Given a large enough integer $n\in\Bbb N$ and a real $r\in\big(0,\frac12\big]$ and $n\_1\in\Bbb N\_{> n}$ is the smallest integer such that $n\_1=AB$ for two coprime integers $A$ bigger than but close to $\lceil n^{1-r}\rceil$ and $B$ smaller than but close to $\lfloor n^{r}\rfloor$ then is there an upper bound and average case (lower bound) estimate for $n\_1-n$? Note if $n\_1$ is prime then we have Cramer's conjecture for upper bound and average case bound given by prime number theorem.


I think something similar to $O((\log n)^c)$ upper bound is true (similar to Cramer's conjecture but due to coprimeness might be easier to show). Or is there an $\Omega(n^r)$ lower bound?



On a coprime generalization of Cramer's conjecture
Put $B=\lfloor n^r\rfloor$, and let $n\_0$ be the smallest integer larger than $n$, which is divisible by $B$. If for an integer $k$ we have $(\frac{n\_0}{B}+k, B)=1$, then $n\_1 = n\_0+kB$ satisfies your condition. You can bound $k$ by Jacobsthal's function, by Iwaniec's estimate we get $n\_1-n\ll n^r\log^2 n$. 


If you rather have a better bound for $n\_1-n$, and are willing to relax the bound for $B$ to $n^r/2\leq B\leq n^r$, say, you can argue as follows. Write
$$
\#\{\nu\in[n, n+\Delta]:\exists p\in[n^r/2, n^r], p\mbox{ prime}\} =
\sum\_{p\in[n^r/2, n^r]} \left[\frac{n+\Delta}{p}\right]-\left[\frac{n}{p}\right].
$$
Now use Exponential sums and Vaughan's identity to give a lower bound for the sum over primes. Then use some combinatorial counting to show that the number of integers $\nu\in[n, n+\Delta]$, which are divisible by the square of a prime $p\in[n^r/2, n^r]$ is small. 


The combinatorial part might become tricky, in case you get stuck there you could have a look at the literature dealing with the difference between squarefree numbers.


