%I
%S 34,339,1760,6293,17598,41677,87328,166677,295898,495745,791734,
%T 1215373,1804320,2602645,3662110,5042499,6811262,9045701,11832586,
%U 15267973,19459520,24526335,30597824,37817343,46341088,56337177,67989404,81496455
%N Number of length 4+2 0..n arrays with the sum of the maximum minus the median of adjacent triples multiplied by some arrangement of +1 equal to zero
%C Row 4 of A251935
%H R. H. Hardin, <a href="/A251938/b251938.txt">Table of n, a(n) for n = 1..94</a>
%F Empirical: a(n) = a(n1) +3*a(n3) a(n4) 2*a(n5) 3*a(n6) 3*a(n7) +5*a(n8) +2*a(n9) +5*a(n10) 3*a(n11) 3*a(n12) 2*a(n13) a(n14) +3*a(n15) +a(n17) a(n18)
%F Empirical for n mod 12 = 0: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2  (73/20)*n + 1
%F Empirical for n mod 12 = 1: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2  (146743/25920)*n + (3073/576)
%F Empirical for n mod 12 = 2: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2  (7913/1620)*n + (169/54)
%F Empirical for n mod 12 = 3: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2  (1423/320)*n + (209/64)
%F Empirical for n mod 12 = 4: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2  (7273/1620)*n + (31/9)
%F Empirical for n mod 12 = 5: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2  (156983/25920)*n + (4355/1728)
%F Empirical for n mod 12 = 6: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2  (73/20)*n + (7/2)
%F Empirical for n mod 12 = 7: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2  (137023/25920)*n + (3289/576)
%F Empirical for n mod 12 = 8: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2  (7913/1620)*n + (17/27)
%F Empirical for n mod 12 = 9: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2  (1543/320)*n + (185/64)
%F Empirical for n mod 12 = 10: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2  (7273/1620)*n + (107/18)
%F Empirical for n mod 12 = 11: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2  (147263/25920)*n + (5003/1728)
%e Some solutions for n=6
%e ..2....2....3....5....6....1....1....4....3....5....4....3....1....4....6....6
%e ..4....4....5....1....2....5....2....5....0....2....2....2....5....0....5....6
%e ..5....6....6....0....0....3....2....2....5....1....5....0....6....4....5....2
%e ..3....1....3....5....0....0....2....6....2....4....6....6....1....3....4....3
%e ..4....3....2....4....4....6....6....6....1....0....5....4....5....4....5....6
%e ..0....6....0....3....6....3....2....4....6....0....3....5....4....5....3....6
%K nonn
%O 1,1
%A _R. H. Hardin_, Dec 11 2014
