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.
[Math] Consecutive numbers with n prime factors
factorizationnt.number-theory
Best Answer
$P(4,1)$ 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.