# Can one enumerate the set partitions of the faces of a regular dodecahedron mathematically?

If we partition the faces of a regular dodecahedron into one nonempty subset, there is only one partition; for twelve nonempty subsets, there is also just one partition; for eleven nonempty subsets there are three partitions--the two faces in one subset are either adjacent, opposite, or adjacent to a third face but not each other. The full sequence of oriented partitions (two partitions are equivalent if one is a rotation of the other; chiral pairs are counted as two) into $$n$$ nonempty subsets is

1, 55, 1561, 10517, 23425, 22510, 10729, 2782, 416, 41, 3, 1

according to my computer program. For unoriented partitions (two partitions are equivalent if one can be obtained from the other by a series of reflections; chiral pairs are counted as one) the full sequence is

1, 45, 911, 5597, 12207, 11690, 5637, 1508, 246, 29, 3, 1

according to my computer program. Is there a way to calculate these sequences mathematically?

The cycle index for the group of rotations of the faces of a regular dodecahedron is $$(x_1^{12}+24x_1^2x_5^2+20x_3^4+15x_2^6)/60$$, which can be transformed into a formula for the number of oriented colorings with up to $$n$$ colors by substituting $$\sum_k k!{j\brace k}{n\choose k}$$ for each $$x_i^j$$ to obtain

$${n\choose1}+94{n\choose2}+8814{n\choose3}+245008{n\choose4}+2759250{n\choose5}+15884004{n\choose6}+52701264{n\choose7}+106866144{n\choose8}+134719200{n\choose9}+103118400{n\choose10}+43908480{n\choose11}+7983360{n\choose12}$$.

The cycle index for the full automorphism group is $$(x_1^{12}+15x_2^6+20x_3^4+24x_1^2x_5^2+15x_1^4x_2^4+x_2^6+20x_6^2+24x_2^1x_{10}^1)/120$$, which can be transformed into a formula for the number of unoriented colorings with up to $$n$$ colors by the same substitution to obtain

$${n\choose1}+80{n\choose2}+5136{n\choose3}+127620{n\choose4}+1395390{n\choose5}+7965948{n\choose6}+26368272{n\choose7}+53438112{n\choose8}+67359600{n\choose9}+51559200{n\choose10}+21954240{n\choose11}+3991680{n\choose12}$$