A262181 a(n) = total number of convex equilateral n-gons with corner angles of m*Pi/n (0 < m <= n).
1, 2, 1, 11, 1, 42, 64, 202, 1, 1557, 1, 5539, 32298, 30666, 1, 405200, 1, 1035642
Offset: 3
Examples
For n = 3 there is one convex n-gon, the equilateral triangle, with m angle factors (3 3 3); so a(3) = 1. For n = 4 there are two convex n-gons, the square and a rhombus, with respective m angle factors (2 2 2 2) and (1 3 1 3); so a(4) = 2. For n = 5, there is the regular pentagon, m factors (3 3 3 3 3); so a(5) = 1. For n = 6 there are 11 convex n-gons; here are the m factors:(1 5 6 1 5 6), (1 6 5 1 6 5), (2 4 6 2 4 6), (2 5 5 2 5 5), (2 6 2 6 2 6), (2 6 4 2 6 4), (3 3 6 3 3 6), (3 4 5 3 4 5), (3 5 3 5 3 5), (3 5 4 3 5 4), (4 4 4 4 4 4); so a(6) = 11.
Links
- Stuart E Anderson, for n=3, 1 solution, the equilateral triangle
- Stuart E Anderson, for n=4, 2 solutions
- Stuart E Anderson, for n=5, 1 solution
- Stuart E Anderson, for n=6, 11 solutions
- Stuart E Anderson, for n=7, 1 solution
- Stuart E Anderson, for n=8, 42 solutions
- Stuart E Anderson, n-gons.cpp produces postscript images of convex polygons with n sides, or alternatively produces an unsorted list of interior angle multiples m for each polygon. One rotationally invariant representative polygon is produced in postscript and there is an option to exclude mirror image reflected polygons.
- Gilbert Labelle and Annie Lacasse, Closed paths whose steps are roots of unity, in FPSAC 2011, Reykjavik, Iceland DMTCS proc. AO, 2011, 599-610.
Crossrefs
Formula
a(n) = A292355(n) for n prime or twice prime. - Andrew Howroyd, Sep 14 2017
a(n) = -(1+(-1)^n)/2 + (1/(2*n))*(A321415(n) - binomial(3*n-1, n) + Sum_{d|n} phi(n/d) * binomial(3*d-1, d)). - Andrew Howroyd, Nov 09 2018
Extensions
a(10) corrected and a(12)-a(17) from Andrew Howroyd, Sep 14 2017
a(18)-a(20) from Andrew Howroyd, Nov 09 2018
Comments