Given $n \in \mathbb{N}$, let $\pi(n)$ denote the number of prime numbers $\leq n$. What is $$ \limsup_{m \rightarrow \infty} \left( \limsup_{n \rightarrow \infty} \frac{\pi(n+m) - \pi(n)}{\pi(m)} \right)? $$ If needed, answers may be conditional under the assumption that a suitable generalization of the Bunyakovsky conjecture holds.

As observed by Hensley and Richards in

*Douglas Hensley and Ian Richards*, **Primes in intervals**, *Acta Arith.* **25** (1973-74), 375--391,

if the prime tuples conjecture is true, then $\limsup_{n \to \infty} \pi(n+m) - \pi(n) = \rho^*(m)$, where $\rho^*(m)$ is the length $k$ of the largest admissible $k$-tuple of diameter less than $m$ (a $k$-tuple is admissible if it avoids a residue class modulo $p$ for every prime $p$). So your quantity would then be equal to $\limsup_{m \to \infty} \rho^*(m)/\pi(m)$, which is also easily seen to be equal to $\limsup_{k \to\infty} k \log k / H(k)$, where $H(k)$ is the minimal diameter of an admissible $k$-tuple.

The best known upper and lower bounds for $H(k)$ can be found in Section 3 of the (unabridged version of) the Polymath8a paper. The best asymptotic upper bound known is $k \log k + k \log\log k - (1+2 \log 2) k + o(k)$ (due to Schinzel), and the best asymptotic lower bound is a bit tricky to state (coming from the sharp Montgomery-Vaughan version of the large sieve) but is roughly of the form $\frac{1}{2} k \log k - o(k \log k )$. So the best that is known about your quantity is (as was pointed out in comments) it is somewhere between $1$ and $2$. (One does not need the best results on $H(k)$ for these bounds: the lower bound comes from observing that the first $k$ primes larger than $k$ are automatically admissible, while the upper bound follows from (the proof of) the Brun-Titchmarsh inequality.) Based on the known numerical values and bounds for $H(k)$ (see this online database, as well as Figures 2 and 3 of the Polymath8a paper referenced above) one can tentatively conjecture that your quantity is equal to $1$, but the evidence is not terribly strong for this conjecture yet. Improving the upper bound of $2$ would be a major breakthrough in sieve theory (closely related to the parity problem), and is one of the motivations of the nascent field of "inverse sieve theory", discussed in

*Ben Green and Adam J. Harper*, **Inverse questions for the large sieve**, *Geom. Funct. Anal.* **24** (2014), no. 4, 1167--1203.

The upper bound of $2$ is unconditional (following directly from Brun-Titchmarsh). No non-trivial unconditional lower bound (beyond the trivial bound of $0$) is known; the best result is by Maynard in

*James Maynard*, **Small gaps between primes**, *Ann. of Math. (2)* **181** (2015), no. 1, 383--413

with slightly better implied constants obtained by the Polymath8b project

*D. H. J. Polymath*, **Variants of the Selberg sieve, and bounded intervals containing many primes**, *Res. Math. Sci.* **1** (2014), Art. 12, 83,

basically showing that $\limsup_{n \to \infty} \pi(n+m) - \pi(n) \gg \log m$. This is known to be the limit of the existing Selberg sieve method to unconditionally produce many primes in short intervals; getting beyond the $\log m$ bound here would be a significant breakthrough (though I think breaking the parity upper bound of $2$ mentioned earlier would be even more revolutionary).

Note that until the famous recent breakthrough of Zhang in

*Yitang Zhang*, **Bounded gaps between primes**, *Ann. of Math. (2)* **179** (2014), no. 3, 1121--1174.

it was not even known unconditionally that $\limsup_{n \to \infty} \pi(n+m) - \pi(n)$ was larger than $1$ for any $m$.

EDIT: Ben Green pointed out to me an observation of Selberg in Vol. I of his "Collected works" (page 610 in the new edition) that one cannot obtain effective improvements to the $1/2$ constant in the lower bound $H(k) \geq \frac{1}{2} k \log k - o(k \log k)$ without also getting effective bounds in Siegel's theorem, since a Siegel zero implies the existence of a medium-sized arithmetic progression with about twice as many primes as one would expect without the zero. This leads to an admissible k-tuple that is also about twice as dense as expected. So the upper bound of 2 in the original question is unlikely to be improved without some sort of breakthrough on the Siegel zero problem (or else some massive extension of the repulsion phenomenon).

shortis of course relative here. $\endgroup$ – Stefan Kohl Apr 6 '16 at 12:44under the assumptionof the $k$-tuple conjecture? $\endgroup$ – Stefan Kohl Apr 6 '16 at 12:52