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, 23671413563491, 36966736685739

Offset: 1

Author

Randy L. Ekl, Feb 21 2003

Keywords

• fini
• nonn

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

a(21) > 2 * 10^15, if it exists. - Toshitaka Suzuki, Jun 23 2025

Examples

			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.
		

Links

• 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. Runs of consecutive numbers such that... (I), The Prime Puzzles & Problems Connection.

Crossrefs

Cf. A064708, A064709, A185032, A048932.

Programs

Mathematica

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

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--,) )

Extensions

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

More terms from Don Reble, Mar 02 2003

a(19)-a(20) from Toshitaka Suzuki, Apr 01 2025