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or the reverse, where a repeating expansion in a variable base to be rational? been trying some trial and error cases without success

by variable base we mean each digit can be in a different base but any digit can occur as many times in the same base or not

Just in case anybody tries to pull a fast one no base $\sqrt 2$, $\pi$ , $e$ etc. can not have the irrational number itself as base, that would make this trivial.

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    $\begingroup$ Can you give an example of what you mean by a variable base? $\endgroup$ – quasi Feb 15 '17 at 10:19
  • $\begingroup$ @quasi bade i or first digit in base 3 , second in base 5 , or first digit in base 1/2 , second in base 5, 3rd in I how is that for variable bars. Can also use Maurice as base as well but I don't think any combination of that will ever end up being a scaler number $\endgroup$ – Arjang Feb 15 '17 at 11:27
  • $\begingroup$ Can you use bases 3,9,27,81,... ? $\endgroup$ – quasi Feb 15 '17 at 11:36
  • $\begingroup$ @quasi : sure, but that is just base 3 and the answer is obvious $\endgroup$ – Arjang Feb 15 '17 at 11:49
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    $\begingroup$ But the digits can vary, non-repeating $\endgroup$ – quasi Feb 15 '17 at 12:14
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Consider $.1141144114441144441144444\dots$, where the 1s are in base 3, and the 4s are in base 9. The expansion is nonrepeating, but it represents $${1\over3}+{1\over9}+{4\over9^2}+{1\over3\cdot9^2}+{1\over9^3}+{4\over9^4}+{4\over9^5}+\cdots$$ which is $${4\over9}+{4\over9^2}+{4\over9^3}+{4\over9^4}+{4\over9^5}+\cdots={1\over2}$$ which is rational.

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This is more a comment than an answet.

If the expansion is of the form $x=\sum_{k=0}^{\infty} d_k/B_k$ where $B_{k+1} >B_k$ and $0\le d_k < B_{k+1}/B_k$ (the standard decimal system has $B_k=10^k$) then the representation is unique if and only if $B_{k+1}/B_k$ is an integer for all $k$.

In the factorial system, with $B_k=k!$, the representation terminates if and only if the number is rational. In this system, $e$ has $d_k=1$ for all $k$, so this transcendental number has a repeating representation.

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Using bases $\,3,3^2,3^3,3^4,\ldots\,$ with corresponding digits $\,1,2,3,4,\ldots\;$ yields

$${\large{\sum_{n=1}^\infty \frac{n}{3^n}}} = \frac{3}{4}$$

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