A249291 Number of length 1+3 0..n arrays with no four consecutive terms having the sum of any three elements equal to three times the fourth.

14, 66, 204, 524, 1098, 2070, 3584, 5808, 8934, 13202, 18828, 26100, 35306, 46758, 60792, 77792, 98118, 122202, 150476, 183396, 221442, 265142, 315000, 371592, 435494, 507306, 587652, 677204, 776610, 886590, 1007864, 1141176, 1287294

Offset: 1

Author

R. H. Hardin, Oct 24 2014

Keywords

• nonn

Examples

			Some solutions for n=10:
  3  3  2  5  3  1  7  2  9  5  9  1  9  6  0  0
  3  7  0  9  5  7 10  0  8  9  2  5  5  7  1  5
  1  3  8 10  6  7  3  1  7  8  1  9  3  0  2  9
  1  1  4  6  0  3 10  9  0  2 10  8  4  4 10  1
		

Links

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

Crossrefs

Row 1 of A249290.

Formula

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

Empirical for n mod 6 = 0: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n

Empirical for n mod 6 = 1: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (7/3)

Empirical for n mod 6 = 2: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (8/3)

Empirical for n mod 6 = 3: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n - 3

Empirical for n mod 6 = 4: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (16/3)

Empirical for n mod 6 = 5: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n - (1/3).

Empirical g.f.: 2*x*(7 + 12*x + 17*x^2 + 29*x^3 + 7*x^5) / ((1 - x)^5*(1 + x)*(1 + x + x^2)). - Colin Barker, Nov 09 2018