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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. 3

%I #24 Oct 21 2017 13:16:22

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

%N 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.

%C Similar to constructions for A002717 (dividing a triangle), A000330 (dividing a square) and sequences pending for dividing other polygons.

%C 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.

%H Michel Marcus, <a href="/A128153/a128153.png">Figure with 1 midpoint on each side</a>

%H Noah Priluck, <a href="http://www.uccs.edu/geombina/2007.html#issue4">On Counting Regular Polygons Formed by Special Families of Parallel Lines</a>, Geombinatorics Quarterly, Vol XVII (4), 2008, pp. 166-171. (Note that there is no document to download; see A128127 for PDF file.)

%F Conjecture: a(n) = (10*n^2 + 16*n + 9 -(-1)^n)/4 for n > 0.

%F From _Chai Wah Wu_, Oct 21 2017: (Start)

%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4 (conjectured).

%F G.f.: (-x^4 + x^3 - 2*x^2 - 7*x - 1)/((x - 1)^3*(x + 1)) (conjectured). (End)

%e 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.

%K more,nonn

%O 0,2

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

%E Edited by _Michel Marcus_, Jul 10 2013

%E a(4)-a(12) from _Giovanni Resta_, Aug 20 2017

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