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%I #30 Oct 19 2017 03:14:12
%S 30,230,644,1308,2664,6850,10280,39693,44360,48919,218972,526095,
%T 526095,526095,17233173,127890362,29138958036,146216247221
%N a(n) is the first number in the first run of at least n successive numbers, all having exactly 3 distinct prime factors.
%C The 19th term, if it exists, is at least 1.1 * 10^12. - _Fred Schneider_, Jan 05 2008
%C 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_, Jan 05 2008
%C Eggleton and MacDougall show that there are no more than 59 terms in this sequence. [From _T. D. Noe_, Oct 13 2008]
%C a(19) > 10^13. - _Donovan Johnson_, Jun 11 2013
%C a(19) <= 7523987244435061. - _Donovan Johnson_, Jul 08 2013
%H Roger B. Eggleton and James A. MacDougall, <a href="http://www.jstor.org/stable/27643119">Consecutive integers with equally many principal divisors</a>, Math. Mag. 81 (2008), 235-248.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_427.htm">Prime Puzzle 427</a>
%e 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 k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]
%o (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 Cf. A064708, A064709, A185032, A048932.
%K fini,nonn
%O 1,1
%A _Randy L. Ekl_, Feb 21 2003
%E Edited and extended by _Robert G. Wilson v_, Feb 22 2003
%E More terms from _Don Reble_, Mar 02 2003
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