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A261891
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Least k>0 such that n AND (k*n) = 0, where AND stands for the binary AND operator.
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4
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2, 2, 4, 2, 2, 4, 8, 2, 2, 2, 12, 4, 10, 8, 16, 2, 2, 2, 4, 2, 2, 12, 24, 4, 4, 10, 12, 8, 10, 16, 32, 2, 2, 2, 4, 2, 2, 4, 40, 2, 2, 2, 12, 12, 10, 24, 48, 4, 4, 4, 4, 10, 34, 12, 56, 8, 18, 10, 12, 16, 42, 32, 64, 2, 2, 2, 4, 2, 2, 4, 8, 2, 2, 2, 12, 4, 10
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OFFSET
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1,1
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COMMENTS
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All terms are even.
a(2n) = a(n) for any n>0.
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LINKS
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EXAMPLE
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For n=7:
+---+-------------+
| k | 7 AND (k*7) |
| | (in binary) |
+---+-------------+
| 1 | 111 |
| 2 | 110 |
| 3 | 101 |
| 4 | 100 |
| 5 | 11 |
| 6 | 10 |
| 7 | 1 |
| 8 | 0 |
+---+-------------+
Hence, a(7) = 8.
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MATHEMATICA
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Table[k = 1; While[BitAnd[k n, n] != 0, k++]; k, {n, 60}] (* Michael De Vlieger, Sep 06 2015 *)
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PROG
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(Perl) sub a {
my $n = shift;
my $k = 1;
while ($n & ($k*$n)) {
$k++;
}
return $k;
}
(PARI) a(n) = {k=1; while (bitand(n, k*n), k++); k; } \\ Michel Marcus, Sep 06 2015
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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