%I A080569
%S A080569 30,230,644,1308,2664,6850,10280,39693,44360,48919,218972,526095,526095,
%T A080569 526095,17233173,127890362,29138958036,146216247221
%N A080569 a(n) is the first number in a run of at least n successive numbers, all
having exactly 3 distinct prime factors.
%C A080569 The 19th term, if it exists, is at least 1.1 * 10^12. - Fred Schneider
(frederick.william.schneider(AT)gmail.com), Jan 05 2008
%C A080569 There can be at most 209 terms in this sequence. Any list of 210 consecutive
numbers must contain a number n which is multiple of 2*3*5*7 = 210.
So omega(n) would be >3. - Fred Schneider (frederick.william.schneider(AT)gmail.com),
Jan 05 2008
%C A080569 Eggleton and MacDougall show that there are no more than 59 terms in
this sequence. [From T. D. Noe (noe(AT)sspectra.com), Oct 13 2008]
%D A080569 Roger B. Eggleton and James A. MacDougall, Consecutive integers with
equally many principal divisors, Math. Mag. 81 (2008), 235-248. [From
T. D. Noe (noe(AT)sspectra.com), Oct 13 2008]
%H A080569 Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_427.htm">
Prime Puzzle 427</a>
%e A080569 a(3) = 644 because 644 = 2^2 * 7 * 23, so omega(644) = 3, 645 = 3*5*43,
so omega(645) = 3 and 646 = 2*17*19, so omega(646) = 3 and no other
number n < 644 has omega(n)=omega(n+1)=omega(n+2)=3.
%t A080569 k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k +
n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]
%o A080569 (PARI) k=1; for(i=1,600000,s=1; for(j=1,k,if(omega(i+j-1)!=3,s=0,));
if(s==1,print1(i,", "); k++; i--,) )
%Y A080569 Cf. A064708 and A064709.
%Y A080569 Sequence in context: A076389 A156372 A064241 this_sequence A081779 A069487
A008385
%Y A080569 Adjacent sequences: A080566 A080567 A080568 this_sequence A080570 A080571
A080572
%K A080569 fini,nonn
%O A080569 1,1
%A A080569 Randy L. Ekl (Randy.Ekl(AT)Motorola.com), Feb 21 2003
%E A080569 Edited and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 22
2003
%E A080569 More terms from Don Reble (djr(AT)nk.ca), Mar 02 2003
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