A342065 Primes p such that p^9 - 1 has 16 divisors.

383, 12227, 44519, 44687, 56003, 97523, 130259, 148727, 160739, 169007, 208799, 258887, 270563, 281783, 331883, 336143, 353099, 364979, 498119, 501707, 550679, 573107, 577667, 716747, 753023, 775367, 781007, 784727, 861299, 887543, 1084247, 1085159, 1099139

Offset: 1

Author

Jon E. Schoenfield, Feb 27 2021

Keywords

• nonn

Comments

Conjecture: sequence is infinite.

The only primes p such that p^9 - 1 has fewer than A309906(9)=16 divisors are p=2 (2^9 - 1 = 511 = 7*73 has 4 divisors) and p=3 (3^9 - 1 = 19682 = 2*13*757 has 8 divisors).

For every term p, p^9 - 1 is of the form 2*q*r*s, where q = (p-1)/2, r = (p^2 + p + 1), and s = (p^6 + p^3 + 1) are primes (see Example section).

The Generalized Dickson's Conjecture implies there are infinitely many p such that p, (p-1)/2, p^2+p+1 and p^6+p^3+1 are prime. - Robert Israel, Feb 28 2021

Examples

			                       factorization of p^9 - 1
    p =   ===================================================
n   a(n)  2 * (p-1)/2 * (p^2+p+1) *      (p^6 + p^3 + 1)
-  -----  ---------------------------------------------------
1    383  2 *   191   *    147073 *          3156404483062657
2  12227  2 *  6113   * 149511757 * 3341330794198073514753973
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000005, A000040, A309906, A341670.

Programs

Maple

R:= NULL: count:= 0: q:= 1:
while count < 100 do
  q:= nextprime(q);
  p:= 2*q+1;
  if isprime(p) and isprime(p^2+p+1) and isprime(p^6+p^3+1) then
    count:= count+1; R:= R, p;
  fi
od:
R; # Robert Israel, Feb 28 2021