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.



Is $\Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$ unique $\forall n \in \Bbb{N}, n>1$
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 $p >m/2$.


