%I #14 May 13 2013 01:54:22

%S 1,0,2,1,0,0,3,2,0,1,1,0,0,0,3,2,0,1,2,0,0,1,2,0,1,1,0,0,0,0,3,2,0,1,

%T 2,0,0,1,3,0,1,2,0,0,0,1,3,0,1,2,0,0,2,1,0,1,1,0,0,0,0,0,3,2,0,1,2,0,

%U 0,1

%N The number of steps it takes, by taking the smallest number so that it minus the number of ones in its binary representation is the previous number, to reach a number which has no others that satisfy this property.

%C A218252 is divided into groups with length the numbers with odd position in this sequence.

%H Charles R Greathouse IV, <a href="/A218253/b218253.txt">Table of n, a(n) for n = 1..10000</a>

%e For n=7, the numbers you get are 7,8,10,12, and because this takes 3 steps, the 7th term is 3.

%o (PARI) f(n)=for(k=n+1,n+log(n)\log(2)+1,if(k-hammingweight(k)==n,return(k)))

%o a(n)=my(k);while(n=f(n),k++);k \\ _Charles R Greathouse IV_, Oct 29 2012

%Y Cf. A218252, A218254.

%K nonn,easy

%O 1,3

%A _Nico Brown_, Oct 24 2012