Range of inverse harmonic mean of two integers

Question

Today I was solving an exercise and one of the things I tried (which later turned out to be useless) involved considering the following:

Is there a simple way to describe in terms of $n$ the range of the inverse harmonic mean of two integers?

Let $m, k \in \mathbb N$, $m,k \le n$. Consider

$$\frac 1m + \frac 1k = \frac 2{\mathcal H(m,k)}$$

For which $f: \mathbb N \mapsto \mathbb Q$ can I find $m,k$ such that

$$\frac 1m + \frac 1k = \frac 2{\mathcal H(m,k)} = f(n)$$?

Answer 1

So you are looking for a function $f:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{Q})$,where $\mathcal{P}(\mathbb{Q})$ is the power set of the rationals, rather than a function $f:\mathbb{N}\rightarrow\mathbb{Q}$. The way you are posing the problem causes some confusion.

$\mathcal H(m,k)$ obtains its min value for $m=k=1$ and then $\mathcal H(1,1)=1$ and its max value for $m=k=n$ and then $\mathcal H(n,n)=n$. So, generally, $$f(n)=\left\{\frac{2mk}{m+k}|k,m=1,2,...n\right\}\subset\mathbb{Q}$$

Answer 2

First, I want to note that this is not a full solution, it just makes finding $a,b$ with $\frac 1a+\frac 1b=\frac pq$ a little easier.

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All solutions to $$a+b|ab$$ are $(a,b)=(\alpha\gamma(\alpha+\beta),\beta\gamma(\alpha+\beta))$ for some $\alpha,\beta,\gamma$.

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So if $$\frac 1a+\frac 1b=\frac 1n$$ then substitution yields $n=\alpha\beta\gamma$. But we wish to find solutions to $\frac 1a+\frac 1b=\frac pq$ or, when dividing by $p$, we need $$\frac 1{pa}+\frac 1{pb}=\frac 1q$$ So $\alpha\beta\gamma=q$, and $ap=\alpha\gamma(\alpha+\beta)$ and $bp=\beta\gamma(\alpha+\beta)$. But since $\gcd(p,q)=1$, we know that $p|\alpha+\beta$.

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We'll try to get $a$ and $b$ as low as possible now too see whether or not they can get less than or equal to $n$. We have a factorization of $q$, namely $\alpha\beta\gamma$, but we also need $p|\alpha+\beta$, where $p\not|\alpha,\beta$, as well as $ap,bp\leq n$. As you can see, the solvability of this highly depends on the factorization of $p$ and $q$, so I don't think there's a closed formula for it. The result does however give you an easier method to seeing whether or not there is a solution to $\frac 1a+\frac 1b=\frac pq$.