A207257 Number of 6 X n 0..1 arrays avoiding 0 0 1 and 1 0 1 horizontally and 0 1 0 and 1 0 1 vertically.

26, 676, 4238, 15652, 43498, 101036, 207298, 388232, 677898, 1119716, 1767766, 2688140, 3960346, 5678764, 7954154, 10915216, 14710202, 19508580, 25502750, 32909812, 41973386, 52965484, 66188434, 81976856, 100699690, 122762276

Offset: 1

Author

R. H. Hardin, Feb 16 2012

Keywords

• nonn

Comments

Row 6 of A207254.

Examples

			Some solutions for n=4:
..1..1..1..1....0..1..1..1....1..0..0..0....0..1..0..0....1..1..0..0
..1..1..1..1....1..1..0..0....1..1..0..0....1..0..0..0....1..0..0..0
..1..1..1..1....1..1..0..0....0..1..0..0....1..0..0..0....1..0..0..0
..1..1..1..0....1..0..0..0....0..0..0..0....1..0..0..0....1..1..1..1
..1..0..0..0....0..0..0..0....0..0..0..0....0..1..1..0....0..1..1..1
..1..0..0..0....0..1..1..0....0..0..0..0....0..1..1..0....0..0..0..0
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210

Crossrefs

Cf. A207254.

Formula

Empirical: a(n) = (13/180)*n^6 + (143/20)*n^5 + (1235/36)*n^4 - (39/4)*n^3 - (377/45)*n^2 + (13/5)*n.

Conjectures from Colin Barker, Jun 21 2018: (Start)

G.f.: 26*x*(1 + 19*x + 2*x^2 - 28*x^3 + 7*x^4 + x^5) / (1 - x)^7.

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>7.

(End)