%I #13 Aug 29 2022 09:40:03
%S 2,1,3,3,7,8,17,26,55,89,170,298,585,1059,1988,3640,6943,12990,24469,
%T 45663,86454,163324,309092,582651,1103457,2092206,3971963,7529743,
%U 14293584,27163872,51678766,98293571,187034535,356132703,678651768
%N Number of nX1 binary arrays with the number of 0-1 adjacencies equal to the number of 0-0 adjacencies
%C Column 1 of A183262
%H R. H. Hardin, <a href="/A183256/b183256.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) = [x^n] f(n,x) where f(1,x) = 2*x, f(2,x) = x^3+x^2+2*x, and f(n,x) = (1-x^3)*f(n-2,x) + (x+x^2)*f(n-1,x) otherwise. - _Robert Israel_, Nov 13 2019
%e All solutions for 5X1
%e ..0....0....1....1....0....1....0
%e ..0....1....0....1....0....1....0
%e ..1....0....0....1....1....1....0
%e ..1....0....0....0....0....1....1
%e ..1....0....1....0....0....1....0
%p fx:= proc(n) option remember; expand((1-x^3)*procname(n-2)+(x+x^2)*procname(n-1)) end proc:
%p fx(0):= 0: fx(1):= 2*x: fx(2):= x^3 + x^2 + 2*x:
%p seq(coeff(fx(n),x,n),n=1..40); # _Robert Israel_, Nov 13 2019
%t fx[n_] := fx[n] = Expand[(1-x^3)*fx[n-2] + (x+x^2)*fx[n-1]];
%t fx[0] = 0; fx[1] = 2x; fx[2] = x^3 + x^2 + 2x;
%t Table[Coefficient[fx[n], x, n], {n, 1, 40}] (* _Jean-François Alcover_, Aug 29 2022, after _Robert Israel_ *)
%Y Cf. A183262.
%K nonn
%O 1,1
%A _R. H. Hardin_, Jan 03 2011