TITLE: Is $\Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$ unique $\forall n \in \Bbb{N}, n>1$
QUESTION [12 upvotes]: If we define $$f(n) = \Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$$
is it true that $f(n) \ne f(m)$ whenever $n \ne m, \forall m,n \in \Bbb{N}$ (where the curly braces denote the fractional part)?
I wanted to explore coming up with a kind of "global residue" concept for a number, based on all the numbers less than it.

REPLY [13 votes]: It follows from Bertrand's postulate that $mH_m - nH_n$ is an integer only in the case $n=m$.  Suppose $n<m$, and assume that $m$ is reasonably large below (say $\ge 20$).
Let $p$ denote the largest prime below $m$, so that by Bertrand's postulate $p>m/2$.   The only term with $p$ in the denominator in $mH_m$ is $m/p$ which is not an integer (since $m/2 <p <m$).  So if there is no term with $p$ in the denominator in $nH_n$, then $mH_m -nH_n$ cannot be an integer.  This takes care of the case when $p>n$.
If $p\le n$, then $nH_n$ has the unique term $n/p$ with $p$ in the denominator.  But the only way $m/p - n/p$ can be an integer is if $n=(m-kp)$ (for some integer $k\ge 1$) and we also have $p\le n$, but this is impossible since $p >m/2$.