Let $\Phi(n)=\{p_1, p_2, ..., p_k\}$ be the set of prime factors of a number $n$. How does
$$ p_1(n) = \prod_{p_i\in\Phi(n) \\ 1 \le i \le k}{p_i} $$
compare to
$$ p_2(n) = \prod_{p_i\in\Phi(n) \\ 1 \le i \le k}{(p_i +1)} \hspace{1em}?? $$
Just computing $p_2(n) - p_1(n)$ for $n$ up to a million, it ranges from $1 \le p_2(n) - p_1(n) \le 2165646$. Computing $\frac{p_2(n)}{p_1(n)}$, gives me $1 \le \frac{p_2(n)}{p_1(n)} \le 3.4119292472233647$. However, the difference between these two functions seem to vary wildly. For example, $p_2(10^6) - p_1(10^6) = 8$ and $\frac{p_2(10^6)}{p_1(10^6)} = 1.8$. My question is, what is the asymptotic behavior of $p_2(n)$ compared to $p_1(n)$? Or, alternatively, is there another way of representing $p_2(n)$ that will make this problem easier?
I know that we can convert the product $p_2(n)$ to a sum as
$$ \prod_{p_i\in\Phi(n) \\ 1 \le i \le k}{(p_i + 1)} = e^{\sum{\log(p_i + 1)}} $$
But this clearly doesn't help in simplifying the product. What other methods am I overlooking, if there are any?
Note that $$ \frac{p_2(n)}{p_1(n)} = \prod_{p\mid n} \biggl( 1+\frac1p \biggr); $$ this function is unbounded (but not by much—it's always smaller than a constant times $\log(\log n)$). I think this is a more natural object then $p_2(n)-p_1(n)$.