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.)

## Comments

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TopNewestHow did you come across such observation? – Julian Poon · 7 months, 3 weeks ago

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– Jake Lai · 7 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.Log in to reply

A good place to start might be to prove this for \(q = \frac{1}{x}\) (\(x\in \mathbb{N}\)), which Euler himself actually proved. – Eli Ross Staff · 7 months, 3 weeks ago

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– Jake Lai · 7 months, 3 weeks ago

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

English translation of Euler's paper. – Eli Ross Staff · 7 months, 3 weeks ago

Here is anLog in to reply

@Jake Lai Have you had a chance to think about this more? (I haven't, but am definitely curious.) – Eli Ross Staff · 7 months, 1 week ago

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– Jake Lai · 7 months, 1 week ago

I have not, sadly. Been too busy with too much :<Log in to reply