While thinking about a different problem, I noticed that half-primorials when decremented (or incremented) by a power of $2$ are frequently primes.
Let $A_m$ be a half-primorial, i.e. the product of the first $(m-1)$ odd primes: $$A_m=\prod_{i=2}^m (p_i)=\frac{p_m\#}{2}$$ and let $k(m)$ be defined such that: $$2^{k(m)}<A_m<2^{k(m)+1}$$ By computation, I found many numbers of the form $A_m \pm 2^j$ where $1\le j \le k(m)$ are prime. I have looked at values of $m$ up to $12$ and found no cases where such numbers did not include several primes, for either decrementing or incrementing. This observation is not astonishing, as such numbers have none of the first $m$ primes as factors.
My question is: Is there a way to prove that among the numbers $A_m - 2^j$ or $A_m + 2^j$, there must always be one or more primes? As a follow on, if no such proof can be provided, can someone identify a value for $m$ where every $A_m - 2^j$ and/or every $A_m + 2^j$ is composite?