Burnside's formula on a hexagon I was trying to use Burnside's formula on a regular hexagon. I believe the answer is $13$, but I am trying to show my work. Here is what I have figured out.
$$\frac{1}{12}(2^6 + 36 + 30 + \cdots?)$$
$2^6$ is $D_{6}$ (at least that is what I think the regular hexagon is called). Then I reflected over the vertices. This gave me $3$ different possible reflections with $12$ different possible ways which is $36$. Then I reflected over the midpoints. This gave me $3$ different possible reflections with $10$ different possible ways which is $30$. Now the rotations are a bit confusing to me. 
 A: The action on the dihedral group on the hexagon is illustrated below:

The number of assignments of $2$ colors to the vertices that are preserved by a group element $\alpha$ is $$2^{\text{Number of vertex orbits under } \langle \alpha \rangle}$$ since each vertex orbit can be assigned any color, and every vertex in any orbit must be colored the same.
The vertex orbits are highlighted below corresponding to the group elements above (vertices in the same orbit are assigned the same color):

Inputting this into Burnside's Lemma gives the number of assignments of $2$ colors (inequivalent under rotations and reflections) as
$$\tfrac{1}{12}(2^6 + 2^1 + 2^2 + 2^3 + 2^2 + 2^1 + 2^3 + 2^4 + 2^3 + 2^4 + 2^3 + 2^4)=13.$$
Precisely two of these inequivalent assignments of $2$ colors have all colours the same: when they're all white, and when they're all black.  That leaves $11$ inequivalent assignments of $2$ colors to the vertices where both colors are used.
A: I upvoted the first answer but I would like to show how to compute the
cycle index $Z(D_6)$  of the dihedral group $D_6$  and apply the Polya
Enumeration Theorem to this problem.
We  need  to  enumerate   and  factor  the  twelve  permutations  that
contribute to $Z(D_6).$
There is the idenity, which contributes
$$a_1^6.$$
The two rotations by a distance of one and five contribute
$$2 a_6.$$
The two rotations by a distance of two or four contribute
$$2 a_3^2.$$
Finally the rotation by a distance of three contributes
$$a_2^3.$$
There  are three reflections  about an  axis passing  through opposite
vertices, giving
$$3 a_1^2 a_2^2$$
and three reflections  about an axis passing through  the midpoints of
opposite edges, giving
$$3 a_2^3.$$
This finally yields the cycle index
$$Z(D_6) = \frac{1}{12}
\left(a_1^6 + 2a_6 + 2a_3^2 + 3a_1^2 a_2^2 + 4a_2^3\right).$$
Coloring the hexagon with at most two colors we get
$$Z(D_6)(A+B)_{A=1, B=1}.$$
This yields
$$\frac{1}{12}
\left(2^6 + 2\times 2 + 2\times 2^2 
+ 3\times 2^2 \times 2^2 + 4 \times 2^3\right)$$
which evaluates to
$$13.$$
There is a list of similar computations at
MSE Meta.
