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A104234
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Number of k >= 1 such that k+n == 0 mod 2^k.
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5
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0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 1, 2, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1
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OFFSET
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0,6
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COMMENTS
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Number of terms in the summation in the formula for A102370(n).
Also, a(n) is the number of 1's in (A103185(n) written in base 2).
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LINKS
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David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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FORMULA
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a(2^k + y) = a(y) + 1 if y = 2^k - k - 1, = a(y) otherwise (where 0 <= y <= 2^k - 1).
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MAPLE
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f:=proc(n) local t1, l; t1:=0; for l from 1 to n do if n+l mod 2^l = 0 then t1:=t1+1; fi; od: t1; end;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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