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The standard proof (or I think that´s a standard one) that we can have as much as we want consecutive composite numbers is to observe the numbers $n!+2,...,n!+n$, which are $n-1$ consecutive composite numbers which also means that there are arbitrarily large gaps between two consecutive primes.

So for, for example, $n=5$ we have that $122,123,124,125$ are four consecutive composites since $5!=120$.

But although this theorem shows us the existence of as much as we want consecutive composite numbers, it does not show us, not even within a good approximation, when the first such set of $k>1$ numbers occur?

Because, very much before $122,123,124,125$ we have four consecutive composites $24,25,26,27$.

Are there some results and/or conjectures about when do we have the first $k>1$ consecutive composite numbers?

Are there some estimates and conjectured functional dependencies of the form $c_k(p_k)$, where $c_k$ is the "first occurence of $k>1$ consecutive composite numbers" function, dependent on the $k$-th prime $p_k$?

What are the main results "in this direction"?

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    To begin with, you can replace $n!$ with the product of all primes $\le n$. This still finds four consecutuve composites beginning at $2\cdot3\cdot 5+2=32$, but it is certainly a lot better. -- Next, except for small $n$, you can subtract instead of add $2,3,\ldots, n$ (which takes us to $25$ for the example of four composites. -- next you can subtract mor ethan $n$, up to some $n'>n$ such that $n'+1$ is the next prime $>n$; this tells us tha we are lucky and get five instead of four composites (and starting at $24$) – Hagen von Eitzen Oct 22 '19 at 21:50
  • @HagenvonEitzen Do you expect that there are some nice regularities in the "first occurence of k>1 consecutive composite numbers" function as a function of k? –  Oct 22 '19 at 22:00

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Let $[k]_n = \{k, k+1, ..., k+n-1\}$ and let $\mathbb{P}$ denote the set of primes. Define $f(n) = \min\{k \in \mathbb{N} : [k]_{n} \cap \mathbb{P} = \varnothing\}$.

Consider that the prime number theorem says that $\frac{n}{\log n}(1+o(1))$ of the positive integers $\leq n$ are prime. By the pigeonhole principle, this implies that we should have a string of $\log n(1-o(1))$ consecutive composites $\leq n$. Therefore, we can expect that the first string of $k$ consecutive composites occurs somewhere in $\left[1, \exp\left(\frac{k}{1-o(1)}\right)\right]$. In other words, $f(n) \leq \exp\left(\frac{k}{1-o(1)}\right)$. Since $\exp\left(\frac{k}{1-o(1)}\right) = o(k!)$, this is certainly better than the usual $k!+2, k!+3...$ argument.

However, this is not the best we can do. It's for example known that $\limsup_{n \to \infty} \frac{p_{n+1} - p_n}{\log p_n} = \infty$, and hence (per the PNT) $\limsup_{n \to \infty} \frac{p_{n+1} - p_n}{\log p_{n+1}} = \infty$. From this, it's not hard to see that for any $M>0$, there are infinitely many $k$ such that $f(\lceil M\log p_{k+1} \rceil) \leq p_{k+1}$. In other words, for any $M>0$, there are infinitely many integers $r$ such that $f(r) \leq \exp\left(\frac{r}{M}\right) $.

The current best known result of this flavor is that there is an absolute constant $C>0$ such that $$p_{n+1} - p_n > \frac{C \log n \log \log n \log \log \log \log n}{\log \log \log n}$$

MathematicsStudent1122
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