A273308 Maximum population of a 2 X n still life in Conway's Game of Life.
0, 4, 4, 6, 8, 8, 10, 12, 12, 14, 16, 16, 18, 20, 20, 22, 24, 24, 26, 28, 28, 30, 32, 32, 34, 36, 36, 38, 40, 40, 42, 44, 44, 46, 48, 48, 50, 52, 52, 54, 56, 56, 58, 60, 60, 62, 64, 64, 66, 68, 68, 70, 72, 72, 74, 76, 76, 78, 80, 80, 82, 84, 84, 86, 88, 88
Offset: 1
Examples
a(2) = 4 because the largest number of alive cells in a 2 X 2 still life is 4, which is attained by the block. a(4) = 6 because the largest number of alive cells in a 2 X 4 still life is 6, which is attained by the snake.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- G. Chu, P. Stuckey, and M.G. de la Banda, Using relaxations in Maximum Density Still Life, In Proc. of Fifteenth Intl. Conf. on Principles and Practice of Constraint Programming, 258-273 (2009).
- LifeWiki, Still life
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Programs
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Maple
seq(4*floor((n+1)*(1/3))+2*floor((1/2)*(`mod`(n+1, 3))), n = 2 .. 110);
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Mathematica
LinearRecurrence[{1,0,1,-1},{0,4,4,6,8},70] (* Harvey P. Dale, Apr 19 2023 *) -
PARI
concat(0, Vec(2*x^2*(2+x^2-x^3)/((1-x)^2*(1+x+x^2)) + O(x^50))) \\ Colin Barker, May 24 2016
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Python
def A273308(n): return n+sum(divmod(n,3)) if n > 1 else 0 # Chai Wah Wu, Jan 29 2023
Formula
For n >= 1, a(3*n) = a(3*n-1) = 4*n and a(3*n+1) = 4*n+2.
From Colin Barker, May 24 2016: (Start)
a(n) = a(n-1)+a(n-3)-a(n-4) for n>5.
G.f.: 2*x^2*(2+x^2-x^3) / ((1-x)^2*(1+x+x^2)). (End)
Comments