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%I #8 Nov 09 2018 07:24:53
%S 2,53,204,585,1326,2817,5028,8789,13970,21601,31512,45353,62194,84725,
%T 111556,145697,185598,235249,291740,360701,438402,530913,634224,
%U 755745,889506,1045157,1215548,1410889,1623630,1865681,2126932,2422133,2739634
%N Number of length 1+4 0..n arrays with every five consecutive terms having two times the sum of some three elements equal to three times the sum of the remaining two.
%H R. H. Hardin, <a href="/A248988/b248988.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 3*a(n-2) + 2*a(n-3) - 3*a(n-4) - 6*a(n-5) + 6*a(n-7) + 3*a(n-8) - 2*a(n-9) - 3*a(n-10) + a(n-12).
%F Empirical for n mod 6 = 0: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (32/3)*n + 1
%F Empirical for n mod 6 = 1: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 - (41/9)*n - (493/72)
%F Empirical for n mod 6 = 2: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (136/9)*n + (29/9)
%F Empirical for n mod 6 = 3: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 + (13/3)*n - (197/8)
%F Empirical for n mod 6 = 4: a(n) = (175/72)*n^4 - (40/9)*n^3 + (115/6)*n^2 - (176/9)*n + (169/9)
%F Empirical for n mod 6 = 5: a(n) = (175/72)*n^4 - (40/9)*n^3 + (185/12)*n^2 - (1/9)*n - (1613/72).
%F Empirical g.f.: x*(2 + 53*x + 198*x^2 + 422*x^3 + 614*x^4 + 825*x^5 + 810*x^6 + 653*x^7 + 416*x^8 + 206*x^9 + x^11) / ((1 - x)^5*(1 + x)^3*(1 + x + x^2)^2). - _Colin Barker_, Nov 09 2018
%e Some solutions for n=6:
%e ..3....1....0....3....6....5....5....6....0....6....3....6....6....2....4....2
%e ..4....2....2....0....5....2....3....6....4....3....1....6....1....1....5....2
%e ..0....0....0....6....2....1....5....5....0....6....6....1....0....4....3....1
%e ..2....1....4....6....6....2....6....1....5....3....5....2....6....3....2....5
%e ..6....1....4....0....1....0....1....2....1....2....0....0....2....0....6....0
%Y Row 1 of A248987.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 18 2014
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