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Is it possible to use the sieve method to solve problems like these?

  1. Count a number of $p_1p_2\cdots p_r$, a product of $r\geqslant 1$ primes such that $p_i\in [P,2P]$ for all $i=1,2,\ldots, r$ in the short range such as $[AP^r,AP^r+Y]$ for any suitable constant $A\in [1,2^r)$ and $Y=P^{1.9}$ for instance. Heuristically, since the probability of getting a prime in $[P,2P]$ is roughly $\frac{1}{\log P}$, it is quite reasonable to expect the count as $\ll \frac{P^{1.9}}{\log^r P}$?

For $r=2$, the bound is possible by the Prime number theorem in short intervals $$\sum_{\substack{AP^2<pq\leqslant AP^2+Y}\\ p,q\in [P,2P]}1=\sum_{\substack{p\leqslant AP^2+Y\\p\in[P,2P]}}\sum_{\substack{\frac{AP^2}{p}<q\leqslant \frac{AP^2+Y}{p}\\q\in [P,2P]}}1\ll \sum_{\substack{p\leqslant AP^2+Y\\p\in[P,2P]}}\frac{Y}{p\log P}\ll \frac{Y}{P\log P}\sum_{p\in [P,2P]}1\ll \frac{P^{1.9}}{\log^2P}$$ but for higher $r$ the size of $Y$ is too small to adapt.

  1. Given the setting as in 1., Can we show any result on counting a number of pairs of products of $r$ primes such that, for a fixed even number $g=O(P)$, their difference is $g$? Similarly as above, I expect the count to be $\ll\frac{P^{1.9}}{(\log P)^{2r}}$.

What can we say about the implicit constant? Can it be bounded by $\ll (\log P)^{o(1)}$?

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  • $\begingroup$ I think for large $r$ even counting $n_1,\dots,n_r$ such that $n_i \in [P,2P]$ and $n_1 n_2\dots n_r \in [A P^r, A P^r +Y]$ is not so easy. $\endgroup$ Commented Nov 12 at 14:52

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