A067813
Start of a record-breaking run of consecutive integers with a number of prime factors (counted with multiplicity) equal to 3.
Original entry on oeis.org
8, 27, 170, 602, 2522, 211673
Offset: 1
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.
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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 *)
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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
A045984
a(n) = smallest number m such that factorizations of n consecutive integers starting at m have same number of primes (counted with multiplicity).
Original entry on oeis.org
1, 2, 33, 602, 602, 2522, 211673, 3405122, 3405122, 49799889, 202536181, 3195380868, 5208143601, 85843948321, 97524222465
Offset: 1
a(4) = 602 as 602 = 2 * 7 * 43, 603 = 3 * 3 * 67, 604 = 2 * 2 * 151, 605 = 5 * 11 * 11 so four consecutive positive integers have the same number of prime factors starting at 602, the first such number. - _David A. Corneth_, Feb 24 2024
A374449
Triangle read by rows: T(m,k) is the first number that starts a sequence of exactly k consecutive numbers with m prime factors, counted with multiplicity, if such a sequence is possible.
Original entry on oeis.org
5, 2, 4, 9, 33, 8, 27, 170, 1083, 602, 2522, 211673, 16, 135, 1274, 4023, 12122, 204323, 355923, 6612470, 3405122, 49799889, 202536181, 3195380868, 5208143601
Offset: 1
Triangle starts
5 2
4 9 33
8 27 170 1083 603 3533 211673
T(3,2) = 27 because 27 = 3^3 and 28 = 2^2 * 7 each have 3 prime factors (counted with multiplicity) while 26 = 2*13 and 29 (prime) do not.
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f:= proc(n)
uses priqueue;
local V,L, count, T, v, j, q, p, TP;
V:= Vector(2^n-1); count:= 0;
L:= [(-1)$(2^n),2^n];
initialize(pq);
insert([-2^(n),2$n],pq);
while count < 2^n-1 do
T:= extract(pq); v:= -T[1];
if L[-1] <> v-1 then
for j from 1 while L[-1]-L[-j] = j-1 do
if L[-j]-L[-j-1] <> 1 and V[j] = 0 then
V[j]:= L[-j]; count:= count+1;
fi od fi;
L:= [op(L[2..-1]),v];
q:= T[-1];
p:= nextprime(q);
for j from n+1 to 2 by -1 do
if T[j] <> q then break fi;
TP:= [T[1]*(p/q)^(n+2-j), op(T[2..j-1]), p$(n+2-j)];
insert(TP,pq);
od od;
op(convert(V,list));
end proc:
f(1):= 5,2:
seq(f(i),i=1..3);
Showing 1-3 of 3 results.
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