A189322 Number of nondecreasing arrangements of n+2 numbers in 0..5 with the last equal to 5 and each after the second equal to the sum of one or two of the preceding four.

8, 12, 21, 33, 54, 84, 119, 157, 195, 233, 271, 309, 347, 385, 423, 461, 499, 537, 575, 613, 651, 689, 727, 765, 803, 841, 879, 917, 955, 993, 1031, 1069, 1107, 1145, 1183, 1221, 1259, 1297, 1335, 1373, 1411, 1449, 1487, 1525, 1563, 1601, 1639, 1677, 1715

Offset: 1

Author

R. H. Hardin, Apr 20 2011

Keywords

• nonn

Comments

Column 5 of A189326.

Examples

			Some solutions for n=3:
..1....1....2....5....1....1....1....1....1....4....1....2....3....2....1....0
..2....2....3....5....3....2....2....1....3....5....3....5....5....3....4....5
..2....3....3....5....3....2....3....2....4....5....4....5....5....3....4....5
..4....3....5....5....4....3....4....3....5....5....4....5....5....3....4....5
..5....5....5....5....5....5....5....5....5....5....5....5....5....5....5....5
		

Links

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

Crossrefs

Cf. A189326.

Formula

Empirical: a(n) = 38*n - 147 for n>6.

Empirical g.f.: x*(8 - 4*x + 5*x^2 + 3*x^3 + 9*x^4 + 9*x^5 + 5*x^6 + 3*x^7) / (1 - x)^2. - Colin Barker, May 02 2018