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

1,1

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

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

Cf. A274230.

Conjectured to be the main diagonal of A274636.

Sequence in context: A118691 A087718 A033476 * A118696 A065856 A136357

Adjacent sequences: A183975 A183976 A183977 * A183979 A183980 A183981

Keyword

nonn,easy

Author

R. H. Hardin, Jan 08 2011

Status

approved