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%I #13 Mar 07 2020 13:50:20
%S 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,
%T 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,
%U 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
%N 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.
%C Similar to A274641, except that here we consider the mex of squares that are a knight's moves rather than queen's moves.
%C Since there are at most 4 earlier cells in the spiral at a knight's move from any square, a(n) <= 4.
%H F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v27i1p52/8039">Queens in exile: non-attacking queens on infinite chess boards</a>, Electronic J. Combin., 27:1 (2020), #P1.52.
%H N. J. A. Sloane, <a href="/A308884/a308884.png">Beginning of the spiral showing the initial values.</a>
%e A knight at square 0 cannot see any numbers, so a(0)=0. Similarly a(1)=a(2)=a(3)=0.
%e 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.
%e A knight at square 7 can see a(2)=0 and a(4)=1, so a(7) = mex{0,1} = 2.
%e And so on. See the illustration for the start of the spiral.
%Y Cf. A274641, A308885-A308895.
%K nonn
%O 0,8
%A _N. J. A. Sloane_, Jul 01 2019
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