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A183982 1/4 the number of (n+1) X 6 binary arrays with all 2 X 2 subblock sums the same.

%I A183982 #10 Aug 31 2024 15:40:42
%S A183982 25,27,30,36,46,66,102,174,310,582,1110,2166,4246,8406,16662,33174,
%T A183982 66070,131862,263190,525846,1050646,2100246,4198422,8394774,16785430,
%U A183982 33566742,67125270,134242326,268468246,536920086,1073807382,2147581974,4295098390
%N A183982 1/4 the number of (n+1) X 6 binary arrays with all 2 X 2 subblock sums the same.
%C A183982 Column 5 of A183986.
%H A183982 R. H. Hardin, <a href="/A183982/b183982.txt">Table of n, a(n) for n = 1..200</a>
%H A183982 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 0, -6, 4).
%F A183982 Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).
%F A183982 Conjectures from _Colin Barker_, Apr 08 2018: (Start)
%F A183982 G.f.: x*(25 - 48*x - 51*x^2 + 96*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).
%F A183982 a(n) = 3*2^(n/2-1) + 2^(n-1) + 22 for n even.
%F A183982 a(n) = 2^(n-1) + 2^((n+1)/2) + 22 for n odd.
%F A183982 (End)
%F A183982 The above empirical formula is correct. See note from Andrew Howroyd in A183986.
%e A183982 Some solutions for 5 X 6:
%e A183982 ..0..1..1..0..1..0....0..1..1..1..0..1....0..1..0..0..1..0....0..1..0..1..1..1
%e A183982 ..1..0..0..1..0..1....1..1..0..1..1..1....1..0..1..1..0..1....1..0..1..0..0..0
%e A183982 ..0..1..1..0..1..0....0..1..1..1..0..1....0..1..0..0..1..0....0..1..0..1..1..1
%e A183982 ..1..0..0..1..0..1....1..1..0..1..1..1....1..0..1..1..0..1....1..0..1..0..0..0
%e A183982 ..0..1..1..0..1..0....0..1..1..1..0..1....0..1..0..0..1..0....0..1..0..1..1..1
%Y A183982 Cf. A183986.
%K A183982 nonn
%O A183982 1,1
%A A183982 _R. H. Hardin_, Jan 08 2011