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Empirical: a(n) = a(n-1) + 4*a(n-2) - 2*a(n-3) - 4*a(n-4).

Number of (n+1) X (n+1) -3..3 symmetric matrices with every 2X2 2 X 2 subblock having sum zero and one, three or four distinct values.

Symmetry and 2X2 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).

Empirical: a(n) = a(n-1) +4*a(n-2) -2*a(n-3) -4*a(n-4).

Conjectures from Colin Barker, Jul 17 2018: (Start)

G.f.: x*(15 + 18*x - 24*x^2 - 28*x^3) / ((1 + x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = (-9*2^(n/2) + 29*2^n + 1)/3 for n even.

a(n) = (-3*2^(n/2+3/2) + 29*2^n - 1)/3 for n odd.

(End)

Some solutions for n=3:

R. H. Hardin , Apr 07 2012

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R. H. Hardin, <a href="/A211327/b211327.txt">Table of n, a(n) for n = 1..210</a>

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Number of (n+1)X(n+1) -3..3 symmetric matrices with every 2X2 subblock having sum zero and one, three or four distinct values

15, 33, 69, 143, 293, 595, 1205, 2427, 4885, 9803, 19669, 39403, 78933, 157995, 316245, 632747, 1266005, 2532523, 5066069, 10133163, 20268373, 40538795, 81081685, 162167467, 324343125, 648694443, 1297405269, 2594826923, 5189686613

1,1

Symmetry and 2X2 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)

Empirical: a(n) = a(n-1) +4*a(n-2) -2*a(n-3) -4*a(n-4)

Some solutions for n=3

.-1..2..1..0....0.-1..0.-1...-2..1..0..1....1.-2..1.-2....0..0..0..0

..2.-3..0.-1...-1..2.-1..2....1..0.-1..0...-2..3.-2..3....0..0..0..0

..1..0..3.-2....0.-1..0.-1....0.-1..2.-1....1.-2..1.-2....0..0..0..0

..0.-1.-2..1...-1..2.-1..2....1..0.-1..0...-2..3.-2..3....0..0..0..0

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R. H. Hardin Apr 07 2012

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