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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?

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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 $\epsilon>0$.

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  • $\begingroup$ The idea is there. I would say that if $x_m$ is $q$-periodic and $C = \frac{1}{q}\sum_{m=1}^q x_m$ then $\sum_{m=1}^\infty (x_m-C)m^{-s}$ is analytic for $\Re(s) > 0$. Here $\sum_{m=1}^\infty (x_m-C)m^{-s} = \frac{(1-2^{-s}) \zeta(s)}{(1-2^{-2s})\zeta(2s)}-C \zeta(s)$ which we know isn't analytic for $\Re(s) > 0$ since $\frac{1}{\zeta(2s)}$ has a pole at $2s = 1/2+i 14.15...$ not cancelled by the other terms $\endgroup$ – reuns Jul 12 '17 at 22:47
  • $\begingroup$ Indeed the function that appears is $\frac{\zeta(s)}{\zeta(2s)}$ rather than $\frac{\zeta(2s)}{\zeta(s)}$ - I have edited the solution accordingly - thanks for the correction. $\endgroup$ – Rodrigo Jul 12 '17 at 23:04

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