Question

Let $n\geq 1$ be an integer, $P$ the set of rational prime numbers. I am interested in upper bounds for $\prod_{p\in P; \ p\mid n}(1+1/p)$ in terms of n.

I would like to find explicit real numbers $a,b$ such that for any integer $n\geq 1$ it holds $\prod_{p\in P; \ p\mid n}(1+1/p)\leq a\log(n)^b$.

By the answer of GH this is possible for any $b>0$ provided $n$ is sufficiently large.

Now, I would like to find explicit real numbers $a,c$ such that for any integer $n\geq 10$ it holds $\prod_{p\in P; \ p\mid n}(1+1/p)\leq a\log(n)+c$.

What are small possible values for $c$ and $a$?

Answer 1

If you want a weaker bound that is completely explicit, the following method is pretty simple. Let $\omega(n)$ denote the number of distinct prime factors of $n$. Since the function $1+1/x$ is a decreasing function of $x$, and the smallest prime equals 2, we certainly have $$ \prod_{p\mid n} \bigg( 1+\frac1p \bigg) \le \prod_{k=2}^{\omega(n)+1} \bigg( 1+\frac1k \bigg) = \frac{\omega(n)+2}2 $$ since the product telescopes. Moreover, $\omega(n) \le (\log n)/(\log 2)$, again since each prime factor of $n$ is at least 2. Therefore $$ \prod_{p\mid n} \bigg( 1+\frac1p \bigg) \le \frac{\log n}{2\log2}+1 $$ (with equality at $n=1$ and $n=2$).

Answer 2

It is easy to verify that for sufficiently large $n$ $$ \sum_{p\mid n}\frac{1}{p}<\sum_{p<2\log n}\frac{1}{p}<\log\log\log n+O(1) $$ whence $$ \prod_{p\mid n} \left( 1+\frac{1}{p} \right) \ll \log\log n.$$ In other words, you can choose any $b>0$. Also, one can further refine the above bound.