# What might the (normalized) pair correlation function of prime numbers look like? [closed]

You might have read about the fortuitous meeting between Montgomery and Dyson. The background is that the nontrivial zeros of the Riemann zeta function, when normalized to have unit spacing on average, (seem to) have the pair correlation function $1-\mathrm{sinc}^2(x)$, where $\mathrm{sinc}$ is the normalized function $\sin(\pi x)/ (\pi x)$. It's still a conjecture but it has good numerical support.

So what about prime numbers? Let $\Sigma(x,u)$ be the number of pairs of primes $p,q\le x$ which satisfy the inequality $0\le p-q\le u\,x/\pi(x)$, where $\pi(x)$ is the prime counting function. This inequality is chosen because multiplying primes by $\pi(x)/x$ will ensure the gaps between consecutives is exactly unity (hence they are normalized). Then what might

$$g(u)=\frac{d}{du}\left(\lim_{x\to\infty}\frac{\Sigma(x,u)}{\pi(x)}\right)$$

end up looking like? This basically asks, "what is the density of normalized primes around so-and-so apart from each other?" (You can see the original Montgomery conjecture as equation 12 here. I've adapted it to prime numbers by essentially changing the asymptotic number of zeta zeros to the prime counting function instead.)

I have also posted this question on MathOverflow here. I figured this question might be at more of a researcher's level than Math.StackExchange is used to and so would be beneficial to cross-post. MO user David Roberts informs me this is something of a malpractice, as it makes collecting answers awkward and a delay should be allowed for one site (presumably SE) to have a go at it. I'll keep this in mind in the future.

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## closed as off topic by Eric Naslund, Qiaochu YuanAug 31 '11 at 16:29

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Also posted to MathOverflow. – Gerry Myerson Aug 17 '11 at 23:38
I'd suggest keeping the MO one open and closing this one, since it already has a lot of good comments, and I think you are going to need someone pretty sophisticated to answer this. – David Speyer Aug 18 '11 at 17:03
I am voting to close since this is answered on Overflow – Eric Naslund Aug 31 '11 at 15:06