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.

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I might be completely wrong on this, but isn't it a consequence of the PNT that $p(n)\sim n\ln(n)$? Could this not give an easy proof? –  Olivier Bégassat Oct 27 '11 at 21:57
@SivaramAmbikasaran I don't understand, $p(n)$ must grow quicker than $n$, and thus quicker than $\frac{n}{\ln(n)}$ –  Olivier Bégassat Oct 27 '11 at 22:01
@SivaramAmbikasaran You must have meant $\pi(n)$ :) –  Olivier Bégassat Oct 27 '11 at 22:02
@Oliver: Right. I thought the op meant $\pi(n)$. Sorry for the previous comment. –  user17762 Oct 27 '11 at 22:03
@Olivier: I don't think it's that easy. You need to cancel terms with $n^2$ in them, leaving terms with $n$, but the asymptotic you quote doesn't exclude terms of order $\sqrt n$ that might contribute. See here for improvements on that asymptotic result, but I don't think any of those entail this result here, since they all allow deviations of order $\sqrt n$. This seems to be a shorter-range correlation that may not be derivable if you treat the two terms in the numerator as independent. –  joriki Oct 27 '11 at 22:03

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 $n\ln n$.
I was wondering about the $0.525$ result: Wikipedia has this as an asymptotic result, but the paper only contains the lower bound; the same for Huxley's result. Perhaps you (or someone else knowledgeable enough) could clean up that section? –  joriki Oct 28 '11 at 0:41
P.S.: It seems the asymptotic claims in that section are correct up to and including the result by Ingham. The $~0.55$ result you refer to doesn't seem to be mentioned there at all; the unconditional asymptotic results only go down to $0.75$, and Ingham's conditional result is said to yield $0.625$. –  joriki Oct 28 '11 at 0:57