# How many primes of this form?

In the famous proof of infinitely many primes, one assumes there are finitely many and then defines $n=p_1p_2\cdots p_k$. Now when we consider $n+1$, it is a possibility that $n+1$ could be a prime. I would like to know how many such number are there where $n+1$ is a prime and study the behavior of all the numbers of this form. Eg: $3, 7, 11$, etc.

I tried re-framing the question to see if it help in any way: "How many primes $p$ exist such that $p-1$ is square free?" I was not able to come to any solutions.

• Is $p_1$ the first prime,$p_2$ the second etc.If it isn't it would be worth adding it into the question. – kingW3 Feb 11 '17 at 18:08

The type of prime you are looking for is a type of primorial prime, so named because it is $\pm 1$ the product of the first $n$ primes (called the primorial).
• Thank you. This a partial answer to my question. Your answer deals with the product of the first $n$ primes $+1$. I would like to know about the product of $k$ arbitrary primes $+1$. – Abhijit A J Feb 11 '17 at 17:55
At first I thought you were talking about primorial primes, but this assumes that $p_1, \ldots, p_k$ consists of the first $k$ primes. However, your inclusion of $11$ as one of the example primes suggests that this is not the case, that the finite sets of primes may be arbitrary so long as no prime is repeated. However, the arbitrary set does need to include $2$ so that $n$ is even and $n + 1$ is odd.
With that in mind, the OEIS sequence that's more relevant to your question is http://oeis.org/A039787, "Primes $p$ such that $p - 1$ is squarefree." The entry is not marked "fini" (OEIS lingo for "finite") but that doesn't prove anything conclusively.
It is believed that there are infinitely many Sophie Germain primes. If that is the case, then that guarantees infinitely many primes $p$ such that $p - 1$ is squarefree, as the corresponding safe prime is a bit more than twice a Sophie Germain prime. Well, I guess this is another partial answer to your question.