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%I #19 Apr 28 2023 16:23:19
%S 0,1,3,4,7,16,17,28,70,85,125,392,379,704,3359,2248,4111,18510,14309,
%T 30820
%N Number of strictly-convex unit-sided polygons with all internal angles equal to a multiple of Pi/n, ignoring rotational and reflectional copies.
%F a(p) = (2^(p-1)-1)/p + 2^((p-1)/2) for odd prime p. - _Andrew Howroyd_, Mar 22 2023
%e For n=3, a(3) is computed as follows: The base angle is Pi/3 (60 degrees). Thus any internal angle can only be either Pi/3 or 2*Pi/3. Call an interior angle with Pi/3 a "1" and with 2*Pi/3 a "2". Since all external angles will add to 2*Pi, we know that the only possible sequences (ignoring rotation and reflection) are {{1, 1, 1}, {1, 1, 2, 2}, {1, 2, 1, 2}, {1, 2, 2, 2, 2}, {2, 2, 2, 2, 2, 2}}. However, neither {1, 1, 2, 2} nor {1, 2, 2, 2, 2} forms a closed polygon. Thus the final set is {{1, 1, 1}, {1, 2, 1, 2}, {2, 2, 2, 2, 2, 2}}, which gives a(3) = 3.
%Y Cf. A164896, A361659.
%K nonn,more
%O 1,3
%A _Roman Mecholsky_, Mar 18 2023
%E a(7) and a(9) corrected and a(11)-a(20) from _Andrew Howroyd_, Mar 22 2023
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