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Mar 7, 2022 at 13:40 vote accept Damn it My Foot
Jul 10, 2020 at 16:07 history closed Alex M.
LSpice
YCor
Adam P. Goucher
Mark Wildon
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Jun 10, 2020 at 12:32 review Close votes
Jul 10, 2020 at 16:07
Jun 10, 2020 at 7:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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May 17, 2019 at 3:48 answer added Damn it My Foot timeline score: 0
May 17, 2019 at 3:29 comment added Damn it My Foot Yes I understood. I mistook the cyclic group for dihedral group. I was kind of confused. Thanks for the comment. I now understand how to apply the formula.
May 17, 2019 at 2:20 comment added Max Alekseyev 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, 2019 at 2:04 comment added Damn it My Foot Then isn't it the same thing as above in the question?
May 17, 2019 at 2:02 comment added Max Alekseyev In your case $a_d=p_d$.
May 17, 2019 at 1:53 comment added Damn it My Foot What is $a_d$ in that? Here $p_d$ is $\Sigma x_n^d$
May 16, 2019 at 17:46 comment added Max Alekseyev Just use the cycle index of the dihedral group rather than that of the cyclic group.
May 16, 2019 at 17:01 review First posts
May 16, 2019 at 17:19
May 16, 2019 at 17:01 history asked Damn it My Foot CC BY-SA 4.0