TITLE: Have all numbers with "sufficiently many zeros" been proven transcendental?
QUESTION [21 upvotes]: 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 equivilantly 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 demostrating this result for the case $g=2$; it can be found here: http://www.escholarship.org/uc/item/44t5s388?display=all

REPLY [19 votes]: I must apologize and correct what I wrote in an earlier paper: the method does extend to base > 2. Using Liouville's inequality, you lose a bit in the constant c, but the great advantage is that everything can be made fully explicit.