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.

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