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

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

Offset: 3

Author

Andrew Howroyd, Sep 14 2017

Keywords

• nonn

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).

Examples

			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.
		

Links

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

Crossrefs

Cf. A262181.

Programs

PARI

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

Formula

a(n) = -(1+(-1)^n)/2 + (1/n)*Sum_{d | n} (phi(n/d)-moebius(n/d)) * binomial(3*d-1, d-1).

a(n) = A262181(n) for n prime or twice prime.

Conjecture: a(2^n) = A262181(2^n).