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%I #19 Feb 24 2024 14:25:11
%S 8,27,170,602,2522,211673
%N Start of a record-breaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 3.
%C 602 is the first number having 4 and 5 consecutive integers with 3 prime factors. - _T. D. Noe_, Mar 19 2014
%e 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.
%t 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]]
%t 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 *)
%o (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
%Y Cf. A007304, A045984, A067814, A067820, A067821, A067822, A117969, A259756.
%K fini,full,nonn
%O 1,1
%A _Shyam Sunder Gupta_, Feb 07 2002
%E Edited by _Dean Hickerson_, Jul 31 2002