TITLE: Consecutive numbers with n prime factors
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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