OFFSET
1,1
COMMENTS
If 4p - 1 is prime then n^2 + n + p = p(4p - 1) for some n = 1, 2, 3, ... [Proof. Let n + 1 = 2p, etc.]
From Alonso del Arte, Jan 14 2024: (Start)
The first six terms correspond to rings of algebraic integers of Q(sqrt(-a(n))) which are unique factorization domains.
In the ring of algebraic integers of Q(sqrt(-a(n))), the corresponding prime p = (a(n) + 1)/4 is divisible by 1/2 - sqrt(-a(n))/2 and 1/2 + sqrt(-a(n))/2, both of those being algebraic integers with minimal polynomial x^2 - x + p. For example, in Q(sqrt(-163)), we see that (1/2 - sqrt(-163)/2)(1/2 + sqrt(-163)/2) = 1/4 + 163/4 = 41, with both of the divisors having the minimal polynomial x^2 - x + 41. (End)
FORMULA
a(n) >> n log^2 n. - Charles R Greathouse IV, Jun 14 2022
MATHEMATICA
Select[Prime[Range[1000]], PrimeQ[(# + 1)/4] &] (* Alonso del Arte, Jan 14 2024 *)
PROG
(PARI) is(n)=isprime(n) && isprime(4*n-1) \\ Charles R Greathouse IV, Jun 14 2022
(Scala) def isPrime(num: Int): Boolean = Math.abs(num) match {
case 0 => false; case 1 => false; case n => (2 to Math.floor(Math.sqrt(n)).toInt) forall (p => n % p != 0)
}
((1 to 2500).map(4 * _ - 1)).filter(n => isPrime(n) && isPrime((n + 1)/4)) // Alonso del Arte, Jan 14 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Daniel Tisdale, Jul 14 2009
EXTENSIONS
More terms from N. J. A. Sloane, Jul 19 2009
STATUS
approved