Do there exist numbers normal in every base except for one?

Question

A number $x$ is called normal in base $b$ if every sequence of base $b$ digits $b_1b_2...b_n$ occurs with natural density $1/b^n$ in the decimal expansion of $x$.

There exist numbers normal in every base (called absolutely normal) and irrational numbers normal in no base (called absolutely non-normal), an example is given here.

Is it known whether there exist numbers that are normal in every base except one or numbers non-normal in every base except one?

The question can be stated rather easily but an answer probably will take a lot of effort, so thanks in advance, also for any reference to literature :).

Answer 1

This is not possible. In Kuipers, L. and Niederreiter, H. Uniform Distribution of Sequences (1974), you can find the following exercise, on page $77$:

If $b_1$ and $b_2$ are integers $≥2$, such that one is a rational power of the other, then $a$ is normal to the base $b_1$ if and only if $a$ is normal to the base $b_2$.

(This is also theorem 2.7. in Bailey, D. H. and Crandall, R. E. On the Random Character of Fundamental Constant Expansions, Experiment. Math., Volume 10, Issue 2 (2001), 175-190.)

Therefore, if $x$ is a real number that is normal in every base but one, say $b_1$, then actually it wouldn't be normal to the base $b_1^2 \neq b_1$, contradicting the assumption. The situation is similar for numbers that are non-normal in every base except one.

Answer 2

Theorem 1

If $x$ is normal in infinitely many bases of the powers of $b^k$ then it is normal in base $b$.

Conclusion: This rules out your first curiosity.

Theorem 2:

If $x$ is not normal in base $b^k$ then it cannot be normal in base $b$.

Conclusion: This rules out your second curiosity.

Source for the theorems: On normal numbers, Veronica Becher, page 20-21.

Answer 3

The most general answer to date for your question comes from a paper of Schmidt (http://matwbn.icm.edu.pl/ksiazki/aa/aa7/aa7311.pdf). Schmidt showed that if $A\subset \mathbb{N}_{\ge 2}$ is a set closed under multiplicative dependence*, then there exist uncountably many numbers that are normal to every base $b\in A$ and not normal to every base $b\not \in A$. And by earlier work of Maxfield and Cassels, if $A$ is not closed under multiplicative dependence, then no such number exists. Since if one element is missing from (resp., included in) a set closed under multiplicative dependence, infinitely many are missing (resp, included). Thus, the answer to your question is no.

*A set is closed under multiplicative dependence if $n\in A$ implies every rational power of $n$ that is an integer $>1$ is also in $A$.

I've heard numbers that are normal to some bases but not normal to others be referred to as selectively normal.

As a curiosity, the problem of characterizing possible sets of simple normal bases was recently solved by Becher, Bugeaud, and Slaman (http://arxiv.org/pdf/1311.0332.pdf).