%I #5 Sep 17 2022 00:56:56
%S 0,1,0,-1,-2,-3,-2,-1,0,-1,0,1,2,1,2,3,4,3,4,5,6,7,6,5,4,5,4,3,2,3,2,
%T 1,0,-1,0,1,2,3,2,1,0,1,0,-1,-2,-1,-2,-3,-4,-5,-4,-3,-2,-1,-2,-3,-4,
%U -3,-4,-5,-6,-5,-6,-7
%N Hilbert's Hamiltonian walk projected onto the second diagonal: M'(3) (difference between sequences A059253 and A059252; their sum is A059261).
%F Initially, M'(0)=0; recursion: M'(2n)=M'(2n-1). (-f(-M'(2n-1), 2n-1)).(-M'(2n-1)).f(M'(2n-1), 2n-1), M'(2n+1)=M'(2n).f(M'(2n), 2n).(-M'(2n)).(-(f(-M'(2n), 2n+1)). f(m, n) is the complementation to 2^n, [example: f(4 3 4 5 6 7 6 5 4 5 4 6 2 3 2 1, 3)=4 5 4 3 2 1 2 3 4 3 4 5 6 5 6 7]; (-m) is the opposite[example: m=4 5 4 3 2 1 2 3 4 3 4 5 6 5 6 7, (-m)=-4 -5 -4 -3 -2 -1 -2 -3 -4 -3 -4 -5 -6 -5 -6 -7] [Corrected by _Sean A. Irvine_, Sep 17 2022]
%e [M'(0)=0, M'(1)=0 1 0 -1, M'(2)=0 1 0 -1 -2 -3 -2 -1 0 -1 0 1 2 1 2 3]
%Y The x-projection m(3) is A059253, the y-projection m(3) is A059252 and the projection onto the first diagonal, M(3), is A059261.
%K sign
%O 0,5
%A Claude Lenormand (claude.lenormand(AT)free.fr), Jan 24 2001