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A069813
Maximum number of triangles in polyiamond with perimeter n.
6
1, 2, 3, 6, 7, 10, 13, 16, 19, 24, 27, 32, 37, 42, 47, 54, 59, 66, 73, 80, 87, 96, 103, 112, 121, 130, 139, 150, 159, 170, 181, 192, 203, 216, 227, 240, 253, 266, 279, 294, 307, 322, 337, 352, 367, 384, 399, 416, 433, 450, 467, 486, 503, 522, 541, 560, 579
OFFSET
3,2
LINKS
FORMULA
a(n) = round(n^2/6) - (0 if n = 0 mod 6, 1 else) = A056829(n)-A097325(n).
From Colin Barker, Jan 18 2015: (Start)
a(n) = round((-25 + 9*(-1)^n + 8*exp(-2/3*i*n*Pi) + 8*exp((2*i*n*Pi)/3) + 6*n^2)/36), where i=sqrt(-1).
G.f.: x^3*(1+x-x^2)*(1+x^2) / ((1-x)^3*(1+x)*(1+x+x^2)). (End)
a(n) = A001399(n-3) + A001399(n-4) + A001399(n-6) - A001399(n-7). - R. J. Mathar, Jul 14 2015
EXAMPLE
a(10) = 16 because the maximum number of triangles in a polyiamond of perimeter 10 is 16.
MAPLE
A069813 := proc(n)
round(n^2/6) ;
if modp(n, 6) <> 0 then
%-1 ;
else
% ;
end if;
end proc: # R. J. Mathar, Jul 14 2015
MATHEMATICA
LinearRecurrence[{1, 1, 0, -1, -1, 1}, {1, 2, 3, 6, 7, 10}, 60] (* Jean-François Alcover, Jan 03 2020 *)
PROG
(PARI) a(n) = round(n^2/6) - (n % 6 != 0) \\ Michel Marcus, Jul 17 2013
(PARI) Vec(x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)) + O(x^60)) \\ Colin Barker, Jan 19 2015
(Magma) R<x>:=PowerSeriesRing(Integers(), 65); Coefficients(R!( x^3*(x^2-x-1)*(x^2+1)/((x-1)^3*(x+1)*(x^2+x+1)))); // Marius A. Burtea, Jan 03 2020
CROSSREFS
Sequence in context: A331075 A378595 A062720 * A266534 A293392 A161336
KEYWORD
nonn,easy
AUTHOR
Winston C. Yang (winston(AT)cs.wisc.edu), Apr 30 2002
STATUS
approved