A141135 - OEIS

%I #24 Jan 05 2025 04:44:04

%S 5,9,13,17,21,24,28,32,36,39,43,47,50,54,58,61,65,69,72,76,80,83,87,

%T 90,94,98,101,105,109,112

%N Minimal number of unit edges required to construct n regular pentagons when allowing edge-sharing.

%H Ralph H. Buchholz, <a href="http://sites.google.com/site/teufelpi/papers/sps.pdf">Spiral polygon series</a>, preprint 1985 SMJ 31, School Mathematics Journal, 1995.

%H Ralph H. Buchholz and Warwick de Launey, <a href="http://sites.google.com/site/teufelpi/papers/edgemin.pdf">Edge minimization</a>, June 1996, (revised June 2008).

%H Ralph H. Buchholz and Warwick de Launey, <a href="https://doi.org/10.37236/179">An edge minimization problem for regular polygons</a>, The Electronic Journal of Combinatorics, Volume 16, Issue 1 (2009), #R90.

%F Conjectures from _Colin Barker_, Apr 05 2019: (Start)

%F G.f.: x*(5 + 4*x + 4*x^2 - x^3 - x^5 + x^8 - x^9) / ((1 - x)^2*(1 + x + x^2)).

%F a(n) = a(n-1) + a(n-3) - a(n-4) for n>10.

%F (End)

%F Conjecture: if n is a term in A121149, a(n) = a(n-1) + 3, otherwise a(n) = a(n-1) + 4. - _Jinyuan Wang_, Apr 05 2019

%e a(6) = 24 since the first pentagon requires 5 edges, the 2nd, 3rd, 4th and 5th pentagons require an additional 4 edges each and the 6th pentagon requires 3 edges since it can share 2 edges (if one tiles via a 6-cycle). Thus 24 = 5 + 4 + 4 + 4 + 4 + 3.

%Y Cf. equilateral triangles A137228, squares A078633, regular hexagons A135708.

%Y Cf. A121149.

%K nonn,more

%O 1,1

%A Ralph H. Buchholz (teufel_pi(AT)yahoo.com), Jun 08 2008

%E a(21)-a(30) from _Jinyuan Wang_, Apr 05 2019