What is the best unconditional upper bound for $p\_{n^2}-p\_{(n-1)^2}$ such that $p\_n$ is the $n$-th prime number?


Asymptotics suggest it's somewhere near $4 n \ln n$, but how to prove this?


Edit: it's known that there is a prime between $[x,x+x^{13/23}]$, so one can bound $p\_{n^2}-p\_{(n-1)^2} \leq 2 n n^{13/23}$, but are there any better bounds?



Upper bound for $p_{n^2} - p_{(n-1)^2}$?
By the results of 


*Baker, R.C.; Harman, G.; Pintz, J.*, [**The difference between consecutive primes. II**](http://dx.doi.org/10.1112/plms/83.3.532), Proc. Lond. Math. Soc., III. Ser. 83, No.3, 532-562 (2001). [ZBL1016.11037](https://zbmath.org/?q=an:1016.11037).


the number of primes in the interval $[x, x+x^{0.525}]$ is $\gg x^{0.525}/\log x$ for $x$ large enough (see the final inequality of the paper); in particular, $p\_{(n-1)^2} + p\_{(n-1)^2}^{0.525} \geq p\_{n^2}$ for $n$ large enough. This gives a bound of the form $O( n^{1.05} \log^{0.525} n )$. Any improvement on this bound would likely improve the Baker-Harman-Pintz bound on large prime gaps, so I doubt one can do much better using what is in the literature. Of course, the situation is much better on RH or even on LH (the Lindelöf Hypothesis). (Also, one can show the expected bound of $O(n \log n)$ for almost all $n$, with a fairly small exceptional set, using the known results on primes in short intervals on average.)


