[Math] Other interesting consequences of $d=163$

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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:

  1. The exact value of the j-function $j(\tau)=-640320^3= -12^3(231^2-1)^3$.
  2. Is prime for $0\leq n \leq 19$. (OEIS link)
  3. 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).
  4. 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 $c<41$.
  5. Where $u=(e^{\pi\sqrt{163}}+24)^{1/24}$.
  6. 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).
  7. The perfect square appears in the pi formula.
  8. By the Chudnovsky brothers (based on Ramanujan's formulas).
  9. 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.
  10. $E_n$ are Eisenstein series. More details in this MO post.
  11. The prime factors of $j(\tau)$ and $j(\tau)-12^3$ .
  12. $\eta(\tau)$ is the Dedekind eta function.
  13. 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.
  14. Based on a formula for the j-function using eta quotients.
  15. $K(k_d)$ is the complete elliptic integral of the first kind.
  16. $\,_2F_1(a,b;c;z)$ is the hypergeometric function.
  17. $\alpha$ is a "singular moduli" of signature 6.
  18. Is a product of 81 gamma function $\Gamma(n)$ given in the link above.
  19. A special case of the Heine continued fraction where as usual $q = e^{2\pi i \tau}$ and x as above.
  20. The general form is by M. Naika(?).
  21. Ramanujan's octic continued fraction and $k_{163}=k$ is the value such that $\frac{K'(k)}{K(k)}=\sqrt{163}$.
  22. As observed by Conway and Norton, the moonshine functions span a linear space of dimension 163. (The relevance of 163 is still speculative though.)

Best Answer

Noam Elkies kindly pointed out that in his preprint "Three Lectures on elliptic surfaces and curves of high rank" (page 9), there is a discriminant $-163$ elliptic surface with torsion group $\mathbb{Z}/4\mathbb{Z}$. Explicitly, this is,

$$y^2 + a x y + a b y = x^3 + b x^2\tag{1}$$

where,

$$a = (8t - 1)(32t + 7),\;\;b = 8(t + 1)(15t - 8)(31t - 7)$$

Elkies gave several parametric solutions to (1). After some fiddling around, I found it can also be solved as,

$$x = \frac{-a^2}{8},\;\;y = \frac{a(a^2-8b)}{16},\;\;t = \frac{3(3-2v)}{4(2+v)}$$

and where $v$ (among others) are the roots of the discriminant $-163$ cubic,

$$v^3-6v^2+4v-2=0$$

mentioned in the post. (Though I don't know why it works.)

P.S: Another version of Elkies' K3 surface can be found at http://www.math.rice.edu/~hassett/conferences/Clay2006/Elkies/CMIPelkies.pdf

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