A249002 Number of length 1+4 0..n arrays with no five consecutive terms having two times the sum of any three elements equal to three times the sum of the remaining two.

30, 190, 820, 2540, 6450, 13990, 27740, 50260, 86030, 139450, 217320, 325940, 475630, 674650, 937020, 1274160, 1703970, 2240850, 2908260, 3723400, 4715230, 5905430, 7328400, 9009880, 10991870, 13303750, 15994820, 19100260, 22676370

Offset

1,1

Links

R. H. Hardin, Table of n, a(n) for n = 1..209

Formula

Empirical: a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 5*a(n-4) - 3*a(n-5) + 6*a(n-6) + 6*a(n-7) - 3*a(n-8) - 5*a(n-9) - a(n-10) + 3*a(n-11) + a(n-12) - a(n-13).

Empirical for n mod 6 = 0: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (55/6)*n^2 + (47/3)*n

Empirical for n mod 6 = 1: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (65/12)*n^2 + (86/9)*n + (565/72)

Empirical for n mod 6 = 2: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (55/6)*n^2 + (181/9)*n - (20/9)

Empirical for n mod 6 = 3: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (65/12)*n^2 + (2/3)*n + (205/8)

Empirical for n mod 6 = 4: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (55/6)*n^2 + (221/9)*n - (160/9)

Empirical for n mod 6 = 5: a(n) = n^5 + (185/72)*n^4 + (130/9)*n^3 - (65/12)*n^2 + (46/9)*n + (1685/72).

Empirical g.f.: 10*x*(3 + 16*x + 54*x^2 + 118*x^3 + 179*x^4 + 178*x^5 + 143*x^6 + 84*x^7 + 44*x^8 + 24*x^9 + 21*x^10) / ((1 - x)^6*(1 + x)^3*(1 + x + x^2)^2). - Colin Barker, Nov 09 2018

Example

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

Crossrefs

Row 1 of A249001.

Sequence in context: A265037 A249001 A249466 * A249467 A309924 A120339

Adjacent sequences: A248999 A249000 A249001 * A249003 A249004 A249005

Keyword

nonn

Author

R. H. Hardin, Oct 18 2014

Status

approved