How many ways can you color the edges of a hexagon in two colors?

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Homework Statement



How many ways can you color the edges of a hexagon in two colors? It is assumed two colorings are identical if there is a way to flip or rotate the hexagon.

Homework Equations



Orbit Stabilizer Lemma and Burnside's Lemma

The Attempt at a Solution



This, implements the Orbit Stabilizer Lemma and Burnside's Lemma (think necklace permutations) however, is there anything special or different to computing this because you are now dealing with the faces/edges rather than the vertices (or beads of a necklace)?

Thanks.
 
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To answer my own question: it does not matter.

Di6 symmetry with order 12.
6 rotation symmetries
6 reflection symmetries

By Burnside's Lemma:

ƒ(n) = \frac{1}{12}⋅(2⋅n + 2⋅n^{2} + 4⋅n^{3} + 3⋅n^{4} + n^{6})

Where
n := # of colors
ƒ(n) := # of unique colorings

n = 2
ƒ(2) = 13


n = 3
ƒ(3) = 92

n = 4
ƒ(4) = 430

n = 5
ƒ(5) = 1505


\cdots
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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