Here is a similar question that asks for $101$ numbers none of them are prime and it is well known that $101!+1,101!+2,101!+3,\cdots,101!+101$ are those numbers. I am interested to know, how we can find exactly $101$(more generally, $n$) consecutive composites, i.e; if the list of composites is $k,k+1,\cdots,k+100$, then neither $k-1$ nor $k+101$ are composite,$($i.e; $k-1,k+101\in \mathbb{P})$.
From this OEIS list, and from wikipedia also, $n$ for which we have primes of the form $n!+1$ are listed. This list doesn't includes $101$, that means, $101!+1$ is composite also. So $101!+1,101!+2,101!+3,\cdots,101!+101$ is not list of exactly $101$ consecutive composite numbers(as, $101!+1$ is also composite). Is there a way to find such list for $101$ consecutive composites? and is it possible to find such list for any given length $n\in\mathbb{N}$ which includes exactly $n$ consecutive composite numbers.
I have tried this for a while and not getting any way out. In other way it is similar to finding primes with gap $102$. Though it is not listed here.
Edit:
I don't think there can be any solution without programming. I have translated this code to Python, one can check here.
Best Answer
Average gap between primes is ~100 around $e^{100}=2.7\times10^{43}$, but it's amazing how frequently we find such gap for much, much smaller numbers:
Gaps of size 100 or 102:
etc.
The following Java code will list all prime pairs with the given gap:
In the following line just replace 102 with the gap you are interested in: