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%I #8 Jul 17 2018 08:21:05
%S 15,33,69,143,293,595,1205,2427,4885,9803,19669,39403,78933,157995,
%T 316245,632747,1266005,2532523,5066069,10133163,20268373,40538795,
%U 81081685,162167467,324343125,648694443,1297405269,2594826923,5189686613
%N Number of (n+1) X (n+1) -3..3 symmetric matrices with every 2 X 2 subblock having sum zero and one, three or four distinct values.
%C Symmetry and 2 X 2 block sums zero implies that the diagonal x(i,i) are equal modulo 2 and x(i,j) = (x(i,i)+x(j,j))/2*(-1)^(i-j).
%H R. H. Hardin, <a href="/A211327/b211327.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = a(n-1) + 4*a(n-2) - 2*a(n-3) - 4*a(n-4).
%F Conjectures from _Colin Barker_, Jul 17 2018: (Start)
%F G.f.: x*(15 + 18*x - 24*x^2 - 28*x^3) / ((1 + x)*(1 - 2*x)*(1 - 2*x^2)).
%F a(n) = (-9*2^(n/2) + 29*2^n + 1)/3 for n even.
%F a(n) = (-3*2^(n/2+3/2) + 29*2^n - 1)/3 for n odd.
%F (End)
%e Some solutions for n=3:
%e .-1..2..1..0....0.-1..0.-1...-2..1..0..1....1.-2..1.-2....0..0..0..0
%e ..2.-3..0.-1...-1..2.-1..2....1..0.-1..0...-2..3.-2..3....0..0..0..0
%e ..1..0..3.-2....0.-1..0.-1....0.-1..2.-1....1.-2..1.-2....0..0..0..0
%e ..0.-1.-2..1...-1..2.-1..2....1..0.-1..0...-2..3.-2..3....0..0..0..0
%K nonn
%O 1,1
%A _R. H. Hardin_, Apr 07 2012
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