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In this 2010 post, I pointed out that given the fundamental unit $U_{29}= \frac{5+\sqrt{29}}2$, and fundamental solutions to Pell equations $x^2-29y^2 = \pm1$,

$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$

$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot\color{blue}{1820}^2=1$$

and the relations,

$$2^6\left((U_{29})^6+(U_{29})^{-6}\right)^2 =\color{blue}{396^4}$$ $$e^{\pi\sqrt{\color{blue}{58}}} = 396^4-104.00000017\dots$$

then we find these six blue numbers all over Ramanujan's famous formula,

$$\frac1{\pi} = \frac{\sqrt2}{\color{blue}{9801}} \sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac{\color{blue}{58}\cdot\color{blue}{70\cdot13}\,n+2206}{\color{blue}{(396^4)}^n}$$

except one, namely $y = \color{blue}{1820}$. I've always been bothered by this.

Revisiting this problem, there may be a solution. We use three fundamental units,

$$U_2 = 1+\sqrt2\\ U_{29}= \frac{5+\sqrt{29}}2\\ U_{58}= 99+13\sqrt{58}$$

since the relevant discriminant is actually $d = -2\cdot29 = -58$ which has class number $h(d) = 2$. A complete elliptic integral of the first kind $K(k_r)$ is also needed. In Mathematica syntax, this is EllipticK[ModularLambda[Sqrt[-58]]]. Then we get,

$$\frac1{\pi}-\left(\frac{12^4\,\big(U_{29}\big)^3}{396\,U_{58}}\right) \left(\frac{2K(k_{58})}{\pi\,\big(U_2)^3}\right)^2 = \frac1{\color{blue}{9801}\sqrt2} \sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac{2\cdot\color{blue}{58}\cdot\color{blue}{70\cdot13}\,n+\color{blue}{1820}}{\color{blue}{(396^4)}^n}$$

and all six blue numbers finally appear!

Question: But how do we prove this though?

P.S. In the Mathworld list above for $K(k_r)$, they give closed-forms in terms of gamma functions $\Gamma(n)$, but don't have $r=58$. Since $4\times58 = 232$, then,

$$\left(\frac{58\,\big(U_{29}\big)^3}{\pi\,U_{58}}\right)^2 \left(\frac{4K(k_{58})}{\big(U_2)^3}\right)^4= \prod_{m=1}^{232}\Gamma\left(\frac{m}{232}\right)^{\big(\frac{-232}{\; m}\big)}$$

where the exponent $\big(\frac{-232}{\; m}\big)$ is the Kronecker symbol.

Tito Piezas III
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  • I think the proof of your final identity would be equivalent to a proof of the Ramanujan's famous series. If you desire I can write a proof for the equivalence mentioned in last sentence, but it will take me some reasonable time to do so. – Paramanand Singh Sep 04 '24 at 17:37
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    One needs to use the identity $(2K/\pi)^2=(1+k^2)^{-1}{}3F{2}(1/4,3/4,1/2;1,1;((g^{12}+g^{-12})/2)^{-2})$ with $g=\sqrt{(5+\sqrt{29})/2}$ and $k=(13\sqrt{58}-99)(\sqrt{2}-1)^6$. Note that $k$ is a unit. – Paramanand Singh Sep 04 '24 at 21:17
  • @ParamanandSingh Feel free to write a proof when you have time. Yes, $k$ is a unit that is a product of units, namely $k = \sqrt{ModularLambda(\sqrt{-58})} = (13\sqrt{58}−99)(\sqrt2−1)^6 = (U_{58})^{-1}(U_2)^{-6}$, verified by Wolfram here. Gotta love how Ramanujan was a master at factoring radicals. – Tito Piezas III Sep 05 '24 at 01:56
  • We have $58\cdot 70\cdot 13=29\cdot 1820$ so 1820 does appear, but multiplied by $29$. – Paramanand Singh Sep 06 '24 at 01:41
  • @ParamanandSingh After all these years, I never checked the factorization was just $1820 = 2\cdot13\cdot70$, hehe. Since that is the case, the formula above greatly simplifies as, $$\frac1{\pi}-\left(\frac{12^4,\big(U_{29}\big)^3}{396,U_{58}}\right) \left(\frac{2K(k_{58})}{\pi,\big(U_2)^3}\right)^2 = \frac{1820}{9801\sqrt2} \sum_{n=0}^\infty \frac{(4n)!}{n!^4} ,\frac{58n+1}{(396^4)^n}$$ – Tito Piezas III Sep 06 '24 at 02:18

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