A115403 Numbers k such that k^3+1 is 3-almost prime (product of three primes).

3, 9, 10, 12, 13, 21, 25, 30, 34, 36, 40, 46, 52, 66, 76, 81, 90, 96, 118, 120, 126, 130, 132, 142, 144, 154, 165, 168, 172, 177, 180, 193, 196, 198, 204, 216, 226, 228, 238, 240, 246, 250, 256, 262, 268, 273, 282, 288, 294, 312, 333, 336, 345, 346, 366, 370

Offset: 1

Author

Zak Seidov, Mar 08 2006

Keywords

• nonn

Comments

It appears that there is only one known example of three consecutive primes p, q, r whose product is 1 more than a perfect cube, namely 7*11*13 = 1001 and that probably no other examples exist. - N. J. A. Sloane, Apr 27 2008

Examples

			9 is a member because 9^3+1=730=2*5*73 (product of three primes).
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001093, A014612 (3-almost primes).

Programs

Mathematica

Select[Range[370],PrimeOmega[#^3+1]==3&] (* James C. McMahon, Aug 10 2024 *)

PARI

isok(n) = bigomega(n^3+1) == 3; \\ Michel Marcus, Oct 10 2013

Formula

k^3+1=p*q*r where p, q, r are primes (not necessarily distinct).