%I #19 Feb 12 2023 10:53:41
%S 126,336,906,2484,7218,21024,61398,182520,542754,1614492,4829706,
%T 14448456,43225326,129555936,388309626,1163860164,3490511778,
%U 10468335024,31395421638,94176681480,282501311634,847417788972,2542167220986
%N Number of (n+2) X 3 binary matrices with every 3 X 3 block having exactly four 1's.
%C Column 1 of A181262.
%C The number of 1s in each row repeats with period 3, and we can divide the matrices into 12 classes (013, 022, 031, 103, 112, 121, 130, 202, 211, 220, 301, or 310) based on the pattern of row sums. The number of matrices in each class satisfies b(n) = 3*b(n-1), 3*b(n-3), or 9*b(n-3), depending on the number of 1s and 2s in the pattern. Therefore, the combined sequence satisfies [(T - 3I)(T^3 - 3I)(T^3 - 9I)](a)(n) = 0, where T is the right shift operator defined by T(a)(n) = a(n+1), and I is the identity operator. This is equivalent to the empirical formula for a(n) given below. - _David Radcliffe_, Jan 12 2023
%H R. H. Hardin, <a href="/A181255/b181255.txt">Table of n, a(n) for n = 1..200</a>
%F Empirical: a(n) = 3*a(n-1) + 12*a(n-3) - 36*a(n-4) - 27*a(n-6) + 81*a(n-7).
%F Empirical g.f.: 6*x*(21 - 7*x - 17*x^2 - 291*x^3 + 45*x^4 + 99*x^5 + 756*x^6) / ((1 - 3*x)*(1 - 3*x^3)*(1 - 9*x^3)). - _Colin Barker_, Mar 26 2018
%e Some solutions for 4 X 3:
%e 1 0 1 0 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 0 0
%e 0 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 0
%e 1 0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1
%e 1 1 0 1 1 0 1 0 0 1 1 0 0 1 0 0 0 1 0 0 1
%Y Cf. A181262.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 10 2010
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