A128153 The number of regular pentagons found by constructing n equally-spaced points on each side of the pentagon and drawing lines parallel to the pentagon sides, as well as lines connecting vertices.

1, 9, 20, 37, 58, 85, 116, 153, 194, 241, 292, 349, 410

Offset: 0

Author

Noah Priluck (npriluck(AT)gmail.com), May 02 2007

Keywords

• more
• nonn

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.

Examples

			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.
		

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)

Extensions

Edited by Michel Marcus, Jul 10 2013

a(4)-a(12) from Giovanni Resta, Aug 20 2017