|
%I #8 Jun 22 2015 19:03:31
%S 2,1,12,187561
%N Least number k such that A258772(k) = n.
%C If a(n) exists, a(n) > 10^6 for n > 3.
%C Excluding k = 12, for n = 2, the two fixed points in the trajectory of k occur at 29 and 34 in the section [... 58, 29, 88, 44, 22, 11, 34, 17, 52 ...]. Excluding k = 12, it appears all of the trajectories of the possible k values have length 47 or 48.
%C For n = 3, the three fixed points occur at position 215, 233, and 251 of the trajectory. It appears all of the trajectories of the possible k values have length 316.
%e T(12) = [12, 6, 3, 10, 5, 16, 8, 4, 2, 1]. The numbers 3 and 5 are in the 3rd and 5th position, respectively. Since 12 is the smallest number to have exactly two fixed points, a(2) = 12. Note that the length of this trajectory is 10. For all other trajectories with exactly 2 fixed points, the length is either 47 or 48.
%o (PARI) Tvect(n)=v=[n]; while(n!=1, if(n%2, k=3*n+1; v=concat(v, k); n=k); if(!(n%2), k=n/2; v=concat(v, k); n=k)); v
%o 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 Cf. A258772, A006370, A070165.
%K nonn,hard,more,bref
%O 0,1
%A _Derek Orr_, Jun 11 2015
|
