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),

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

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

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

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


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

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

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

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

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



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

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

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

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

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

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

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

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

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

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

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

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

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

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



  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.)
  • $\begingroup$ I'm sorry, but what $545140134$ mean? Wolfram says its square root is $3 \sqrt{60571126}$. Typo or I'm missing something? $\endgroup$ – Ian Mateus Dec 17 '13 at 0:53
  • 2
    $\begingroup$ $163(12^3+640320^3) = 12^2\cdot 545140134^2$. I meant that the square root of the RHS appears in the Chudnovsky brothers’ pi formula. $\endgroup$ – Tito Piezas III Dec 17 '13 at 1:00
  • $\begingroup$ Very nice list... and $5$ is just great! $\endgroup$ – CODE Dec 17 '13 at 18:49
  • $\begingroup$ Oh just remember to change the numbers in notes, because you changed the numbers on the list itself. $\endgroup$ – CODE Dec 17 '13 at 18:56
  • 1
    $\begingroup$ I don't know if it is a consequence of class number 1, but $$\pi\approx 3+\frac{1}{(2^{14}+2^{10}+163)^\frac{1}{5}}\approx3.1415926$$ $\endgroup$ – Jaume Oliver Lafont Apr 8 '16 at 20:56

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,

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


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

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,


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


$\left(\frac{-163}{n}\right) = \mu(n)$ for squarefree $n \leq 40$, where the LHS is the Kronecker symbol and the RHS is the Mobius function. (Iwaniec & Kowalski pg. 520)

  • $\begingroup$ Interesting. The formula in the post for $K(k_{163})$ is, $$K(k_{163}) = \frac{x^2}{2} \sqrt{ \frac{\pi}{2d} \prod_{n=1}^{d}\Big[\Gamma\big(\tfrac{n}{d}\big)\Big]^{\Big(\frac{-d}{n}\Big)}} $$ with $d=163$ and Kronecker symbol $\Big(\frac{-d}{n}\Big)$ which, after simplification, can be reduced to just a product of 81 $\Gamma(m)$. $\endgroup$ – Tito Piezas III Dec 27 '13 at 19:57

We also have the real series given above,

$$\big(K(k_{163})\big)^2 = \frac{\pi^2}{4} \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \left(\frac{1}{x_1}\right)^{24n} =2.467401\dots\tag{1}$$

and the complex series,

$$\big(K(k_{163})\big)^2 = \frac{\pi^2}{\pm\,4\sqrt{-1}}\left(\frac{x_2}{x_3}\right)^4 \,\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \left(\frac{x_2}{\sqrt{2}}\right)^{24n} =2.467401\dots\tag{2}$$

where $x_1$ being the real root, and $x_2,x_3$ as the complex conjugates of,


and the sign of $\pm\sqrt{-1}$ chosen appropriately.


H.H.Chan found that given the fundamental solution to the Pell equation,


for $d=163$, hence the unit,

$$u = 7592629975+343350596\sqrt{489} = \big(35573\sqrt{3}+4826\sqrt{163}\big)^2$$


$$3\sqrt{3u}-3\sqrt{\tfrac{3}{u}}+6 = 640320$$

Similarly for $d=19,43,67$.

P.S. Incidentally, let $\tau = \tfrac{3+\sqrt{-163}}{6}$, then the eta quotient,

$$\Big(\tfrac{\eta^2(3\tau)}{\eta(\tau)\,\eta(9\tau)}\Big)^6 = -3\sqrt{3u} = -640314.000042\dots$$

can appear in a level 9 Ramanujan-type pi formula.


As pointed out by Greg Martin in the context of the abc conjecture,

$$3^37^211^219^2127^2 163- 2^{12}5^323^329^3=1\tag{1}$$

Note this equality has the form,


which solves the Pell equation,

$$(p r^2 + q s^2)^2 - p q(2r s)^2 = 1$$

In fact, the fundamental solution to,


is given by the $p,q,r,s$ derivable from $(1)$. See also the related post https://mathoverflow.net/questions/154655/.

  • $\begingroup$ This is true but how is it relevant? If $N=2286831727304144$, $N$ and $N+1$ are 73-smooth, and $N$ is greater than the numbers quoted here. So there's probably a pair of consecutive integers much bigger than these which are both 163-smooth. $\endgroup$ – Rosie F Apr 21 at 15:02

In Traces of Singular Moduli (p.2), Zagier defines the modular form of weight 3/2,

$$g(\tau) = \frac{\eta^2(\tau)}{\eta(2\tau)}\frac{E_4(4\tau)}{\eta^6(4\tau)}=\vartheta_4(\tau)\, \eta^2(4\tau)\,\sqrt[3]{j(4\tau)}$$

which has the nice q-expansion (A027652, negated terms),

$$g(\tau) = \tfrac{1}{q} - 2 + 248q^3 - 492q^4 +(15^3+744)q^7 + \dots + (5280^3+744)q^{67} + \dots + (640320^3+744)q^{163}+\dots$$

See also "Monstrous Moonshine for Thompson group Th?".


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