A386915 Numbers k such that k^3 - 1 is a triprime.

4, 5, 15, 27, 32, 42, 44, 48, 50, 59, 60, 66, 72, 75, 78, 84, 98, 104, 108, 114, 119, 132, 140, 143, 147, 152, 162, 167, 174, 180, 182, 188, 200, 203, 206, 212, 215, 218, 224, 228, 234, 236, 240, 252, 258, 264, 266, 270, 279, 288, 290, 294, 308, 318, 336, 338, 342, 350, 374, 378, 383, 384, 390

Offset

1,1

Comments

Numbers k such that either k-1 is prime and k^2 + k + 1 is a semiprime, or k-1 is a semiprime and k^2 + k + 1 is prime.

If k is odd, k-1 = 2*p for a prime p such that 4*p^2 + 6*p + 3 is prime. The Generalized Bunyakovsky conjecture implies that there are infinitely many of these.

The Generalized Bunyakovsky conjecture also implies that there are infinitely many j such that 14*j + 3, 28*j^2 + 18*j + 3, 7*j + 2 and 196*j^2 + 154*j + 31 are all prime. This implies that both k = 14*j + 4 and k + 1 are terms of the sequence.

There are no k where k, k + 1 and k + 2 are all terms of the sequence, since there are no terms == 1 (mod 3) except 4 (if k == 1 (mod 3), then k^3 == 1 (mod 9)).

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

a(3) = 15 is a term because 15^3 - 1 = 3374 = 2 * 7 * 241 is the product of three primes.

Maple

select(t -> numtheory:-bigomega(t^3-1)=3, [$1..1000]);

Mathematica

Select[Range[400], PrimeOmega[#^3 - 1] == 3 &] (* Amiram Eldar, Aug 08 2025 *)

Prog

(PARI) isok(k) = bigomega(k^3-1) == 3; \\ Michel Marcus, Aug 08 2025

Crossrefs

Cf. A001358, A014612, A068601, A115403.

Sequence in context: A026634 A026656 A184244 * A006491 A321174 A304921

Adjacent sequences: A386912 A386913 A386914 * A386916 A386917 A386918

Keyword

nonn

Author

Robert Israel, Aug 07 2025

Status

approved