User darrell plank - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-25T01:53:27Zhttp://mathoverflow.net/feeds/user/12295http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/52417/consecutive-numbers-with-n-prime-factorsConsecutive numbers with n prime factorsDarrell Plank2011-01-18T17:57:46Z2011-01-19T22:35:00Z
<p>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.</p>
http://mathoverflow.net/questions/52417/consecutive-numbers-with-n-prime-factorsComment by Darrell PlankDarrell Plank2011-01-20T11:27:51Z2011-01-20T11:27:51ZI gave the answer to Aaron though it's hard to pick. I'm not sure what order the answers come out here - they're all just marked "yesterday". I thought the primorial was a good general point. Anyway, thanks to all for some thought provoking insights!http://mathoverflow.net/questions/52417/consecutive-numbers-with-n-prime-factors/52432#52432Comment by Darrell PlankDarrell Plank2011-01-20T06:50:30Z2011-01-20T06:50:30ZAaron is, strictly speaking, correct, but it's a really interesting paper anyway. Thanks for the link!