A partir del artículo de la Wikipedia tenemos
$\displaystyle \begin{aligned}e^{\pi\sqrt{43}} &\approx 884736743.999777466 \\ e^{\pi\sqrt{67}} &\approx 147197952743.999998662454\\ e^{\pi\sqrt{163}}&\approx 262537412640768743.99999999999925007 \end{aligned}$
En su papel de "Modular ecuaciones y aproximaciones a $\pi$" Ramanujan le da un enfoque muy simple para encontrar estas aproximaciones y demostrar que son casi enteros. Cito del artículo:
Desde $G_{n}, g_{n}$ puede ser expresado como raíces algebraico de ecuaciones con coeficientes racionales, lo mismo es cierto de $G_{n}^{24}$ o $g_{n}^{24}$. Así que vamos a suponer que los $$1 = ag_{n}^{-24} - bg_{n}^{-48} + \cdots\\ \text{o }g_{n}^{24} = a - bg_{n}^{-24} + \cdots$$ But we know that $$64e^{-\pi\sqrt{n}}g_{n}^{24} = 1 - 24e^{-\pi\sqrt{n}} + 276e^{-2\pi\sqrt{n}} + \cdots\\ 64g_{n}^{24} = e^{\pi\sqrt{n}} - 24 + 276e^{-\pi\sqrt{n}} + \cdots\\ 64a - 64bg_{n}^{-24} + \cdots = e^{\pi\sqrt{n}} - 24 + 276e^{-\pi\sqrt{n}} + \cdots\\ 64a - 4096be^{-\pi\sqrt{n}} + \cdots = e^{\pi\sqrt{n}} - 24 + 276e^{-\pi\sqrt{n}} + \cdots\\ \text{es }e^{\pi\sqrt{n}} = (64a + 24) - (4096b + 276)e^{-\pi\sqrt{n}} + \cdots$$
Aquí tenemos definiciones estándar $$G_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots$$ and $$g_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots$$ From the theory of modular equations I understand that the values $g_{n}, G_{n}$ are algebraic numbers if $n > 0$ is rational. But this does not make them make algebraic integers. I am therefore not able to understand how Ramanujan uses the numbers $a, b$ in the derivations above as integers and is able to conclude that the value of $e^{\pi\sqrt{n}}$ is close to the integer $(64a + 24)$. I believe the approach provided above works only when $g_{n}, G_{n}$ are algebraic integers. Please let me know if we can conclude $e^{\pi\sqrt{n}}$ is an almost integer even if $g_{n}$ es algebraico, pero no algebraicas entero.
