**Conjecture:**


For every irrational algebraic number $q$ and natural number $b$, the representation of $q$ on base $b$ contains all the digits $[0,\dots,b-1]$.


**Questions:**


1. Has this conjecture been proved, refuted or neither?
2. If proved:


Is there an estimate of the minimum length of $q\_b$ containing all the digits?


For example, I would expect something like $2b$ or $b^2$ for any given $q\_b$.
3. If not refuted:


I suppose that it is not true for transcendental numbers. Is that correct?


How can we construct a transcendental number $q\_b$ which does not contain all the digits?


Thanks



Conjecture on irrational algebraic numbers
The conjecture has been neither refuted nor proved. The state of the art, as far as I know, is contained in the papers of [Adamczewski and Bugeaud](http://annals.math.princeton.edu/2007/165-2/p04), in which they show that anything with a very low complexity decimal expansion cannot be an algebraic irrational. The complexity is the function $c\_x(n)$ giving the number of blocks of length $n$ in the decimal expansion of $x$ (or any base). They show that if there exists a $k$ such that $c\_x(n)\le kn$ for all $n$, then $x$ is either rational or transcendental. Of course, it's conjectured that $c\_x(n)=10^n$ for all algebraic irrationals $x$. Your condition would be implied by the conjecture $c\_x(n)>9^n$ for all algebraic irrationals $x$.


