A161789 a(n) is the largest integer k such that 2^k - 1 divides n.

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

The sums of the first 10^k terms, for k = 1, 2, ..., are 15, 183, 1898, 19219, 192464, 1924900, 19249110, 192491275, 1924913468, 19249135108, ... . Apparently, the asymptotic mean of this sequence is 1.924913... . - Amiram Eldar, Jun 30 2025

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Formula

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

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

Cf. A000225, A161788, A161790.

Sequence in context: A180262 A395698 A394540 * A109671 A141289 A368878

Adjacent sequences: A161786 A161787 A161788 * A161790 A161791 A161792

Keyword

nonn,easy

Author

Leroy Quet, Jun 19 2009

Extensions

Extended by R. J. Mathar, Jun 27 2009

Status

approved