Count of bracelets with no adjacent colours the same I already know how to calculate the amount of bracelets of length $n$ and $k$ colours.
I'd like to add a condition: only count bracelets with no adjacent colours the same.
For context, this is a follow-on to this question.
 A: We use the data from the following MSE 
link as pointed out 
in the comments. Now with bracelets we have dihedral symmetry so we 
need the two cycle indices for the cyclic group and the dihedral group.
We have for the former
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
and the latter
$$Z(D_n) = 
\frac{1}{2} Z(C_n) +
\begin{cases} 
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
Observe that  when $n$ odd  and $n\ge 3$  the reflections map  the two
vertices opposite  the fixed vertex  to each other. Burnside  says the
coloring must be constant  on the pair, but this is  not possible in a
proper coloring because  they are adjacent. Therefore when  $n$ is odd
we get
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2n} \sum_{d|n} \varphi(n/d) P_d(k).}$$
The  same  phenomenon occurs  when  $n$  is  even and  the  reflection
partitions everything into two-cycles.  (The coloring must be constant
on the two  two-cycles that are intersected by the  axis of reflection
and  consist of  adjacent beads,  which is  not possible  in a  proper
coloring.) On  the other hand when  there are two fixed  points we can
use  any  proper coloring  connecting  them  say counterclockwise  and
reflect  its interior  points  across the  axis  connecting the  fixed
points to get the  colors on the right side so  that the assignment is
fixed as required by Burnside. The length of this path is $(n-2)/2+2 =
n/2+1$ so we get for $n$ even
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2n} \sum_{d|n} \varphi(n/d) P_d(k)
+ \frac{1}{4} k (k-1)^{n/2}.}$$
We can implement this in Maple, as follows.

with(numtheory);

P := (d,k) -> (k-1)^d+(-1)^d*(k-1);

C :=
proc(n, k)
local d;

    1/2/n*add(phi(n/d)*P(d,k), d in divisors(n))
    + `if`(type(n,even), 1/4*k*(k-1)^(n/2), 0);
end;

We get for $n\ge 2$ and three colors the sequence
$$3, 1, 6, 3, 13, 9, 30, 29, 78, 93, 224, 315, \ldots$$
which  points us  to  OEIS A208539,  which
looks to be a match. Four colors yields
$$6, 4, 21, 24, 92, 156, 498, 1096, 3210,
8052, 22913, 61320,\ldots$$
which points to  OEIS A208540, which looks
correct as well. The OEIS assumes that a singleton can have $k$ proper
colorings while the  above formulae produce zero  colorings for $n=1$,
regarding a singleton cycle having  the singleton connected to itself,
for no proper colorings.
 Remark.  Note that  the term $\sum_{d|n}  \varphi(n/d) (-1)^d$
appearing in the cyclic component
$$P_d(k) = (k-1)^d + (-1)^d\times (k-1)$$ 
can be simplified  as follows. We have $\sum_{d|n} \varphi(d)  = n$ so
that
$$\sum_{n\ge   1} \frac{\varphi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}.$$
Moreover  $\sum_{n\ge 1} (-1)^n/n^s  = (2/2^s-1) \zeta(s)$ so that at
last
$$L(s) = \sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} \varphi(n/d)  (-1)^d
= (2/2^s-1) \zeta(s-1).$$ 
It follows that for  $n$ odd we have $[n^{-s}] L(s) =  -n$ and for $n$
even $[n^{-s}] L(s) = 0.$ This gives the closed forms for $n$ odd
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2n} \sum_{d|n} \varphi(n/d) (k-1)^d
- \frac{1}{2} (k-1)}$$
and for $n$ even
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2n} \sum_{d|n} \varphi(n/d) (k-1)^d
+ \frac{1}{4} k (k-1)^{n/2}.}$$
