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.

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

R. H. Hardin, Table of n, a(n) for n = 1..198

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

Adjacent sequences: A199124 A199125 A199126 * A199128 A199129 A199130

Keyword

nonn

Author

R. H. Hardin, Nov 03 2011

Status

approved