A308884 Follow along the squares in the square spiral (as in A274641); in each square write the smallest nonnegative number that a knight placed at that square cannot see.

0, 0, 0, 0, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2, 3, 3, 2, 0, 1, 3, 2, 1, 0, 3, 3, 3, 2, 0, 1, 3, 3, 1, 0, 0, 0, 3, 3, 0, 0, 0, 2, 1, 3, 3, 0, 0, 0, 2, 2, 3, 2, 0, 0, 0, 2, 1, 3, 3, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 2, 2, 1, 3, 2, 1, 0, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 2, 1

Offset: 0

Author

N. J. A. Sloane, Jul 01 2019

Keywords

• nonn

Comments

Similar to A274641, except that here we consider the mex of squares that are a knight's moves rather than queen's moves.

Since there are at most 4 earlier cells in the spiral at a knight's move from any square, a(n) <= 4.

Examples

			A knight at square 0 cannot see any numbers, so a(0)=0. Similarly a(1)=a(2)=a(3)=0.
A knight at square 4 in the spiral can see the 0 in square 1 (because square 1 is a knight's move from square 4), so a(4) = 1. Similarly a(5)=a(6)=1.
A knight at square 7 can see a(2)=0 and a(4)=1, so a(7) = mex{0,1} = 2.
And so on. See the illustration for the start of the spiral.
		

Links

• F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
• N. J. A. Sloane, Beginning of the spiral showing the initial values.

Crossrefs

Cf. A274641, A308885-A308895.