A183898 Number of nondecreasing arrangements of n+3 numbers in 0..3 with each number being the sum mod 4 of three others.

5, 14, 38, 70, 111, 162, 224, 298, 385, 486, 602, 734, 883, 1050, 1236, 1442, 1669, 1918, 2190, 2486, 2807, 3154, 3528, 3930, 4361, 4822, 5314, 5838, 6395, 6986, 7612, 8274, 8973, 9710, 10486, 11302, 12159, 13058, 14000, 14986, 16017, 17094, 18218, 19390

Offset: 1

Author

R. H. Hardin, Jan 07 2011

Keywords

• nonn

Comments

Column 3 of A183904.

Examples

			All solutions for n=2:
..0....2....0....1....0....0....1....1....0....0....0....1....0....0
..2....2....0....3....1....0....1....1....0....0....1....1....1....0
..2....2....2....3....1....0....3....1....2....0....2....1....1....1
..3....2....3....3....2....0....3....3....2....2....3....1....2....1
..3....2....3....3....3....0....3....3....2....2....3....3....2....2
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..62

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/6)*n^3 + (5/2)*n^2 + (25/3)*n - 14 for n>1.

Conjectures from Colin Barker, Apr 05 2018: (Start)

G.f.: x*(5 - 6*x + 12*x^2 - 18*x^3 + 8*x^4) / (1 - x)^4.

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

(End)