A248988 Number of length 1+4 0..n arrays with every five consecutive terms having two times the sum of some three elements equal to three times the sum of the remaining two.

2, 53, 204, 585, 1326, 2817, 5028, 8789, 13970, 21601, 31512, 45353, 62194, 84725, 111556, 145697, 185598, 235249, 291740, 360701, 438402, 530913, 634224, 755745, 889506, 1045157, 1215548, 1410889, 1623630, 1865681, 2126932, 2422133, 2739634

Offset

1,1

Links

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

Formula

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

Empirical for n mod 6 = 0: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (32/3)*n + 1

Empirical for n mod 6 = 1: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 - (41/9)*n - (493/72)

Empirical for n mod 6 = 2: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (136/9)*n + (29/9)

Empirical for n mod 6 = 3: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 + (13/3)*n - (197/8)

Empirical for n mod 6 = 4: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (176/9)*n + (169/9)

Empirical for n mod 6 = 5: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 - (1/9)*n - (1613/72).

Empirical g.f.: x*(2 + 53*x + 198*x^2 + 422*x^3 + 614*x^4 + 825*x^5 + 810*x^6 + 653*x^7 + 416*x^8 + 206*x^9 + x^11) / ((1 - x)^5*(1 + x)^3*(1 + x + x^2)^2). - Colin Barker, Nov 09 2018

Example

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

Crossrefs

Row 1 of A248987.

Sequence in context: A249663 A130698 A364080 * A248989 A248990 A248991

Adjacent sequences: A248985 A248986 A248987 * A248989 A248990 A248991

Keyword

nonn

Author

R. H. Hardin, Oct 18 2014

Status

approved