考虑公式
$$
\pi =\sum_{k=0}^{\infty }\left[{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)\right]
$$
两边乘以 $16^d$ 并取小数, 则:
$$
\{16^d\pi\} = \left\{\sum _{k=0}^{\infty }\left(
{\frac {16^{d-k} 4}{8k+1}}-
{\frac {16^{d-k} 2}{8k+4}}-
{\frac {16^{d-k}}{8k+5}}-
{\frac {16^{d-k}}{8k+6}}
\right)\right\}
$$
举个例子, $d = 4$ 时有:
$$
\begin{aligned}
\pi &≈ \texttt{3.243f6a8885a368d31\ldots} \\
16^4 \pi &≈ \texttt{3243f.6a8885a368d31\ldots} \\
\left\{16^4 \pi\right\} &≈ \phantom{0000\,}\texttt{0.6a8885a368d31\ldots}
\end{aligned}
$$