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

%I A261892 #22 Jul 19 2024 19:08:52
%S A261892 2,4,12,8,10,24,56,16,18,20,132,48,130,112,240,32,34,36,76,40,42,264,
%T A261892 552,96,100,260,324,224,290,480,992,64,66,68,140,72,74,152,1560,80,82,
%U A261892 84,516,528,450,1104,2256,192,196,200,204,520,1802,648,3080,448
%N A261892 Least multiple m of n such that n and m have no common one bit in their binary representations.
%C A261892 All terms are even.
%C A261892 a(A003714(n)) = 2*A003714(n) for any n>0.
%C A261892 a(2n) = 2*a(n) for any n>0.
%C A261892 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
%H A261892 Paul Tek, <a href="/A261892/b261892.txt">Table of n, a(n) for n = 1..16384</a>
%F A261892 a(n) = n*A261891(n) for any n>0.
%e A261892 For n=7:
%e A261892 +---+-----+---------------+-----------------+
%e A261892 | k | 7*k | 7*k in binary | Common one bits |
%e A261892 +---+-----+---------------+-----------------+
%e A261892 | 1 |   7 |           111 |             111 |
%e A261892 | 2 |  14 |          1110 |             110 |
%e A261892 | 3 |  21 |         10101 |             101 |
%e A261892 | 4 |  28 |         11100 |             100 |
%e A261892 | 5 |  35 |        100011 |              11 |
%e A261892 | 6 |  42 |        101010 |              10 |
%e A261892 | 7 |  49 |        110001 |               1 |
%e A261892 | 8 |  56 |        111000 |               0 |
%e A261892 +---+-----+---------------+-----------------+
%e A261892 Hence, a(7) = 56.
%t A261892 Table[k = 1; While[BitAnd[k n, n] != 0, k++]; k n, {n, 60}] (* _Michael De Vlieger_, Sep 06 2015 *)
%o A261892 (Perl) sub a {
%o A261892     my $n = shift;
%o A261892     my $m = $n;
%o A261892     while ($n & $m) {
%o A261892         $m += $n;
%o A261892     }
%o A261892     return $m;
%o A261892 }
%o A261892 (PARI) a(n) = {k=1; while (bitand(n, k*n), k++); k*n;} \\ _Michel Marcus_, Sep 06 2015
%o A261892 (Python)
%o A261892 from itertools import count
%o A261892 def A261892(n): return next(k for k in count(n<<1,n) if not n&k) # _Chai Wah Wu_, Jul 19 2024
%Y A261892 Cf. A003714, A261891.
%K A261892 nonn,base,look
%O A261892 1,1
%A A261892 _Paul Tek_, Sep 05 2015