A253391 Number of (n+1) X (2+1) 0..1 arrays with every 2 X 2 subblock antidiagonal maximum minus diagonal minimum nondecreasing horizontally and diagonal maximum minus antidiagonal minimum nondecreasing vertically.

44, 102, 143, 197, 250, 320, 391, 477, 564, 666, 769, 887, 1006, 1140, 1275, 1425, 1576, 1742, 1909, 2091, 2274, 2472, 2671, 2885, 3100, 3330, 3561, 3807, 4054, 4316, 4579, 4857, 5136, 5430, 5725, 6035, 6346, 6672, 6999, 7341, 7684, 8042, 8401, 8775, 9150

Offset: 1

Author

R. H. Hardin, Dec 31 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Column 2 of A253397.

Formula

Empirical: a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>8.

Empirical for n mod 2 = 0: a(n) = 4*n^2 + (45/2)*n + 41 for n>4.

Empirical for n mod 2 = 1: a(n) = 4*n^2 + (45/2)*n + (75/2) for n>4.

Empirical g.f.: x*(44 + 14*x - 61*x^2 - x^3 + 16*x^4 + 4*x^5 + 2*x^6 - 2*x^7) / ((1 - x)^3*(1 + x)). - Colin Barker, Dec 11 2018