The sequence is now https://oeis.org/A305382 Comments welcomed. 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 10:24 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Yup, I get the same thing as Edwin.
Among the first 1M primes, the largest number of distinct primes that ever appear when doing Jim's expansion is 37, a maximum first attained by 9453569 (the 630705th prime). The set of primes that appear, closed under the factors-of-2p+1 operation, is {3, 5, 7, 11, 13, 17, 19, 23, 29, 41, 43, 47, 53, 59, 67, 71, 79, 83, 107, 167, 179, 359, 443, 719, 809, 887, 1163, 1439, 1619, 1889, 2833, 2879, 3779, 7559, 15581, 23371, 9453569}.
For Neil, the first 100 terms of the sequence of total-number-of-primes-seen is: 9, 8, 8, 8, 8, 8, 9, 8, 8, 11, 9, 9, 12, 12, 8, 14, 10, 13, 9, 9, 9, 15, 11, 17, 9, 12, 9, 13, 10, 10, 10, 12, 9, 10, 9, 13, 9, 11, 10, 12, 16, 9, 12, 13, 16, 9, 9, 10, 9, 10, 11, 11, 9, 16, 10, 11, 9, 10, 10, 10, 9, 10, 13, 18, 9, 11, 10, 9, 11, 12, 13, 15, 9, 12, 9, 11, 13, 15, 10, 9, 11, 11, 11, 10, 11, 11, 13, 14, 10, 10, 10, 10, 9, 12, 10, 15, 17, 10, 13, 9
Quickie mma code to compute this stuff:
propp1[p_] := propp1[p] = #[[1]]& /@ FactorInteger[2*p+1] propp[p_Integer] := propp[{p}] propp[s_List] := propp[s, Union[s, Union @@ propp1 /@ s]] propp[s_, t_] := If[s==t, s, If[Length[t]>1000, OVERFLOW[t], propp[t]]]
Then propp[9453569] is the length=37 set of primes above, and Table[Length[propp[Prime[n]]], {n,100}] is the sequence.
--Michael
On Sat, Jun 16, 2018 at 9:57 PM W. Edwin Clark <wclark@mail.usf.edu> wrote:
If my program is correct, starting with the n-th prime (n<=10^5) the procedure always ends with {3, 5, 7, 11, 13, 19, 23, 47} which is sequence A020575: (In particular the number of primes obtained will always be finite.)
Example starting with the the 10000th prime:
{104729} {209459} {251, 1669} {3, 7, 53, 503} {3, 5, 7, 19, 53, 107} {3, 5, 7, 11, 13, 43, 107} {3, 5, 7, 11, 23, 29, 43} {3, 5, 7, 11, 23, 29, 47, 59} {3, 5, 7, 11, 17, 19, 23, 47, 59} {3, 5, 7, 11, 13, 17, 19, 23, 47} {3, 5, 7, 11, 13, 19, 23, 47}
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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