A Question Inspired by Pi Han Goh

Can there be more than 5 positive integers such that they are in a harmonic progression ie there reciprocals are in arithmetic progression? If yes, the find some with more than 5 terms and show your working, and if not, prove your observation.

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Note that $\frac{1}{n!}, \frac{2}{n!}, \frac{3}{n!}, \ldots, \frac{n}{n!}$ form an arithmetic progression, and the numerator divides the denominator so they all simplify to unit fractions. Thus their reciprocals are positive integers that form a harmonic progression. Take $n$ as large as you want.

Ivan Koswara - 5 years, 2 months ago

That's a cool solution. Do other solutions exist?

Kushagra Sahni - 5 years, 2 months ago

Of course. Let $a_1, a_2, \ldots, a_n$ be an arithmetic progression of positive integers, and let $P$ be their least common multiple. Then $\frac{kP}{a_1}, \frac{kP}{a_2}, \ldots, \frac{kP}{a_n}$ is a solution, for any positive integer $k$ . (The above is when you put $a_i = i$ and $k$ is such so $kP = n!$ .)

Whether the above gives all the solutions, I haven't proved it yet, although I think it does.

Ivan Koswara - 5 years, 2 months ago

@Ivan Koswara – Yes I guess that covers all.

Kushagra Sahni - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$