This question is inspired by [Project Euler's Problem 47](http://projecteuler.net/index.php?section=problems&id=47).


Let *m* and *n* be positive integers. Consider the following four (related) statements:


1. There exists *m* consecutive integers, each of which has at least *n* distinct prime factors.
2. There exists *m* consecutive integers, each of which has precisely *n* distinct prime factors.
3. There exists *m* consecutive integers, each of which has at least *n* prime factors (counted with multiplicity)
4. There exists *m* consecutive integers, each of which has precisely *n* prime factors (counted with multiplicity).


**Question** $i$: For which pairs $(m,n)$ is statement $i$ true?


For $n = 2$ and any $m$, statement 3 is obviously true (it is just the elementary fact that there are prime gaps of arbitrary length). I suppose it is not too hard to avoid prime powers, so that statement 1. is also true for $n=2$ and arbitrary $m$ (although at the moment I cannot see if this follows by an easy modification of the usual $(m+1)!+2, \ldots, (m+1)!+m+1$ argument). Are there other elementary cases? In particular, is there an easy proof that Problem 47 (which is to find the first instance of Statement 1 for $m=n=4$) is actually solvable? The actual answer is small enough that it is found rather quickly by a naive brute force approach, but it would be nice to know in advance that there is an answer (perhaps even having an upper bound, so that one also knows that one may find the answer before the heat death of the universe).


Statement $i$ can be strengthened by requiring the existence of infinitely many $m$-tuples; call this Statement $i'$.


Statements 2 and 4 are probably too hard to say anything about in general (but maybe not?), but for $m=2$, Statements $2'$ and $4'$ are both weaker versions of [this question](https://mathoverflow.net/questions/32412/question-on-consecutive-integers-with-similar-prime-factorizations).



Consecutive integers with many prime factors
Answer for the first question. Let $w\_k$ be the product $p\_{nk+1} p\_{nk+2} \cdots p\_{nk+n}$ where $p\_i$ is $i$-th prime. So each $w\_k$ is equal to product of $n$ distinct primes and $w\_k$ and $w\_q$ are coprime for $k \ne q$. Now take $t$ such that $t \equiv -k \pmod {w\_k}$ for $k = 1, 2, \cdots, m$. Such $t$ exists by Chinese Remainder Theorem. Now $t + k$ has at least $n$ primes because $t + k$ is a multiple of $w\_k$. Hence numbers $t + 1, t + 2, \cdots, t + m$ satisfy requirements in the first question.


