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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?2011-10-27
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    @SivaramAmbikasaran I don't understand, $p(n)$ must grow quicker than $n$, and thus quicker than $\frac{n}{\ln(n)}$2011-10-27
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    @SivaramAmbikasaran You must have meant $\pi(n)$ :)2011-10-27
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    @Oliver: Right. I thought the op meant $\pi(n)$. Sorry for the previous comment.2011-10-27
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    @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](http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number) 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.2011-10-27
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    @joriki I tried it my method and you're obviously right.2011-10-27
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    @OlivierBégassat: Thank you for pointing out my error.2011-10-27
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    @daniel: We get a lot of, let's say, suboptimally posed questions here that have redundant words like "question" in the title. They tend to be much less interesting than yours, so I could imagine people who might be interested in this question might not look at it when they see that title -- I'd suggest replacing it by something more descriptive and appealing.2011-10-27
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    To answer the question - it is not currently known whether the above is true or not. Current methods cannot give square root accuracy, however using zero density theorems and other tricks we can actually get a range of $n^{0.525+\epsilon}$. In the other direction, Maier proved that we cannot have such results in a range of $(\log n)^k$ for any $k$.2011-10-27
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    joriki--Sure. I'm open to suggestions. If you have the ability, feel free to replace. I'm bad at titles.2011-10-27
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    @Eric, can you prove the equation in the question on the assumption that there's always a prime between $n$ and $n+\sqrt n$? can you prove the converse?2011-10-27
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    Note that the equation follows from the special case $m=0$ by additivity of limits.2011-10-27

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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 $n\ln n$.

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    I was wondering about the $0.525$ result: [Wikipedia](http://en.wikipedia.org/wiki/Prime_gap#Upper_bounds) has this as an asymptotic result, but [the paper](http://www.cs.umd.edu/~gasarch/BLOGPAPERS/BakerHarmanPintz.pdf) only contains the lower bound; the same for [Huxley's result](http://www.kryakin.com/files/Invent_mat_(2_8)/15/15_11.pdf). Perhaps you (or someone else knowledgeable enough) could clean up that section?2011-10-28
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    P.S.: It seems the asymptotic claims in that section are correct up to and including [the result by Ingham](http://qjmath.oxfordjournals.org/content/os-8/1/255.extract). 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 $0.625$.2011-10-28
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    I just plucked the 0.55 number out of the air - I remember the 0.525 offhand, but not the exponent for the asymptotic formula.2011-10-28