%I
%S 4,6,9,15,25,45,81,153,289,561,1089,2145,4225,8385,16641,33153,66049,
%T 131841,263169,525825,1050625,2100225,4198401,8394753,16785409,
%U 33566721,67125249,134242305,268468225,536920065,1073807361,2147581953,4295098369
%N 1/4 the number of (n+1) X 2 binary arrays with all 2 X 2 subblock sums the same.
%C Column 1 of A183986
%C Based on the conjectured recursion formula, it is also the number of notches in a sheet of paper when you fold it n times and cut off the four corners (see A274230).  _Philippe Gibone_, Jul 06 2016
%H R. H. Hardin, <a href="/A183978/b183978.txt">Table of n, a(n) for n = 1..46</a>
%H Robert Israel, <a href="/A183978/a183978.pdf">Proof of empirical formulas for A183978</a>
%F Empirical: a(n) = 3*a(n1)  6*a(n3) + 4*a(n4)
%F Based on the conjectured recursion formula, we may prove (by a tedious induction) that a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) = A051032(n) * A051032(n1) for n >= 1.  _Philippe Gibone_, Jul 06 2016, corrected by _Robert Israel_, May 21 2019
%F Empirical: G.f.: x*(46*x9*x^2+12*x^3) / ( (x1)*(2*x1)*(2*x^21) ).  _R. J. Mathar_, Jul 15 2016
%F Empirical formulas verified (see link): _Robert Israel_, May 21 2019.
%e Some solutions for 5X2
%e ..0..1....1..0....1..0....1..1....0..1....1..0....1..0....0..1....0..1....0..1
%e ..0..0....1..0....1..0....1..0....1..0....1..0....1..0....0..1....1..0....0..1
%e ..1..0....1..0....0..1....1..1....0..1....0..1....0..1....1..0....0..1....1..0
%e ..0..0....1..0....1..0....0..1....1..0....1..0....0..1....1..0....0..1....0..1
%e ..1..0....1..0....1..0....1..1....1..0....0..1....0..1....1..0....1..0....0..1
%p seq((1+2^floor((n1)/2))*(1+2^ceil((n1)/2)), n=1..20); # _Robert Israel_, May 21 2019
%Y Cf. A274230.
%Y Conjectured to be the main diagonal of A274636.
%K nonn
%O 1,1
%A _R. H. Hardin_, Jan 08 2011
