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 = 3p = 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!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: