A249234 Number of length 1+5 0..n arrays with no six consecutive terms having two times the sum of any two elements equal to the sum of the remaining four.

42, 486, 2772, 10620, 32070, 81402, 183696, 376752, 718530, 1289430, 2201892, 3603396, 5688582, 8702250, 12954360, 18823392, 26773506, 37358622, 51242100, 69200820, 92147742, 121136346, 157385352, 202283880, 257418690, 324579342

Offset

1,1

Links

R. H. Hardin, Table of n, a(n) for n = 1..92

Formula

Empirical: a(n) = 4*a(n-1) - 4*a(n-2) - 3*a(n-3) + 6*a(n-4) - 6*a(n-7) + 3*a(n-8) + 4*a(n-9) - 4*a(n-10) + a(n-11).

Empirical for n mod 6 = 0: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n

Empirical for n mod 6 = 1: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (445/12)

Empirical for n mod 6 = 2: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (40/3)

Empirical for n mod 6 = 3: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (255/4)

Empirical for n mod 6 = 4: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (80/3)

Empirical for n mod 6 = 5: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (605/12).

Empirical g.f.: 6*x*(7 + 53*x + 166*x^2 + 267*x^3 + 314*x^4 + 167*x^5 + 266*x^6 + 53*x^7 + 147*x^8) / ((1 - x)^7*(1 + x)^2*(1 + x + x^2)). - Colin Barker, Nov 09 2018

Example

Some solutions for n=7:
  6  4  6  2  2  0  6  2  4  0  6  4  6  2  1  4
  0  2  2  4  5  6  2  5  7  2  6  4  6  1  4  0
  1  6  6  2  4  7  4  0  1  2  6  7  1  5  2  0
  4  5  5  7  1  0  2  0  2  6  6  0  4  1  0  5
  1  7  2  2  0  4  6  6  3  6  4  6  2  6  7  3
  5  2  5  6  2  7  2  3  3  3  6  7  7  2  0  2

Crossrefs

Row 1 of A249233.

Sequence in context: A204566 A348833 A249233 * A250324 A086944 A263289

Adjacent sequences: A249231 A249232 A249233 * A249235 A249236 A249237

Keyword

nonn

Author

R. H. Hardin, Oct 23 2014

Status

approved