OFFSET
1,1
COMMENTS
Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - Charles R Greathouse IV, Sep 14 2015
See A321503 for numbers n such that n & n+1 have at least 3 prime divisors, disjoint union of this and A321493, the terms of A321503 which are not in this sequence. A321493 has A140078 as a subsequence, which in turn is subsequence of A321504, and so on. Since n and n+1 can't share a prime factor, we have a(1) > sqrt(p(3+3)#) > A000196(A002110(3+3)). Note that A000196(A002110(3+4)) = A321493(1) exactly! - M. F. Hasler, Nov 13 2018
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..10000
D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim, Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.
FORMULA
MATHEMATICA
a = {}; Do[If[Length[FactorInteger[n]] == 3 && Length[FactorInteger[n + 1]] == 3, AppendTo[a, n]], {n, 1, 100000}]; a (*Artur Jasinski*)
SequencePosition[PrimeNu[Range[1250]], {3, 3}][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 27 2017 *)
PROG
(PARI) is(n)=omega(n)==3&&omega(n+1)==3 \\ Charles R Greathouse IV, Sep 14 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, May 07 2008
STATUS
approved