Let a(n) = number of distinct primes produced by starting with the n-th prime p and iterating Jim's process of looking at all the prime factors of 2p+1, and then performing the same process (double, add 1, find all prime factors) with those primes. Let a(n) = -1 if this produces infinitely many primes. Jim's example, starting with 2, seems to produce nine primes, 2 3 5 7 11 13 19 23 47, so a(1)=9. How does this sequence begin? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sat, Jun 16, 2018 at 6:14 PM, James Propp <jamespropp@gmail.com> wrote:
Given any prime p, we can look at all the prime factors of 2p+1, and then perform the same process (double, add 1, find all prime factors) with those primes, and so on; must the process eventually stabilize?
E.g.: 2 -> 5 -> 11 -> 23 -> 47 -> 95=5*19 19 -> 39=3*13 3 -> 7 -> 15=3*5 13 -> 27=3*3*3 Done.
I looked at a variant where one uses 2p-1 instead of 2p+1; it appears to lead to closure a lot more rapidly. For the 2p-1 version, I was able to check in my head that all primes under 100 lead to closure; for the 2p+1 version, I wasn’t able to do this.
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