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
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
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