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

1,1

Links

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

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

Example

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

Crossrefs

Row 2 of A248433.

Sequence in context: A090398 A091941 A294540 * A272211 A259035 A093835

Adjacent sequences: A248432 A248433 A248434 * A248436 A248437 A248438

Keyword

nonn

Author

R. H. Hardin, Oct 06 2014

Status

approved