%I #20 Oct 13 2021 10:50:01
%S 1,2,2,12,30,30,210,560,560,4200,11550,11550,90090,252252,252252,
%T 2018016,5717712,5717712,46558512,133024320,133024320,1097450640,
%U 3155170590,3155170590,26293088250,75957810500,75957810500,638045608200
%N Number of n X 2 0..2 arrays with values 0..2 introduced in row major order, the number of instances of each value within one of each other, and no element equal to any horizontal or vertical neighbor.
%C Column 2 of A199133.
%C a(n) is the last term in row n of triangle in A286030 (see also formulas below). _Bob Selcoe_, Sep 26 2021
%H R. H. Hardin, <a href="/A199127/b199127.txt">Table of n, a(n) for n = 1..198</a>
%F Conjecture: a(3n+2) = a(3n+3) = A208881(n+1). - _R. J. Mathar_, Nov 01 2015
%F Conjecture: -(458*n-1205) *(n+2) *(n+1)*a(n) +(-208*n^3+2578*n^2-4613*n-2410) *a(n-1) +9*(-339*n-638) *a(n-2) +27*(n-2) *(458*n^2-289*n-1146) *a(n-3) +54*(n-2) *(n-3) *(104*n-1081) *a(n-4)=0. - _R. J. Mathar_, Nov 01 2015
%F Conjecture: (n+2)*(n+1)*a(n) +(5*n^2-2)*a(n-1) +3*(5*n^2-15*n+3) *a(n-2) +3*(n^2 -60*n +81)*a(n-3) +135*(-n^2+3*n-1)*a(n-4) -405*(n-2)*(n-4) *a(n-5) -810*(n-4) *(n-5) *a(n-6)=0. - _R. J. Mathar_, Nov 01 2015
%F From _Bob Selcoe_, Sep 26 2021: (Start)
%F When n == 0 (mod 3), a(n) = n!/(3*(n/3)!^3);
%F when n == 1 (mod 3), a(n) = n!/(((n+2)/3)!*((n-1)/3)!^2);
%F when n == 2 (mod 3), a(n) = n!/(((n-2)/3)!*((n+1)/3)!^2).
%F (End)
%e Some solutions for n=5:
%e 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
%e 1 0 1 2 1 2 1 2 1 0 1 2 1 0 1 2 1 2 1 0
%e 0 2 2 0 0 1 2 0 0 2 2 1 0 2 0 1 2 0 2 1
%e 2 1 0 2 2 0 0 1 2 1 1 0 2 0 1 2 0 1 1 2
%e 0 2 2 1 0 2 2 0 1 2 0 2 1 2 2 0 1 2 2 0
%Y Cf. A286030.
%K nonn
%O 1,2
%A _R. H. Hardin_, Nov 03 2011