## Confusion regarding the Dihedral Group and Burnside's Lemma

Hello everyone

I am trying to solve a coloring problem i.e find the number of distinct coloring's of a $n-$beaded necklace with $c$ colors. In doing so I want to derive an equation regarding the number of distinct orbits of a group G which are more applicable than:

$Distinct \: orbits = \frac{1}{\left| G \right|} \sum \limits_{i=1}^{n} \left| Fix(g) \right|, \hspace{0.5cm} g \in G$

In doing so I am considering the coloring of the vertices of a $n-$gon when two coloring's are regard the same if reachable by:

1. The Cyclic Group $C_n$
2. The Dihedral Group $D_n$

Acting on the $n-$gon.
So far I have figured out that for the Cyclic group the above (Burnside's Lemma) can be calculated easier as:

$Distinct \: orbits = \frac{1}{\left| G \right|} \sum \limits_{i=0}^{n-1}c^{gcd(i,n)}$

Or put in words the number of colors raised to the number of disjoint-cycles for each permutation. However when I consider the Dihedral Group I find that for the first $\frac{n}{2}$ elements, which are basically the $C_n$ group, I do the same, but for all the elements which can be expressed as an action $r^is$ for $r$ an rotation counterclockwise and $s$ a reflection, I need to add one to the number of disjoint cycles. As an example using $C_6$ and $D_6$ I find the two polynomials.

$P_{C_6}(c) = \frac{1}{6}(c^6+2c+2c^2+c^3)$ for $c=3$ this gives 130

Which makes sense as it fits the number of disjoint cycles, however I also find that:

$P_{D_6}(c) = \frac{1}{12}(c^6+2c+2c^2+c^3+3c^4+3c^3)$ for $c=3$ this gives 92

But for the elements $r^is \in D_6$, I find that the elements are $(1, 3)(4, 6)$, $(1, 5)(2, 4)$, $(2, 6)(3, 5)$, $(1, 2)(3, 6)(4, 5)$, $(1, 4)(2, 3)(5, 6)$ and $(1, 6)(2, 5)(3, 4)$.

So my question is why do I have to add and extra to the number of disjoint cycles to elements $g \in D_n \setminus C_n$.

Thank you