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%I #10 Nov 09 2018 21:54:36
%S 22,243,1324,5005,14586,36247,78448,154689,281470,482131,784092,
%T 1223413,1840954,2688375,3822856,5314177,7238718,9687259,12757900,
%U 16565301,21232162,26899543,33717624,41856745,51497086,62841147,76101988,91516789
%N Number of length 1+5 0..n arrays with every six consecutive terms having two times the sum of some two elements equal to the sum of the remaining four.
%H R. H. Hardin, <a href="/A249086/b249086.txt">Table of n, a(n) for n = 1..96</a>
%F Empirical: a(n) = 3*a(n-1) - a(n-2) - 4*a(n-3) + 2*a(n-4) + 2*a(n-5) + 2*a(n-6) - 4*a(n-7) - a(n-8) + 3*a(n-9) - a(n-10).
%F Empirical for n mod 6 = 0: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n + 1
%F Empirical for n mod 6 = 1: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (457/12)
%F Empirical for n mod 6 = 2: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (37/3)
%F Empirical for n mod 6 = 3: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (259/4)
%F Empirical for n mod 6 = 4: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (77/3)
%F Empirical for n mod 6 = 5: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (617/12).
%F Empirical g.f.: x*(22 + 177*x + 617*x^2 + 1364*x^3 + 1823*x^4 + 2260*x^5 + 1135*x^6 + 880*x^7 + 3*x^8 - x^9) / ((1 - x)^6*(1 + x)^2*(1 + x + x^2)). - _Colin Barker_, Nov 09 2018
%e Some solutions for n=6:
%e 4 3 1 3 4 4 6 3 6 1 1 1 1 3 4 6
%e 6 2 2 6 1 3 1 1 0 2 5 4 6 1 0 2
%e 5 1 0 6 6 6 4 6 6 0 6 0 5 0 4 0
%e 0 5 5 5 1 4 5 2 3 5 0 1 3 2 3 4
%e 5 6 1 5 3 2 0 0 5 2 1 3 4 5 6 2
%e 1 1 6 2 3 2 5 3 4 2 5 0 5 4 1 4
%Y Row 1 of A249085.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 20 2014