A250648 Number of length 4+1 0..n arrays with the sum of the maximum of each adjacent pair multiplied by some arrangement of +-1 equal to zero.

20, 125, 476, 1293, 2954, 5901, 10766, 18305, 29478, 45361, 67364, 96961, 135976, 186445, 250688, 331213, 431054, 553277, 701474, 879553, 1091754, 1342593, 1637320, 1981153, 2380028, 2840317, 3368740, 3972397, 4659346, 5437517, 6315766

Offset

1,1

Comments

Row 4 of A250646.

Links

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

Formula

Empirical: a(n) = a(n-1) +2*a(n-2) +a(n-3) -4*a(n-4) -5*a(n-5) +3*a(n-6) +6*a(n-7) +3*a(n-8) -5*a(n-9) -4*a(n-10) +a(n-11) +2*a(n-12) +a(n-13) -a(n-14).

Empirical for n mod 6 = 0: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (17/3)*n^2 + (124/45)*n + 1.

Empirical for n mod 6 = 1: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (151/27)*n^2 + (1361/405)*n + (301/162).

Empirical for n mod 6 = 2: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (17/3)*n^2 + (1036/405)*n + (49/81).

Empirical for n mod 6 = 3: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (17/3)*n^2 + (169/45)*n + (5/2).

Empirical for n mod 6 = 4: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (151/27)*n^2 + (956/405)*n + (29/81).

Empirical for n mod 6 = 5: a(n) = (2/15)*n^5 + (403/162)*n^4 + (532/81)*n^3 + (17/3)*n^2 + (1441/405)*n + (341/162).

Example

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

Crossrefs

Cf. A250646.

Sequence in context: A227057 A263543 A249709 * A250421 A073968 A219710

Adjacent sequences: A250645 A250646 A250647 * A250649 A250650 A250651

Keyword

nonn

Author

R. H. Hardin, Nov 26 2014

Status

approved