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

1,2

Links

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

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

Example

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

Maple

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

Crossrefs

Column 2 of A183884.

Sequence in context: A184497 A184489 A140540 * A107953 A085249 A272130

Adjacent sequences: A183874 A183875 A183876 * A183878 A183879 A183880

Keyword

nonn

Author

R. H. Hardin, Jan 07 2011

Status

approved