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

Author

R. H. Hardin, Oct 18 2014

Keywords

• nonn

Examples

			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
		

Links

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

Crossrefs

Row 1 of A249001.

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