Is the sequence of composite numbers aperiodic? 
Let $a_1,a_2,\ldots$ be the composite numbers in strictly increasing order and consider the sequence $x_i \equiv a_i \pmod{2}$ where $x_i \in \{0,1\}$. Is the sequence $\{x_i\}$ aperiodic?  That is, does there not exist $T$ and $n_0$ such that for all $n \geq n_0$, $x_{n+T} = x_n$?

The sequence starts out as 
$$0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,1,0,0,0,\\1,0,1,0,0,1,0,0,0,1,0,0,1,0,1,0,0,1,0,1,\\0,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,\\1,0,1,0,0,1,0,1,0,1,0,0,1,0$$
and I couldn't find any points where it could be periodic. How could we show, if it is, aperiodic?
 A: It is aperiodic. First notice that there exist arbitrarily long sequences of consecutive composite positive integers: $n!+2,n!+3,\dots,n!+n$ are all composite. Therefore, if the sequence $(x_n)$, after some point, keeps repeating a cycle, then it must be that the cycle is alternating $1$'s and $0$'s. But the sequence can't be alternating $1$'s and $0$'s after some point since there are arbitrarily large (odd) primes.
A: Recall that we can find consecutive strings of composite numbers of arbitrary length.  It's easy to see that, in fact, we can find such strings wherein the least element is larger than any preset level.  Now, suppose we have produced $n_0,T$ as desired.  Find a consecutive string of composites of length greater than $T$ far enough out so as to be in the periodic range.  That is to say, we have produced a consecutive string $x_i,x_{i+1},\cdots, x_{i+T}$ where $i>n_0$.  Of course these alternate $\pmod 2$.  But every larger $j$ is congruent $\pmod T$ to exactly one of those subscripts $i+k$ for suitable $k\in [0,T-1]$.  It would follow that the composites alternate in parity beyond a certain point.  But this is absurd.
