Denote with $p_n$ the $n$-th prime number and let

$$ h_N(d) = |\{ n : p_{n+1} < N, p_{n+1} - p_n = d \}| $$

be the number of times that prime gap $d$ happens for primes less than $N$. Let $H = \sum_{d \in {\mathbb Z}} h_N(d)$ and let

$$ f_N(d) = \frac{h_N(d)}{H} $$

be the relative frequency of gap $d$. (Yes, actually $H=\pi(N)-1$, where $\pi(N)$ is the number of primes less than $N$.) I would like to have an estimate for the sum of squares of $f_N(d)$

$$ S = \sum_{d \in {\mathbb Z}} f_N^2(d) $$ when $N$ gets large (very large, say $N=2^{512}, 2^{1024}$). Alternatively, I could use an estimate of the Renyi entropy $-\log_2 S$. Are you aware of some result about this problem?

I looked around a bit and I found a work by Marek Wolf where it is conjectured that $h_N(d)$ can be approximated by

$$ 2 c_2 \frac{\pi^2(N)}{N} \prod_{p | d} \frac{p-1}{p-2} e^{-d\pi(N)/N} $$

I tried to use this approximation, ignoring the product of (p-1)/(p-2) terms for simplicity, but the results are not totally convincing. I am not sure if it is because I discarded the product or because of some problem that comes out at very large $N$.

Assad Ebrahim
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Riccardo B.
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    I did not see this question until today. Your $S$ and mine appear to be almost the same (I do not square $H$ in the denominator of $f$). While I have an estimate it is only a guess based on a few calculations. Now that I see our sums are not identical maybe it's not useful to you. http://math.stackexchange.com/questions/893875/sorting-of-prime-gaps. – daniel Sep 11 '14 at 04:10

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