cp's OEIS Frontend

A096303 Number of iterations of n -> n + (number of 1's in binary representation of n) needed for the trajectory of n to join the trajectory of A010062.

%I A096303 #13 Apr 05 2020 15:07:53
%S A096303 0,0,0,1,0,4,0,3,2,0,1,0,2,0,1,1,0,2,0,1,6,0,2,5,0,4,1,0,3,2,0,3,2,1,
%T A096303 1,0,5,0,2,4,0,3,1,0,2,7,0,7,1,6,1,0,5,3,0,2,4,2,1,0,3,1,6,0,0,2,0,1,
%U A096303 5,0,2,4,0,3,1,0,2,5,0,5,1,4,1,0,3,10,0,2,2,9,1,0,1,8,1,0,8,0,7,7,0,6,6,6,0
%N A096303 Number of iterations of n -> n + (number of 1's in binary representation of n) needed for the trajectory of n to join the trajectory of A010062.
%C A096303 Conjecture: For any positive integer starting value n, iterations of n -> n + (number of 1's in binary representation of n) will eventually join A010062.
%H A096303 Rémy Sigrist, <a href="/A096303/b096303.txt">Table of n, a(n) for n = 1..10000</a>
%H A096303 <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>
%e A096303 a(6)=4 because the trajectory for 1 (sequence A010062) starts
%e A096303 1->2->3->5->7->10->12->14->17->19->22->25...
%e A096303 and the trajectory for 6 starts
%e A096303 6->8->9->11->14->17->19->22->25->28->31->36...
%e A096303 so the sequence beginning with 6 joins A010062 after 4 steps.
%o A096303 (PARI) a(n) = { my (o=1); for (k=0, oo, while (o<n, o=o+hammingweight(o)); if (o==n, return (k), n=n+hammingweight(n))) } \\ _Rémy Sigrist_, Apr 05 2020
%Y A096303 Cf. A010062.
%Y A096303 For records see A229743, A229744.
%K A096303 base,nonn
%O A096303 1,6
%A A096303 _Jason Earls_, Jun 25 2004