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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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    We can factor out an $r^c$ term from the sum. For special values of $a$ and $b$, the sum is expressible in terms of Jacobi theta functions, but I know of no closed form for the general form you have.2011-07-31
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    let's say $r=2^{-1/2}$, $a=1$, $b=5$, $c=4$, which I think makes the series equal to $N$ in binary. Also, thanks for fixing my TeX.2011-07-31
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    The $b=5$ portion would be troublesome in figuring a closed form for what you say you have...2011-07-31
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    @J.M. I tried to complete the squares hoping to kill the linear term, but it didn't work because $5$ "happened" to be an odd number. Did you anticipate this in your comment? Also, if I gave you just $an^2+c$, then do closed-form solutions exist?2011-07-31
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    Yeah, something like that @Srivatsan. ;) For your much simpler case, you have a representation in terms of the third Jacobi theta function: $\frac{r^c}{2}(1+\vartheta_3(0,r^a))$2011-07-31
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    "their applicability would depend on the quadratic not containing the linear term." - not necessarily, @Srivatsan. You'll notice for instance that the series for the second Jacobi theta function involves triangular number exponents...2011-07-31
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    @J.M. Grrr, I don't know why I am getting all muddled up with these formulas. Thanks for pointing it to me. I think I will look at them carefully after I get some sleep. :-) (Btw I guess I'll just delete my previous redundant comments tomorrow.)2011-07-31
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    No need @Sri; at least you've provided the opportunity for me to mention the "cream puff" cases. ;)2011-07-31
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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

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