A188777 T(n,k) = Number of n-turn bishop's tours on a k X k board summed over all starting positions.
1, 4, 0, 9, 4, 0, 16, 20, 0, 0, 25, 56, 28, 0, 0, 36, 120, 152, 24, 0, 0, 49, 220, 488, 328, 8, 0, 0, 64, 364, 1192, 1720, 584, 0, 0, 0, 81, 560, 2468, 5816, 5464, 840, 0, 0, 0, 100, 816, 4560, 15424, 26360, 15824, 784, 0, 0, 0, 121, 1140, 7760, 34736, 91120, 112680, 40496
Offset: 1
Examples
Some n=4 solutions for 4 X 4 ..0..0..0..0....0..4..0..1....0..0..0..4....3..0..0..0....0..1..0..0 ..0..3..0..0....0..0..3..0....0..0..0..0....0..2..0..0....4..0..2..0 ..2..0..4..0....0..2..0..0....0..3..0..1....0..0..4..0....0..3..0..0 ..0..1..0..0....0..0..0..0....0..0..2..0....0..0..0..1....0..0..0..0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..113
Crossrefs
Row 2 is A002492(n-1).
Formula
Empirical: T(1,k) = k^2.
Empirical: T(2,k) = (4/3)*k^3 - 2*k^2 + (2/3)*k.
Empirical: T(3,k) = 4*T(3,k-1)-5*T(3,k-2)+5*T(3,k-4)-4*T(3,k-5)+T(3,k-6).
Empirical: T(4,k) = 4*T(4,k-1)-4*T(4,k-2)-4*T(4,k-3)+10*T(4,k-4)-4*T(4,k-5)-4*T(4,k-6)+4*T(4,k-7)-T(4,k-8).
Empirical: T(5,k) = 4*T(5,k-1)-3*T(5,k-2)-8*T(5,k-3)+14*T(5,k-4)-14*T(5,k-6)+8*T(5,k-7)+3*T(5,k-8)-4*T(5,k-9)+T(5,k-10).
Empirical: T(6,k) = 4*T(6,k-1)-2*T(6,k-2)-12*T(6,k-3)+17*T(6,k-4)+8*T(6,k-5)-28*T(6,k-6)+8*T(6,k-7)+17*T(6,k-8)-12*T(6,k-9)-2*T(6,k-10)+4*T(6,k-11)-T(6,k-12).
Comments