limit of a prime sequence Let $p[n]$ be the $n$-th prime. Let $0 \leq m < k$.
Prove 
$$\lim_{n\rightarrow\infty}\frac{   p[(n+k)^2] - p[(n+m)^2] }{ p[n]} = 4(k-m)\;.$$
This is a generalization of something I looked at a while ago. I have some empirical evidence for it but cannot prove it. I think it is hard (and interesting). I do not think the PNT helps. 
 A: This would follow from unproved hypotheses on the distribution of prime numbers in short intervals. Therefore it's surely correct, but nobody knows how to prove it.
It is conjectured that the number of primes in a short interval of the form $(x,x+x^\theta)$ is asymptotic to $x^\theta/\ln x$, for any fixed $1\ge\theta>0$. (This is only known for $\theta>0.55$ or something like that, I can't remember. The Baker-Harman-Pintz result related to $\theta=0.525$ is not an asymptotic but only a lower bound.)
This short-interval conjecture is equivalent to saying that $p[n+f(n)] - p[n]$ is asymptotic to $f(n) \ln n$ for any nice function $f(n)$ satisfying $n \ge f(n) > n^\epsilon$ for some $\epsilon>0$. That in turn implies that $p[n^2+f(n)] - p[n^2+g(n)]$ is asymptotic to $(f(n)-g(n))\ln n^2$ for two such nice functions.
Your quotient is the case $f(n) = 2kn+k^2$, $g(n) = 2mn+m^2$. If we knew the short interval conjecture for some $\theta<\frac12$, say (since $f$ and $g$ are about the square root of $n^2$), it would follow that the numerator of your quotient is asymptotic to $4(k-m)n\ln n$, while the denominator is known to be asymptotic to $n\ln n$.
