[seqfan] Re: Collatz-like problem with prime iterations

[franktaw at netscape.net]

The sequence is almost certainly unbounded. Just as there are believed 
to be arbitrarily long sequences of primes with p_(n+1) = 2*p_n + 1 
(the same for 2*p_n - 1), there are almost certainly arbitrarily long 
sequences of primes with p_(n+1) = (3*p_n + 1) / 2.

Franklin T. Adams-Watters

-----Original Message-----
From: Vladimir Shevelev <shevelev at bgu.ac.il>

 Let P(n) be the maximal prime divisor of 3*n+1. For a given n, 
consider
iterations p_1=P(n), p_2=P(p_1),

..., p_(k+1)=P(p_(k)),... I believe that, for every n, some iterate 
equals 2,
when we stop the process.
E.g., for n=27, we have the following primes: 
41,31,47,71,107,23,7,11,17,13,5,2.
To every n>=1corresponds a set of primes.
The sequence of  products of primes over these sets begins with  2, 
170170, 10,
130, 2, 13394810, 24310,10,170170, etc.
Let t=t(n) be the smallest number of iterates such that P^t(n)<n 
(n>=3). The
first numbers t(n) are t(3)=2, t(4)=3, t(5)=1, t(6)=6, t(7)=4, t(8)=1, 
t(9)=1,
t(10)=6,...(I submitted A198724).
       Is this sequence bounded? Even if the answer is in negative, the 
sequence
of places  of records of this sequence, most likely,  grows very fast.

Regards,
Vladimir



 Shevelev Vladimir‎

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