I have seen people use the polya enumeration theorem/burnsides lemma to derive the formula $$\frac{1}{n}\sum_{d|n} \phi(d)2^{n/d}$$ that counts the number of length $n$ necklaces with two beads (binary necklaces). Now I came across another formula in a book I was reading which talked about these necklace like structures of length $2n$ that were of the form $\alpha \overline{\alpha}$ where $\alpha$ is some binary string and the overline denotes bitwise complement. These are basically length $2n$ necklaces where the first half is the bitwise complement of the second half. Now he gave a formula to count them, and this was it $$\frac{1}{2n}\sum_{\text{odd }d|n}\phi(d) 2^{n/d}$$
The way he derived it was basically showing there was some sort of correspondence between length $n+1$ necklaces with an odd number of ones and these necklace like objects. I have been trying to use the traditional method of using burnsides lemma to enumerate them but I have no idea what to do, the group action is obviously similar to that of necklaces but how do you account for the first half being the complement of the second half.
The Burnside argument is pretty much exactly the same; one has a cyclic group of order $2n$ acting on the necklaces, of which there are $2^n$ overall. A necklace is admissible if "antipodal" beads have opposite colours. The only wrinkle is counting the necklaces fixed by each element of the group. Let $g$ be the natural generator and consider the necklaces fixed by $g^r$. Write $k=\gcd(2n,r)$. Then $g^r$ has $k$ orbits of beads, each of length $2n/k$. If $2n/k$ is even, then each orbit contains a bead and its antipodal bead, which must have opposite colours so $g^r$ fixes no necklaces. But if $2n/k$ is odd we get $k/2$ pairs of "antipodal orbits" and so $2^{k/2}$ admissible colourings.
To get your formula you set $d=2n/k$ etc.