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

0,2

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

Table of n, a(n) for n=0..12.

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

Sequence in context: A147479 A146680 A143704 * A249044 A109805 A345727

Adjacent sequences: A128150 A128151 A128152 * A128154 A128155 A128156

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