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I am looking for references regarding the following statement.

For any two natural numbers x and y there must be a prime k-tuple (a, b, ...) corresponding to x consecutive primes (n+a, n+b, ...) for at least y different n.

I am not familiar with much of the literature and would like to know whether it is a known result. I offer my idea of a proof here:

https://math.stackexchange.com/questions/3373121/abundance-of-prime-k-tuples

It is detailed in my answer under this (my) question. The idea/structure of my (attempt at a) proof is given under Approach. The proof is written in terms of prime patterns rather than k-tuples. The term prime pattern is defined and linked to k-tuple in the section Glossary.

I would much appreciate any kind of answer. I am an amateur mathematician with fairly little formal training and limited, eclectic knowledge of the literature, but have been reading and thinking about prime numbers for 20 years.

Since the first commenter (Gerhard Paseman) has mentioned the twin primes conjecture and asks for a list of consequences, these are my own thoughts on both.

Consequences:

If the above is a theorem, then my understanding is that:

1.) there must either be patterns (k-tuples) of any size and inifnite abundance or be infinitely many different patterns (k-tuples) of any size and a given abundance

2.) the result also holds for non-overlapping patterns (k-tuples). That is, for any x and y one can find k-tuples (a, b, ...) with arbitrarily many different n such that none of the resulting primes (n+a, n+b, ...) for one n is equal to any prime given by one of the other n. (To be clear: I am not saying there are no two n with overlap, just that one can find arbitrarily many n without any overlap)

Regarding the twin primes conjecture, the Polignac conjecture, Dickson's conjecture, the prime k-tuples conjecture and Zhang's theorem:

I don't think this result implies any of these. It is obviously far more elementary than one would expect a proof of any of these to be. Unlike them, it does not state that there are infinitely abundant k-tuples, only that there are some k-tuples of any finite abundance. It allows only an either-or-conclusion as to what is inifinitely abundant (see Consequences).

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    $\begingroup$ A better way to do this on this forum is as follows, where I use single capital consonant letters to abbreviate long character strings of appropriate and meaningful content: "I am looking for references for the following theorem : S . The idea of proof for this theorem is this: P . I am not familiar with the literature, but I believe the following is a partial list of consequences: L. Are there references to this theorem in the literature?" There are even better ways, but this way is an improvement over your current presentation. Gerhard "Don't Ask To Read Much" Paseman, 2019.10.03. $\endgroup$ – Gerhard Paseman Oct 3 '19 at 19:58
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    $\begingroup$ Also, my brief skim suggests that you are considering something harder than the twin prime conjecture, and that your reasoning at best might help understand how an interesting statement might hold in some statistical model of the primes, but does not actually apply to the model of the primes. If you fill in S and P properly above, I may read more. Gerhard "Sometimes Prefers To Read Less" Paseman, 2019.10.03. $\endgroup$ – Gerhard Paseman Oct 3 '19 at 20:02
  • $\begingroup$ Thank you for your help. I have now restated my question to fill in S, P (and L) as asked (though I refer the reader to the "Approach" section in my proof attempt for P). $\endgroup$ – Thomas Traill Oct 4 '19 at 8:09
  • $\begingroup$ Thanks also for the other thoughts in your second comment. I don't think this is harder than the twin primes and have added a section to the end of my question explaining why. I also don't see why my work is not applicable to the primes. My hunch is that the "rectangular distribution" under 1.2 in the "Approach" section is the cause of confusion. I am using this to show that something must be true for all possible abundance distributions and therefore also for the actual abundance distribution of pattens of size X in the primes up to N. I have expanded 1.2 in the approach to clarify this. $\endgroup$ – Thomas Traill Oct 4 '19 at 8:55
  • $\begingroup$ Theorem 1.2 in arxiv.org/abs/1311.4600 seems relevant $\endgroup$ – Ville Salo Oct 4 '19 at 14:26
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The following is Theorem 1.2 in [Maynard, J. (2015). Small gaps between primes. Annals of Mathematics, 181(1), second series, 383-413.]:

Let $m \in \mathbb{N}$. Let $r \in \mathbb{N}$ be sufficiently large depending on $m$ and let $\mathcal{A} = \{a_1,a_2,...,a_r\}$ be a set of $r$ distinct integers. Then we have $$ \frac{\# \{\{h_1,...,h_m\} \subseteq \mathcal{A} \;:\; \mbox{for infinitely many $n$ all of $n + h_1, ..., n + h_m$ are prime}\}}{\#\{\{h_1,...,h_m\} \subseteq \mathcal{A}\}} \gg_m 1. $$

Now, I interpret your question as follows: you want an $m$-tuple of primes such that infinitely many translates are sets of consecutive primes. This indeed follows:

For every $m$, there exists an $m$-tuple of primes $(p_1,...,p_m)$ such that for infinitely many $n$, $(n+p_1,...,n+p_m)$ are an $m$-tuple of consecutive primes.

Proof. Consider the subshift $X \subset \{0,1\}^{\mathbb{Z}}$ obtained as the $\omega$-limit set of the characteristic function of the primes, i.e. $$ \bigcap_N \overline{\{\sigma^n(x) \;:\; n \geq N\}}. $$ where $x \in \{0,1\}^{\mathbb{Z}}$, $x_i = 1 \iff i \geq 1 \wedge i \mbox{ is prime}$, and $\sigma$ is the left shift $\sigma(y)_i = y_{i+1}$.

Maynard's theorem shows (much more than) that the language of this subshift (I mean the set of finite subwords $\{y_{[0,n-1]}\;|\; y \in X\}$) contains a word $w$ with at least $m$ many $1$s. Namely, a tuple $(h_1,...,h_m)$ whose translates are simultaneously prime infinitely often implies that translates of $x$ visit the clopen (cylinder) set $C = \bigcap_{i = 1}^m \sigma^{-h_i}([1])$ infinitely many times. If $\ell \geq \max_i h_i$, then $C$ is a finite union of clopen sets of the form $[w]$ for words $w$ with length $|w| \leq \ell$, thus one of them is visited infinitely many times by the pigeonhole principle.

But if there's a word with at least $m$ many $1$s, there is of course also a word $v$ with exactly $m$ many $1$s. And when this word $v$ appears in the characteristic function of the primes (and it does: that's exactly what the language of $X$ is), we have exactly an $m$-tuple of consecutive primes. So then for any $n$ such that $(i+n \;|\; v_i = 1)$ is a prime $m$-tuple, it is in fact an $m$-tuple of primes such that infinitely many translates of it are consecutive primes.

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