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%I #18 Mar 09 2019 00:24:32
%S 0,0,1,1,2,2,1,0,3,3,2,1,1,-1,0,0,4,4,3,2,2,0,1,1,1,-2,-1,-2,0,1,0,0,
%T 5,5,4,3,3,1,2,2,2,-1,0,-1,1,2,1,1,1,-3,-2,-4,-1,-1,-2,-3,0,2,1,3,0,
%U -1,0,0,6,6,5,4,4,2,3,3,3,0,1,0,2,3,2,2,2,-2
%N The bottom entry in the difference table of the positions of the ones in the binary representation of n.
%C By convention, a(0) = 0.
%C For any n > 0:
%C - let (b_1, b_2, ..., b_h) be the positions of the ones in the binary representation of n,
%C - h = A000120(n) and 0 <= b_1 < b_2 < ... < b_h,
%C - n = Sum_{k = 1..h} 2^b_k,
%C - a(n) is the unique value remaining after taking successively the first differences of (b_1, ..., b_h) h-1 times.
%H Rémy Sigrist, <a href="/A306754/b306754.txt">Table of n, a(n) for n = 0..16384</a>
%F a(2^k) = k for any k >= 0.
%F a(2^k-1) = [k=2].
%F a(2*n) = a(n) + A209229(n).
%e For n = 59:
%e - the binary representation of 59 is "111011",
%e - so h = 5 and b_1 = 0, b_2 = 1, b_3 = 3, b_4 = 4, b_5 = 5,
%e - the corresponding difference table is:
%e 0 1 3 4 5
%e 1 2 1 1
%e 1 -1 0
%e -2 1
%e 3
%e - hence a(59) = 3.
%o (PARI) a(n) = { my (h=hammingweight(n), o=0, v=0); forstep (k=h-1, 0, -1, my (w=valuation(n, 2)); o += w; v += (-1)^k * binomial(h-1, k) * o; o++; n\=2^(1+w)); v };
%Y See A306607 for a similar sequence.
%Y Cf. A000120, A133457, A209229.
%K sign,base
%O 0,5
%A _Rémy Sigrist_, Mar 08 2019
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