The theorem of the Chinese mathematician Chen Jingrun is currently one of the best results with respect to the binary Goldbach problem. It asserts that every even integer $n$ is in the sumset $P + P_2$, where $P$ denotes the set of primes and $P_2$ denotes the set of positive integers with at most two prime factors. However, I am not aware of this result asserting any control over the potential prime factors arising from the summand in $P_2$. In particular, I am interested in whether the following variant of Chen's theorem can be obtained:

Let $\mathcal{R}_f(N)$ denote the number of ways that $N$ can be written in the form $p + q_1 q_2$, where $p, q_1$ are primes and $q_2$ is either a prime or 1, and further $q_2 \ll f(N)$. Here $f : \mathbb{N} \rightarrow \mathbb{R}^+$ is a slowly increasing function that tends to infinity. Then for what choice of $f$ can we prove $\mathcal{R}_f(N) \rightarrow \infty$?

Chen's theorem would amount to the choice $f = N$ say and the result would be $r(N) = \mathcal{R}_f(N) \gg \mathfrak{S}(N) \frac{2N}{(\log N)^2}$, where $\mathfrak{S}(N) = \displaystyle \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right) \prod_{\substack{p | N \\ p > 2}} \frac{p-1}{p-2}.$

For instance, can we prove something analogous to Chen's theorem if we pick say $f = N^\theta$ for some $\theta \leq 1/3 - \epsilon$ for some $\epsilon > 0$? (See the edit for why the choice of $1/3 - \epsilon$ is chosen)

The motivation of course is if we can choose $f(N)$ so that $q_2 \leq 1$ for all sufficiently large $N$, then we would have proven the asymptotic binary Goldbach conjecture. If $q_2$ can be chosen to be bounded for all $N$ sufficiently large, that would still be a very tantalizing result.

Edit: I looked at the proof of Chen's theorem in Nathanson's book "Additive Number Theory: The Classical Bases" again and in that argument he (actually, the proof in Nathanon's book is due to Iwaniec) basically gives a lower bound for $r(N)$ by considering the case when $N = q_1 + q_2 q_3$, where $q_2, q_3$ are primes at least $N^{1/3}$. In light of this, it seems that my question is trivial when $f(N) \gg N^{1/3}$. Hence I modified the question by forcing $f$ to be smaller.

The theorem of the Chinese mathematician Chen Jingrun is currently one of the best results with respect to the binary Goldbach problem. It asserts that every even integer $n$ is in the sumset $P + P_2$, where $P$ denotes the set of primes and $P_2$ denotes the set of positive integers with at most two prime factors. However, I am not aware of this result asserting any control over the potential prime factors arising from the summand in $P_2$. In particular, I am interested in whether the following variant of Chen's theorem can be obtained:

Let $\mathcal{R}_f(N)$ denote the number of ways that $N$ can be written in the form $p + q_1 q_2$, where $p, q_1$ are primes and $q_2$ is either a prime or 1, and further $q_2 \ll f(N)$. Here $f : \mathbb{N} \rightarrow \mathbb{R}^+$ is a slowly increasing function that tends to infinity. Then for what choice of $f$ can we prove $\mathcal{R}_f(N) \rightarrow \infty$?

Chen's theorem would amount to the choice $f = N$ say and the result would be $r(N) = \mathcal{R}_f(N) \gg \mathfrak{S}(N) \frac{2N}{(\log N)^2}$, where $\mathfrak{S}(N) = \displaystyle \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right) \prod_{\substack{p | N \\ p > 2}} \frac{p-1}{p-2}.$

For instance, can we prove something analogous to Chen's theorem if we pick say $f = N^\theta$ for some $\theta \leq 1/8 - \epsilon$ for some $\epsilon > 0$? (See the edit for why the choice of $1/8 - \epsilon$ is chosen)

The motivation of course is if we can choose $f(N)$ so that $q_2 \leq 1$ for all sufficiently large $N$, then we would have proven the asymptotic binary Goldbach conjecture. If $q_2$ can be chosen to be bounded for all $N$ sufficiently large, that would still be a very tantalizing result.

