Theorem 3.2 in <http://arxiv.org/pdf/1405.2593.pdf> shows that for any $x$ there are $\gg x\exp(-\sqrt{\log x})$ integers $x\_0 \in [x; 2x]$ such that $\pi(x\_0 + \log x) - \pi(x\_0) \gg \log\log x$. 


Is there an upper bound for number of such $x\_0$? I think it must be $<x(\log x)^{-c}$ for any $C$.


UPD: It is interesting to find such upper bound for *prime* numbers $x\_0$ such that $\pi(x\_0 + \log x) - \pi(x\_0) \gg \log\log x$



Upper bound for number of prime numbers in a range
If $x\_0<x$ satisfies that $[x\_0, x\_0+\log x]$ contains $\log\log x$ primes, then for a parameter $r$ we have that this interval contains $\binom{\log\log x}{r}$ different $r$-tuples $p, p+d\_1, p+d\_2, \ldots, p+d\_{r-1}$ of primes, such that $0<d\_1<\dots<d\_{r-1}\leq\log x$. The number of possible choices for $d\_1, \ldots, d\_{r-1}$ is $\binom{\log x}{r-1}$. Apply Selberg's sieve to each of them, and take the sum over all tuples. This will lead to some lengthy computation involving singular series, but on average the singular series will be of magnitude $\mathcal{O}(1)$. From this you obtain that the number of short $r$-tuples is bounded above by $$\ll C(r)\binom{\log x}{r-1} \frac{x}{\log^r x} \ll \frac{C(r)}{r!}\frac{x}{\log x},$$ where $C(r)$ is the coefficient of Selberg's sieve. Each tuple belongs to $\leq\log x$ values of $x\_0$, hence 
comparing with the $\binom{\log\log x}{r}$ tuples produced by a single $x\_0$ and restricting $x\_0$ you obtain that there are $\ll C(r)\frac{x}{(\log\log x)^r}$ possibilities for $x\_0$. Optimizing for $r$ should give an upper bound of magnitude $\frac{x}{(\log x)^c}$ for some $c>0$, which falls somewhat short of your expectation. 


Addendum: Since $r$ is large, it is better to use the large sieve in place of Selberg's sieve. The details will become more complicated, but the results should be better. For a model you can look at the proof of Lemma 2 in Elsholtz, On cluster primes, Acta Arith. 109, 281-284.


