I wondered, inspired in a result mentioned from [1] (page 45), what should be the asymptotic behaviour of the sequence on assumption of the First Hardy–Littlewood conjecture

$$\sum_{\substack{\text{primes }p_n\leq x\\\text{such that }p_n+2\text{ is prime}}}(p_{n+1}-p_n)^2$$

as $x\to\infty$. Thus the summation is over the terms of the sequence A001359 from the OEIS, lesser of twin primes $$p_1=3, p_2= 5,p_3= 11, p_4=17,p_5= 29,\ldots$$

and we assume that the First Hardy–Littlewood conjecture is true.

A reference for the first Hardy–Littlewood conjecture is this section of Wikipedia.

I don't know if this exercise is in the literature.

Question. Deduce under the assumption that the First Hardy–Littlewood conjecture is true, what should be the asymptotic behaviour of $$\sum_{\substack{\text{primes }p_n\leq x\\\text{such that }p_n+2\text{ is prime}}}(p_{n+1}-p_n)^2$$ as $x\to\infty$. If it is in the literature, feel free to refer the reference and I try to search and read the result from the literature. Many thanks.


[1] Richard Crandall and Carl Pomerance, Prime Numbers: A Computational Perspective, Second Edition, Springer (2005).

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    $\begingroup$ Isn't the sum just equal to 4 times the number of twin primes $\le x$? $\endgroup$
    – tj_
    Oct 3, 2019 at 9:07
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    $\begingroup$ @tj_ I was confused by the notation too, but I think $p_n$ enumerates not all primes, but rather just twin primes (or rather the smaller of the two in each pair) $\endgroup$
    – Wojowu
    Oct 3, 2019 at 9:09
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    $\begingroup$ @tj_ If you read my previous comment again, then no - if I understand OP correctly, $p_{n+1}$ is the next prime after $p_n$ such that $p_{n+1}+2$ is prime. $\endgroup$
    – Wojowu
    Oct 3, 2019 at 9:15
  • $\begingroup$ @Wojowu: I see, thanks for clarification. $\endgroup$
    – tj_
    Oct 3, 2019 at 9:18
  • $\begingroup$ @tj_ many thanks to you and the other user in comments. The sequence $p_n$ is the sequence A001359 from the OEIS: lesser of twin primes. $\endgroup$
    – user142929
    Oct 3, 2019 at 9:36

1 Answer 1


You want to estimate $x \to +\infty$: $$\sum_{\substack{\text{primes }p_n\leq x\\\text{such that }p_{n+1}+2\text{ is prime}}}(p_{n+1}-p_n)^2$$ Let $n\in 2\mathbb{N}$, and consider the 3 tuple $\mathcal{H}_3 = (0,n,n+2)$.

The 3-tuple $(0,n,n+2)$ is admissible iff $n = 1 \pmod 3$ or $n = 0 \pmod 3$.

Let $\pi_{\mathcal{H}_3}(x) = \#\{(p,p+n,p+n+2) \in \mathbb{P}^3 \, | \, p+n+2 \leq x\}$

The Hardy-Littlewood conjecture states that : $$\pi_{\mathcal{H}_3}(x) \sim \left(\displaystyle\prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_3, p)}{p}}{(1-\frac1p)^{3}}\right) \, \dfrac{x}{\log(x)^3}$$

Where $w(\mathcal{H}_3, p)$ is the number of distinct residues $\pmod p$ in $\mathcal{H}_3$.

We have $w(\mathcal{H}_3, 2) = 1$ and $w(\mathcal{H}_3, 3)=2$ and $p \geq 5 \implies w(\mathcal{H}_3, p)=3$

Then: $$\pi_{\mathcal{H}_3}(x) \sim \dfrac{9}{2} \left(\displaystyle\prod_{\substack{5 \leq p \\ \text{p prime}}}\frac{1-\frac{3}{p}}{(1-\frac1p)^{3}}\right) \, \dfrac{x}{\log(x)^3}$$

See that The k-tuple conjecture holds also for consecutive primes.

Then if $\pi_{\mathcal{H}_3}(x,n)=\#\{(p_i,p_{i+1},p_{i+2}) \, | \, p_{i+2}-p_{i+1}=2, p_{i+1}-p_i=n,p_{i+2}\leq x\}$ we have:

$$\pi_{\mathcal{H}_3}(x, n) \sim \dfrac{9}{2} \left(\displaystyle\prod_{\substack{5 \leq p \\ \text{p prime}}}\frac{1-\frac{3}{p}}{(1-\frac1p)^{3}}\right) \, \dfrac{x}{\log(x)^3}$$

We have $n = 0 \bmod 2$ and $\big( n = 0 \bmod 3 \text{ or } n = 1 \bmod 3 \big)$, then: $$n = 0 \bmod 6 \text{ or } n = 4 \bmod 6 \quad , n \neq 0$$

Suppose that $n \leq N(x)$.

($N(x)$ is the maximum gap that $p_{i+2}-p_{i+1}=2, p_{i+1}-p_i=n$ hold in the interval $[1,x]$)

Then we have:

$$\displaystyle\sum_{\substack{\text{primes }p_k\leq x\\\text{such that }p_{k+1}+2\text{ is prime}}}(p_{k+1}-p_k)^2 = \displaystyle\sum_{\substack{n \leq N(x) \\ n = 6k, k \in \mathbb{N}^*}}n^2 \cdot \pi_{\mathcal{H}_3}(x, n) + \displaystyle\sum_{\substack{n \leq N(x) \\ n = 6k+4, k \in \mathbb{N}}}n^2 \cdot \pi_{\mathcal{H}_3}(x, n)$$

Using Granville's conjecture $N(x) \leq (\alpha+o(1))\log(x)^2$ with $\alpha \geq 2 e^{-\gamma}$.

We have the asymptotic formula for $\pi_{\mathcal{H}_3}(x, n)$ not depending of $n$, Then: $$\displaystyle\sum_{\substack{\text{primes }p_k\leq x\\\text{such that }p_{k+1}+2\text{ is prime}}}(p_{k+1}-p_k)^2 \sim \pi^{\prime}_{\mathcal{H}_3}(x) \left( \displaystyle\sum_{\substack{n \leq N(x) \\ n = 6k, k \in \mathbb{N}^*}}n^2 + \displaystyle\sum_{\substack{n \leq N(x) \\ n = 6k+4, k \in \mathbb{N}}}n^2 \right)$$

With $\pi^{\prime}_{\mathcal{H}_3}(x) = \pi_{\mathcal{H}_3}(x, 4)$ as example.

  • $\begingroup$ Many thanks for your attention, I'm going to study your answer. $\endgroup$
    – user142929
    Oct 3, 2019 at 11:02
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    $\begingroup$ @user142929, i edit my post to answer all questions. good luck .. $\endgroup$ Oct 3, 2019 at 13:04
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    $\begingroup$ No problem about the edits, and I prefer the detailed answers as yours. My knowledge about these techniques is not the best, therefore there may be a delay of a few days before I am accepting an answer, many thanks again. $\endgroup$
    – user142929
    Oct 3, 2019 at 17:52

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