%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