ラマヌジャン型円周率公式
モジュラー形式をバックにする円周率の公式たち。私用メモです。
n: 非負整数
\begin{align}
h_1 (n) &= \frac{(6n)!}{(3n)! n!^3} \\
h_2 (n) &= \frac{(4n)!}{n!^4} \\
h_3 (n) &= \frac{(2n)! (3n)!}{n!^5} \\
h_4 (n) &= \frac{(2n)! ^3}{n!^6} \\
\end{align}
\begin{align}
\frac{1}{\pi} &= 3 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{63n+8}{(15^3) ^{n+1/2}} \\
&= 4 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{154n+15}{(32^3) ^{n+1/2}} \\
&= 12 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{342n+25}{(96^3) ^{n+1/2}} \\
&= 36 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{506n+31}{(3 \times 160^3) ^{n+1/2}} \\
&= 12 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{16254n+789}{(960^3) ^{n+1/2}} \\
&= 12 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{261702n+10177}{(5280^3) ^{n+1/2}} \\
&= 12 \sum_{n=0}^{\infty} h_1(n) (-1)^n \frac{545140134n+13591409}{(640320^3) ^{n+1/2}}
\end{align}
\begin{align}
\frac{1}{\pi} &= 8 \sum_{n=0}^{\infty} h_1(n) \frac{28n+3}{(20^3) ^{n+1/2}} \\
&= 72 \sum_{n=0}^{\infty} h_1(n) \frac{11n+1}{(2 \times 30^3) ^{n+1/2}} \\
&= 24 \sqrt{2} \sum_{n=0}^{\infty} h_1(n) \frac{63n+5}{(66^3) ^{n+1/2}} \\
&= 162 \sum_{n=0}^{\infty} h_1(n) \frac{133n+8}{(255^3) ^{n+1/2}}
\end{align}
\begin{align}
\frac{1}{\pi} &= 4 \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{20n + 3}{(32 ^ 2)^{n+1/2}} \\
&= 4 \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{260n + 23}{(288 ^ 2)^{n+1/2}} \\
&= 4 \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{21460n + 1123}{(14112 ^ 2)^{n+1/2}}
\end{align}
\begin{align}
\frac{1}{\pi} &= \sqrt{7} \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{65n + 8}{(63 ^ 2)^{n+1/2}} \\
&= 12 \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{28n + 3}{(3 \times 64 ^ 2)^{n+1/2}} \\
&= 20 \sum_{n=0}^{\infty} h_2(n) (-1)^n \frac{644n + 41}{(5 \times 1152 ^ 2)^{n+1/2}}
\end{align}
\begin{align}
\frac{1}{\pi} &= 4 \sqrt{2} \sum_{n=0}^{\infty} h_2(n) \frac{7n + 1}{(2 \times 18^2) ^ {n+1/2}} \\
&= 8 \sqrt{3} \sum_{n=0}^{\infty} h_2(n) \frac{8n + 1}{(48^2) ^ {n+1/2}} \\
&=32 \sqrt{2} \sum_{n=0}^{\infty} h_2(n) \frac{10n + 1}{(12^4) ^ {n+1/2}} \\
&= 48 \sqrt{3} \sum_{n=0}^{\infty} h_2(n) \frac{40n + 3}{(784^2) ^ {n+1/2}} \\
&= 8 \sqrt{11} \sum_{n=0}^{\infty} h_2(n) \frac{280n + 19}{(1584^2) ^ {n+1/2}} \\
&= 32 \sqrt{2} \sum_{n=0}^{\infty} h_2(n) \frac{26390n + 1103}{(396^4) ^ {n+1/2}}
\end{align}
加筆中
