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# 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{align} 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{align}

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
11 months, 3 weeks ago

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How did you come across such observation? · 11 months, 3 weeks ago

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. · 11 months, 3 weeks ago

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 · 11 months, 3 weeks ago

Where is this result found? It would be great to look at the proof for insight :) · 11 months, 2 weeks ago

Here is an English translation of Euler's paper. Staff · 11 months, 2 weeks ago

@Jake Lai Have you had a chance to think about this more? (I haven't, but am definitely curious.) Staff · 11 months, 1 week ago

I have not, sadly. Been too busy with too much :< · 11 months, 1 week ago

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