Proposal for Brilliant Polymath Project

Classify all (rational) qq such that, if eqe^q has the continued fraction expansion [a0;a1,a2,][a_0;a_1,a_2,\ldots], {amn+c}n=0\{ a_{mn+c} \}_{n=0}^\infty is an arithmetic progression for some mm and every c>0c>0.

(Conjectured: All qq rational have this property.)


As examples, observe the following expansions:

q=1,eq=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,],mmin=3q=2,eq=[7;2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,11,1,1,12,54,],mmin=5q=1/2,eq=[1;1,1,1,5,1,1,9,1,1,13,1,1,17,1,1,],mmin=3\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:

ez=11z1+zz2+z2z3+z3z4+ze^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
3 years, 8 months ago

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How did you come across such observation?

Julian Poon - 3 years, 8 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, exe^x might display similar regularity as well.

Jake Lai - 3 years, 8 months ago

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

Eli Ross Staff - 3 years, 8 months ago

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

Jake Lai - 3 years, 8 months ago

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Here is an English translation of Euler's paper.

Eli Ross Staff - 3 years, 8 months ago

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

Eli Ross Staff - 3 years, 8 months ago

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

Jake Lai - 3 years, 8 months ago

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