另外,根据上述讨论,可以把不整除的 `k` 具体列出来。
比如情况(1)中,`p=\text{*000*000**0}_2`,则 `k=2p=\text{*000*000**00}_2`,即
\begin{align*}
k\in\{ & 0,100_2,1000_2,1100_2, 10000000_2,10000100_2,10001000_2,10001100_2, \\
& 100000000000_2,100000000100_2,100000001000_2,100000001100_2, \\
& 100010000000_2,100010000100_2,100010001000_2,100010001100_2 \},
\end{align*}
化回十进制之后就是
\[k\in\{0, 4, 8, 12, 128, 132, 136, 140, 2048, 2052, 2056, 2060, 2176, 2180, 2184, 2188\};\]
同样的方法列情况(2)(3)(4)分别为
\begin{align*}
k&\in\{2,10,130,138,2050,2058,2178,2186\};\\
k&\in\{64,68,72,76,2112,2116,2120,2124\};\\
k&\in\{1024,1028,1032,1036,1152,1156,1160,1164\}.
\end{align*}
将以上四个集合与前面的 `k\in\{0,547,2\times547,3\times547,4\times547\}` 并起来之后,就得到了不整除的所有 `k` 为
0, 2, 4, 8, 10, 12, 64, 68, 72, 76, 128, 130, 132, 136, 138, 140, 547, 1024, 1028, 1032, 1036, 1094, 1152, 1156, 1160, 1164, 1641, 2048, 2050, 2052, 2056, 2058, 2060, 2112, 2116, 2120, 2124, 2176, 2178, 2180, 2184, 2186, 2188。 |