A261892 Least multiple m of n such that n and m have no common one bit in their binary representations.

2, 4, 12, 8, 10, 24, 56, 16, 18, 20, 132, 48, 130, 112, 240, 32, 34, 36, 76, 40, 42, 264, 552, 96, 100, 260, 324, 224, 290, 480, 992, 64, 66, 68, 140, 72, 74, 152, 1560, 80, 82, 84, 516, 528, 450, 1104, 2256, 192, 196, 200, 204, 520, 1802, 648, 3080, 448

Offset

1,1

Comments

All terms are even.

a(A003714(n)) = 2*A003714(n) for any n>0.

a(2n) = 2*a(n) for any n>0.

Without the condition on m being a multiple of n, then we'd get another sequence b(n) that seems to be A006519(n+1). - Michel Marcus, Sep 06 2015

Links

Paul Tek, Table of n, a(n) for n = 1..16384

Formula

a(n) = n*A261891(n) for any n>0.

Example

For n=7:
+---+-----+---------------+-----------------+
| k | 7*k | 7*k in binary | Common one bits |
+---+-----+---------------+-----------------+
| 1 |   7 |           111 |             111 |
| 2 |  14 |          1110 |             110 |
| 3 |  21 |         10101 |             101 |
| 4 |  28 |         11100 |             100 |
| 5 |  35 |        100011 |              11 |
| 6 |  42 |        101010 |              10 |
| 7 |  49 |        110001 |               1 |
| 8 |  56 |        111000 |               0 |
+---+-----+---------------+-----------------+
Hence, a(7) = 56.

Mathematica

Table[k = 1; While[BitAnd[k n, n] != 0, k++]; k n, {n, 60}] (* Michael De Vlieger, Sep 06 2015 *)

Prog

(Perl) sub a {
    my $n = shift;
    my $m = $n;
    while ($n & $m) {
        $m += $n;
    }
    return $m;
}
(PARI) a(n) = {k=1; while (bitand(n, k*n), k++); k*n; } \\ Michel Marcus, Sep 06 2015
(Python)
from itertools import count
def A261892(n): return next(k for k in count(n<<1, n) if not n&k) # Chai Wah Wu, Jul 19 2024

Crossrefs

Cf. A003714, A261891.

Sequence in context: A338118 A294063 A290096 * A278225 A186118 A352993

Adjacent sequences: A261889 A261890 A261891 * A261893 A261894 A261895

Keyword

nonn,base,look

Author

Paul Tek, Sep 05 2015

Status

approved