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A253132 Number of length 4+2 0..n arrays with the sum of medians of adjacent triples multiplied by some arrangement of +-1 equal to zero 1

%I #4 Dec 27 2014 09:39:18

%S 40,393,2058,7257,19990,46945,98124,187593,335106,566597,915078,

%T 1423385,2144128,3140689,4491642,6289609,8642310,11679353,15549300,

%U 20420721,26492194,33986741,43152838,54278609,67682648,83714937,102776362

%N Number of length 4+2 0..n arrays with the sum of medians of adjacent triples multiplied by some arrangement of +-1 equal to zero

%C Row 4 of A253129

%H R. H. Hardin, <a href="/A253132/b253132.txt">Table of n, a(n) for n = 1..64</a>

%F Empirical: a(n) = 3*a(n-1) -3*a(n-2) +5*a(n-3) -12*a(n-4) +12*a(n-5) -10*a(n-6) +18*a(n-7) -18*a(n-8) +10*a(n-9) -12*a(n-10) +12*a(n-11) -5*a(n-12) +3*a(n-13) -3*a(n-14) +a(n-15)

%F Empirical for n mod 3 = 0: a(n) = (2/15)*n^6 + (1273/405)*n^5 + (298/27)*n^4 + (601/81)*n^3 + (143/15)*n^2 + (238/45)*n + 1

%F Empirical for n mod 3 = 1: a(n) = (2/15)*n^6 + (1273/405)*n^5 + (298/27)*n^4 + (203/27)*n^3 + (4001/405)*n^2 + (32/5)*n + (17/9)

%F Empirical for n mod 3 = 2: a(n) = (2/15)*n^6 + (1273/405)*n^5 + (298/27)*n^4 + (601/81)*n^3 + (3781/405)*n^2 + (218/45)*n + (73/81)

%e Some solutions for n=8

%e ..8....0....8....3....6....1....3....8....3....1....2....0....1....7....3....6

%e ..2....2....1....5....2....1....3....0....1....2....3....1....4....3....3....2

%e ..4....8....3....4....3....7....2....0....0....0....3....6....2....1....0....3

%e ..2....0....1....5....8....1....4....6....3....0....5....7....6....7....5....7

%e ..6....2....0....8....8....4....5....5....8....1....7....1....2....1....6....5

%e ..6....7....8....6....1....4....2....0....3....8....2....0....4....8....4....3

%K nonn

%O 1,1

%A _R. H. Hardin_, Dec 27 2014

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