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A116361
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Smallest k such that n XOR n*2^k = n*(2^k + 1).
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9
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0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 4, 3, 4, 1, 1, 1, 2, 1, 1, 4, 5, 2, 2, 4, 5, 3, 5, 4, 5, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 6, 4, 4, 5, 6, 2, 2, 2, 2, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 5, 6, 7, 1, 1, 1, 7, 1, 1, 6, 7, 4, 5, 4, 7, 5, 5, 6, 7, 2, 2, 2, 2, 2, 7, 2, 7, 4
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OFFSET
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0,4
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COMMENTS
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LINKS
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MATHEMATICA
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a[n_] := Module[{k}, For[k = 0, True, k++,
If[BitXor[n, n*2^k] == n*(2^k+1), Return[k]]]];
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PROG
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(Python)
from itertools import count
def A116361(n): return next(k for k in count(0) if n^(m:=n<<k)==m+n) # Chai Wah Wu, Jul 19 2024
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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