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The idea is a more convenient form for $N = 0.01001000100001000001...$ in base $r$, hopefully to show whether it is transcendental.

Sorry for brevity.

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    In the title, it's natural to take $r=1/10$, but in the body it's evident that $r$ is meant to be 10. We can square these by having $r=10$ and $a\lt0$, but maybe it's more natural to write $\sum r^{-(an^2+bn+c)}$ with$a$quadratic that's non-negative for non-negative $n$.2011-08-01

3 Answers 3

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Of course $N$ is transcendental. [I said it is, I didn't say I can prove it.] It is conjectured that all irrational algebraic numbers are normal in all bases. If this were not transcendental, it would be a spectacular counterexample to that conjecture.

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This may come under the heading of "Siegel E-functions" or "Siegel G-functions", for which transcendence results are known.

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Here's a technique for this sort of sum (though this doesn't address the transcendentality question) I found a while ago which involves incomplete theta functions:

\begin{equation} \sum_{n=0}^{\infty} q^{an^{2}+2akn+ak^{2}+p} = q^{p}\left[\frac{\theta_{3}(0,q^{a})+1}{2} - \sum_{m=0}^{k-1} q^{am^{2}}\right] \end{equation}

for example:

\begin{equation} \sum_{n=0}^{\infty} \frac{1}{e^{7n^{2}+70n+173}} = e^{2}\left[\frac{\theta_{3}(0,1/e^{7})+1}{2} - \sum_{m=0}^{4} (1/e^{7})^{m^2}\right] \end{equation}

I'm not sure it works for arbitrary $a$, $b$, and $c$, though.