More sensible heuristics

If you’re not familiar with the problem I’m semipublicly trying to solve, please start reading here. Coming to think about it, this is basically just another polymath-project. With the only difference that not a lot of people participate 😛 Hopefully that’ll change though. Anyway, the heuristics and assumptions in the previous post may seem a bit weird, so I try to make things more senseful  here; let $n$ be given. Assume $a_ip_i^{k_i} \le n < (a_i+1)p_i^{k_i}$ with $1 \le a_i < p_i$ and where the $p_i$ run over all primes such that $2 < p_i < n$. Then $g_n = 1$ if, and only if $p_i$ does not divide $X_{a_i}$ for all $i$. Let $f(p)$ be the amount of $a$ for which $a < p$ and $p | X_a$ (so $f(p)$ is the function we want an decent upper bound on). Then the ‘chance’ that $p$ doesn’t divide $X_n$ (assuming for simplicity that $P(a_i = x) = \dfrac{1}{p-1}$ for all $x$, $1 \le x \le p-1$), equals $\dfrac{p - 1 - f(p)}{p - 1}$. The reasonable assumption (and possibly not even that hard to make rigorous) seems that $a_i$ and $a_j$ are independent for $i \neq j$. So the ‘chance’ that $p_i$ does not divide $X_{a_i}$ for all $i$, should be $\displaystyle \prod_{2 < p_i < n} \left(1 - \dfrac{f(p)}{p-1}\right)$.  And if we have a good upper bound on $f(p)$, say $f(p) = O(\log p)$, then we should be able to show that this product doesn’t go to $0$ fast enough to prevent $g_n = 1$ from happening.