-q^2}{1+\cfrac{q^5-q^3}{1+\cfrac{q^7-q^4}{1+ \ddots}}}}\tag{19}$$

$$\frac{(e^{\pi\sqrt{163}})^{1/8}}{x^2}\frac{\eta(\tau)}{e^{\pi i/24}} = \cfrac{1}{1-\cfrac{q}{1+q-\cfrac{q^2}{1+q^2-\cfrac{q^3}{1+q^3-\ddots}}}}\tag{20}$$

$$\frac{(e^{\pi\sqrt{163}})^{1/8}}{\sqrt{2}\,\big(1/k_{163}-1\big)^{1/8}} = \cfrac{1}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}\tag{21}$$

$$\text{Moonshine}_{\,d}=163\tag{22}$$

*Notes*:

- The exact value of the j-function $j(\tau)=-640320^3= -12^3(231^2-1)^3$.
- Is prime for $0\leq n \leq 19$. (OEIS link)
- Euler's polynomial (a) is prime for $0\leq n \leq 39$, while (b) is prime for $1\leq n \leq 40$ (but yields different values from the former).
- As observed by Adam Bailey and Mercio in this comment, Euler's polynomial implies the smallest positive integer $c$ to the Diophantine equation $a^2+b^2=c^2-163$ can't be $c<41$.
- Where $u=(e^{\pi\sqrt{163}}+24)^{1/24}$.
- Where $y\approx\frac{1}{5}(e^{\pi\sqrt{163}}+6)^{1/3}$. (The sextic factors over $\sqrt{5}$ hence has a solvable Galois group).
- The perfect square appears in the pi formula.
- By the Chudnovsky brothers (based on Ramanujan's formulas).
- By J. Guillera in this paper. Let $G_1 = 12^{3n} (\frac{1}{2}+m)_n (\frac{1}{6}+m)_n (\frac{5}{6}+m)_n (m+1)_n^{-3}$ where $m = \frac{1}{2}$ and $(x)_n$ is the Pochhammer symbol.
- $E_n$ are Eisenstein series. More details in this MO post.
- The prime factors of $j(\tau)$ and $j(\tau)-12^3$ .
- $\eta(\tau)$ is the Dedekind eta function.
- Expressing
*x*as infinite nested cube roots. The continued fraction of $x-2$ was also described by H. M. Stark as*exotic*. See OEIS link. - Based on a formula for the j-function using eta quotients.
- $K(k_d)$ is the complete elliptic integral of the first kind.
- $\,_2F_1(a,b;c;z)$ is the hypergeometric function.
- $\alpha$ is a "singular moduli" of signature 6.
- Is a product of 81 gamma function $\Gamma(n)$ given in the link above.
- A special case of the
*Heine continued fraction*where as usual $q = e^{2\pi i \tau}$ and*x*as above. - The general form is by M. Naika(?).
- Ramanujan's
*octic continued fraction*and $k_{163}=k$ is the value such that $\frac{K'(k)}{K(k)}=\sqrt{163}$. - As observed by Conway and Norton, the moonshine functions
*span a linear space of dimension 163*. (The relevance of 163 is still speculative though.)