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

Author

Leroy Quet, Jun 19 2009

Keywords

• nonn
• easy

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

Crossrefs

Cf. A000225, A161788, A161790.

Programs

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 *)

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

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

Extensions

Extended by R. J. Mathar, Jun 27 2009