A249086 Number of length 1+5 0..n arrays with every six consecutive terms having two times the sum of some two elements equal to the sum of the remaining four.

22, 243, 1324, 5005, 14586, 36247, 78448, 154689, 281470, 482131, 784092, 1223413, 1840954, 2688375, 3822856, 5314177, 7238718, 9687259, 12757900, 16565301, 21232162, 26899543, 33717624, 41856745, 51497086, 62841147, 76101988, 91516789

Offset

1,1

Links

R. H. Hardin, Table of n, a(n) for n = 1..96

Formula

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

Empirical for n mod 6 = 0: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n + 1

Empirical for n mod 6 = 1: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (457/12)

Empirical for n mod 6 = 2: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (37/3)

Empirical for n mod 6 = 3: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (259/4)

Empirical for n mod 6 = 4: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (77/3)

Empirical for n mod 6 = 5: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (617/12).

Empirical g.f.: x*(22 + 177*x + 617*x^2 + 1364*x^3 + 1823*x^4 + 2260*x^5 + 1135*x^6 + 880*x^7 + 3*x^8 - x^9) / ((1 - x)^6*(1 + x)^2*(1 + x + x^2)). - Colin Barker, Nov 09 2018

Example

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

Crossrefs

Row 1 of A249085.

Sequence in context: A126903 A035706 A249085 * A300624 A249087 A348707

Adjacent sequences: A249083 A249084 A249085 * A249087 A249088 A249089

Keyword

nonn

Author

R. H. Hardin, Oct 20 2014

Status

approved