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%I #7 Jun 22 2015 19:03:55
%S 1,2,105,305
%N Least number k such that A258825(k) = n.
%C If it exists, a(n) > 10^6 for n > 3.
%C For n = 2, after 17 and 20 iterations, you arrive at 17 and 20, respectively. It appears the total number of iterations of the possible k values is either 26 or 33.
%C For n = 3, after 14, 17, and 20 iterations, you arrive at 14, 17, and 20, respectively. It appears the total number of iterations of the possible k values is 26.
%e For n = 105, the Collatz function does the following: 105 -> 158 -> 79 -> 119 -> 179 -> 269 -> 404 -> 202 -> 101 -> 152 -> 76 -> 38 -> 19 -> 29 -> 44 -> 22 -> 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1. After the 17th and 20th iteration, we can see we reach 17 and 20, respectively. Since 105 is the smallest number to have exactly two occurrences, a(2) = 105. Note that there are 26 iterations before you reach 1. It appears that all numbers with exactly two occurrences have either 26 or 33 total iterations to get to 1.
%e For n = 305, the Collatz function does the following: 305 -> 458 -> 229 -> 344 -> 172 -> 86 -> 43 -> 65 -> 98 -> 49 -> 74 -> 37 -> 56 -> 28 -> 14 -> 7 -> 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1. After the 14th, 17th, and 20th iteration, we can see we reach 14, 17, and 20, respectively. Since 305 is the smallest number to have exactly 3 occurrences, a(3) = 305. Note that there are 26 iterations before you reach 1. It appears that all numbers with exactly three occurrences have 26 total iterations to get to 1.
%o (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 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. A258819, A258824, A258825, A014682, A070168.
%K nonn,more,hard,bref
%O 0,2
%A _Derek Orr_, Jun 11 2015
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