

A183978


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


4



4, 6, 9, 15, 25, 45, 81, 153, 289, 561, 1089, 2145, 4225, 8385, 16641, 33153, 66049, 131841, 263169, 525825, 1050625, 2100225, 4198401, 8394753, 16785409, 33566721, 67125249, 134242305, 268468225, 536920065, 1073807361, 2147581953, 4295098369
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OFFSET

1,1


COMMENTS

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


LINKS



FORMULA

Empirical: a(n) = 3*a(n1)  6*a(n3) + 4*a(n4)
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
Empirical: G.f.: x*(46*x9*x^2+12*x^3) / ( (x1)*(2*x1)*(2*x^21) ).  R. J. Mathar, Jul 15 2016
Empirical formulas verified (see link): Robert Israel, May 21 2019.


EXAMPLE

Some solutions for 5X2
..0..1....1..0....1..0....1..1....0..1....1..0....1..0....0..1....0..1....0..1
..0..0....1..0....1..0....1..0....1..0....1..0....1..0....0..1....1..0....0..1
..1..0....1..0....0..1....1..1....0..1....0..1....0..1....1..0....0..1....1..0
..0..0....1..0....1..0....0..1....1..0....1..0....0..1....1..0....0..1....0..1
..1..0....1..0....1..0....1..1....1..0....0..1....0..1....1..0....1..0....0..1


MAPLE

seq((1+2^floor((n1)/2))*(1+2^ceil((n1)/2)), n=1..20); # Robert Israel, May 21 2019


CROSSREFS

Conjectured to be the main diagonal of A274636.


KEYWORD

nonn


AUTHOR



STATUS

approved



