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A116361
Smallest k such that n XOR n*2^k = n*(2^k + 1).
9
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
0,4
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
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 *)
PROG
(PARI) a(n)=my(k); while(bitxor(n, n<<k)-n!=n<<k, k++); k \\ 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<<k)==m+n) # Chai Wah Wu, Jul 19 2024
CROSSREFS
Sequence in context: A136277 A133233 A174430 * A375571 A106796 A265743
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Feb 04 2006
EXTENSIONS
Offset corrected by Charles R Greathouse IV, Mar 07 2013
STATUS
approved