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%I #28 May 26 2020 06:27:25
%S 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,
%T 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,
%U 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
%N a(n) is the largest integer k such that 2^k - 1 divides n.
%C A161788(n) = 2^a(n) - 1. a(A161790(n)) = 1.
%C Conjecture: gcd(n, m) = a(2^n + 2^m - 2) for n > 0 and m > 0. - _Velin Yanev_, Aug 24 2017
%H Robert Israel, <a href="/A161789/b161789.txt">Table of n, a(n) for n = 1..10000</a>
%p 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
%p # Alternative:
%p N:= 200: # for a(1)..a(N)
%p V:= Vector(N,1):
%p for k from 2 to ilog2(N) do
%p t:= 2^k-1;
%p V[[seq(i,i=t..N,t)]]:= k
%p od:
%p convert(V,list); # _Robert Israel_, May 12 2020
%t 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 *)
%o (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
%Y Cf. A000225, A161788, A161790.
%K nonn,easy
%O 1,3
%A _Leroy Quet_, Jun 19 2009
%E Extended by _R. J. Mathar_, Jun 27 2009
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