%I #18 Dec 11 2022 12:31:29
%S 3,8,19,48,120,299,747,1865,4656,11625,29024,72464,180921,451705,
%T 1127771,2815704,7029963,17551696,43821288,109408531,273160083,
%U 681998289,1702743904,4251237649,10614057408,26500098080,66162747313,165188412465
%N Number of n X 2 0..1 arrays with each 1 horizontally or vertically adjacent to 0 or 2 1s.
%C Column 2 of A295051.
%C From  _Mikhail Lavrov_, Oct 31 2022: (Start)
%C If the first row is 0 0, the array can continue with anything; there are a(n1) such arrays for n > 1.
%C If the first row is 1 1, then the second row must be 1 1, and the third row must be 0 0; there are a(n3) such arrays for n > 3.
%C If the first row is 1 0 or 0 1, then the rows after that can alternate between 0 1 and 1 0 for any number of steps before going to 0 0 (or reaching the end); there are 2(a(n2) + a(n3) + ...) such arrays for n > 2. This gives a(n) = a(n1) + a(n3) + 2(a(n2) + a(n3) + ...); to simplify the recurrence, apply it to a(n)  a(n1), getting a(n)  a(n1) = a(n1)  a(n2) + a(n3)  a(n4) + 2a(n2). This proves _Colin Barker_'s recurrence relation below for n > 4. (End)
%H R. H. Hardin, <a href="/A295045/b295045.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 2*a(n1) + a(n2) + a(n3)  a(n4).
%F Empirical g.f.: x*(3 + 2*x  x^3) / (1  2*x  x^2  x^3 + x^4).  _Colin Barker_, Feb 21 2018
%e Some solutions for n=7:
%e 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 0
%e 1 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0
%e 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 1
%e 1 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 1 0 0
%e 0 1 1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0
%e 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0
%e 0 1 1 0 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 0
%t LinearRecurrence[{2, 1, 1, 1}, {3, 8, 19, 48}, 30] (* _Mikhail Lavrov_, Nov 01 2022 *)
%Y Cf. A295051.
%K nonn
%O 1,1
%A _R. H. Hardin_, Nov 13 2017
