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Numbers n such that k iterations of n under the modified Collatz function yield k for some k.
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%I #6 Jun 18 2015 14:52:09

%S 2,6,7,8,17,20,21,22,23,24,36,37,38,43,48,52,53,64,68,69,70,75,86,87,

%T 89,97,98,99,105,111,112,116,117,120,122,130,131,132,133,134,137,160,

%U 169,192,208,212,213,226,227,242,243,260,261,262,264,266,268,269,273,288,290,291,292,293,294,296,298,299,305

%N Numbers n such that k iterations of n under the modified Collatz function yield k for some k.

%C Numbers n such that A258825(n) > 0.

%e For n = 5, the Collatz function does the following: 5 -> 8 -> 4 -> 2 -> 1. Here, for k = 1, 2, 3, 4, applying k iterations to 5 does not yield k. So 5 is not a member of this sequence.

%e For n = 6, the Collatz function does the following: 6 -> 3 -> 5 -> 8 -> 4 -> 2 -> 1. After the 4th iteration, you arrive at 4. So 6 is a member of this sequence.

%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=1; while(n<10^3, d=Tvect(n); c=0; for(i=1, #d, if(d[i]==i-1, print1(n, ", "); break)); n++)

%Y Cf. A258825, A066861, A014682, A070168.

%K nonn

%O 1,1

%A _Derek Orr_, Jun 11 2015