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%I #8 Nov 08 2018 19:09:09
%S 10,36,148,380,862,1652,2956,4860,7642,11400,16488,23044,31482,41952,
%T 54956,70672,89662,112128,138708,169632,205610,246884,294240,347960,
%U 408890,477324,554196,639828,735214,840700,957356,1085556,1226442
%N Number of length 2+2 0..n arrays with no three consecutive terms having the sum of any two elements equal to twice the third.
%H R. H. Hardin, <a href="/A248463/b248463.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 2*a(n-1) - a(n-3) - 2*a(n-5) + 2*a(n-6) + a(n-8) - 2*a(n-10) + a(n-11).
%F Empirical for n mod 12 = 0: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
%F Empirical for n mod 12 = 1: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (5/2).
%F Empirical for n mod 12 = 2: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n - (4/3).
%F Empirical for n mod 12 = 3: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (3/2).
%F Empirical for n mod 12 = 4: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
%F Empirical for n mod 12 = 5: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (7/6).
%F Empirical for n mod 12 = 6: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
%F Empirical for n mod 12 = 7: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (3/2).
%F Empirical for n mod 12 = 8: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n - (4/3).
%F Empirical for n mod 12 = 9: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (5/2).
%F Empirical for n mod 12 = 10: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
%F Empirical for n mod 12 = 11: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (17/6).
%F Empirical g.f.: 2*x*(5 + 8*x + 38*x^2 + 47*x^3 + 69*x^4 + 48*x^5 + 42*x^6 + 17*x^7 + 14*x^8) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)). - _Colin Barker_, Nov 08 2018
%e Some solutions for n=6:
%e ..2....1....0....2....3....2....2....4....4....2....1....6....0....4....1....1
%e ..0....4....0....6....2....3....5....4....5....5....0....3....6....6....3....5
%e ..6....4....3....2....0....5....6....1....5....0....1....6....1....1....4....5
%e ..4....1....4....3....6....2....6....0....0....6....1....2....4....2....6....1
%Y Row 2 of A248461.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 06 2014
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