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A180938
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Smallest k such that k*n has an even number of 1's in its base-2 expansion.
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3
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3, 3, 1, 3, 1, 1, 9, 3, 1, 1, 3, 1, 3, 9, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 9, 1, 1, 33, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 3, 9, 1, 1, 3, 1, 3, 33, 1, 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 1, 5, 1, 11, 3, 1, 1, 3, 3, 1, 3, 1, 1, 9, 3, 1
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OFFSET
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1,1
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COMMENTS
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k must always be odd.
First occurrence of odd k: 3, 1, 87, 109, 7, 93, 457, 1143, 5501, 7921, 889, 12775, 11753, 635, 111209, 6093, 31, 33823, 7665, ..., .
(End)
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LINKS
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EXAMPLE
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For n = 7, a(n) = 9, since the smallest multiple of 7 with an even number of 1's in its base-2 expansion is 9*7 = 63.
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MATHEMATICA
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a[n_] := Block[{k = 1}, While[OddQ@ DigitCount[k*n, 2, 1], k++ ]; k]; Array[a, 100] (* Robert G. Wilson v, Sep 29 2010 *)
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PROG
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(Python)
def a(n):
k=1
while True:
if not bin(k*n)[2:].count('1')%2: return k
k+=1
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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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