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%I #8 Nov 07 2018 11:29:21
%S 8,61,220,561,1124,2009,3220,4901,7016,9737,13000,17025,21688,27245,
%T 33572,40929,49140,58553,68924,80613,93400,107641,123056,140113,
%U 158440,178509,200012,223393,248260,275209,303748,334453,366920,401689,438264
%N Number of length 3+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum.
%H R. H. Hardin, <a href="/A247535/b247535.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = a(n-2) + 2*a(n-3) + a(n-4) - 2*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8) + 2*a(n-9) + a(n-10) - a(n-12).
%F Also as a cubic plus a linear quasipolynomial with period 12:
%F Empirical for n mod 12 = 0: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n + 1
%F Empirical for n mod 12 = 1: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)
%F Empirical for n mod 12 = 2: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (97/27)
%F Empirical for n mod 12 = 3: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)
%F Empirical for n mod 12 = 4: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (37/27)
%F Empirical for n mod 12 = 5: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54)
%F Empirical for n mod 12 = 6: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - 1
%F Empirical for n mod 12 = 7: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)
%F Empirical for n mod 12 = 8: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (151/27)
%F Empirical for n mod 12 = 9: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)
%F Empirical for n mod 12 = 10: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (91/27)
%F Empirical for n mod 12 = 11: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54).
%F Empirical g.f.: x*(8 + 61*x + 212*x^2 + 484*x^3 + 774*x^4 + 963*x^5 + 892*x^6 + 661*x^7 + 330*x^8 + 120*x^9 - x^11) / ((1 - x)^4*(1 + x)^2*(1 + x^2)*(1 + x + x^2)^2). - _Colin Barker_, Nov 07 2018
%e Some solutions for n=6:
%e ..2....1....1....3....5....1....1....3....5....3....3....5....6....2....3....4
%e ..2....5....4....4....5....0....3....0....4....1....2....6....4....5....2....2
%e ..0....3....5....2....6....5....5....3....2....3....0....4....3....1....1....3
%e ..4....3....2....3....4....6....3....6....3....5....1....5....1....4....2....3
%e ..2....5....3....1....5....1....5....3....3....3....3....3....2....2....1....2
%e ..6....5....6....2....5....0....3....6....2....5....4....4....0....5....0....4
%Y Row 3 of A247533.
%K nonn
%O 1,1
%A _R. H. Hardin_, Sep 18 2014
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