# Circular (bracelets) permutations with alike things(reflections are equivalent) using polya enumeration [closed]

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.

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

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.