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A342065
Primes p such that p^9 - 1 has 16 divisors.
1
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
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
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Feb 27 2021
STATUS
approved