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

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)
    return len(reach)
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Jun 29 2022

Crossrefs

Cf. A020575, A306035, A316226.

Sequence in context: A363704 A195477 A157680 * A347199 A011228 A175572

Adjacent sequences: A305379 A305380 A305381 * A305383 A305384 A305385

Keyword

nonn

Author

N. J. A. Sloane, Jun 17 2018

Status

approved