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%I #17 Sep 30 2018 20:22:52
%S 1,31,171,631,2059,6399,19483,58807,176859,531103,1593931,4782519,
%T 14348395,43046143,129139515,387419767,1162260667,3486783519,
%U 10460352235,31381058551,94143177675,282429535231,847288608091,2541865826871
%N Number of arrangements of n+2 numbers in 0..2 with each number being the sum mod 3 of two others.
%H R. H. Hardin, <a href="/A183877/b183877.txt">Table of n, a(n) for n = 1..200</a>
%F Empirical (for n>=2): 3^(n+2) - 2*(n+3)^2. - _Vaclav Kotesovec_, Nov 27 2012
%F Conjectures from _Colin Barker_, Apr 05 2018: (Start)
%F G.f.: x*(1 + 25*x - 3*x^2 - 33*x^3 + 18*x^4) / ((1 - x)^3*(1 - 3*x)).
%F a(n) = 6*a(n-1) - 12*a(n-2) + 10*a(n-3) - 3*a(n-4) for n>5.
%F (End)
%F Conjecture is true. The complement consists of arrangements of the forms
%F 1*, 2*, 01*, 02*, 10*, 12*, 20*, 21*, 001*, 002* and 120*. _Robert Israel_, Sep 30 2018
%e Some solutions for n=4:
%e ..2....1....1....2....1....2....2....2....0....1....2....1....2....1....0....2
%e ..2....0....2....1....0....2....0....2....2....0....1....1....2....0....1....1
%e ..1....0....2....0....0....1....0....1....1....0....1....2....0....0....1....0
%e ..1....2....2....0....2....1....0....0....2....2....0....1....1....2....1....1
%e ..1....1....1....2....2....1....0....0....0....2....0....2....0....0....2....1
%e ..2....2....1....1....2....0....2....0....0....1....1....0....0....1....0....2
%p 1, seq(3^(n+2)-2*(n+3)^2, n=2..30); # _Robert Israel_, Sep 30 2018
%Y Column 2 of A183884.
%K nonn
%O 1,2
%A _R. H. Hardin_, Jan 07 2011
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