User darrell plank - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T01:53:27Z http://mathoverflow.net/feeds/user/12295 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52417/consecutive-numbers-with-n-prime-factors Consecutive numbers with n prime factors Darrell Plank 2011-01-18T17:57:46Z 2011-01-19T22:35:00Z <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-factors Comment by Darrell Plank Darrell Plank 2011-01-20T11:27:51Z 2011-01-20T11:27:51Z I 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 &quot;yesterday&quot;. 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#52432 Comment by Darrell Plank Darrell Plank 2011-01-20T06:50:30Z 2011-01-20T06:50:30Z Aaron is, strictly speaking, correct, but it's a really interesting paper anyway. Thanks for the link!