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%I #6 May 05 2014 15:02:12
%S 0,1,1,0,1,2,0,1,1,0,2,1,0,1,1,0,1,2,0,1,2,0,1,1,0,1,1,0,1,1,0,1,1,0,
%T 2,1,0,1,1,0,2,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,2,0,1,
%U 2,0,1,1,0,1,1,0,1,1,0,1,2,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1
%N Fix e = 3; a(n) = minimal Hamming distance between the binary representation of n and the binary representation of any multiple ke (0 <= k <= n/e) which is a child of n.
%C A number m is a child of n if the binary representation of n has a 1 in every position where the binary representation of m has a 1.
%C In other words, this tells us how closely (in Hamming weight) we can approximate n "from below" by a multiple of e.
%H Nadia Heninger and N. J. A. Sloane, <a href="/A140080/b140080.txt">Table of n, a(n) for n = 0..5000</a>
%H N. J. A. Sloane, <a href="/A140080/a140080.f.txt">Fortran program for this and related sequences</a>
%e If n = 14 = 1110_2, take k=2, ke = 6 = 110_2, which is Hamming distance 1 from n. This is the best we can do, so a(14) = 1.
%o (Fortran) See Sloane link.
%Y For e=2 and 4 through 9 see A000035 and A140081 through A140086.
%Y Cf. A140137, A140138, A140200-A140206.
%K nonn
%O 0,6
%A _Nadia Heninger_ and _N. J. A. Sloane_, Jun 03 2008
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