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Let $n \in \mathbb{N}$ be a positive integer.

It is possible to prove that the equation $F_1(n)=m$ where $m \in \mathbb{Z}$ and $$ F_1(n)=(1+2+\ldots+n)\cdot\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}\right) $$ is solvable only for $n=3$ and $m=11$. I've tried to generalize the result for the function $$ F_k(n)=\big(1^k+2^k+\ldots+n^k\big)\cdot\left(\frac{1}{1^k}+\frac{1}{2^k}+\dots+\frac{1}{n^k}\right) $$ where $k \geq 2$, i.e. asking for which $n$ the expression $F_k(n)$ is an integer, but I'm not succeeding. Does some of you know if this equation is solvable in general or for some particular values of $k$, for example for $k=2$?

I've tried with the basic elementary number theory techniques. Thanks in advance.

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    $\begingroup$ It should be noted that $n=m=1$ is also a solution. $\endgroup$
    – Aeryk
    Commented Nov 8, 2022 at 19:44
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    $\begingroup$ for $k=2$ one needs that $n(n+1)(2n+1)/6$ cancels out all $1/p^2$ for $n/2 <p \le n$ prime, so in particular $n^2/4 \le p^2 \le 2n+1$ for at least one such $p$ by Bertrand since $p$ can divide only one of the three factors, or $n \le 8$ etc; similar arguments work for higher $k$ given that we know that $1^k+2^k+\ldots+n^k$ is a polynomial of degree $k+1$ in $n$ $\endgroup$
    – Conrad
    Commented Nov 8, 2022 at 20:26
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    $\begingroup$ $F_1(n)=\frac{n\ (n+1)}{2} H_n$ and $F_k(n)=H_n^{(-k)}\ H_n^{(k)}$ where $H_n$ is the harmonic number and $H_n^{(k)}$ is the harmonic number of order $k$. $\endgroup$
    – Steven Clark
    Commented Nov 8, 2022 at 20:48

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