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

Author

R. H. Hardin, Nov 26 2014

Keywords

• nonn

Comments

Row 4 of A250646.

Examples

			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
		

Links

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

Crossrefs

Cf. A250646.

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).