cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A209723 1/4 the number of (n+1) X 5 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

6, 7, 8, 10, 12, 16, 20, 28, 36, 52, 68, 100, 132, 196, 260, 388, 516, 772, 1028, 1540, 2052, 3076, 4100, 6148, 8196, 12292, 16388, 24580, 32772, 49156, 65540, 98308, 131076, 196612, 262148, 393220, 524292, 786436, 1048580, 1572868, 2097156
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 4 of A209727.

Examples

			Some solutions for n=4:
..2..1..2..0..2....0..2..0..1..0....0..1..0..1..0....0..1..0..1..0
..0..2..0..1..0....2..1..2..0..2....2..0..2..0..2....2..0..2..0..2
..2..1..2..0..2....0..2..0..1..0....0..1..0..1..0....0..1..0..1..0
..0..2..0..1..0....2..1..2..0..2....2..0..2..0..2....2..0..2..0..2
..2..1..2..0..2....0..2..0..1..0....1..2..1..2..1....0..1..0..1..0
		

Crossrefs

Cf. A209727.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
Conjectures from Colin Barker, Jul 12 2018: (Start)
G.f.: x*(6 + x - 11*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2 - 1) + 4 for n even.
a(n) = 2^((n + 1)/2) + 4 for n odd.
(End)