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This question generalizes Symmetry in Hardy-Littlewood k-tuple conjecture. Say two prime constellations are equivalent up to permutation if they consist of the same multiset of prime gaps. One can order such constellations lexicographically according to the considered prime gaps, a smaller prime gap playing a role analogous to a letter occurring earlier in the alphabet.

That way, one can uniquely define a permutation $\sigma$ such that the action of $\sigma$ on the multiset of prime gaps contained in the constellation $C$ gives rise to the constellation $C_{\sigma}$.

My question is: do any two prime constellations equivalent up to permutation have the same distribution? Can a link with Chebotarev's theorem be established, viewing the set of prime constellations equivalent to $C$ up to permutation as an analogue of a conjugacy class?

I'm interested in both conditional and unconditional results or heuristics.

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    $\begingroup$ The constellation of primes $11,13,17,19$ has gaps $2,4,2$ but $4,2,2$ never happens and $2,2,4$ only once. That is a minor issue, but you need to tighten up the question. Have tried looking at data? $\endgroup$
    – Aaron Meyerowitz
    Commented Sep 8, 2021 at 8:26
  • $\begingroup$ Indeed a permutation should preserve the admissibility, just like not every permutation of the letters of a given word makes sense or can be pronounced. I looked at the relevant wikipedia article and it seems the expression of the asymptotics conjectured by Hardy and Littlewood does not depend on the order of the prime gaps (assuming the admissibility). $\endgroup$
    – Sylvain JULIEN
    Commented Sep 8, 2021 at 9:08
  • $\begingroup$ So one may try to study the "Hardy-Littlewood-Galois group" of permutations of a given constellation preserving the admissibility and see if it leaves the distribution invariant. $\endgroup$
    – Sylvain JULIEN
    Commented Sep 8, 2021 at 9:13

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Let $a_x$ be the number of primes $p<x$ starting a constellation $18,18,12$, i.e. $p,p+18,p+36,p+48$ are consecutive primes. Similarly, let $b_x$ and $c_x$ count primes starting a constellation $12,18,18$ and $18,12,18$ respectively. I think that there is strong reason to expect that as $x \rightarrow \infty$ we have $$ \frac{b_x}{a_x} \rightarrow 1$$ $$ \frac{c_x}{a_x} \rightarrow \frac32.$$

I will give my heuristic reasoning for expecting this and some limited computational support. Of course we don't know that the gap $12$ even occurs infinitely often.

Heuristic:

  • For $p,p+18,p+36,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3\}$

  • For $p,p+12,p+30,p+48$ to contain no multiples of $5$ also requires $p \bmod 5 \in \{1,3\}$

  • For $p,p+18,p+30,p+48$ to contain no multiples of $5$ requires $p \bmod 5 \in \{1,3,4\}$

The added condition that the four primes be consecutive seems to be equally restrictive in all three cases.

Computation: Here is a graph,

enter image description here

The top curve is $$\frac{c'_x}{a'_x}$$ and the one below is $$\frac{b'_x}{a'_x}$$ up to $x=3\cdot 10^7$ where $a'_x$ is the number of primes $p<x$ so that $p,p+18,p+36,p+48$ are all primes (but not required to be consecutive.)

Each curve actually has $60$ data points, those for $x$ a multiple of $500,000.$

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  • $\begingroup$ I don't think those make prime constellations as the diameter of the related $k$-tuples is not minimal. But your argument looks sound and there seems to be some subtlety I can't pinpoint yet. $\endgroup$
    – Sylvain JULIEN
    Commented Sep 8, 2021 at 9:29
  • $\begingroup$ Actually I think we should require that all iterates of a "good" permutation $\sigma$ preserve the admissibility, but it may not be enough. $\endgroup$
    – Sylvain JULIEN
    Commented Sep 8, 2021 at 9:42
  • $\begingroup$ I décided to accept your answer, as it shows we need the sets of possible residue classes of $p$ mod the relevant prime to have the same cardinal. Those sets of residue classes with same cardinal may then play the role of analogues of conjugacy classes in Chebotarev's theorem. $\endgroup$
    – Sylvain JULIEN
    Commented Sep 9, 2021 at 9:24
  • $\begingroup$ That is probably all that matters. gaps $2,4,8$ should occur equally often but only half as frequency as gaps $6,12,18$ due to residue classes $\bmod 3.$ $\endgroup$
    – Aaron Meyerowitz
    Commented Sep 9, 2021 at 15:43

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