Is there a conjectured gap between semiprimes?


There is a conjectured gap between primes in form of Cramer's conjecture. Using this we have $p\_1\leq p\_0+c(\log p\_0)^2$ for consecutive primes $p\_0$ and $p\_1$.


Then we have $s\_0=p\_0p\_1\leq p\_0p\_2=s\_1$ and $p\_2p\_0-p\_1p\_0\geq c(\log^2p\_0)p\_0$. So is the gap between consecutive semiprimes at least $\sqrt{s\_0}$.


Could there always or never be a prime smaller than $s\_0$ and larger than $s\_1$ whose product falls in the interval $[s\_0, s\_0+\sqrt{s\_0}]$?


In general is gap between product of $m$-primes at least $s^{\frac{m-1}m}\log^2s$?



Gap between semiprimes
I guess by "semiprime" you mean a product of two primes. If this is the case, then your conclusion that the gap between consecutive semiprimes is at least $\sqrt{s\_0}$ is incorrect, because in your reasoning you restrict to semiprimes divisible by the same prime $p\_0$. As a rule of thumb, the products of two (or more) primes form a denser set than the primes themselves. In particular, the gaps between them are expected to be only smaller, not bigger, than the prime gaps. Here is a [recent nice preprint](http://arxiv.org/abs/1510.06005) dealing with this question.


**Added.** It is straightforward to derive from Cramer's conjecture that for any $k$ there are numbers with exactly $k$ prime factors in $(x,x+c\_k(\log x)^2)$, where $c\_k>0$ is a suitable constant. Indeed, if $P\_{k-1}$ denotes the product of the first $k-1$ primes, then by Cramer's conjecture there is a prime $p$ in $(x/P\_{k-1},x/P\_{k-1}+c(\log x)^2)$, whence the product $pP\_{k-1}$ has exactly $k$ prime factors and lies in $(x,x+cP\_{k-1}(\log x)^2)$. That is, $c\_k:=cP\_{k-1}$ does the job. One can even achieve that $p$ is coprime with $P\_{k-1}$ by slightly enlarging this $c\_k$.


