Prepending strings and primes? we all know that 31,331,3331,33331,333331,3333331,33333331 all are primes. Here we prepend the digit 3 to 31, to get a list of 7 primes.This gives me the following thought:
Let $D = \{\text{all possible nonnull finite digit strings}\}$,  $D' = \{\text{all things in D that do not start with "0"}\}$.  Define a function $m: D' \times D -> N \cup {\infty}$  by: $m(A,B)= |${all prime members of the list AB, AAB, AAAB, 
...up until but not including the first composite member}| (the size of the set).Then:
Does m ever take the value $\infty$ ?  If not, is it an unbounded function?
 A: I think it's easy to show that $m$ is finite. Assume $AB$ is prime (otherwise, $m=0$, which is extremely finite). I'll prove there are later terms in the sequence divisible by $AB$. Subtract $AB$ from each of the later terms in the sequence (that won't affect divisibility by $AB$), then all I have to do is show $AB$ divides $AA\dots A00\dots0$, indeed, $AA\dots A$, for some number of repetitions of $A$. If $A$ is of length $s$, then $AA\dots A=A\times(10^{ns}-1)/(10^s-1)$ where $n$ is the number of repetitions of $A$. If we take $n=AB-1$, then by Fermat, $AB$ divides $(10^s)^n-1$, and we're done. 
Well, there are a few little things I've swept under the rug here, but I think they're all easily handled. 
Proving $m$ is unbounded looks harder. It's not a million miles from proving that for every $m$ there's a prime $p$ such that $2p+1,4p+3,\dots 2^mp+2^m-1$ are all prime (in binary, these are the numbers you get from $p$ by tacking ones onto the right end of it), and that's a notorious unsolved problem. 
