A305382 - OEIS

A305382

a(n) is the number of distinct primes produced by starting with the n-th prime p and repeatedly 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; a(n) = -1 if this produces infinitely many primes.

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

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OFFSET

1,1

COMMENTS

Based on a question posed by James Propp. Terms computed by Michael Kleber.

W. Edwin Clark observes (Jun 16 2018) that, based on analysis of the first 10^5 primes, the procedure always ends with {3, 5, 7, 11, 13, 19, 23, 47}, which is sequence A020575. In particular, it appears that the total number of primes obtained is always finite.

First occurrences are in A316226.

REFERENCES

James Propp, Posting to Math Fun Mailing List, Jun 16 2018

LINKS

Hans Havermann, Table of n, a(n) for n = 1..10000

EXAMPLE

a(1)=9: Starting with the first prime, 2, we see that:

2 -> 5 -> 11 -> 23 -> 47 -> 95=5*19,

19 -> 39=3*13,

3 -> 7 -> 15=3*5,

13 -> 27=3*3*3,

which produces 9 different primes, 2 3 5 7 11 13 19 23 47.

MATHEMATICA

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]]];

Table[Length[propp[Prime[n]]], {n, 100}] (* Michael Kleber, Jun 16 2018 *)

g[lst_List] := Union@ Join[lst, First@# & /@ Flatten[FactorInteger[2 lst + 1], 1]]; f[n_] := Length@ NestWhile[g@# &, {Prime@ n}, UnsameQ, All]; Table[ f[n], {n, 100}] (* Robert G. Wilson v, Jun 17 2018 *)

PROG

(Python)

from sympy import prime, primefactors

def a(n):

pn = prime(n)

reach, expand = {pn}, [pn]

while len(expand) > 0:

p = expand.pop()

for q in primefactors(2*p+1):

if q not in reach:

expand.append(q)

reach.add(q)