A183877 Number of arrangements of n+2 numbers in 0..2 with each number being the sum mod 3 of two others.

1, 31, 171, 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

Examples

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

Links

• R. H. Hardin, Table of n, a(n) for n = 1..200

Crossrefs

Column 2 of A183884.

Programs

Maple

1, seq(3^(n+2)-2*(n+3)^2, n=2..30); # Robert Israel, Sep 30 2018

Formula

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

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

G.f.: x*(1 + 25*x - 3*x^2 - 33*x^3 + 18*x^4) / ((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>5.

(End)

Conjecture is true. The complement consists of arrangements of the forms

1*, 2*, 01*, 02*, 10*, 12*, 20*, 21*, 001*, 002* and 120*. Robert Israel, Sep 30 2018