Circular permutations of N objects of n1 are identical of one type, n2 are identical of another type and so on, such that n1+n2+n3+..... = N? A similar question exists but it doesn't address the case where reflections are under the same equivalent class.$$\frac{1}{N}\sum_{d | N} \phi(d) p_d^{N/d}$$ This is when reflections are not the same. How does the equation change under this new restriction.

Note: I couldn't comment on that question due to my low reputation, so I made this question.

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    $\begingroup$ Just use the cycle index of the dihedral group rather than that of the cyclic group. $\endgroup$ May 16 '19 at 17:46
  • $\begingroup$ What is $a_d$ in that? Here $p_d$ is $\Sigma x_n^d$ $\endgroup$
    – Karthik
    May 17 '19 at 1:53
  • $\begingroup$ In your case $a_d=p_d$. $\endgroup$ May 17 '19 at 2:02
  • $\begingroup$ Then isn't it the same thing as above in the question? $\endgroup$
    – Karthik
    May 17 '19 at 2:04
  • $\begingroup$ What thing? Cyclic indices are different for the cyclic and dihedral groups, but their arguments (ie. $a_d=p_d$ here) are the same. $\endgroup$ May 17 '19 at 2:20

Answer based on Max Alekseyev's helpful comments. Unlike the cyclic group, using which all rotations are equivalent, under the dihedral group all rotations as well as reflections are equivalent.
Intuitively it seems right is use dihedral group because dihedral group is group of symmetic transformations of regular polygons,where as cyclic groups consider only rotational symmetries.
Thus we must use the cyclic index of the dihedral group. That is $$\frac{1}{2N}(\sum_{d|N}\phi(d)p_d^\frac{N}{d})+\frac{1}{2}p_1p_2^{(N-1)/2} $$ If n is odd.

$$\frac{1}{2N}(\sum_{d|N}\phi(d)p_d^\frac{N}{d})+\frac{1}{4}(p_1^2p_2^{(N-2)/2}+p_2^{N/2} ) $$ If n is even.


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