A190035 Number of nondecreasing arrangements of n+2 numbers in 0..3 with the last equal to 3 and each after the second equal to the sum of one or two of the preceding three.

5, 7, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 222, 226, 230, 234, 238

Offset: 1

Author

R. H. Hardin, May 04 2011

Keywords

• nonn
• changed

Comments

Column 3 of A190041.

Examples

			All solutions for n=4:
..1....1....0....3....2....0....1....0....1....1....0....1....1....1
..1....1....3....3....3....1....1....1....2....1....1....2....3....1
..2....1....3....3....3....1....2....1....3....1....1....2....3....1
..2....1....3....3....3....2....3....2....3....2....1....3....3....2
..3....2....3....3....3....2....3....3....3....3....2....3....3....2
..3....3....3....3....3....3....3....3....3....3....3....3....3....3
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A190041, A073760, A016825.

Formula

Empirical: a(n) = 4*n - 2 for n>2.

Conjectures from Colin Barker, May 03 2018: (Start)

G.f.: x*(5 - 3*x + x^2 + x^3) / (1 - x)^2.

a(n) = 2*a(n-1) - a(n-2) for n>4.

(End)