TITLE: $p$-adic valuation of harmonic numbers
QUESTION [8 upvotes]: For an integer $m$ let $\nu_p(m)$ be its $p$-valuation i.e.  the greatest non-negative integer such that $p^{\nu_p(m)}$ divides $m$. Let now 
$H_n=1+\dfrac{1}{2}+ \cdots+ \dfrac{1}{n}$. 
If $H_n=\dfrac{a_n}{b_n}$  then $\nu_p(H_n)=\nu_p(a_n)-\nu_p(b_n).$
It is known that if $2^k \leq n < 2^{k+1}$  then $\nu_2(H_n)=-k.$   
Question   For what $n$ and arbitrary $p$  we may be sure that $\nu_p(H_n)<0?$
Any estimates and reference, please.

REPLY [2 votes]: 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)