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A258824 Least number k such that A258822(k) = n. 2

%I #8 Jun 22 2015 19:03:42

%S 1,2,24,63105

%N Least number k such that A258822(k) = n.

%C If a(n) exists, a(n) > 10^6 for n > 3.

%C Excluding k = 24, for n = 2, after 29 and 34 iterations, you arrive at 29 and 34, respectively. Excluding k = 24, it appears all of the trajectories of the possible k values have length 48 or 49.

%C For n = 3, after 216, 234, and 252 iterations, you arrive at 216, 234, and 252, respectively. It appears all of the trajectories of the possible k values have length 317.

%e For n = 24, the '3x+1' map is as follows: 24 -> 12 -> 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. After the 3rd iteration, we reach 3 and after the 5 iteration, we reach 5. Since 12 is the smallest number to have exactly two occurrences, a(2) = 24. Note that the length of this trajectory is 11. For all other trajectories with exactly two occurrences, the length is either 48 or 49.

%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-1, c++)); if(c==n, print1(m, ", "); m=0; n++); m++)

%Y Cf. A258822, A006370, A070165.

%K nonn,hard,more,bref

%O 0,2

%A _Derek Orr_, Jun 11 2015

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