A252179 Number of length 3+2 0..n arrays with the sum of the maximum minus the minimum of adjacent triples multiplied by some arrangement of +-1 equal to zero.

12, 83, 264, 687, 1428, 2729, 4680, 7661, 11764, 17535, 25056, 35067, 47628, 63701, 83312, 107673, 136764, 172075, 213528, 262919, 320100, 387201, 463992, 552965, 653796, 769367, 899248, 1046739, 1211292, 1396653, 1602144, 1831985, 2085356

Offset: 1

Author

R. H. Hardin, Dec 15 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 3 of A252177.

Formula

Empirical: a(n) = 3*a(n-1) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 8*a(n-6) + 3*a(n-8) - a(n-9).

Empirical for n mod 2 = 0: a(n) = (1/60)*n^5 + (17/16)*n^4 + (14/3)*n^3 + (11/2)*n^2 + (77/30)*n + 1.

Empirical for n mod 2 = 1: a(n) = (1/60)*n^5 + (17/16)*n^4 + (14/3)*n^3 + (39/8)*n^2 + (79/60)*n + (1/16).

Empirical g.f.: x*(12 + 47*x + 15*x^2 - 9*x^3 - 41*x^4 - 13*x^5 + 3*x^6 + 3*x^7 - x^8) / ((1 - x)^6*(1 + x)^3). - Colin Barker, Dec 01 2018