# Gap between semiprimes

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 $s^{\frac{m-1}m}\log^2s$?

• Frank Thorne's paper "Bounded gaps between products of primes with applications to elliptic curves and ideal class groups" (people.math.sc.edu/thornef/bounded-gaps.pdf) contains some conditional results on bounded gaps between square-free products of primes lying in certain subsets of positive density which may be of interest to you. – Ben Linowitz Oct 26 '15 at 1:42
• The desnity of semiprimes is $\log \log n/ log \n$, so maybe $c \log n^2/ \log log n$ i the right gap? – Will Sawin Oct 26 '15 at 2:24

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 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 $c_k$.
• Ok I gave $\sqrt{s_0}$ as a highly trivial calculation I suspected better but had no clue how to show – user76479 Oct 26 '15 at 1:59
• Your $\sqrt{s_0}$ is much larger than the expected gap size. You concentrate on gaps between special semiprimes, not all semiprimes. This is the point. – GH from MO Oct 26 '15 at 2:00
• @Arul: See my added section. It shows that the original Cramer conjecture implies a variant for numbers with exactly $k$ prime factors. All you need to do is adjust the constant $c$ in terms of $k$. – GH from MO Oct 26 '15 at 2:19