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1/16 the number of (n+1)X5 X 5 0..3 arrays with all 2X2 2 X 2 subblocks having the same four values.
Column 4 of A184039.
Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).
Conjectures from Colin Barker, Apr 10 2018: (Start)
G.f.: x*(91 - 170*x - 185*x^2 + 340*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).
a(n) = 9*2^(n/2-1) + 9*2^(n-1) + 76 for n even.
a(n) = 9*2^(n-1) + 3*2^((n+1)/2) + 76 for n odd.
(End)
Some solutions for 3X53 X 5:
Cf. A184039.
R. H. Hardin , Jan 08 2011
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_R. H. Hardin (rhhardin(AT)att.net) _ Jan 08 2011
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R. H. Hardin, <a href="/A184034/b184034.txt">Table of n, a(n) for n = 1..200</a>
allocated for Ron Hardin1/16 the number of (n+1)X5 0..3 arrays with all 2X2 subblocks having the same four values
91, 103, 124, 166, 244, 400, 700, 1300, 2476, 4828, 9484, 18796, 37324, 74380, 148300, 296140, 591436, 1182028, 2362444, 4723276, 9443404, 18883660, 37761100, 75515980, 151019596, 302026828, 604029004, 1208033356, 2416017484, 4831985740
1,1
Column 4 of A184039
Empirical: a(n)=3*a(n-1)-6*a(n-3)+4*a(n-4)
Some solutions for 3X5
..3..1..0..1..3....2..3..3..3..3....2..1..2..0..2....1..1..3..2..1
..0..2..3..2..0....3..0..2..0..2....3..0..3..1..3....3..2..1..1..3
..3..1..0..1..3....2..3..3..3..3....2..1..2..0..2....1..1..3..2..1
allocated
nonn
R. H. Hardin (rhhardin(AT)att.net) Jan 08 2011
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allocated for Ron Hardin
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