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Let's define a function: $$f:\mathbb{N}\rightarrow \mathbb{R}$$ $$f(n) = \prod_{p|n}(1+\frac{2}{p})$$ Where $p\in \mathbb{P}$

Find a differentiable function $g:\mathbb{R}\rightarrow\mathbb{R}$ such $f(n) = O(g(n))$ as $n$ goes to infinity. Where $O$ is a big O.

I have an intuition that $g(x) = \log(\log (x))$. I have calculated $\frac{f(n)}{\log(\log (n))}$ on my computer up to $4000$ numbers, and it seems that this quotient is bounded from above by $18$ and from below by $0.4$. However i would like to be sure, whether $\log(\log(x))$ is a good function.

Regards.

-Edit-

As far i understand, i can do the following: $$f(n)=\prod_{p|n,p\in \mathbb{P}}(1+\frac{2}{p})\le \prod_{1\le k\le \omega(n)}(1+\frac{2}{p_k})$$ Where $p_k$ is the $k$-th prime number and $\omega(n)$ is the number of prime divisors of $n$. Let's use a logarithm here. Notice that $\log(1+x)\le x$. $$\log f(n) \le \sum_{k=1}^{\omega(n)}\log(1+\frac{2}{p_k})\le 2\sum_{k=1}^{\omega(n)}\frac{1}{p_k}$$ $$2\sum_{k=1}^{\omega(n)}\frac{1}{p_k}=2\log\log\omega(n)+O(1)$$ Hence: $f(n) = \exp(\log(f(n)))\le \exp(2\log\log\omega(n)+O(1))=e^{O(1)}(\log\omega(n))^2$. Because $\omega(n) = O(\log(n))$, now we see that $f(n)=O((\log\log(n))^2)$.

Am i right? Please correct me if i am wrong.

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    $\begingroup$ It will be more like this, I mean Mertens' third theorem. $\endgroup$
    – rtybase
    Commented Mar 14, 2021 at 13:27

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Anything you don't like with $3^n$ ? The better result is $f(n)\le f(\prod_{p\le k_n} p)$ with $ \prod_{p\le k_n} p$ the largest primorial $\le n$ so that (from the $\theta(2m)-\theta(m)\le\log {2m\choose m}\le \psi(2m)$ argument) $k_n =O(\log n)$ and $\log n=O(k_n)$ ie. $\log k_n\sim \log \log n$.

Mertens theorem gives $\log f(\prod_{p\le k_n} p)=2 \log \log k_n+C+o(1)$. Whence $f(n)\sim e^C (\log\log n)^2$.

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