考虑公式
$$
π=\frac{1}{16}\sum_{n=0}^∞\frac{1}{256^n} \left(\frac{256}{16 n+1}-\frac{128}{16 n+4}-\frac{64}{16 n+5}-\frac{64}{16 n+6}+\frac{16}{16 n+9}-\frac{8}{16 n+12}-\frac{4}{16 n+13}-\frac{4}{16 n+14}\right)
$$
$$
\{256^dπ\}=\frac{1}{16}\sum_{n=0}^∞\left(\frac{256}{16 n+1}-\frac{128}{16 n+4}-\frac{64}{16 n+5}-\frac{64}{16 n+6}+\frac{16}{16 n+9}-\frac{8}{16 n+12}-\frac{4}{16 n+13}-\frac{4}{16 n+14}\right)
$$