Definition: In the gap between any two consecutive odd primes we have one or more composite numbers. One of these composite number will have a prime factor which is greater than that of any other number in the gap. E.g. $43$ is the largest prime in the gap between the consecutive primes $83$ and $89$. I am interested in the largest prime factor in the gap between two consecutive primes.
Claim: Every prime is the largest prime factor in some prime gap.
I am looking for a proof or disproof.
Update, 21 Dec 2019: Conjecture verified for $p \le 10^{10}.$
Note: This question was posted in MSE and got many upvotes but no answer hence posting in MO.
Best Answer
Heuristically this should be the case. For any prime p greater than 5, consider the set of numbers of the form $2^a3^b5^c p \pm 1$. The "probability" that one of of these is prime should be about $$\frac{1}{\log (2^a 3^b 5^c)} = \frac{1}{a \log 2 + b \log 3 + c \log 5} .$$ So the probability that both $2^a3^b5^c p + 1$ and $2^a3^b5^c p - 1$ are prime should be about $$\frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}.$$ Now, note that the product $$\prod_{a,b,c} \left(1-\frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}\right)$$ diverges to $0$, it represents the probability that none of those $2^a3^b5^c p \pm 1$ is a prime pair. So we should expect for a given $p$ there should be such an $a$,$b$ and $c$. So after noting the twin prime pairs $(3,5)$, $(5,7)$ and $(29,31)$ it seems like we should expect a much stronger statement. For any prime $p$, there should be a positive integer $n$ such that $p$ is the largest prime divisor of $n$ and $n+1$ and $n-1$ are both prime.
Obviously, proving something like this is well beyond current technology. I'd also say that it is highly likely that even your weaker statement is well beyond what is currently doable.