OFFSET
0,2
COMMENTS
Similar to constructions for A002717 (dividing a triangle), A000330 (dividing a square) and sequences pending for dividing other polygons.
Use 1 midpoint (resp. 2 points) on each side placed to divide each side into 2 (resp. 3) equally-sized segments or so on, do the same construction for every side of the pentagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least 1 side of the polygon. Also connect all vertices of the pentagon with lines that are parallel to at least 1 side of the pentagon.
LINKS
Michel Marcus, Figure with 1 midpoint on each side
Noah Priluck, On Counting Regular Polygons Formed by Special Families of Parallel Lines, Geombinatorics Quarterly, Vol XVII (4), 2008, pp. 166-171. (Note that there is no document to download; see A128127 for PDF file.)
FORMULA
Conjecture: a(n) = (10*n^2 + 16*n + 9 -(-1)^n)/4 for n > 0.
From Chai Wah Wu, Oct 21 2017: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4 (conjectured).
G.f.: (-x^4 + x^3 - 2*x^2 - 7*x - 1)/((x - 1)^3*(x + 1)) (conjectured). (End)
EXAMPLE
With 0 points, there is only 1 pentagon. With 1 point (a midpoint on each side), 9 regular pentagons are found. With 2 points, 20 regular pentagons are found in total.
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Noah Priluck (npriluck(AT)gmail.com), May 02 2007
EXTENSIONS
Edited by Michel Marcus, Jul 10 2013
a(4)-a(12) from Giovanni Resta, Aug 20 2017
STATUS
approved