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A115403
Numbers k such that k^3+1 is 3-almost prime (product of three primes).
3
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
KEYWORD
nonn
AUTHOR
Zak Seidov, Mar 08 2006
STATUS
approved