# Generalization of Wilson's theorem for prime tuples

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 )

• 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 Sep 3, 2019 at 18:29
• So maybe raising each "Wilson gamma factor" to a positive power could help rule them out. Sep 3, 2019 at 18:30
• 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! Sep 3, 2019 at 22:44
• @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 ?
– TPC
Sep 3, 2019 at 22:59
• @Gerry Myerson you commented in support on that MSE post of mine .
– TPC
Sep 3, 2019 at 23:10

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 $$n(n+2)\mid (4(n-1)!+n+4)$$.

• Thank you very much , sir .
– TPC
Sep 4, 2019 at 6:14
• What about the main question ?
– TPC
Sep 4, 2019 at 6:16
• Typo is corrected. I doubt this approach can help to prove infinitude of prime twins or tuples. Sep 4, 2019 at 11:13
• @ Max Alekseyev The doubt is mutual . But I don't have very strong reason to firmly believe it .
– TPC
Sep 4, 2019 at 11:56