cp's OEIS Frontend

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.

%I A249291 #10 Nov 09 2018 21:55:00
%S A249291 14,66,204,524,1098,2070,3584,5808,8934,13202,18828,26100,35306,46758,
%T A249291 60792,77792,98118,122202,150476,183396,221442,265142,315000,371592,
%U A249291 435494,507306,587652,677204,776610,886590,1007864,1141176,1287294
%N 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.
%H A249291 R. H. Hardin, <a href="/A249291/b249291.txt">Table of n, a(n) for n = 1..210</a>
%F A249291 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).
%F A249291 Empirical for n mod 6 = 0: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n
%F A249291 Empirical for n mod 6 = 1: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (7/3)
%F A249291 Empirical for n mod 6 = 2: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (8/3)
%F A249291 Empirical for n mod 6 = 3: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n - 3
%F A249291 Empirical for n mod 6 = 4: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n + (16/3)
%F A249291 Empirical for n mod 6 = 5: a(n) = n^4 + (8/3)*n^3 + 5*n^2 + 3*n - (1/3).
%F A249291 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
%e A249291 Some solutions for n=10:
%e A249291   3  3  2  5  3  1  7  2  9  5  9  1  9  6  0  0
%e A249291   3  7  0  9  5  7 10  0  8  9  2  5  5  7  1  5
%e A249291   1  3  8 10  6  7  3  1  7  8  1  9  3  0  2  9
%e A249291   1  1  4  6  0  3 10  9  0  2 10  8  4  4 10  1
%Y A249291 Row 1 of A249290.
%K A249291 nonn
%O A249291 1,1
%A A249291 _R. H. Hardin_, Oct 24 2014