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

1,1

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

Links

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

Example

                       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

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

Crossrefs

Cf. A000005, A000040, A309906, A341670.

Sequence in context: A064721 A252060 A046014 * A046491 A046492 A110296

Adjacent sequences: A342062 A342063 A342064 * A342066 A342067 A342068

Keyword

nonn

Author

Jon E. Schoenfield, Feb 27 2021

Status

approved