# 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

2
– joriki Oct 27 '11 at 22:03
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## 1 Answer

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
– 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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