A187859 Number of 4-step one space for components leftwards or up, two space for components rightwards or down asymmetric quasi-queen's tours (antidiagonal moves become knight moves) on an n X n board summed over all starting positions.

0, 0, 216, 968, 2754, 5428, 9237, 14040, 19837, 26628, 34413, 43192, 52965, 63732, 75493, 88248, 101997, 116740, 132477, 149208, 166933, 185652, 205365, 226072, 247773, 270468, 294157, 318840, 344517, 371188, 398853, 427512, 457165, 487812

Offset: 1

Author

R. H. Hardin, Mar 14 2011

Keywords

• nonn

Comments

Row 4 of A187857.

Examples

			Some solutions for 4 X 4:
..0..0..0..0....3..2..4..0....0..0..0..0....0..4..0..0....0..0..0..4
..0..0..0..0....0..1..0..0....0..4..3..0....0..0..3..0....0..3..2..0
..4..3..2..0....0..0..0..0....2..1..0..0....1..0..2..0....0..0..0..1
..0..0..1..0....0..0..0..0....0..0..0..0....0..0..0..0....0..0..0..0
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..50

Crossrefs

Cf. A187857.

Formula

Empirical: a(n) = 497*n^2 - 2652*n + 3448 for n>5.

Conjectures from Colin Barker, Apr 26 2018: (Start)

G.f.: x^3*(216 + 320*x + 498*x^2 - 146*x^3 + 247*x^4 - 141*x^5) / (1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>8.

(End)