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

Author

Paul Tek, Sep 05 2015

Keywords

• nonn
• base
• look

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A003714, A261891.

Programs

Mathematica

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

PARI

a(n) = {k=1; while (bitand(n, k*n), k++); k*n;} \\ Michel Marcus, Sep 06 2015

Perl

sub a {
    my $n = shift;
    my $m = $n;
    while ($n & $m) {
        $m += $n;
    }
    return $m;
}

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

Formula

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