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Let $k \in\mathbb{N}, k \geq 2$, and $\mathbb{P}$ represent the set of prime numbers.

Consider the k-tuple $(0,h_1,h_2,\cdots,h_{k-1})$ with $0 < h_1 < \cdots < h_{k-1}$.

The well known k-tuple conjecture states:

If the k-tuple $(0,h_1,h_2,\cdots,h_{k-1})$ is admissible then they are infinitly many k-tuples $(p,p+h_1,\cdots,p+h_{k-1})\in\mathbb{P}^{k}$

(The k-tuple $(0,h_1,h_2,\cdots,h_{k-1})$ is admissible iff $\forall p \in \mathbb{P} \,\, ,p(p+h_1)\cdots(p+h_{k-1})$ haven't a fix prime divisor)

I ask if we can prove:

If the k-tuple $(0,h_1,h_2,\cdots,h_{k-1})$ is admissible then exists at least a k-tuple that $(p,p+h_1,\cdots,p+h_{k-1})\in\mathbb{P}^{k}$

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    $\begingroup$ No. Using larger tuples, you can prove if that for any tuple there is at least one k-tuple, then there are infinitely many. $\endgroup$
    – Wojowu
    Commented Sep 20, 2019 at 21:08
  • $\begingroup$ @Wojow, you see the idea .. we move from k-tuple to (k+1)-tuple, (k+2)-tuple,. then the existence of at least prime k-tuple implies infinitly many k-tuple primes .. $\endgroup$
    – Lagrida Yassine
    Commented Sep 20, 2019 at 21:21
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    $\begingroup$ What you ask is not even known for $k=2$. That is, we don't know whether every even integer is a difference of two primes. $\endgroup$
    – GH from MO
    Commented Sep 21, 2019 at 1:08
  • $\begingroup$ @GH: what is the best $\delta>0$ so far so that the proportion of even integers that are the difference of two primes is at least $\delta$? $\endgroup$
    – Sylvain JULIEN
    Commented Sep 21, 2019 at 8:25
  • $\begingroup$ @SylvainJULIEN: I think the best known result is recorded in (1.2) of the following paper: arxiv.org/abs/1410.8198v2 $\endgroup$
    – GH from MO
    Commented Sep 21, 2019 at 8:45

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