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%I #10 Nov 09 2018 21:54:46
%S 42,486,2772,10620,32070,81402,183696,376752,718530,1289430,2201892,
%T 3603396,5688582,8702250,12954360,18823392,26773506,37358622,51242100,
%U 69200820,92147742,121136346,157385352,202283880,257418690,324579342
%N Number of length 1+5 0..n arrays with no six consecutive terms having two times the sum of any two elements equal to the sum of the remaining four.
%H R. H. Hardin, <a href="/A249234/b249234.txt">Table of n, a(n) for n = 1..92</a>
%F Empirical: a(n) = 4*a(n-1) - 4*a(n-2) - 3*a(n-3) + 6*a(n-4) - 6*a(n-7) + 3*a(n-8) + 4*a(n-9) - 4*a(n-10) + a(n-11).
%F Empirical for n mod 6 = 0: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n
%F Empirical for n mod 6 = 1: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (445/12)
%F Empirical for n mod 6 = 2: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (40/3)
%F Empirical for n mod 6 = 3: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (255/4)
%F Empirical for n mod 6 = 4: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (80/3)
%F Empirical for n mod 6 = 5: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (605/12).
%F Empirical g.f.: 6*x*(7 + 53*x + 166*x^2 + 267*x^3 + 314*x^4 + 167*x^5 + 266*x^6 + 53*x^7 + 147*x^8) / ((1 - x)^7*(1 + x)^2*(1 + x + x^2)). - _Colin Barker_, Nov 09 2018
%e Some solutions for n=7:
%e 6 4 6 2 2 0 6 2 4 0 6 4 6 2 1 4
%e 0 2 2 4 5 6 2 5 7 2 6 4 6 1 4 0
%e 1 6 6 2 4 7 4 0 1 2 6 7 1 5 2 0
%e 4 5 5 7 1 0 2 0 2 6 6 0 4 1 0 5
%e 1 7 2 2 0 4 6 6 3 6 4 6 2 6 7 3
%e 5 2 5 6 2 7 2 3 3 3 6 7 7 2 0 2
%Y Row 1 of A249233.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 23 2014
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