

A141135


Minimal number of unit edges required to construct n regular pentagons when allowing edgesharing.


0



5, 9, 13, 17, 21, 24, 28, 32, 36, 39, 43, 47, 50, 54, 58, 61, 65, 69, 72, 76, 80, 83, 87, 90, 94, 98, 101, 105, 109, 112
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OFFSET

1,1


LINKS

Ralph H. Buchholz and Warwick de Launey, Edge minimization, June 1996, (revised June 2008).


FORMULA

G.f.: x*(5 + 4*x + 4*x^2  x^3  x^5 + x^8  x^9) / ((1  x)^2*(1 + x + x^2)).
a(n) = a(n1) + a(n3)  a(n4) for n>10.
(End)
Conjecture: if n is a term in A121149, a(n) = a(n1) + 3, otherwise a(n) = a(n1) + 4.  Jinyuan Wang, Apr 05 2019


EXAMPLE

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 6cycle). Thus 24 = 5 + 4 + 4 + 4 + 4 + 3.


CROSSREFS



KEYWORD

nonn,more


AUTHOR

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


EXTENSIONS



STATUS

approved



