Have all numbers with "sufficiently many zeros" been proven transcendental? Any number less than 1 can be expressed in base g as $\sum _{k=1}^\infty {\frac {D_k}{g^k}}$, where $D_k$ is the value of the $k^{th}$ digit. If we were interested in only the non-zero digits of this number, we could equivalently express it as $\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}}$, where $Z(k)$ is the position of the $k^{th}$ non-zero digit base $g$ and $C_k$ is the value of that digit (i.e. $C_k = D_{Z(k)}$).
Now, consider all the numbers of this form $(\sum _{k=1}^\infty {\frac {C_k}{g^{Z(k)}}})$ where the function $Z(k)$ eventually dominates any polynomial. Is there a proof that any number of this form is transcendental?
So far, I have found a paper demonstrating this result for the case $g=2$; it can be found here.
 A: Unless I am confused (and I hope someone will correct me!) you are describing a subset of the Liouville numbers and they are all transcendental.  The point is - the numbers you describe are not rational, but are very very well approximated by rational numbers.  I think, also, that it suffices for $Z(k)$ to be super-quadratic (so its difference $Z(k+1) - Z(k)$ is super-linear).  The canonical example is discussed on the same webpage.
A: The answer to your question is yes. All numbers of the form $x=\sum_{n\ge0}\frac{C_k}{g^{Z(k)}}$ for Z(k) eventually dominating any polynomial are indeed transcendental. As in the question, g and Ck are integers with 1 ≤ Ck ≤ g-1. In fact, the methods used by the paper linked in the question generalize in a quite straightforward way to handle this situation. I don't know of this result appearing in any published paper, but note the following points.


*

*We can say straight-away that x is irrational. This follows from Z(n) eventually dominating any linear function of n, so its base-g expansion is not eventually periodic. 

*If Z(n+1)/Z(n) is unbounded then, as noted in the comments, x will be a Liouville number so, by Liouville's theorem, it is transcendental. For any N > 0, Z(n+1) ≥ NZ(n) for infinitely many n. The fact that it is a Liouville number follows from taking $p=\sum_{k=1}^nC_kg^{Z(n)-Z(k)}$ and $q=g^{Z(n)}$, giving the rational approximation $|x-p/q|< g^{1+Z(n)-Z(n+1)}\le gq^{-N}$.

*By the Thue–Siegel–Roth theorem, if Z(n+1)/Z(n) ≥ 2+ε infinitely often (any ε > 0) then x will be transcendental. The theorem says that an irrational algebraic number has only finitely many rational approximations $\vert x-p/q\vert\le cq^{-2-\epsilon}$ for any fixed c,ε > 0. That x has infinitely many such rational approximations follows in the same way as for point 2 above. Every Z(n+1) ≥ (2+ε)Z(n) gives a rational approximation $\vert x-p/q\vert< gq^{-2-\epsilon}$, so x cannot be algebraic. This covers the case where Z(n) grows exponentially of rate an for any a > 2, but is not strong enough to cover cases such as Z(n) = 2n.

*If Z(n+1)/Z(n) > 1+ε infinitely often (any ε > 0) then x will be transcendental. This is a consequence of the Roth-Ridout theorem, from the 1957 paper Rational approximations to algebraic numbers (not free access, but is also quoted in the freely available paper An explicit version of the theorem of Roth-Ridout, Theorem 2). The Roth-Ridout theorem strengthens the Thue-Siegel-Roth theorem implying, in particular, for irrational and algebraic x, there are only finitely many rational approximations $\vert x-p/q\vert\le cq^{-1-\epsilon}$ when the prime factors of q all belong to some fixed finite set P. In our case, we can let P be the set of prime factors of g and the result follows in the same way as for point 3 above. This shows that x is transcendental if Z(n) grows exponentially. (thanks to Mike Bennett over at mathoverflow for pointing out the Roth-Ridout theorem).

*A paper by Bugeaud, On the b-ary expansion of an algebraic number shows that, if x is irrational and algebraic then for large enough n, there are at least (log n)1+1/(ω+4) (loglog n)-1/4 nonzero digits among the first n digits of the base g expansion. Here, ω is the the number of prime divisors of g. This shows that, if Z(n) ≥ exp(cnα) for large n and any fixed c > 0, α > 1/(1+1/(ω+4)) then x is transcendental.

*After reading through the details of the paper linked in the original question, I note that they do generalize to the base g ≥ 2 case. So x is transcendental as long as Z(n) eventually dominates any polynomial. I don't know of any published paper proving this, but posted my proof on mathoverflow where this question was also asked. I have re-read through this proof a few times to be sure, and am now confident that it is correct (modulo small typos, etc). Also, Bugeaud posted an answer to the question agreeing that the method generalizes. Using #(x,n) to denote the number of non-zero digits in the first n digits of the base g expansion of x, the precise statement is as follows.

If x is irrational and satisfies a rational polynomial of degree D then #(x,n) ≥ cn1/D for a positive constant c and all large enough n.

In fact, you can easily remove the "large enough" from this statement, although I find it convenient stated in this way. The proof I wrote out is a generalization of the methods used in the paper linked in the question. There is one change worth noting though. Whereas the paper made use of the Thue-Siegel-Roth theorem at one point (Theorem 3.1), I used Liouville's theorem. This means that the constant c appearing in the statement above is not quite as good (if you go through the proof and work it out explicitly) although, in any case, the paper linked in the question could have obtained a better value by using the Roth-Ridout theorem instead. Using Liouville's theorem does have two advantages though. Firstly, it is elementary. A proof of Liouville's theorem is given in the linked Wikipedia article. Secondly, it is effective. That is, not only can the constant c be calculated but you can also work out exactly what "large enough" means for n in the statement above (which will depend on the polynomial satisfied by x). The strengthened versions of Liouville's theorem such as Thue-Siegel-Roth and Roth-Ridout are not effective.
