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A183978 1/4 the number of (n+1) X 2 binary arrays with all 2 X 2 subblock sums the same. 4

%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(n-1) - 6*a(n-3) + 4*a(n-4)

%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(n-1) for n >= 1. - _Philippe Gibone_, Jul 06 2016, corrected by _Robert Israel_, May 21 2019

%F Empirical: G.f.: -x*(4-6*x-9*x^2+12*x^3) / ( (x-1)*(2*x-1)*(2*x^2-1) ). - _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((n-1)/2))*(1+2^ceil((n-1)/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

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Last modified April 5 08:04 EDT 2020. Contains 333238 sequences. (Running on oeis4.)