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Chen's Theorem with congruence conditions.

Chen's Theorem with congruence conditions.

The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape. However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the argument of the last paragraph that $\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$. The latter is comparable to the number of primes less than $x$, where the former has an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated by numbers with (a few) small prime factors. In fact, as Kowalski pointed out to me, it is not even obvious that one can easily obtain the correct upper bound for $\pi_r(x)$ simply by sieving over primes. From the asymptotic for $\pi_{r}(x)$, one expects that $$(*): \qquad \pi_{r,2}(x) =^{?} \ O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$ (I expect that such a result, perhaps slightly weaker, does exist in the literature, I will update this answer when I find the relevant links - actual experts are being consulted as we speak.) All one needs to answer 3) is that the exponent of $\log(x)$ in the denominator is $> 1$.

The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape. However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the argument of the last paragraph that $\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$. The latter is comparable to the number of primes less than $x$, where the former has an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated by numbers with (a few) small prime factors. In fact, as Kowalski pointed out to me, it is not even obvious that one can easily obtain the correct upper bound for $\pi_r(x)$ simply by sieving over primes. From the asymptotic for $\pi_{r}(x)$, one expects that $$(*): \qquad \pi_{r,2}(x) =^{?} \ O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$ (EDIT: My resident expert reports that this is known. Here is a sketch of the idea in the simpler case where we want to count pairs $n$ and $n+2$ where $n$ is a $2$-almost prime and $n+2$ is prime. First, for a small prime $p$, we want to find an upper bound for the number of $n < x$ such that $n$ is divisible by $p$ and both $n/p$ and $n+2$ are prime. This is a similar problem to counting twin primes, and in a similar way one obtains a bound of the form $O(x/\log x)$ (key point: the implied constant does not depend on $p$). If we wish to bound the number of pairs $(n,n+2)$ such that $n+2$ is prime and $n$ is a $2$-almost prime, we may instead count the triples $(p,n,n+2)$ where $p < x$ is prime, $n < x$ is divisible by $p$, $n/p$ is prime, and $n+2$ is prime. If for each $p < x$ we have a upper bound of $Ax/\log x$ (for the same $A$), in total we obtain the upper bound: $$ \frac{Ax}{\log x} \cdot \sum_{p < x} \frac{1}{p} \sim \frac{Ax \log \log x}{\log x}.$$ Of course, the devil is in the details! END EDIT) All one needs to answer 3) is that the exponent of $\log(x)$ in the denominator is $> 1$.

(Caveat: normally I wouldn't answer a question which such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.)

(Caveat: normally I wouldn't answer a question with such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.)