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I'm learning Selburg Sieve in an Additive Number Theory book. Let $N$ be a positive even integer and $r(N)$ denote the number of primes $p$ less than $N$ such that $N-p$ is also a prime, in other words, $r(n)$ counts the representation of $N$ as a sum of two primes. Then $$r(N)=\mathcal O\bigg(\dfrac{N}{(\log N)^2}\prod_{p|N}\bigg(1+\dfrac 1p\bigg)\bigg).$$ The smaller terms as stated in the book are up to $N^{9/10}$, so I guess the order of $\prod_{p|N}(1+p^{-1})$ shouldn't exceed $(\log N)^2$, and then I have do some calculation on estimating the order:

Let $N=p_1^{k_1}\cdots p_r^{k_r}$ where each $k_i\geq 1$, then let $M=p_1\cdots p_r$, the "square-free kernel" of $N$, then the product is the same when calculating with respect to $M$:

\begin{align*}\prod_{p|N}\bigg(1+\dfrac1p\bigg)&=\prod_{p|M}\bigg(1+\dfrac1p\bigg)\\ &=\sum_{d|M}\dfrac1d\\ &\leq \sum_{d\leq M}\dfrac1d\\ &=\mathcal O(\log M)\\ &=\mathcal O(\log N).\end{align*} which fulfill my expectation. But is there any well-known smaller estimate about this product? For example I wish it could be estimated as $\mathcal O(\log\log N)$ or other function smaller than $\log N$.

(Sorry if this post is duplicated because I don't know the name of this product. )

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Let $f(N)=\prod_{p\mid N}(1+1/p)$. For $N=\prod_{1\leqslant i\leqslant r}p_i^{k_i}$ (like above, $p_i$ are distinct primes and $k_i>0$), let $m(N)$ be the product of the first $r$ primes, and let $l(N)$ be the largest (i.e., $r$-th) of these primes.

From $m(N)\leqslant N$ and $\lim\limits_{n\to\infty}(1/n)\sum_{p\leqslant n}\ln p=1$ ($\approx$ PNT) we have $l(N)=\mathcal{O}(\ln N)$.

Now $f(N)\leqslant f(m(N))=\mathcal{O}(\ln l(N))$ because Mertens' results give $$\lim_{n\to\infty}\left[\ln n\prod_{p\leqslant n}\Big(1-\frac{1}{p}\Big)\right]=e^{-\gamma}\implies\lim_{n\to\infty}\left[\frac{1}{\ln n}\prod_{p\leqslant n}\Big(1+\frac{1}{p}\Big)\right]=\frac{6e^{\gamma}}{\pi^2}.$$ Finally, $f(N)=\mathcal{O}(\ln\ln N)$ as expected.

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