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Initial segment of A045934 is identical to this sequence but in A045934 the 12th term is divisible by 3 prime factors. Is the present sequence complete?

No more terms < 3*10^8. - David Wasserman, Aug 29 2005

a(12) > 10^40, if it exists. - Giovanni Resta, May 10 2017

From David A. Corneth, May 14 2017: (Start)

We're looking for at least 6 consecutive positive integers that each have exactly two distinct prime divisors. I.e. 6 consecutive positive integers m with omega(m) = 2. Now of exactly 6 consecutive integers, exactly one of them is divisible by 6, i.e. m is of the form 2*3*k. However m has exactly 2 distinct prime divisors, so k can only have prime divisors 2 or 3. Now, suppose m ends in 6 or higher. Then one of the consecutive integers is divisible by 10 = 2*5. I.e. it's of the form 2*5*t. Then t can only have prime divisors 2 and 5. (End)

This sequence has no run of four consecutive integers, since Eggleton and MacDougall prove that there are no more than 9 consecutive integers with A001221(k) = 2. They conjecture that A007774 contains no runs of 9 consecutive integers, and has only two runs of size 8 (at 141 and 212) and two maximal runs of size 7 (at 323 and 2302); they add that the maximal run of size 6 at 91 might be the only such run, so A088983 might be complete. - Roger Eggleton via Jason Kimberley, Jul 12 2017