TITLE: Do there exist numbers normal in every base except for one?
QUESTION [18 upvotes]: 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 :).

REPLY [8 votes]: 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).