A080569 a(n) is the first number in the first run of at least n successive numbers, all having exactly 3 distinct prime factors.

30, 230, 644, 1308, 2664, 6850, 10280, 39693, 44360, 48919, 218972, 526095, 526095, 526095, 17233173, 127890362, 29138958036, 146216247221

Offset

1,1

Comments

The 19th term, if it exists, is at least 1.1 * 10^12. - Fred Schneider, 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, Jan 05 2008

Eggleton and MacDougall show that there are no more than 59 terms in this sequence. [From T. D. Noe, Oct 13 2008]

a(19) > 10^13. - Donovan Johnson, Jun 11 2013

a(19) <= 7523987244435061. - Donovan Johnson, Jul 08 2013

Links

Table of n, a(n) for n=1..18.

Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248.

Carlos Rivera, Prime Puzzle 427

Example

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.

Mathematica

k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]

Prog

(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--, ) )

Crossrefs

Cf. A064708, A064709, A185032, A048932.

Sequence in context: A210101 A156372 A064241 * A185032 A291466 A246892

Adjacent sequences: A080566 A080567 A080568 * A080570 A080571 A080572

Keyword

fini,nonn

Author

Randy L. Ekl, Feb 21 2003

Extensions

Edited and extended by Robert G. Wilson v, Feb 22 2003

More terms from Don Reble, Mar 02 2003

Status

approved