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A199127
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.
2
1, 2, 2, 12, 30, 30, 210, 560, 560, 4200, 11550, 11550, 90090, 252252, 252252, 2018016, 5717712, 5717712, 46558512, 133024320, 133024320, 1097450640, 3155170590, 3155170590, 26293088250, 75957810500, 75957810500, 638045608200
OFFSET
1,2
COMMENTS
Column 2 of A199133.
a(n) is the last term in row n of triangle in A286030 (see also formulas below). Bob Selcoe, Sep 26 2021
LINKS
FORMULA
Conjecture: a(3n+2) = a(3n+3) = A208881(n+1). - R. J. Mathar, Nov 01 2015
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
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
From Bob Selcoe, Sep 26 2021: (Start)
When n == 0 (mod 3), a(n) = n!/(3*(n/3)!^3);
when n == 1 (mod 3), a(n) = n!/(((n+2)/3)!*((n-1)/3)!^2);
when n == 2 (mod 3), a(n) = n!/(((n-2)/3)!*((n+1)/3)!^2).
(End)
EXAMPLE
Some solutions for n=5:
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 2 1 2 1 2 1 0 1 2 1 0 1 2 1 2 1 0
0 2 2 0 0 1 2 0 0 2 2 1 0 2 0 1 2 0 2 1
2 1 0 2 2 0 0 1 2 1 1 0 2 0 1 2 0 1 1 2
0 2 2 1 0 2 2 0 1 2 0 2 1 2 2 0 1 2 2 0
CROSSREFS
Cf. A286030.
Sequence in context: A287629 A324919 A130306 * A093044 A151366 A184944
KEYWORD
nonn
AUTHOR
R. H. Hardin, Nov 03 2011
STATUS
approved