Are some numbers more equidistributed than others?

Question

Weyl's theorem says that $n\alpha$ is equidistributed mod 1 for any irrational $\alpha$. One corollary is that, if I consider the fractional part $\{n\alpha\}$ for $n \leq N$, and look at the indices $n_k$ where it's small ($\{n_k \alpha\} \leq \epsilon$), the number of such indices $K$ will be approximately equal $N\epsilon$ as $N \to \infty$. In other words, if I look at the spacings between the consecutive integers $n_k$ where $n_k \cdot \alpha$ is nearly an integer, the mean spacing will be close to $N\epsilon$.

Can anything interesting be said about the variance of the spacing? For example, if $\alpha \approx 1/N$, the numbers $n_k$ will be $1, 2, \dotsc, N\epsilon$, so the spacings are all 1 except for one enormous spacing of size $N-\epsilon$, and the variance of the spacings will be $O(N)$. But this is cheating, since I picked $\alpha$ to depend on $N$. Are there any results on the asymptotics of this for fixed irrational $\alpha$? Are some values clumpier than others, or does it all wash out as you let $N \to \infty$?

Answer 1

Given $\alpha$ and $\varepsilon = 1/N$, fix $k\gg N$ such that $ \lbrace k\alpha \rbrace \ll \varepsilon $. Now if we consider the sequence $\bar s_0 := (kn\alpha:n\in \mathbb N)$, which is a subsequence of the original sequence along an arithmetic sequence of indices, this new sequence behaves just like the example you gave for a small $\alpha$.

If we let $\bar s_i := ((kn+i)\alpha:n\in \mathbb N)$ for $i=0,\dots,k-1$, then each of these sequences is just a shifted copy of $\bar s_0$ and has therefore the same behavior. (e.g. same variance)

We can divide the original sequence into blocks of length $k$. For each such block $(nk\alpha, (nk+1)\alpha, \dots, (nk+k-1)\alpha)$, consider the interval $[a,b)$, where $a\in [0,1)$ is $nk\alpha$ mod 1, and $b\in[1,2)$ is $(n+1)k\alpha $ mod 1. Then $b-a\approx 1$, and the $k$ entries in our block divide the interval [a,b) into $k$ equal pieces of length $1/k$ each. (almost)

The whole sequence $(n\alpha)$ is now a finite union of sequences $\bar s_i:=((kn+i)\alpha:n\in \mathbb N)$, for $i=0,\dots, k-1$.

So for each $n$, you will get a set $M_n$ of (approximately) $2k\varepsilon$ distinct indices $i_1(n),\dots, i_{2k\varepsilon}$ in $\{1,\dots, k\}$ such that all the values $(kn+i_j(n))\alpha$ are close to $0$, and the others are not. I think that any behavior (such as the variance) of the sequence $n\alpha $ mod 1 can now be reduced to the behavior of the sequence $nk\alpha$, up to a factor depending on $k$, which is constant (depending on $\varepsilon$, of course).