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

1,1

Links

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

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

Example

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

Crossrefs

Row 4 of A250419.

Sequence in context: A263543 A249709 A250648 * A073968 A219710 A263229

Adjacent sequences: A250418 A250419 A250420 * A250422 A250423 A250424

Keyword

nonn

Author

R. H. Hardin, Nov 22 2014

Status

approved