Checking for prime constellation

Question

How does one systematically check if a given configuration of prime numbers $p_1, p_2, ... p_n$ is the densest possible configuration of primes in the range $[p_1, p_n]$? (The densest configurations seem to be called “prime constellations”).

I am writing a paper for my high school science class (not math). It deals with interesting aspects of prime numbers. In this context I have two questions:

1) It is fairly clear that the configuration of a prime triplet must either be (p, p+2, p+6) or (p, p+4, p+6), as the „obvious“ shorter choice (p, p+2, p+4) is an equally spaced sequence of three elements and therefore one of the elements will be divisible by three (the exception of course is a sequence with “3” as its first element). While this simple example of three primes is clear I would like to know how to continue? What is the rule/algorithm for identifying longer prime constellations and how many constellations of the same length exist?

2) If a configuration of primes is given as $p, p+k_1, p+k_2, … p+k_n$, how can I check if this is the densest possible configuration, a so-called “prime constellation“ in the light of the above?

Thank you for your help. I would appreciate if the explanations are hands-on with examples that are fairly easy to understand.

Answer 1

$n,n+2,n+4$ can't be all primes more than once because whatever is $n \bmod 3$ one of $n,n+2,n+4$ will be $\equiv 0 \bmod 3$.

Given $0<b_1< \ldots < b_k$ there is no obstruction for $n,n+b_1,\ldots,n+b_k$ being all primes (more than once) iff for every prime $p$ there is some $a_p \in 1\ldots p-1$ such that $a_p,a_p+b_1,\ldots,a_p+b_k$ are all coprime with $p$.

If $p > b_k+1$ there is the solution $a_p=1$.

Thus it suffices to look at the primes $p\le b_k+1$ which means there is no obstruction exactly when for some $A$ all of $A,A+b_1,\ldots,A+b_k$ are coprime with $(b_k+1)!$

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The natural generalization of the twin prime conjecture is that when there is no obstruction $n,n+b_1,\ldots,n+b_k$ are all primes infinitely many times.

The conjectured asymptotic (from the random model for the primes) for the number of such $n$ is $$\pi_b(x) \sim \sum_{n \le x} \prod_{p \le n^r} \frac{\sum_{a =1,\ p\, \nmid\, a (a+b_1)\ldots (a+b_k)}^{p-1} 1}{p} \sim C(b) \frac{x}{\ln^{k+1} x}$$ where for $p > b_k+1, \frac{\sum_{a =1,\ p\, \nmid\, a (a+b_1)\ldots (a+b_k)}^{p-1} 1}{p} = 1-\frac{b_k+1}{p}$

and $r =e^{-\gamma} \approx 0.56$ is the constant such that $\prod_{p \le n^r} \frac{p-1}{p} \sim \frac{1}{\ln n}$ (making the predicted $C(b)$ constants compatible with the PNT)

The random model for the primes is merely saying that when choosing $n$ randomly in $[1,x]$, for $p \le n^r$, then $n \bmod p$ can be considered uniformly distributed and more important the events $n \bmod p$ and $n \bmod q$ can be considered independent. Thus, under this model, the probability that $n$ is prime can be considered to be $\prod_{p \le n^r} \frac{p-1}{p}$ which by Mertens's theorem is $\sim \frac1{\ln n}$.

Answer 2

Consider this k-tuple : $\mathcal{H}_k = (0,h_1,h_2,\cdots,h_{k-1})$, with $0 < h_1 < \cdots < h_{k-1}$ evens numbers.

$H_k$ is admissible iff $p(p+h_1)\cdots(p+h_{k-1})$ has no fix divisor.

For example $p(p+2)(p+4)$ is always divisible by $3$ then $(0, 2, 4)$ not admissible.

Consider the constant : $$\mathcal{G}_k = \left(\prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_k, p)}{p}}{(1-\frac1p)^{k}} \right)$$

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

The k-tuple conjecture predicts that the number of primes $(p,p+h_1,\cdots,p+h_{k-1})\in \mathbb{P}^k$ with $p+h_{k-1} \leq x$ is:

$$\pi_{\mathcal{H}_k}(x) \sim \mathcal{G}_k \dfrac{x}{\log(x)^k}$$

And we can prove if $\mathcal{H}_k$ is not admissible then $\mathcal{G}_k = 0$.

Proof: If $q$ is a prime number always dividing $p(p+k_1)\cdots(p+k_n), p \in \mathbb{P}$, then $w(\mathcal{H}_k, q) = q$, then $\displaystyle\mathcal{G}_k = \left(\prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_k, p)}{p}}{(1-\frac1p)^{k}} \right) = 0$

And you can see for $p > h_{k-1}$ we have $w(\mathcal{H}_k, p) = k$, then we check only primes less than $h_{k-1}$ for admissibility.

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Example1 : $\mathcal{H}_k=(0,2,4)$, we have $w(\mathcal{H}_k, 3)=3$ then $(0,2,4)$ not admissible.

Example2 : $\mathcal{H}_k=(0,2,6)$, we have $w(\mathcal{H}_k, 3) = 2$ and $w(\mathcal{H}_k, 5) = 3$ then $(0,2,6)$ is admissible.