A248435 Number of length 2+2 0..n arrays with every three consecutive terms having the sum of some two elements equal to twice the third.

2, 9, 20, 45, 70, 105, 140, 189, 242, 301, 360, 437, 514, 597, 684, 785, 886, 997, 1108, 1233, 1362, 1497, 1632, 1785, 1938, 2097, 2260, 2437, 2614, 2801, 2988, 3189, 3394, 3605, 3816, 4045, 4274, 4509, 4748, 5001, 5254, 5517, 5780, 6057, 6338, 6625, 6912

Offset: 1

Author

R. H. Hardin, Oct 06 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 2 of A248433.

Formula

Empirical: a(n) = a(n-1) + a(n-3) - a(n-5) - a(n-7) + a(n-8).

Empirical for n mod 12 = 0: a(n) = (19/6)*n^2 - (5/3)*n + 1

Empirical for n mod 12 = 1: a(n) = (19/6)*n^2 - (5/3)*n + (1/2)

Empirical for n mod 12 = 2: a(n) = (19/6)*n^2 - (5/3)*n - (1/3)

Empirical for n mod 12 = 3: a(n) = (19/6)*n^2 - (5/3)*n - (7/2)

Empirical for n mod 12 = 4: a(n) = (19/6)*n^2 - (5/3)*n + 1

Empirical for n mod 12 = 5: a(n) = (19/6)*n^2 - (5/3)*n - (5/6)

Empirical for n mod 12 = 6: a(n) = (19/6)*n^2 - (5/3)*n + 1

Empirical for n mod 12 = 7: a(n) = (19/6)*n^2 - (5/3)*n - (7/2)

Empirical for n mod 12 = 8: a(n) = (19/6)*n^2 - (5/3)*n - (1/3)

Empirical for n mod 12 = 9: a(n) = (19/6)*n^2 - (5/3)*n + (1/2)

Empirical for n mod 12 = 10: a(n) = (19/6)*n^2 - (5/3)*n + 1

Empirical for n mod 12 = 11: a(n) = (19/6)*n^2 - (5/3)*n - (29/6).

Empirical g.f.: x*(2 + 7*x + 11*x^2 + 23*x^3 + 16*x^4 + 17*x^5 - x^6 + x^7) / ((1 - x)^3*(1 + x)*(1 + x^2)*(1 + x + x^2)). - Colin Barker, Nov 08 2018