Is this theorem on the abundance of prime patterns/k-tuples known?

Question

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).

Answer 1

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.