Edit: I looked at the proof of Chen's theorem in Nathanson's book "Additive Number Theory: The Classical Bases" again and in that argument he (actually, the proof in Nathanson's book is due to Iwaniec) basically gives a lower bound for $r(N)$ by considering the case when $N = q_1 + q_2 q_3$, where $q_2, q_3$ are primes at least $N^{1/3}$. In light of this, it seems that my question is trivial when $f(N) \gg N^{1/3}$. Hence I modified the question by forcing $f$ to be smaller.

Edit 2: I misread the proof, the lower bound is actually $N^{1/8}$ instead of $N^{1/3}$.

The theorem of the Chinese mathematician Chen Jingrun is currently one of the best results with respect to the binary Goldbach problem. It asserts that every even integer $n$ is in the sumset $P + P_2$, where $P$ denotes the set of primes and $P_2$ denotes the set of positive integers with at most two prime factors. However, I am not aware of this result asserting any control over the potential prime factors arising from the summand in $P_2$. In particular, I am interested in whether the following variant of Chen's theorem can be obtained:

Let $\mathcal{R}_f(N)$ denote the number of ways that $N$ can be written in the form $p + q_1 q_2$, where $p, q_1$ are primes and $q_2$ is either a prime or 1, and further $q_2 \ll f(N)$. Here $f : \mathbb{N} \rightarrow \mathbb{R}^+$ is a slowly increasing function that tends to infinity. Then for what choice of $f$ can we prove $\mathcal{R}_f(N) \rightarrow \infty$?

Chen's theorem would amount to the choice $f = N$ say and the result would be $r(N) = \mathcal{R}_f(N) \gg \mathfrak{S}(N) \frac{2N}{(\log N)^2}$, where $\mathfrak{S}(N) = \displaystyle \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right) \prod_{\substack{p | N \\ p > 2}} \frac{p-1}{p-2}.$

For instance, can we prove something analogous to Chen's theorem if we pick say $f = N^\theta$ for some $\theta < 1$?

The motivation of course is if we can choose $f(N)$ so that $q_2 \leq 1$ for all sufficiently large $N$, then we would have proven the asymptotic binary Goldbach conjecture. If $q_2$ can be chosen to be bounded for all $N$ sufficiently large, that would still be a very tantalizing result.

The theorem of the Chinese mathematician Chen Jingrun is currently one of the best results with respect to the binary Goldbach problem. It asserts that every even integer $n$ is in the sumset $P + P_2$, where $P$ denotes the set of primes and $P_2$ denotes the set of positive integers with at most two prime factors. However, I am not aware of this result asserting any control over the potential prime factors arising from the summand in $P_2$. In particular, I am interested in whether the following variant of Chen's theorem can be obtained:

Let $\mathcal{R}_f(N)$ denote the number of ways that $N$ can be written in the form $p + q_1 q_2$, where $p, q_1$ are primes and $q_2$ is either a prime or 1, and further $q_2 \ll f(N)$. Here $f : \mathbb{N} \rightarrow \mathbb{R}^+$ is a slowly increasing function that tends to infinity. Then for what choice of $f$ can we prove $\mathcal{R}_f(N) \rightarrow \infty$?

Chen's theorem would amount to the choice $f = N$ say and the result would be $r(N) = \mathcal{R}_f(N) \gg \mathfrak{S}(N) \frac{2N}{(\log N)^2}$, where $\mathfrak{S}(N) = \displaystyle \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right) \prod_{\substack{p | N \\ p > 2}} \frac{p-1}{p-2}.$

For instance, can we prove something analogous to Chen's theorem if we pick say $f = N^\theta$ for some $\theta \leq 1/3 - \epsilon$ for some $\epsilon > 0$? (See the edit for why the choice of $1/3 - \epsilon$ is chosen)

The motivation of course is if we can choose $f(N)$ so that $q_2 \leq 1$ for all sufficiently large $N$, then we would have proven the asymptotic binary Goldbach conjecture. If $q_2$ can be chosen to be bounded for all $N$ sufficiently large, that would still be a very tantalizing result.

Edit: I looked at the proof of Chen's theorem in Nathanson's book "Additive Number Theory: The Classical Bases" again and in that argument he (actually, the proof in Nathanon's book is due to Iwaniec) basically gives a lower bound for $r(N)$ by considering the case when $N = q_1 + q_2 q_3$, where $q_2, q_3$ are primes at least $N^{1/3}$. In light of this, it seems that my question is trivial when $f(N) \gg N^{1/3}$. Hence I modified the question by forcing $f$ to be smaller.