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摘自书籍：Groups and Symmetry, Amstrong, 第98页
⋯试错法不是解决这类问题的方法。我们不太可能通过实验发现所有9,099种将正十二面体的每个面涂成红色、白色或蓝色的不同方法！ 
Pólya枚举定理
math.stackexchange.com/questions/380807/in-how-many-different-wa ... ecahedron-be-colored
已知循环指数为$$Z(G) = \frac{1}{60}
\left( a_1^{12} + 24 a_1^2 a_5^2 + 20 a_3^4 + 15 a_2^6\right).$$这允许推导出一个最多使用$n$种颜色的上色公式，该公式为$${\frac {1}{60}}\,{n}^{12}+\frac{1}{4}\,{n}^{6}+{\frac {11}{15}}\,{n}^{4}.$$将$n=3$代入此公式，我们得到最多使用3种颜色的上色数量确实为$9099$。
多面体上色
| solid | polynomial | OEIS | colorings for $n=1, 2, \dots$
| | cube | $\frac13n^2+\frac12n^3+\frac18n^4+\frac1{24}n^6$ | A047780 | $1, 10, 57, 240, 800, 2226, 5390, \dots$
| | dodecahedron | $\frac{11}{15}n^4+\frac14n^6+\frac1{60}n^{12}$ | A000545 | $1, 96, 9099, 280832, 4073375, 36292320,\dots$
| | icosahedron | $\frac25n^4+\frac13n^8+\frac14n^{10}+\frac1{60}n^{20}$ | A054472 | $1, 17824, 58130055, 18325477888, 1589459765875, \dots$
| | octahedron | $\frac14n^2+\frac{17}{24}n^4+\frac1{24}n^8$ | A000543 | $1, 23, 333, 2916, 16725, 70911, 241913, \dots$
| | tetrahedron | $\frac{11}{12}n^2+\frac1{12}n^4$ | A006008 | $1, 5, 15, 36, 75, 141, 245, 400, 621, \dots$ |
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