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

%I #14 Aug 11 2024 07:13:14

%S 3,9,10,12,13,21,25,30,34,36,40,46,52,66,76,81,90,96,118,120,126,130,

%T 132,142,144,154,165,168,172,177,180,193,196,198,204,216,226,228,238,

%U 240,246,250,256,262,268,273,282,288,294,312,333,336,345,346,366,370

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

%C 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

%H Charles R Greathouse IV, <a href="/A115403/b115403.txt">Table of n, a(n) for n = 1..10000</a>

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

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

%t Select[Range[370],PrimeOmega[#^3+1]==3&] (* _James C. McMahon_, Aug 10 2024 *)

%o (PARI) isok(n) = bigomega(n^3+1) == 3; \\ _Michel Marcus_, Oct 10 2013

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

%K nonn

%O 1,1

%A _Zak Seidov_, Mar 08 2006