## A proof of the aforementioned theorem and ideas about some sort of probabilistic proof of the infinitude of n for which g_n = 1

If you already have no idea what I’m talking about; please read the last $4$ posts.

Let $n = ap^k$ with $2 \le a \le p-1$. Then we have $p | X_n$ if, and only if $X_{n-1} + \dfrac{L_n}{n} \equiv 0 \pmod{p}$;

$X_{n-1} + \dfrac{L_n}{n} \equiv L_n \displaystyle \sum_{i=1}^{n-1} \dfrac{1}{i} + \dfrac{L_n}{n} \pmod{p}$
$\equiv L_n \displaystyle \sum_{i=1}^n \dfrac{1}{i}$ $\equiv L_n \displaystyle \sum_{i=1}^a \dfrac{1}{ip^k} \pmod{p}$
$\equiv \dfrac{L_n}{p^k} \displaystyle \sum_{i=1}^a \dfrac{1}{i} \equiv \dfrac{L_n}{p^k}\dfrac{X_a}{L_a} \pmod{p}$

And this is equal to $0 \pmod{p}$ if, and only if $X_a \equiv 0 \pmod{p}$. And this suffices to prove the theorem I mentioned in the previous post. And, by Wolstenholme´s theorem, we know that $p | X_{p-1}$ for all $p \ge 3$. So for every number $n$, for which $(p-1)p^{k-1} \le n < p^k$ holds, we have that $p | X_n$. Assume for the moment that the following converse holds: $a = p-1$ is the only $a$ for which $p | X_a$.  While this is not true ($a = 3$$p = 11$ is a counterexample), there should be a relatively low upper bound for the amount of $a$ with $a < p$ and $p | X_a$. In any case, bear with me for the moment. Let’s pick some ‘random’ number $m$ and let’s assume that it lies pretty much in the middle of two consecutive powers of $p$. Then, the amount of numbers $n$, with $n$ $< m$ and for which $p | X_n$, is roughly equal to $\dfrac{2m}{(p-1)^2}$.  Now the sweet part: if we sum over all primes $< m$, then what we obtain is not more than the number of $n$ such that $X_n$ is divisible by any prime $< m$. And $\displaystyle \sum_{p=3}^m \dfrac{2m}{(p-1)^2}$ $< \displaystyle \sum_{p=3}^\infty \dfrac{2m}{(p-1)^2} \approx 0.75m$. So at least about 25% of all the $X_n$ with $n < m$ is not divisible by any prime $< m$, which implies $g_n = (X_n, L_n) \le (X_n, L_m) = 1$.

Obviously, this doesn’t come very close to a real proof. But it might be a good heurstic start. Anyway, it certainly seems necessary to have some bounds on the $a$ for a given $p$, such that $a < p$ and $p | X_a$. Let me know what you think!