3 Coloring a Baton of n Bands Using, Burnside's Lemma, how many ways are there to $3$-color the $n$ bands of a baton if adjacent bands must be a different color?
Workings:
The first band has $3$ choices for color, the next bands has only $2$ choices since it can not be the same color as adjacent bands.
So $3 \times 2^{n-1}$ is the number of ways to chose these rings of a fixed baton.
If the baton revolves $180^o$ then
$3 \times 2^{\frac{n-1}{2}}$
So $N = \frac{1}{2} (3 \times 2^{\frac{n-1}{2}})$
I'm not sure if what I said is right. Any help will be appreciated
 A: These  batons  represent  the  flip  permutation group  $F_n$  on  $n$
elements that contain two permutations. The cycle indices are
$$Z(F_n) = \frac{1}{2}(a_1^n + a_2^{n/2})$$
for $n$ even and
$$Z(F_n) = \frac{1}{2}(a_1^n + a_1 a_2^{(n-1)/2})$$
for $n$ odd.
In order  to apply  Burnside we  ask how many  colorings are  fixed by
these two permutations. This gives for the first case
$$\frac{1}{2}(3\times 2^{n-1} + 0)
= 3\times 2^{n-2}$$
when  $n$  even  and  $n\ge  2$,  because in  order  to  be  fixed  by
$a_2^{n/2}$ the  two middle  slots would have  to have the  same color
which is not admissible, and for the second case
$$\frac{1}{2}(3\times 2^{n-1} + 2\times 3\times 2^{(n-1)/2-1})
= 3\times 2^{n-2} + 3\times 2^{(n-1)/2-1}.$$
where $n\gt 1$.  There are three colorings when  $n=1$ and the formula
also applies here.

This gives the following sequence $\{A_n\}$:
$$3, 3, 9, 12, 30, 48, 108, 192, 408, 768, 1584, 3072,\ldots$$

Admittedly this problem  is extremely simple but since  there was no
OEIS entry for the sequence I  wrote a Maple program to compute it for
small $n$ by  total enumeration.  The program confirmed  the values of
the sequence. This is the Maple code.

A :=
proc(n)
    if type(n, even) then return 3*2^(n-2) fi;

    3*2^(n-2)+3*2^((n-1)/2-1);
end;

verif :=
proc(n)
    option remember;
    local ind, res, ent, digits, str1, str2, pos;

    res := {};

    for ind from 3^n to 2*3^n-1 do
        digits := convert(ind, base, 3);

        for pos to n-1 do
            if digits[pos] = digits[pos+1] then
                break;
            fi;
        od;

        if pos = n then
            str1 := [seq(digits[pos], pos=1..n)];
            str2 := [seq(digits[n+1-pos], pos=1..n)];

            res := {op(res), {str1, str2}};
        fi;
    od;

    nops(res);
end;

A: The problem is that some colourings are invariant under reversal, so dividing by $2$ will under-count them.  This is where Burnside comes in.
