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We know that Wilson's theorem states the following :

$x$ is a prime if $(\frac {\Gamma(x)+1}{x})$ is an integer .

We can extend this to Twin primes as :

$x$ and $x+2$ is prime if $(\frac {4(\Gamma(x)+1)+x}{x(x+2)})$ is an integer.

Now my question :

what is an equivalent result to the tuples of primes ?

i.e.

for given $k_0$

${\mathcal H} = (h_1, \ldots ,h_{k_0})$

and $n+ {\mathcal H } $ consists entirely of primes.

What is equivalence in terms of analog of Wilson's theorem to State the above condition ( I know it's relatively easy ) ?

(For twin primes ${\mathcal H} = (0,2)$)

Can we first prove the infinity of primes by proving the statement there are infinitely many integers of the form $(\frac {\Gamma(x)+1}{x})$ if x is an integer ; and then proceed in similar direction for tuples ? -- actually this is my main question .

( Forgive my English and type of typing I'm new here )

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    $\begingroup$ I think you have to pay attention to false positives such as $2=4×1/2$, so maybe the naive approach won't be enough $\endgroup$ – Sylvain JULIEN Sep 3 '19 at 18:29
  • $\begingroup$ So maybe raising each "Wilson gamma factor" to a positive power could help rule them out. $\endgroup$ – Sylvain JULIEN Sep 3 '19 at 18:30
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    $\begingroup$ You want to prove there's an infinity of twin primes, based on there being an infinity of integers of the form $(\Gamma(x)+1)/x$? Good luck! $\endgroup$ – Gerry Myerson Sep 3 '19 at 22:44
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    $\begingroup$ @Gerry Myerson No ,sir that's not my main concern ( that's why I added " in some sense" i.e. equivalent type) . I'm just curious is it possible or not with some additional Analysis ? $\endgroup$ – TPC Sep 3 '19 at 22:59
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    $\begingroup$ @Gerry Myerson you commented in support on that MSE post of mine . $\endgroup$ – TPC Sep 3 '19 at 23:10
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I assume that all $h_i$ are even. Then $(n+h_1,n+h_2,\dots,n+h_k)$ is a prime tuple iff $$\begin{cases} h_1!m \equiv -1\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -1\pmod{n+h_k}, \end{cases} $$ where $m=(n-1)!$. The system implies $$\begin{cases} h_k!m \equiv -\frac{h_k!}{h_1!}\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -\frac{h_k!}{h_k!}\pmod{n+h_k}, \end{cases} $$ which further combines into $$h_k!m \equiv - \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}\pmod{(n+h_1)\cdots (n+h_k)}.$$ That is, $(n+h_1)\cdots (n+h_k)$ divides the numerator of $$h_k!(n-1)! + \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}.$$

Example. For twin primes $(n,n+2)$, we have $k=2$ with $h_1=0$ and $h_2=2$. Then the last expression becomes $$2(n-1)!+2\frac{n+2}2 + \frac{n}{-2} = \frac{4(n-1)!+n+4}{2},$$ and thus we want $n(n+2)\mid (4(n-1)!+n+4)$.

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  • $\begingroup$ Thank you very much , sir . $\endgroup$ – TPC Sep 4 '19 at 6:14
  • $\begingroup$ What about the main question ? $\endgroup$ – TPC Sep 4 '19 at 6:16
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    $\begingroup$ Typo is corrected. I doubt this approach can help to prove infinitude of prime twins or tuples. $\endgroup$ – Max Alekseyev Sep 4 '19 at 11:13
  • $\begingroup$ @ Max Alekseyev The doubt is mutual . But I don't have very strong reason to firmly believe it . $\endgroup$ – TPC Sep 4 '19 at 11:56

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