$\frac{2\cdot3\cdot5\cdot7\cdot11\cdots p_k}{5}+1$

Question

Let be $n$ equal $$n=\frac{2\cdot3\cdot5\cdot7\cdot11\cdots p_k}{5}+1$$ for some $k$ where $p_k$ is the $k$th prime number. Is it true that if $n$ is prime, then $n-p_k-1$ is either prime or a multiple of 5?

Can that be proven?

Answer 1

No

$\tfrac{2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83}5 + 1 $ $= 53412903137855170271124803598559$ which is prime

but subtracting $2$

$\tfrac{2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83}5 - 1 $ $= 53412903137855170271124803598557 = 13399153 \times 9500816407 \times 419573365701467 $ which is not prime nor a multiple of $5$

Answer 2

I suppose that $n$ in your question is the result of your formula.

Denote $m\#:=2\times 3\times 5\times\ldots\times p_m$ (where $p_m$ is the $m$th prime). Then $m\#+1,m\#-1$ cannot be a multiple of any of the primes $2,\ldots,p_m$. So any prime dividing $m\#-1$ must be greater than $p_m$. If you divide by $5$, then any prime dividing $n-2=m\#/5-1$ must be either $5$ or greater than $p_m$. But $m\#/5-1$ does not need to be prime itself, nor a multiple of 5. For example, $$9\#/5-1=2\times 3\times 7\times\ldots\times 23-1=59\times 756247.$$

Answer 3

The number:

$$n_{75}=\tfrac{2\times 3\times 5\times \dots\times 379}5+1$$ is prime.

$n_{75}-2$ is neither prime nor divisible by $5$: $n_{75}\mod 5 =3 $.