cp's OEIS Frontend

A258819 Least number k such that A258769(k) = n.

%I A258819 #22 Oct 06 2020 02:07:16
%S A258819 2,1,156,153
%N A258819 Least number k such that A258769(k) = n.
%C A258819 If a(n) exists, a(n) > 4*10^6 for n > 3.
%C A258819 For n = 2, the two fixed points occur at 17 and 20 in the section [... 44, 22, 11, 17, 26, 13, 20, 10 ...] of the trajectory. It appears that all of the trajectories of the possible k values have length 26 or 33.
%C A258819 For n = 3, the three fixed points occur at 14, 17, and 20 in the section [... 56, 28, 14, 7, 11, 17, 26, 13, 20, 10 ...] of the trajectory. It appears that all of the trajectories of the possible k values have length 26.
%C A258819 If a(n) exists for n > 3, a(n) > 5*10^9. For n = 2, besides 26 and 33, the possible k values can also have trajectories of length 182, 294, 323, and 343. - _Jud McCranie_, Oct 05 2020
%e A258819 T(156) = [156, 78, 39, 59, 89, 134, 67, 101, 152, 76, 38, 19, 29, 44, 22, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1], In the 17th and 20th entry, there is a 17 and a 20, respectively. Since 156 is the smallest number of have exactly this, a(2) = 156. Note that T(156) has a length 26. It appears that all candidates with 2 fixed points have a trajectory with length 26 or 33.
%e A258819 T(153) = [153, 230, 115, 173, 260, 130, 65, 98, 49, 74, 37, 56, 28, 14, 7, 11, 17, 26, 13, 20, 10, 5, 8, 4, 2, 1]. In the 14th, 17th, and 20th entry, there is a 14, 17 and 20, respectively. Since 153 is the smallest number to have exactly this, a(3) = 153. Note that T(153) has length 26. It appears that all candidates with 3 fixed points have a trajectory with length 26.
%o A258819 (PARI) Tvect(n)=v=[n];while(n!=1,if(n%2,k=(3*n+1)/2;v=concat(v,k);n=k);if(!(n%2),k=n/2;v=concat(v,k);n=k));v
%o A258819 n=0;m=1;while(m<10^3,d=Tvect(m);c=0;for(i=1,#d,if(d[i]==i,c++));if(c==n,print1(m,", ");m=0;n++);m++)
%Y A258819 Cf. A258769, A014682, A070168.
%K A258819 nonn,hard,more,bref
%O A258819 0,1
%A A258819 _Derek Orr_, Jun 11 2015