Proving prime gaps not monotonic

Question

I want to prove (or see a proof) that the prime gap function is not monotonic. (but not using extremely difficult theory like Zhangs proof please)

The idea for proof is to first suppose it is monotonic and find the slowest possible asymptotic growth it could have.

The get a contradiction by showing that even this is still "too fast" compared to say the chebychev bound $x/\log(x)$.

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Lemma: A sequence of primes with gap $d$ can't be longer than $d+2$.

So we suppose the prime gap function $g(n)$ grows as slow as possible, like this:

1, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 6, ...

Then I found that we would have $\pi(\sum g(n)) = n$, so $\pi(\sum_{d=2k} (d+2)^2) = \sum_{d=2k} d+2$. The first sum being $O(n^3)$ and the second $O(n^2)$. Inserting $n = x^{1/3}$ gives $\pi(x) = O(n^{2/3})$.

This gives us $x/x^{1/3}$ which sadly doesn't beat $x/\log(x)$.

Is this proof strategy doomed then? Or is there a way to repair it? Alternatively does anyone have an accessable reference which has this proof?

Answer 1

This gives us $x/x^{1/3}$ which sadly doesn't beat $x/\log(x)$.

Is this proof strategy doomed then? Or is there a way to repair it?

There is no problem as the inequality holds the other way round. One should combine a hypothetical upper bound of order $x/x^{1/3}$ on the prime counting function, obtained under the assumption of monotone gaps, with a lower bound on the prime counting function to get a contradiction.

The point is if the gaps would grow monotonically they would be large so it is natural there would be too few primes then.

Answer 2

Henry's observation solves the problem relatively simply.

A gap of length $d$ cannot repeat more than $p$ times where $p$ is the smallest prime not dividing $d$.

Let $f(d)$ be any increasing upper bound function for maximum number of consecutive gaps of length $d$. By the previous paragraph, $f$ can be taken to be $O((\log d)^c)$. For a logarithmic choice of $f$ let $g$ be its sum, the function with $g(n+1) - g(n) = f(n)$. Then asymptotically $g(n) \sim nf(n)$.

An increasing integer sequence $P_n$ whose gap lengths are bounded by $f$ has a subsequence $P_{g(n)}$ with strictly increasing differences and thus bounded below by a quadratic function $\frac{n(n+1)}{2}$. Composing this with $g^{-1}(n) \sim \frac{n}{f(n)}$ we get a nearly-quadratic lower bound on $P_n$ which is far larger than the $n$-th prime for large $n$.