This post Density of prime pairs whose gap is less than the average gap says that it is conjectured https://arxiv.org/abs/1103.5886 that (in case # means the cardinal of the set) $$ \#\{p_n<x|\alpha < \frac{p_{n+1} - p_n}{\log(p_n)}<\beta \} \sim \pi(x) \int_{\alpha}^{\beta} \frac{1}{e^t}dt$$ for $0 \leq \alpha < \beta$. Let $p_n$ denote the $n$'th prime number. Define the multiset $$A_1(p_N) = \{g_n = p_{n+1} - p_n | p_{n+1} < p_N\}$$ We make the following notations: $|A_1(p_N)|$ is the number of elements in $A_1(p_N)$ counted with multiplicity and $\sum A_1(p_N)$ is the sum thereof. It is known from the prime number theorem that $|A_1(p_N)| \sim \frac{p_N}{\log(p_N)}$ and obviously $\sum A_1(p_N) \sim p_N$. Define $$ A_2(p_N) = \left\{g_n \in A_1(p_N)| g_n < \frac{\sum A_1(p_n)}{|A_1(p_n)|}\right\}$$ The quantity on the right of the inequality $\frac{\sum A_1(p_n)}{|A_1(p_n)|} \sim \log(p_n)$ It is conjectured (I think GPY applies here too ) that $|A_2(p_N)| \sim N \int_{0}^1 \frac{1}{e^t}dt = c_2\cdot |A_1(p_N)|$ where $c_2$ is a strictly positive constant. Further more define in a similar fashion: $$ A_3(p_N) = \left\{g_n \in A_{2}(p_N)| g_n < \frac{\sum A_{2}(p_n)}{|A_{2}(p_n)|}\right\} $$ the process can continue of course.

Can someone make an estimation of $\frac{\sum A_2(p_N)}{|A_2(p_N)|}$ ? Is it $\sim \log(p_N)$ ?

Also one can see that $$ |A_1(p_N)| \sim \int_1^N \left( \frac{1}{\log(x)} - \frac{1}{\log^2(x)}\right)dx \hspace{0.5cm} |A_2(p_N)| \sim \int_{\frac{N}{e}}^N \left( \frac{1}{\log(x)} - \frac{1}{\log^2(x)}\right)dx$$ then how is intuitively $|A_k(p_N)| \sim \int_{?}^{N} \left( \frac{1}{\log(x)} - \frac{1}{\log^2(x)}\right)dx$ Where does intuitively $e$ arise from, in that integral?

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    $\begingroup$ Can somebody explain the down votes? Why no explanation? $\endgroup$ – C Marius Jun 8 '17 at 14:04
  • $\begingroup$ Do you consider $ A_{1}(p_{N}) $ as a set or as a multiset? $\endgroup$ – Sylvain JULIEN Jun 8 '17 at 16:04
  • $\begingroup$ $A_1(p_N)$ is a set $A_1(p_M)$ another set $\endgroup$ – C Marius Jun 8 '17 at 16:06
  • $\begingroup$ So if $g$ is a given prime gap that appears several times, you only count it once? $\endgroup$ – Sylvain JULIEN Jun 8 '17 at 16:13
  • $\begingroup$ No, it is counted every time it appears ... please edit the post to make a better exposition... $\endgroup$ – C Marius Jun 8 '17 at 16:17

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