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%I #22 Jun 08 2021 13:54:29
%S 0,1,1,0,0,0,1,1,2,2,3,3,3,2,2,3,4,4,5,5,6,7,7,6,6,7,7,6,5,5,4,4,4,4,
%T 5,5,6,7,7,6,6,7,7,6,5,5,4,4,3,2,2,3,3,3,2,2,1,1,0,0,0,1,1,0,0,0,1,1,
%U 2,3,3,2,2,3,3,2,1,1,0,0,0,1,1,0,0,0,1,1,2,2,3,3,3,2,2,3,4,5,5,4,4,4
%N Hilbert's Hamiltonian walk on N X N projected onto y axis: m'(3).
%C This is the Y-coordinate of the n-th term in the type I Hilbert's Hamiltonian walk A163359 and the X-coordinate of its transpose A163357.
%H Antti Karttunen, <a href="/A059253/b059253.txt">Table of n, a(n) for n = 0..65535</a>
%H J. Shallit, <a href="https://arxiv.org/abs/2106.01062">Hilbert's spacefilling curve described by automatic, regular, and synchronized sequences</a>, arXiv:2106.01062 [cs.FL], June 2 2021.
%H <a href="/index/Con#coordinates_2D_curves">Index entries for sequences related to coordinates of 2D curves</a>
%F Initially [m(0) = 0, m'(0) = 0]; recursion: m(2n + 1) = m(2n).m'(2n).f(m'(2n), 2n).c(m(2n), 2n + 1); m'(2n + 1) = m'(2n).f(m(2n), 2n).f(m(2n), 2n).mir(m'(2n)); m(2n) = m(2n - 1).f(m'(2n - 1), 2n - 1).f(m'(2n - 1), 2n - 1).mir(m(2n - 1)); m'(2n) = m'(2n - 1).m(2n - 1).f(m(2n - 1), 2n - 1).c(m'(2n - 1), 2n); where f(m, n) is the alphabetic morphism i := i + 2^n [example: f(0 0 1 1 2 3 3 2 2 3 3 2 1 1 0 0, 2) = 4 4 5 5 6 7 7 6 6 7 7 6 5 5 4 4]; c(m, n) is the complementation to 2^n - 1 alphabetic morphism [example: c(0 0 1 1 2 3 3 2 2 3 3 2 1 1 0 0, 3) = 7 7 6 6 5 4 4 5 5 4 4 5 6 6 7 7]; and mir(m) is the mirror operator [example: mir(0 1 1 0 0 0 1 1 2 2 3 3 3 2 2 3) = 3 2 2 3 3 3 2 2 1 1 0 0 0 1 1 0].
%F a(n) = A025581(A163358(n)) = A002262(A163360(n)) = A059905(A163356(n)).
%o (C) See A059252.
%Y See also the y-projection, m(3), A059252 as well as A163538, A163540, A163542, A059261, A059285, A163547 and A163528.
%K nonn
%O 0,9
%A Claude Lenormand (claude.lenormand(AT)free.fr), Jan 23 2001
%E Extended by _Antti Karttunen_, Aug 01 2009
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