I am trying to determine elementarily can there be an infinite number of composite numbers that have at least the two same numbers as two minimal gaps with respect to other composites both from left and from right.
I started with the construction of gap of size $n-1$ from the right by considering the numbers $n!+2,...,n!+n$, as is customary.
Since $n!-n,...,n!-2,n!,n!+2,...n!+n$ are all composites I could finish my task if I could now show that there is an infinite number of prime-quadruples $(n!-n-1,n!-1,n!+1,n!+n+1)$, which I can´t.
Of course for this to hold $n$ must be even because if $n$ is odd then $n!-n-1$ and $n!+n+1$ are even.
So a question would be settled if there is an infinite number of prime-quadruples of the form $((2k)!-2k-1,(2k)!-1,(2k)!+1,(2k)!+2k+1)$ and I seriously doubt in that there is an infinite number of them.
Because if $(2k)!-2k-1,(2k)!-1,(2k)!+1,(2k)!+2k+1$ are all primes then composite number $(2k)!$ has gap of size $1$ from $(2k)!-2$ from the left and gap of size $1$ from $(2k)!+2$ from the right and also gap of size $2k$ from $(2k)!-2k-1$ from the left and also gap of size $2k$ from $(2k)!+2k+1$ from the right, so it has two same immediate gaps both from the left and from the right.
If there is an infinite number of prime-quadruples of the form $((2k)!-2k-1,(2k)!-1,(2k)!+1,(2k)!+2k+1)$ then also there is an infinite number of even numbers that can be written in at least two ways as difference of two primes since we have $2k=((2k)!-1)-((2k)!-2k-1)=((2k)!+2k+1)-((2k)!+1)$.
But if I want to determine this stuff about gaps I think that this approach is doomed to fail because it seems too much to expect that there are an infinite number of prime-quadruples of this special form.
Is there an infinite number of them?
Edit: If someone will search for those prime-quadruples by extremely elementary considerations we have that if $k$ ends in $2$ or $7$ then $(2k)!+2k+1$ ends in $5$ so is composite, so $20$% of numbers need not be included in search at all.