A215504 Prime numbers p such that p^2 - 1 has exactly one distinct prime factor other than 2 and 3.

11, 13, 19, 23, 31, 37, 47, 53, 73, 97, 107, 127, 163, 193, 257, 383, 487, 577, 863, 1153, 2593, 2917, 4373, 8747, 995327, 1492993, 1990657, 5308417, 28311553, 86093443, 6879707137, 1761205026817, 2348273369087, 5566277615617, 7421703487487, 21422803359743

Offset: 1

Author

Lei Zhou, Aug 13 2012

Keywords

• nonn
• hard

Comments

a) Only primes 2, 3, 5, 7 and 17 are known with the property that p^2 - 1 has no prime factors other than 2 and/or 3.

b) These prime numbers have p+-1 in the form of 2^m*3^n and 2^j*q^k respectively, where q is a prime number other than 2 and 3.

c) Up to the 100000000th prime number (2038074743), first 30 terms were confirmed using first mathematica program.

d) Based on comment b), the second mathematica program checked up to the 12000th term of A003586, 54 terms were found.

e) So far only 97 and 577 have their prime factor repeated.

Item b) above is false for members 31 = 2^5-1; 127 = 2^7-1; and 257 = 2^8+1. Here one of the neighbors to p is of form 2^m, and the other one is of form 2*3^j*q^k. Call such members exceptional terms. The exceptional terms constitute the only difference between this sequence and A284037. The exceptional terms are either Fermat primes (A019434) or Mersenne primes (A000668). - Jeppe Stig Nielsen, Dec 01 2020

a(71) >= 10^300. - David A. Corneth, Dec 21 2020

Examples

			For any primes less than 11, p^2 - 1 does not have factors other than 2 or 3.
11^2 - 1 = 120 = 2^3*3*5, 5 is the only prime factor other than 2 and 3, so a(1) = 11;
13^2 - 1 = 168 = 2^3*3*5, so a(2) = 13;
17^2 - 1 = 288 = 2^5*3^2, so 17 is a not a term;
19^2 - 1 = 360 = 2^3*3^2*5, so a(3) = 19.
97^2 - 1 = 9408 = 2^6 * 3 * 7^2 which shows the other prime (7 here) can have multiplicity > 1. - _David A. Corneth_, Dec 21 2020
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..70 (first 53 terms from Lei Zhou)

Crossrefs

Cf. A003586, A284037.

Programs

Mathematica

(* Method a *)
Table[While[p = Prime[i++]; Length[Delete[Delete[FactorInteger[p^2 - 1], 1], 1]] != 1]; p, {k, 1, 30}]
(* Method b *)
f[n_] := Block[{p2, p3 = 3^Range[0, Floor@ Log[3, n] + 1]}, p2 = 2^Floor[ Log[2, n/p3] + 1]; Min[ Select[ p2*p3, IntegerQ]]]; fQ[n_] := Block[{pq = n}, While[ Mod[pq, 2] == 0, pq /= 2]; While[ Mod[pq, 3] == 0, pq /= 3]; PrimeNu@ pq == 1]; k = 1; lst = {}; While[k < 10^50, If[ PrimeQ[k - 1] && fQ[k - 2], AppendTo[lst, k - 1]]; If[ PrimeQ[k + 1] && fQ[k + 2], AppendTo[lst, k + 1]]; k = f@ k]; lst (* Robert G. Wilson v, Aug 23 2012 *)