|
|
A161789
|
|
a(n) is the largest integer k such that 2^k - 1 divides n.
|
|
3
|
|
|
1, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 3, 1, 4, 5, 1, 2, 1, 3, 2, 1, 1, 2, 1, 1, 3, 1, 1, 4, 1, 1, 2, 3, 1, 2, 1, 1, 2, 1, 3, 2, 1, 1, 4, 1, 5, 6, 1, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 4, 1, 3, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 4, 3, 1, 5, 1, 1, 2, 1, 3, 2, 1, 1, 2, 1, 1, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
Conjecture: gcd(n, m) = a(2^n + 2^m - 2) for n > 0 and m > 0. - Velin Yanev, Aug 24 2017
|
|
LINKS
|
|
|
MAPLE
|
A161789 := proc(n) for k from ilog2(n+1) to 0 by -1 do if n mod (2^k-1) = 0 then RETURN(k); fi; od: end: seq(A161789(n), n=1..120) ; # R. J. Mathar, Jun 27 2009
# Alternative:
N:= 200: # for a(1)..a(N)
V:= Vector(N, 1):
for k from 2 to ilog2(N) do
t:= 2^k-1;
V[[seq(i, i=t..N, t)]]:= k
od:
|
|
MATHEMATICA
|
kn[n_]:=Module[{k=Floor[Log[2, n]]+1}, While[!Divisible[n, 2^k-1], k--]; k]; Array[kn, 110] (* Harvey P. Dale, Mar 26 2012 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|