A140077 Numbers n such that n and n+1 have 3 distinct prime factors.

230, 285, 429, 434, 455, 494, 560, 594, 609, 615, 644, 645, 650, 665, 740, 741, 759, 804, 805, 819, 825, 854, 860, 884, 902, 935, 945, 969, 986, 987, 1001, 1014, 1022, 1034, 1035, 1044, 1064, 1065, 1070, 1085, 1104, 1105, 1130, 1196, 1209, 1220, 1221

Offset: 1

Author

Artur Jasinski, May 07 2008

Keywords

• nonn

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.

Crossrefs

Cf. A074851, A140078, A140079.

Equals A321503 \ A321493.

Programs

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 *)

PARI

is(n)=omega(n)==3&&omega(n+1)==3 \\ Charles R Greathouse IV, Sep 14 2015

Formula

{k: k in A033992 and k+1 in A033992}. - R. J. Mathar, Jul 19 2023