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The start of a record-breaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 5.
7

%I #29 Sep 03 2022 20:01:38

%S 32,944,15470,57967,632148,14845324,69921004,888781058,2674685524,

%T 10077383364,21117216104,393370860205,3157222675953,5509463413255,

%U 24819420480104

%N The start of a record-breaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 5.

%C a(16) > 3*10^13. - _Brian Trial_, May 13 2017

%C All multiples of 32 greater than 32 are of form 2^5*m and have at least 6 factors. Thus this sequence will be limited to a run of at most 31 integers. - _Brian Trial_, May 13 2017

%e a(3)=15470 because 15470 is the start of a record breaking run of 3 consecutive integers (15470 to 15472) each having 5 prime factors; i.e. bigomega(n)=A001222(n)=5 for n = 15470, ..., 15472.

%t bigomega[n_] := Plus@@Last/@FactorInteger[n]; For[n=1; m=l=0, True, n++, If[bigomega[n]==5, l++, If[l>m, m=l; Print[n-l, " ", l]]; l=0]]

%t Table[SequencePosition[PrimeOmega[Range[15*10^6]],PadRight[{},n,5],1][[All,1]],{n,6}]//Flatten (* The program generates the first six terms of the sequence. *) (* _Harvey P. Dale_, Sep 03 2022 *)

%Y Cf. A067813, A067814, A067821, A067822.

%Y Subsequence of A014614.

%K fini,more,nonn

%O 1,1

%A _Shyam Sunder Gupta_, Feb 07 2002

%E Edited by _Dean Hickerson_, Jul 31 2002

%E More terms from _Jens Kruse Andersen_, Aug 23 2003

%E a(13)-a(14) from _Donovan Johnson_, Jan 31 2009

%E a(15) from _Brian Trial_, May 13 2017