Consecutive numbers with n prime factors Let $P(m,n)$ mean that there is a number, $M$, such that starting with $M$ there are $m$ consecutive numbers each having exactly $n$ distinct prime factors.  Is it obvious that $P(m,n)$ is true for all $m$ and $n$?  My gut says "obviously" and $P(4,4)$ and $P(5,5)$ are definitely true (for 134043 and 129963314 respectively).  It seems like some sort of pigeonhole proof based on the number of factors available might work, but upon reflection, I'm not so sure.  Maybe I'm missing something obvious.
 A: Tony Forbes' paper (and its references) is a good place to start.
Fifteen Consecutive Integers with Exactly Four Prime Factors
A: It would be interesting to know what generalization might be true about the starting numbers of such sequence, or at least the smallest such.  
For instance, $P(7,2)$ and $P(8,2)$ are both true at 141 (and 212), so the smallest such number for only 7 is 323.  The next smallest is 2302 (also for 7 only), and there are no others under a million.
Sorry for putting this in an 'answer'; it seems odd to me that one needs additional rep to put in a comment.  Though I think that bounds on such numbers would be quite interesting.
A: Check out the related question Happy New Prime Year! .  I have some code posted there which tracks constant sequences as well as increasing and decreasing sequences. (Check out 2302 to 2308.)
A comment made by someone else and then deleted contained the observation that multiples m of the nth primorial had s(m) >= n, so that runs of values less than n must have length less than the nth primorial.  Also, if you look at multiples of 6, you get that s(m)=2 for at most 11 consecutive values instead of at most 29, so there is room for improvement in
the upper bound to such lengths.
Gerhard "Reduce, Reuse, Recycle for Rep" Paseman, 2011.01.18
A: $P(4,1)$  is true because of $2,3,4,5$ but I strongly doubt that $P(5,1)$ is true. Indeed $P(4,1)$ is only true that once and even $P(3,1)$ happens for the last time at $7,8,9$.
Of any $210$ consecutive integers one is divisible by $2,3,5,7$ so $P(210,3)$ fails. I'm sure that can be improved. It does generalize .
As noted in the comments, there are two cases (at least) of $8$ consecutive integers each with two distinct prime factors and a run of $12$ can be ruled out. Actually it seems extremely unlikely that there are any cases of $9$, let alone $10$ or $11$: Such a run of $9$ would have to include $7$ integers $3(a-1),2(b-1),*,2^i3^j,*,2(b+1),3(a+1)$ where  $a\pm1=2^{i}3^{j-1}\pm1$ and $b\pm1=2^{i-1}3^{j}\pm1$ are two pairs of twin primes (or merely twin prime powers). I would expect only finitely many cases where $2^u3^v\pm1$ are both prime powers (I find $35$ such up to $10^{50}$ with $u,v  \ge 1$, let alone getting two such pairs as above ( I find $4$ cases, $36,144,216$ and $2^{33}3^9$.) This leaves out the requirements that we would also need $2^{i}3^{j}\pm1$ to each have two distinct prime factors (ruling out the smallest and largest) and have at least one of $2^{i}3^{j}\pm4$ to be 4 times a prime power, and more.
