A067813 Start of a record-breaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 3.

8, 27, 170, 602, 2522, 211673

Offset

1,1

Comments

602 is the first number having 4 and 5 consecutive integers with 3 prime factors. - T. D. Noe, Mar 19 2014

Links

Table of n, a(n) for n=1..6.

Example

a(4)=602 because 602 is the start of a record breaking run of 5 consecutive integers (602 to 606) each having 3 prime factors; i.e. bigomega(n)=A001222(n)=3 for n = 602, ..., 606.

Mathematica

bigomega[n_] := Plus@@Last/@FactorInteger[n]; For[n=1; m=l=0, True, n++, If[bigomega[n]==3, l++, If[l>m, m=l; Print[n-l, " ", l]]; l=0]]
Module[{nn=8, po}, po=PrimeOmega[Range[5000000]]; Flatten[Table[ SequencePosition[ po, PadRight[{}, n, 3], 1], {n, nn}], 1]][[All, 1]]//Union (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 14 2019 *)

Prog

(PARI) show(lim)=my(was, r, ct); forfactored(n=2, lim\1+1, is=vecsum(n[2][, 2])==3; if(is, ct++; if(ct>r, r=ct; print(r" "n[1]-r+1)), ct=0)) \\ Charles R Greathouse IV, Jun 26 2019

Crossrefs

Cf. A007304, A045984, A067814, A067820, A067821, A067822, A117969, A259756.

Sequence in context: A343318 A165048 A066963 * A304769 A118720 A197620

Adjacent sequences: A067810 A067811 A067812 * A067814 A067815 A067816

Keyword

fini,full,nonn

Author

Shyam Sunder Gupta, Feb 07 2002

Extensions

Edited by Dean Hickerson, Jul 31 2002

Status

approved