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

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

Offset: 1

Author

R. H. Hardin, Jan 08 2011

Keywords

• nonn
• easy

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

Examples

			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
		

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).

Crossrefs

Cf. A274230.

Conjectured to be the main diagonal of A274636.

Programs

Maple

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

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