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Let $p[n]$ be the $n$-th prime. Let $0 \leq m < k$.


$$\lim_{n\rightarrow\infty}\frac{ p[(n+k)^2] - p[(n+m)^2] }{ p[n]} = 4(k-m)\;.$$

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
up vote 5 down vote accepted

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 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
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. – Greg Martin Oct 28 '11 at 6:41

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