A183885 Number of arrangements of n+3 numbers in 0..2 with each number being the sum mod 3 of three others.

1, 121, 631, 2059, 6399, 19483, 58807, 176859, 531103, 1593931, 4782519, 14348395, 43046143, 129139515, 387419767, 1162260667, 3486783519, 10460352235, 31381058551, 94143177675, 282429535231, 847288608091, 2541865826871

Offset: 1

Author

R. H. Hardin, Jan 07 2011

Keywords

• nonn

Comments

Column 2 of A183892.

Examples

			Some solutions for n=4:
..0....2....2....2....2....1....2....0....2....1....2....1....0....0....2....2
..0....1....0....2....2....2....0....1....1....2....1....2....1....2....2....1
..1....1....2....0....1....0....2....1....1....2....2....1....1....0....0....1
..2....0....1....0....1....1....2....1....2....1....0....0....2....2....2....1
..1....1....1....1....1....0....1....0....1....2....2....0....0....2....0....0
..2....1....0....1....1....1....2....2....1....1....2....1....0....2....2....1
..0....0....2....2....0....0....1....0....0....1....2....2....0....1....0....1
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..62

Crossrefs

Cf. A183892.

Formula

Empirical (for n>=3): 3^(n+3) - 2*(n+4)^2. - Vaclav Kotesovec, Nov 27 2012

Conjectures from Colin Barker, Apr 05 2018: (Start)

G.f.: x*(1 + 115*x - 83*x^2 - 285*x^3 + 410*x^4 - 150*x^5) / ((1 - x)^3*(1 - 3*x)).

a(n) = 6*a(n-1) - 12*a(n-2) + 10*a(n-3) - 3*a(n-4) for n>6.

(End)