A closed-form irrational number with few distinct digits I'm looking for an irrational number which does not have all $b$ distinct digits in its base-$b$ representation and which can be expressed in "closed form" for some reasonable definition thereof.
Acceptable would be +, -, *, /, ^, sin, $\pi$, $e$, and anything else 'reasonable'. Unacceptable would be algorithms, unbound $\sum$ or $\prod$, "write in base 2 and interpret in base $b$", etc.
The problem may be hard, in which case I would be willing to accept an answer explaining that this is so (ideally with a reference).
 A: Regarding hard: it is conjectured that each irrational element of the ring of periods is a normal number. So, they are excluded. This already excludes a lot. $e$ is also conjectured to be normal. Basically, since everything but a set of (Lebesgue) measure zero is normal, you should really have to resort to constructions (that are in some sense artificial) to avoid normal numbers. 
So, this excludes most notions of closed form. 
A: Consider $b = 3$ where we have the digits $0$, $1$, and $2$. We are looking for an irrational number (in closed form) which has ternary expansion which uses at most two of the three digits. In fact, it has to use exactly two digits, otherwise the result would be rational.
To simplify the problem somewhat, let's try to find such a number in $[0, 1]$. Now we have to consider which two digits will appear in the ternary expansion; let's try for $0$ and $2$. So the question is:

For which $x \in [0, 1]$ does $x$ have a ternary representation consisting only of $0$ and $2$? 

The answer is:

Elements of the Cantor set.

As the Cantor set is uncountable and $\mathbb{Q}$ is countable, there are uncountably many irrational elements of the Cantor set. If you can find one which matches your criteria, you're done. Unfortunately, I don't know whether such elements have ever been explicity found
