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A008687
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Number of 1's in 2's complement representation of -n.
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17
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0, 1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 7, 6, 6, 5, 6, 5, 5, 4, 6, 5, 5, 4, 5, 4, 4, 3, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3
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OFFSET
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0,4
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COMMENTS
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Conjecture: a(n)+1 is the length of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
Terms a(n); n >= 2 can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, let S(k) = {S(k-1)+1, S(k-1)}, where +1 means +1 on every term of S(k-1); see Example. Each step of the recursion gives the next 2^k terms of the sequence. - David James Sycamore, Jul 15 2024
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LINKS
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FORMULA
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a(n) = if n<=1 then n else (n mod 2) + a((n mod 2) + floor(n/2)). - Reinhard Zumkeller, Feb 05 2007
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EXAMPLE
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Using the above recursion for a(n); n >= 2, we have:
S(0) = {1} so a(2) = 1;
S(1) = {2,1} so a(3,4) = 2,1;
S(2) = {3,2,2,1}, so a(5,6,7,8) = 3,2,2,1;
As irregular table the sequence for n >= 2 begins:
1;
2,1;
3,2,2,1;
4,3,3,2,3,2,2,1;
5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
6,5,5,4,5,4,4,3,5,4,3,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
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MATHEMATICA
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PROG
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(Haskell)
a008687 n = a008687_list !! n
a008687_list = 0 : 1 : c [1] where c (e:es) = e : c (es ++ [e+1, e])
(PARI) a(n) = if(n<2, n, n--; logint(n, 2) - hammingweight(n) + 2); \\ Kevin Ryde, Apr 14 2021
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CROSSREFS
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This is Guy Steele's sequence GS(4, 3) (see A135416).
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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