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 A181255 Number of (n+2) X 3 binary matrices with every 3 X 3 block having exactly four 1's. 1
 126, 336, 906, 2484, 7218, 21024, 61398, 182520, 542754, 1614492, 4829706, 14448456, 43225326, 129555936, 388309626, 1163860164, 3490511778, 10468335024, 31395421638, 94176681480, 282501311634, 847417788972, 2542167220986 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A181262. Sequence in context: A254465 A063334 A181262 * A329807 A322542 A323759 Adjacent sequences: A181252 A181253 A181254 * A181256 A181257 A181258 KEYWORD nonn AUTHOR R. H. Hardin, Oct 10 2010 STATUS approved

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Last modified September 19 11:33 EDT 2024. Contains 376010 sequences. (Running on oeis4.)