OFFSET
1,1
COMMENTS
Column 1 of A181262.
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
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..200
FORMULA
Empirical: a(n) = 3*a(n-1) + 12*a(n-3) - 36*a(n-4) - 27*a(n-6) + 81*a(n-7).
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
EXAMPLE
Some solutions for 4 X 3:
1 0 1 0 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 0 0
0 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 0
1 0 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1
1 1 0 1 1 0 1 0 0 1 1 0 0 1 0 0 0 1 0 0 1
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Oct 10 2010
STATUS
approved