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
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's 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.  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
Eggleton and MacDougall prove that no terms exist beyond a(9) and conjecture that a(9) does not exist.  Jason Kimberley, Jul 08 2017
