A249712 Number of length 6+3 0..n arrays with every four consecutive terms having the maximum of some two terms equal to the minimum of the remaining two terms.

52, 659, 3440, 11925, 32500, 75495, 156416, 297321, 528340, 889339, 1431728, 2220413, 3335892, 4876495, 6960768, 9730001, 13350900, 18018403, 23958640, 31432037, 40736564, 52211127, 66239104, 83252025, 103733396, 128222667

Offset: 1

Author

R. H. Hardin, Nov 04 2014

Keywords

• nonn

Examples

			Some solutions for n=6:
..4....2....3....2....4....2....4....5....3....0....5....0....3....5....6....6
..3....6....3....0....5....5....2....4....0....3....6....5....3....4....2....5
..6....4....1....2....3....2....2....4....2....3....5....5....3....6....3....2
..4....4....5....6....4....1....0....1....2....6....5....5....3....5....3....5
..4....0....3....2....4....2....2....4....2....3....4....5....2....5....3....5
..1....4....3....2....4....6....2....4....5....3....5....5....3....5....3....5
..4....4....3....0....5....2....2....4....1....2....5....5....3....5....5....6
..4....4....1....5....4....1....2....5....2....4....6....6....4....2....3....5
..6....4....5....2....2....2....2....3....2....3....2....5....3....5....3....1
		

Links

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

Crossrefs

Row 6 of A249707.

Formula

Empirical: a(n) = (1/5)*n^6 + (73/15)*n^5 + 18*n^4 + 22*n^3 + (34/5)*n^2 - (13/15)*n + 1.

Conjectures from Colin Barker, Nov 10 2018: (Start)

G.f.: x*(52 + 295*x - 81*x^2 - 136*x^3 + 20*x^4 - 7*x^5 + x^6) / (1 - x)^7.

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>7.

(End)