Is $\Bigl\{ n \sum_{k=2}^{n-1} \frac{1}{k}\Bigr\}$ unique $\forall n \in \Bbb{N}, n>1$ 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.
 A: Here is a proof that $nH_n-mH_m$ cannot be an integer for $n>m$ in the case that $n$ is odd while $m$ is even. The argument in that case follows directly from the proof that the difference of two distinct harmonic numbers cannot be an integer.
Write the expression out as
$$nH_n-mH_m=\frac{n-m}{1}+\frac{n-m}{2}+\cdots \frac{n-m}{m}+\frac{n}{m+1}+\frac{n}{m+2}\cdots +\frac{n}{n}.$$
According to lemma 1 in the cited proof one and only one number in the sequence $1,2,\ldots n$ has the highest power of two. Denote this number by $c=2^p q$ with $q$ an odd integer. The lowest common multiple of all the $n$ denominators equals $d=2^p q'$ with $q'$ odd.
I now bring each term in the sum under the common denominator $d$, and ask whether the numerator is even or odd. By construction all numerators are even except possibly the numerator of the special term $c$. If I can prove that that numerator is odd I am done, because then the sum will equal an odd integer divided by the even integer $d$, which cannot be an integer.
Let me distinguish the case that $c\leq m$ from the case that $c>m$. In the first case $c\leq m$, the numerator of that special term is equal to an odd number times $n-m$. In the second case $c>m$ the numerator equals an odd number times $n$. In both cases the numerator is odd, because of our assumption that $n$ is odd while $m$ is even.
A: Only a partial answer for now, as it is too long for a comment. From my comment and Carlo Beenakker's answer, it suffices to consider the case where $n$ and $m$ have different radicals but the same parity.
Write $n=\prod_{l=1}^{r_{n}}p_{l}^{\alpha_{l}}$ and $m=\prod_{l=1}^{r_{m}}p_{l}^{\beta_{l}}$.
Let $g:=gcd(n,m)$ where $n<m$, $n'=n/g$ and $m'=m/g$ and write $g=\prod_{l=1}^{r_{g}}p_{l}^{\gamma_{l}}$.
The idea is to consider that there are three different classes of integers greater than $1$ and below $n$: non trivial divisors of $n$ making the class $D_{n}$, integers coprime with $n$ and greater than $1$ making the class $C_{n}$ and "mixed" integers that can be uniquely written as a product of elements of the two previous classes, making the class $M_{n}$.
Denote by $F(a):=\{a\sum_{k=2}^{a-1}\frac{1}{k}\}$. The goal is to prove that the sum over the $k$ that are greater than $g-1$ of the $\frac{1}{k}$ give rise to different denominators depending on the value of $a$, namely $n$ or $m$.
A: 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 $p >m/2$.
