The given terms up to a(8) = 141 are the only terms less than 10^18. To speed the search, note that any string of 6 or more consecutive numbers contains a multiple of 6 and hence must contain a number of the form 2^a * 3^b. Conjecture: 141 is the last term, because numbers with only two different prime factors get pretty rare, so having several in a row near a number of the form 2^a * 3^b is pretty unlikely.  Joshua Zucker, May 05 2006
Sequence cannot have any terms for n > 29, since a run of 30 or more consecutive numbers must contain a multiple of 30, divisible by at least 3 primes.  Franklin T. AdamsWatters, Oct 23 2006
I searched numbers of the form n=2^a * 3^b through 10^700 and could not find any solution where even 4 numbers (n+2, n2, n+3, n3) had omega=2. The last such number through 10^700 is only 169075682574336=2^33 * 3^9. So a full set of 9 numbers seems quite unlikely  Fred Schneider, Jan 05 2008
Comments from Vim Wenders, Apr 02 2008: (Start) The sequence is complete. The argument of Franklin T. AdamsWatters is easily extended: if 2^a.3^b, a,b, >=1 is a term then omega(2^a.3^b+6) > 2 (because the exponents of 2 and 3 follow a ruler like sequence). So the last possible term would be a(11).
Also, if 2.p, p prime, is in the run of an initial value to check, then p+2, p+4, ... has to be prime too, (for the values 2p+4=2(p+2),2p+8=2(p+4) ...), which is impossible for obvious reason.
The two arguments limit the maximum length of a run to 8. (End)
Wenders' argument is incomplete because the consecutive even numbers can have the form 2^a p^b. As stated in the paper by Eggleton and MacDougall, it is still a conjecture that 9 consecutive omega2 numbers do not exist. [From T. D. Noe, Oct 13 2008]
For a(10) to exist, one of the consecutive terms must be in A033846. Also, the sequence cannot have any terms for n > 14. If a(15) exists, one term has to be of the form A = 2^n*5^m. Then, there also must be two other terms divisible by 5, excluding A. Call them B and C.
Case 1: If A is the smallest of these terms, then B = 5*(2^n*5^(m1)+1) and C = 10*(2^(n1)*5^(m1)+1). However, for large enough values of m and n, either B/5 or C/10 is divisible by 3 and another prime > 5.
Case 2: If A is the middle term, then B = 5*(2^n*5^(m1)1) and C = 5*(2^n*5^(m1)+1). For large enough values of m and n, either B/5 or C/5 is a multiple of 3 and another prime > 5.
Case 3: If A is the highest term, B = 5*(2^n*5^(m1)1) and C = 10*(2^(n1)*5^(m1)1). Again, for large enough values of m and n, either B/5 or C/10 is a multiple of 3 and another prime > 5.
Thus, if there are more terms in the sequence, they can only be a(9), a(10), a(11), a(12), a(13) or a(14).
Other comments: The only numbers that satisfy a(8) are 141 and 212. The reason a(9) wasn't satisfied in either of these is because one end of the run of numbers was a prime and the other end stopped at a multiple of 10. I believe it is possible to show that a(10) can never exist because there cannot be a multiple of 10 in the run of consecutive numbers, perhaps because there cannot be two multiples of 5.  Derek Orr, May 24 2014
