Are the squarefree numbers periodic? 
Let $a_1,a_2,\ldots$ be the positive squarefree 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 $$1,0,1,1,0,1,0,1,1,0,1,1,1,1,0,1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,1,1,1,1,1,0,\ldots.$$ How could we show, if it is, aperiodic?
 A: Let $Q_1(x)$ and $Q_2(x)$ count the number of odd and even squarefree numbers up to $x$ respectively. Notice that if the sequence you proposed is periodic, then at every period $T$ we always get the same number of odd and of even squarefree numbers. Assume that the ratio between the count of odd and even numbers in the period is $q$. Then it will follow that $|Q_1(x) - qQ_2(x)|$ is bounded (since at every period we get exactly q times as many odd squarefrees as evens). We will show that is too regular to be true.
First remember that:
$$f(s) = \sum_{squarefree \,n} \frac{1}{n^s} = \prod_p(1+\frac{1}{p^s}) = \prod_p\frac{(1-\frac{1}{p^{2s}})}{(1-\frac{1}{p^s})} = \frac{\zeta(s)}{\zeta(2s)}$$
Now define analogous functions $f_1$, $f_2$,  but restricted only to sums of odd and eben squarefree numbers respectively. 
Notice that:
$$f_2(s) = \sum_{even\,squarefree \,n} \frac{1}{n^s} = \sum_{odd\,squarefree \,n} \frac{1}{(2n)^s} = 2^{-s} f_1(s)$$
While $f_1(s)+f_2(s) = f(s)$. From that it is easy to get:
$$f_1(s) = \frac{2^s}{1+2^s}\frac{\zeta(s)}{\zeta(2s)}\,\,\,,\,\,\, f_2(s) = \frac{1}{1+2^s}\frac{\zeta(s)}{\zeta(2s)}$$
Now we consider: 
$$f_1(s) - qf_2(s) = \frac{2^s-q}{1+2^s}\frac{\zeta(s)}{\zeta(2s)}$$
We can write this as a series:
$$f_1(s) - qf_2(s) = \sum_{n\, squarefree} \frac{A(n)}{n^s}$$
Where $A(n)$ is $1$ if $n$ is odd or $-q$ if $n$ is even. But notice that the partial sums of $A$ satisfy: 
$$A(1)+...+A(x) = Q_1(x)-qQ_2(x)$$
Hence from Abel's lemma (or summing by parts) we obtain that the series $\frac{A(n)}{n^s}$ converges and defines an analytic function for $\Re(s) > 0$. 
This provides an analytic extension for:
$$\frac{2^s-q}{1+2^s}\frac{\zeta(s)}{\zeta(2s)}$$
All the way to $\Re(s)> 0$. But this contradicts the fact that $\zeta(s)$ has at least one zero with real part $\frac{1}{2}$ (so our function must have a pole at that zero). So we get a pole for or function for some $s$ with real part $1/4$ - one can verify that the numerator doesn't vanish.
Indeed, one can show that $Q_1(x) \sim 2Q_2(x)$, and this same method shows that 
$$|Q_1(x) - 2Q_2(x)| = O(x^{1/4-\epsilon})$$
cannot hold for $\epsilon>0$. 
