OFFSET
1,1
COMMENTS
Column 1 of A183986
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
R. H. Hardin, Table of n, a(n) for n = 1..46
Robert Israel, Proof of empirical formulas for A183978
Index entries for linear recurrences with constant coefficients, signature (3, 0, -6, 4).
FORMULA
Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4)
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
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
Empirical formulas verified (see link): Robert Israel, May 21 2019.
2*a(n) = 2+2^n+A029744(n+3). - R. J. Mathar, Jul 19 2024
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((n-1)/2))*(1+2^ceil((n-1)/2)), n=1..20); # Robert Israel, May 21 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. H. Hardin, Jan 08 2011
STATUS
approved