A201271 Number of n X 2 0..2 arrays with every row and column nondecreasing rightwards and downwards, and the number of instances of each value within one of each other.

1, 3, 5, 4, 12, 16, 9, 27, 33, 16, 48, 56, 25, 75, 85, 36, 108, 120, 49, 147, 161, 64, 192, 208, 81, 243, 261, 100, 300, 320, 121, 363, 385, 144, 432, 456, 169, 507, 533, 196, 588, 616, 225, 675, 705, 256, 768, 800, 289, 867, 901, 324, 972, 1008, 361, 1083, 1121, 400, 1200

Offset: 0

Author

R. H. Hardin, Nov 29 2011

Keywords

• nonn
• easy

Examples

			Some solutions for n=5:
..0..1....0..1....0..0....0..0....0..0....0..0....0..0....0..1....0..1....0..0
..0..1....0..1....0..0....0..1....0..1....0..1....0..0....0..2....0..1....0..2
..0..1....0..1....1..2....1..1....1..1....1..2....1..1....0..2....0..2....1..2
..0..2....1..2....1..2....2..2....1..2....1..2....1..2....1..2....1..2....1..2
..2..2....2..2....1..2....2..2....2..2....2..2....2..2....1..2....2..2....1..2
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (terms n = 1..210 from R. H. Hardin)

Crossrefs

Column 2 of A201277.

Formula

a(n) = 3*a(n-3) -3*a(n-6) +a(n-9).

Subsequences for n modulo 3 = 1,2,0:

p=(n+2)/3: a(n) = 3*p^2

q=(n+1)/3: a(n) = 3*q^2 + 2*q

r=(n+0)/3: a(n) = r^2 + 2*r + 1.

G.f.: 1+x*(3 + 5*x + 4*x^2 + 3*x^3 + x^4 - 3*x^5 + x^8) / ((1 - x)^3*(1 + x + x^2)^3). - Colin Barker, May 22 2018

Extensions

a(0)=1 prepended by Alois P. Heinz, Mar 18 2024