$p$-adic valuation of harmonic numbers

As stated by Elaqqad, the question if any fixed prime $p$ divides only a finite number of numerators of $H_n$ is a (probably difficult) open problem, see [1] [2].

Anyway, I can prove the following lemma. You can use it to show that $\nu_p(H_n) < 0$ for any sufficienly large $n$, for small primes like $p=2,3,5$.

Lemma 1. If there exists a positive integer $k$ such that $\nu_p(H_m) \leq 0$ for all integers $m \in [p^k,p^{k+1}[$, then $\nu_p(H_n) < 0$ for any positive integer $n \geq p^{k+1}$.

Proof. We have $$H_n = \sum_{j \, = \, 1}^n \frac1{j} = \sum_{\substack{1 \, \leq \, j \, \leq \, n \\ p \,\nmid\, j}} \frac1{j} + \sum_{d \, \leq \, n / p} \frac1{p d} = \sum_{\substack{1 \, \leq \, j \, \leq \, n \\ p \,\nmid\, j}} \frac1{j} + \frac1{p} \cdot H_{\lfloor n / p \rfloor} = A + B ,$$ where $\nu_p(A) \geq 0$. Hence, if $\nu_p(H_{\lfloor n / p \rfloor}) \leq 0$ then $\nu_p(B) \leq -1 < \nu_p(A)$ and therefore $\nu_p(H_n) = \nu_p(B) < 0$. With this last consideration, the claim follows easily by induction on the number of digits of $n$ in base $p$. []

[1] A. Eswarathasan and E. Levine, p-integral harmonic sums, Discrete Math. 91 (1991), 249–257.

[2] D. W. Boyd, A p-adic study of the partial sums of the harmonic series, Experiment. Math 3 (1994), 287--302.

Question For what $n$ and arbitrary $p$ we may be sure that $v_p(H_n)<0$ ?

Any estimates and reference, please.

In $[1]$ it is proved that for all $x \geq 1$ it holds $v_p(H_n) \leq 0$ for all positive integers $n \leq x$ but at most $129 p^{2/3} x^{0.765}$ exceptions. Hence $v_p(H_n) \leq 0$ for a 100% of the positive integers (respect to asymptotic density) and I guess that this is true even for the strict inequality $v_p(H_n) < 0$.

$[1]$ C. Sanna, On the p-adic valuation of harmonic numbers, J. Number Theory (2016) 166, 41-46. (http://dx.doi.org/10.1016/j.jnt.2016.02.020)