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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
OFFSET
1,3
COMMENTS
A161788(n) = 2^a(n) - 1. a(A161790(n)) = 1.
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:
convert(V, list); # Robert Israel, May 12 2020
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
(PARI) a(n)=forstep(k=logint(n+1, 2), 1, -1, if(n%(2^k-1)==0, return(k))) \\ Charles R Greathouse IV, Aug 25 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Leroy Quet, Jun 19 2009
EXTENSIONS
Extended by R. J. Mathar, Jun 27 2009
STATUS
approved