

A292355


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


4



1, 2, 1, 11, 1, 42, 10, 202, 1, 1077, 1, 5539, 210, 30666, 1, 174620, 1, 1001642, 5547, 5864751, 1, 34799997, 201, 208267321, 173593, 1258579693, 1, 7664723137, 1, 46976034378, 5864759, 289628805624, 5738, 1794967236906, 1, 11175157356523, 208267329
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OFFSET

3,2


COMMENTS

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).


LINKS

Andrew Howroyd, Table of n, a(n) for n = 3..1000


FORMULA

a(n) = (1+(1)^n)/2 + (1/n)*Sum_{d  n} (phi(n/d)moebius(n/d)) * binomial(3*d1, d1).
a(n) = A262181(n) for n prime or twice prime.
Conjecture: a(2^n) = A262181(2^n).


EXAMPLE

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:
d = 1: {2}
d = 2: {04, 13}
d = 3: {015, 024, 033, 042, 051, 114, 123, 132}
In total there are 11 possibilities, so a(6) = 11.
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.
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.


PROG

(PARI) a(n) = (1+(1)^n)/2 + (1/n)*sumdiv(n, d, (eulerphi(n/d)moebius(n/d)) * binomial(3*d1, d1));


CROSSREFS

Cf. A262181.
Sequence in context: A139393 A037916 A320390 * A262181 A309497 A281350
Adjacent sequences: A292352 A292353 A292354 * A292356 A292357 A292358


KEYWORD

nonn


AUTHOR

Andrew Howroyd, Sep 14 2017


STATUS

approved



