Conjecture on irrational algebraic numbers 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:


*

*Has this conjecture been proved, refuted or neither?

*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 $q_b$.

*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
 A: What about the number in base 10 that has decimal expansion using only digits 1 and 2 in the following pattern: 0.121221222122221 ... That is the digit 1 occurs always alone: 2's occur in blocks of increasing length. This has no periodicity and is not a rational number (and misses many digits of the base 10 system)
A: 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, 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 $x$.
