Consecutive identical values of omega

[franktaw at netscape.net]

I have been looking at the question of how many consective integers 
there are with omega (the number of distinct prime factors) equal to n. 
  Trivially, for n = 0 the maximum is 1 (for 1), and for n=1, I think 
current results are sufficient to show the maximum is 4 (2 - 5).  
Checking up to 10^11, the maximum for n=2 is 8 (141 - 148, and again 
212 - 219); this is probably the maximum.  For n=3, the maximum in this 
range is 15 (found 3 times: 17233173 - 17233187, 28926285 - 28926299, 
and 83590082 - 83590096).  I don't have much confidence that this is 
actually the limit.  For n=4, I didn't even bother looking.

Perhaps somebody with a faster machine and a better factorization 
program (I used PARI) can extend these results.

Actual upper bounds are much higher.  The best easily provable upper 
bound is p_{n+1}# - 1 (the product of the first n+1 primes, minus 1).  
This is 29 for n=2, 209 for n=3.  It seems very unlikely that the value 
for n=2 is greater than 9, since any sequence of 10 consecutive 
integers must contain both a multiple of 6 and a multiple of 10; 
getting the cofactors of each to be limited to the prime divisors 
thereof seems very unlikely.
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