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%I #21 Jul 19 2023 07:52:33
%S 230,285,429,434,455,494,560,594,609,615,644,645,650,665,740,741,759,
%T 804,805,819,825,854,860,884,902,935,945,969,986,987,1001,1014,1022,
%U 1034,1035,1044,1064,1065,1070,1085,1104,1105,1130,1196,1209,1220,1221
%N Numbers n such that n and n+1 have 3 distinct prime factors.
%C Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - _Charles R Greathouse IV_, Sep 14 2015
%C 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
%H Harvey P. Dale, <a href="/A140077/b140077.txt">Table of n, a(n) for n = 1..10000</a>
%H D. A. Goldston, S. W. Graham, J. Pintz, C. Y. Yildirim, <a href="http://arxiv.org/abs/0803.2636">Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers</a>, arXiv:0803.2636 [math.NT], 2008.
%F {k: k in A033992 and k+1 in A033992}. - _R. J. Mathar_, Jul 19 2023
%t a = {}; Do[If[Length[FactorInteger[n]] == 3 && Length[FactorInteger[n + 1]] == 3, AppendTo[a, n]], {n, 1, 100000}]; a (*Artur Jasinski*)
%t SequencePosition[PrimeNu[Range[1250]],{3,3}][[All,1]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Feb 27 2017 *)
%o (PARI) is(n)=omega(n)==3&&omega(n+1)==3 \\ _Charles R Greathouse IV_, Sep 14 2015
%Y Cf. A074851, A140078, A140079.
%Y Equals A321503 \ A321493.
%K nonn
%O 1,1
%A _Artur Jasinski_, May 07 2008
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