

A183877


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


1



1, 31, 171, 631, 2059, 6399, 19483, 58807, 176859, 531103, 1593931, 4782519, 14348395, 43046143, 129139515, 387419767, 1162260667, 3486783519, 10460352235, 31381058551, 94143177675, 282429535231, 847288608091, 2541865826871
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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(n1)  12*a(n2) + 10*a(n3)  3*a(n4) 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



