# Proposal for Brilliant Polymath Project

Classify all (rational) $q$ such that, if $e^q$ has the continued fraction expansion $[a_0;a_1,a_2,\ldots]$, $\{ a_{mn+c} \}_{n=0}^\infty$ is an arithmetic progression for some $m$ and every $c>0$.

(Conjectured: All $q$ rational have this property.)

As examples, observe the following expansions:

\begin{aligned} q = 1, & e^q = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, \ldots], & m_{\min} = 3 \\ q = 2, & e^q = [7; 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12, 54, \ldots], & m_{\min} = 5 \\ q = 1/2, & e^q = [1; 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, \ldots], & m_{\min} = 3 \\ \end{aligned}

Known facts:

$e^z = \cfrac{1}{1 - \cfrac{z}{1 + z - \cfrac{z}{2 + z - \cfrac{2z}{3 + z - \cfrac{3z}{4 + z - \ddots}}}}}$

(Due to Euler.)

Note by Jake Lai
4 years, 2 months ago

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## Comments

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Top Newest

How did you come across such observation?

- 4 years, 2 months ago

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I noted the regularity of e's expansion in a YouTube lecture and thought that, since e is intimately linked with exponentiation, $e^x$ might display similar regularity as well.

- 4 years, 2 months ago

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A good place to start might be to prove this for $q = \frac{1}{x}$ ($x\in \mathbb{N}$), which Euler himself actually proved.

Staff - 4 years, 1 month ago

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Where is this result found? It would be great to look at the proof for insight :)

- 4 years, 1 month ago

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Here is an English translation of Euler's paper.

Staff - 4 years, 1 month ago

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@Jake Lai Have you had a chance to think about this more? (I haven't, but am definitely curious.)

Staff - 4 years, 1 month ago

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I have not, sadly. Been too busy with too much :<

- 4 years, 1 month ago

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