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A048735
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a(n) = (n AND floor(n/2)), where AND is bitwise and-operator (A004198).
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14
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0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 0, 0, 0, 1, 0, 0, 2, 3, 8, 8, 8, 9, 12, 12, 14, 15, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 16, 16, 16, 17, 16, 16, 18, 19, 24, 24, 24, 25, 28, 28, 30, 31, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 0
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OFFSET
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0,7
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COMMENTS
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To prove that (n AND floor(n/2)) = (3n-(n XOR 2n))/4 (= A048728(n)/4), we first multiply both sides by 4, to get 2*(n AND 2n) = (3n - (n XOR 2n)) and then rearrange terms: 3n = (n XOR 2n) + 2*(n AND 2n), which fits perfectly to the identity A+B = (A XOR B) + 2*(A AND B) (given by Schroeppel in HAKMEM link).
The number of 1's through 4*2^n appears to yield A000045(n+1). - Ben Burns, Jun 12 2017
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LINKS
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FORMULA
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a(n) = A048728(n)/4. (This was the original definition. AND-formula found Jan 01 2007).
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MAPLE
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seq(Bits:-And(n, floor(n/2)), n=0..200); # Robert Israel, Feb 29 2016
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MATHEMATICA
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Table[BitAnd[n, Floor[n/2]], {n, 0, 127}] (* T. D. Noe, Aug 13 2012 *)
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PROG
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(Python)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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