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Search: id:A080569
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| A080569 |
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a(n) is the first number in a run of at least n successive numbers, all having exactly 3 distinct prime factors. |
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+0
3
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| 30, 230, 644, 1308, 2664, 6850, 10280, 39693, 44360, 48919, 218972, 526095, 526095, 526095, 17233173, 127890362, 29138958036, 146216247221
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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The 19th term, if it exists, is at least 1.1 * 10^12. - Fred Schneider (frederick.william.schneider(AT)gmail.com), Jan 05 2008
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
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]
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REFERENCES
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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]
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LINKS
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Carlos Rivera, Prime Puzzle 427
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EXAMPLE
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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.
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MATHEMATICA
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k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]
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PROGRAM
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(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--, ) )
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CROSSREFS
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Cf. A064708 and A064709.
Sequence in context: A076389 A156372 A064241 this_sequence A081779 A069487 A008385
Adjacent sequences: A080566 A080567 A080568 this_sequence A080570 A080571 A080572
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KEYWORD
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fini,nonn
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AUTHOR
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Randy L. Ekl (Randy.Ekl(AT)Motorola.com), Feb 21 2003
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EXTENSIONS
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Edited and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 22 2003
More terms from Don Reble (djr(AT)nk.ca), Mar 02 2003
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