A116361 Smallest k such that n XOR n*2^k = n*(2^k + 1).

0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 4, 3, 4, 1, 1, 1, 2, 1, 1, 4, 5, 2, 2, 4, 5, 3, 5, 4, 5, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 6, 4, 4, 5, 6, 2, 2, 2, 2, 4, 6, 5, 6, 3, 6, 5, 6, 4, 6, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 2, 5, 6, 7, 1, 1, 1, 7, 1, 1, 6, 7, 4, 5, 4, 7, 5, 5, 6, 7, 2, 2, 2, 2, 2, 7, 2, 7, 4

Offset: 0

Author

Reinhard Zumkeller, Feb 04 2006

Keywords

• nonn
• easy

Comments

a(A003714(n)) <= 1;

a(A048716(n)) <= 2;

a(A115845(n)) <= 3;

a(A115847(n)) <= 4;

a(A114086(n)) <= 5;

a(A116362(n)) = n and a(m) < n for m < A116362(n).

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Programs

Mathematica

a[n_] := Module[{k}, For[k = 0, True, k++,
     If[BitXor[n, n*2^k] == n*(2^k+1), Return[k]]]];
Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Nov 19 2021 *)

PARI

a(n)=my(k);while(bitxor(n,n<Charles R Greathouse IV, Mar 07 2013

Python

from itertools import count
def A116361(n): return next(k for k in count(0) if n^(m:=n<Chai Wah Wu, Jul 19 2024

Extensions

Offset corrected by Charles R Greathouse IV, Mar 07 2013