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%I #9 Nov 12 2018 14:36:33
%S 4,20,96,436,1880,7836,32032,129572,521256,2091052,8376368,33529908,
%T 134168632,536772668,2147287104,8589541444,34358952008,137437380684,
%U 549752668240,2199016964180,8796080439384,35184346923100,140737438023776
%N Number of length n+1 0..3 arrays with the sum of adjacent differences multiplied by some arrangement of +-1 equal to zero.
%H R. H. Hardin, <a href="/A250162/b250162.txt">Table of n, a(n) for n = 1..59</a>
%F Empirical: a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4).
%F Conjectures from _Colin Barker_, Nov 12 2018: (Start)
%F G.f.: 4*x*(1 - 3*x + 5*x^2) / ((1 - x)^2*(1 - 2*x)*(1 - 4*x)).
%F a(n) = 2*(2 - 3*2^n + 4^n + 2*n).
%F (End)
%e Some solutions for n=6:
%e ..1....3....3....2....3....2....2....2....2....1....1....2....1....0....2....1
%e ..1....0....2....3....3....0....2....2....2....1....0....0....1....2....0....0
%e ..0....3....1....3....3....2....3....3....2....3....3....0....1....2....3....2
%e ..3....1....0....0....0....3....1....1....2....2....3....2....3....3....0....1
%e ..0....2....3....3....3....2....3....2....1....0....1....3....3....1....2....2
%e ..1....1....0....1....3....0....2....0....0....1....2....2....0....3....0....2
%e ..3....3....1....2....3....2....2....0....2....3....1....0....1....2....2....3
%Y Column 3 of A250167.
%K nonn
%O 1,1
%A _R. H. Hardin_, Nov 13 2014
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