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Factorial primes are are primes of the form $n! \pm 1$ and primorial primes are primes of the form $p\#\pm 1$, where $p\#$ is the product of all primes $\leq p$.

To cite http://www.ams.org/journals/mcom/2002-71-237/S0025-5718-01-01315-1/: "Careful checks over the last half-century have turned up relatively few such primes". However, they also made the following two conjectures:

The expected numbers of factorial primes of each of the forms $n! \pm 1$ with $n \leq N$ are both approximately $e^\gamma \log N$.

The expected numbers of primorial primes of each of the forms $p\# \pm 1$ with $p \leq N$ are both approximately $e^\gamma \log N$.

I'm somewhat curious about twin primes of the forms $(n!-1,n!+1)$ and $(p\#-1,p\#+1)$. If the conjectures are right, we could ask ourselves if there are infinitely many twin primes of these forms. But clearly, as neither these conjectures are proven and neither is the twin prime conjecture, a positive answer is impossible.

However, what about the converse? Is it possible to prove there can only be finitely many twin primes of these form(s)?

Note: twin primes of both forms exist: $$(3!-1,3!+1) = (5,7)$$ $$(3\#-1,3\#+1) = (5,7)$$ $$(5\#-1,5\#+1) = (29,31)$$ $$(11\#-1,11\#+1) = (2309,2311)$$ There are the only ones known (so far).

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Questions such as these are still considered open problems, hence the heuristic arguments in the paper you provided. We have no idea how to approach such problems currently.

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Without representing the following to constitute a proof of anything, I offer the following observations (too long to be included as comments) and a conjecture about twin primes of the form $p\# \pm1$. It is known (see A twin prime theorem, and a reformulation of the twin prime conjecture) that twin primes have the form $6N\pm 1$ iff $N\ne 6ab\pm a\pm b$.

Since all primes greater than $3$ have the form $6k\pm 1$, all primorials greater than $6$ will have the form $6\cdot (6K\pm 1)$, where the sign between $6K$ and $1$ depends on whether the primes up to $p$ feature an odd or even number of primes of the form $6k-1$. Combining this result with that of the previous paragraph, twin primes of the form $p\# \pm 1$ will occur when $(6K\pm 1)=\frac{p\#}{6}=N$ and $N\ne 6ab\pm a\pm b$.

Note that $(6K\pm 1)$ is divisible (once) by every prime up to $p$ other than $2$ and $3$, and no others. Note further that $K$ itself is not divisible by any prime $\le p$ with the possible exceptions of $2$ and $3$. (For $5\le q\le p, q\mid 6K\pm 1$. $q\mid K \Rightarrow q\mid 5K \Rightarrow q\mid (6K\pm 1 -5K)\Rightarrow q\mid (K\pm 1)$ which is impossible.) So the possible prime factors of $K$ are $2,3,q_j>p$. For several primorials ($p\ge 5$) I have computed values of $K$ and looked at the factors of $K$, and they conform to the previous statement regarding possible prime factors of $K$. $3\#$ contains no primes of the form $6k\pm 1$ and is left out of this analysis.

$$\begin{array}{ccc} \text{Primorial}&K&\text{Factors} \\ 5\#&1&1 \\ 7\#&6&2\cdot 3 \\ 11\#&64&2^6 \\ 13\#&834&2\cdot 3\cdot 139 \\ 17\#&14181&3\cdot 29\cdot 163 \\ 19\#&269436&2^3\cdot 131\cdot 257 \\ 23\#&6197024&2^5\cdot 31\cdot 6247 \\ \end{array}$$

For the two primorials that generate twin primes ($5\#$ and $11\#$), $K$ is a perfect power of $2$ (either $0$ or $6$). The suggestion is (and I understand just how very weak two examples out of seven is) that it might be the case that $p\# \pm 1$ are twin primes iff $\frac{p\#}{6}\pm 1=6\cdot 2^m$. Further pursuit of this line of thought by calculation of more primorials and more values of $K$ is unwieldy, and in any event would never constitute a proof. Can anyone out there think of a mathematically rigorous way to analyze these observations? It might be helpful to prove for what values of $m$ $6\cdot 2^m \pm 1 \ne 6ab\pm a\pm b$, if that can be done.

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