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A292355 Number of distinct convex equilateral n-gons having rotational symmetry and with corner angles of m*Pi/n (0 < m <= n).

%I A292355 #13 Nov 11 2018 00:43:00
%S A292355 1,2,1,11,1,42,10,202,1,1077,1,5539,210,30666,1,174620,1,1001642,5547,
%T A292355 5864751,1,34799997,201,208267321,173593,1258579693,1,7664723137,1,
%U A292355 46976034378,5864759,289628805624,5738,1794967236906,1,11175157356523,208267329
%N A292355 Number of distinct convex equilateral n-gons having rotational symmetry and with corner angles of m*Pi/n (0 < m <= n).
%C A292355 Subset of polygons of A262181 having rotational symmetry. Polygons that differ only by rotation are not considered as distinct. See A262181 for illustrations of initial terms. The first difference between this sequence and A262181 is at a(9).
%H A292355 Andrew Howroyd, <a href="/A292355/b292355.txt">Table of n, a(n) for n = 3..1000</a>
%F A292355 a(n) = -(1+(-1)^n)/2 + (1/n)*Sum_{d | n} (phi(n/d)-moebius(n/d)) * binomial(3*d-1, d-1).
%F A292355 a(n) = A262181(n) for n prime or twice prime.
%F A292355 Conjecture: a(2^n) = A262181(2^n).
%e A292355 Case n=6: The ways to select d angles that are multiples of Pi/n and sum to 2*d which are nonequivalent up to rotation and d is a proper factor of 6 are:
%e A292355 d = 1: {2}
%e A292355 d = 2: {04, 13}
%e A292355 d = 3: {015, 024, 033, 042, 051, 114, 123, 132}
%e A292355 In total there are 11 possibilities, so a(6) = 11.
%e A292355 In the above, 22 and 222 are excluded from the possibilities for d = 2 and 3 because they correspond to the regular hexagon that is covered by d = 1.
%e A292355 Also, 006 has been excluded from d = 3 since 6 corresponds to an angle of 180 degrees which is disallowed by this sequence. This would be the flattened polygon of three sides in one direction and then three back in the opposite.
%o A292355 (PARI) a(n) = -(1+(-1)^n)/2 + (1/n)*sumdiv(n,d, (eulerphi(n/d)-moebius(n/d)) * binomial(3*d-1, d-1));
%Y A292355 Cf. A262181.
%K A292355 nonn
%O A292355 3,2
%A A292355 _Andrew Howroyd_, Sep 14 2017