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Consider the following number:

$R=\frac{1}{9}\sum^\infty_{n=1} 10^{-\frac{n\left(n+1\right)}{2}}\left(10^n-1\right)\left(n\left(\operatorname{mod}10\right)\right)$

=0.122333444455555666666777777788888888999999999000000000011111111111222222222222333333333333344444444444444555555555555555666666666666666677777777777777777888888888888888888999999999999999999900000000000000000000...

which is formed by concatenating $n$ copies of $n\left(\operatorname{mod}10\right)$ after $0$.

The long sequences of repeating digits allow better and better rational approximations as the lengths of repeating digit blocks grow. Can this be a basis to prove this number transcendental?

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    $\begingroup$ Maybe, Liouville's theorem or something similar can help here. $\endgroup$
    – Peter
    May 29, 2019 at 16:52
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    $\begingroup$ The approximation doesn’t look nearly sharp enough to do this by irrationality measure. However, the exact value looks like it could be written down as a combination of theta values. There are transcendence results available for those, but I’m not aware of one that would handle combinations thereof. Maybe the binary version of this question is answerable though. $\endgroup$
    – Erick Wong
    May 29, 2019 at 17:08

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