

A067813


Start of a recordbreaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 3.


11




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[nl, " ", l]]; l=0]]


CROSSREFS

Cf. A067814, A067820, A067821, A067822.
Sequence in context: A056570 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



