A250421 Number of length 4+1 0..n arrays with the sum of the minimum of each adjacent pair multiplied by some arrangement of +-1 equal to zero.

20, 125, 476, 1351, 3154, 6433, 11906, 20461, 33178, 51359, 76520, 110417, 155080, 212797, 286144, 378023, 491638, 630529, 798614, 1000157, 1239806, 1522639, 1854124, 2240161, 2687132, 3201853, 3791620, 4464263, 5228090, 6091937, 7065226

Offset: 1

Author

R. H. Hardin, Nov 22 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 4 of A250419.

Formula

Empirical: a(n) = 4*a(n-1) - 6*a(n-2) + 6*a(n-3) - 9*a(n-4) + 12*a(n-5) - 9*a(n-6) + 6*a(n-7) - 6*a(n-8) + 4*a(n-9) - a(n-10).

Empirical for n mod 3 = 0: a(n) = (2/15)*n^5 + (92/27)*n^4 + (85/27)*n^3 + (68/9)*n^2 + (68/15)*n + 1.

Empirical for n mod 3 = 1: a(n) = (2/15)*n^5 + (92/27)*n^4 + (85/27)*n^3 + (68/9)*n^2 + (632/135)*n + (29/27).

Empirical for n mod 3 = 2: a(n) = (2/15)*n^5 + (92/27)*n^4 + (85/27)*n^3 + (68/9)*n^2 + (652/135)*n + (31/27).

Empirical g.f.: x*(20 + 45*x + 96*x^2 + 77*x^3 + 36*x^4 - 48*x^5 - 44*x^6 - 37*x^7 - x^9) / ((1 - x)^6*(1 + x + x^2)^2). - Colin Barker, Nov 14 2018