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

1,1

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

Links

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

Formula

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

Example

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

Mathematica

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

Prog

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

Crossrefs

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

Sequence in context: A309349 A024574 A119203 * A282409 A059012 A352317

Adjacent sequences: A115400 A115401 A115402 * A115404 A115405 A115406

Keyword

nonn

Author

Zak Seidov, Mar 08 2006

Status

approved