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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

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

Ant
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  • See http://math.stackexchange.com/questions/840720/qm-am-gm-hm-proof-help – lab bhattacharjee Feb 02 '16 at 13:28
  • @labbhattacharjee I am aware of the result, but I fail to see how to apply it to my problem. I don't want to just find the maximum and minimum of my expression – Ant Feb 02 '16 at 13:29
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    Your question is unclear. Are you looking for a way to describe all values assumed by $\frac1m +\frac1k$ as the integers $m$ and $k$ range from $1$ to $n$? – MPW Feb 02 '16 at 13:56
  • @MPW Indeed! :) to describe them as a function of $n$. – Ant Feb 02 '16 at 14:10

2 Answers2

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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, $$f(n)=\left\{\frac{2mk}{m+k}|k,m=1,2,...n\right\}\subset\mathbb{Q}$$

mathreadler
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KonKan
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  • well yeah, that's what I wrote :) I want to find an easy way to see if a particular fraction can be written that way. For example: can $\frac 1n$ be written that way? What about $\frac {1}{n^2}$? and so on – Ant Feb 12 '16 at 21:28
  • I am not sure I'm getting where the problem is: just check whether the given fraction is inside the range given in the answer above. If you have a well defined set and want to see whether a given value is inside it, then it is straightforward to check. For example: if you wanna check whether $\frac{\alpha}{\beta}=\frac{1}{m}+\frac{1}{k}$ for given $\alpha, \beta$ just solve the Diophantine equation $\alpha mk=\beta(m+k)$ for $m,k=1,2,...,n$. – KonKan Feb 12 '16 at 22:17
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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.

All solutions to $$a+b|ab$$ are $(a,b)=(\alpha\gamma(\alpha+\beta),\beta\gamma(\alpha+\beta))$ for some $\alpha,\beta,\gamma$.
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 $p|\alpha+\beta$.
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 $\frac 1a+\frac 1b=\frac pq$.
  • I'll probably can't read this seriously for about 2 days, but since the bounty expires today and the answer seems interesting I decided to award it to you :-) I'll come back later with some remarks :) – Ant Feb 13 '16 at 16:22