Sequences similar to $\{n\alpha\}$ that are both equidistributed and truly random-like

For any $p>0$ the sequence of fractional parts $x_n=\{n^p\alpha\}$ cannot be random-like in the sense defined in the appendix. The case of integer $p$ was already discussed in the comment by Goldstern. Suppose that $k-1<p \le k$. Then some fixed linear combination of $x_n,x_{n+1}\ldots,x_{n+k}$ will approach zero as $n \to \infty$, so that the vectors $(x_n,x_{n+1}\ldots,x_{n+k})$ will asymptotically (almost) lie on a finite union of hyperplanes. For instance, if $1<p<2$ then using the Taylor expansion $(1+u)^p=1+pu+O(u^2)$ as $ u\to 0$, we find that as $n \to \infty$, $$(n+2)^p-2(n+1)^p+n^p=n^p[(1+2/n)^p-2(1+1/n)^p+1]=n^p\cdot O(n^{-2}) \to 0 \,.$$ Thus $x_{n+2}-2x_{n+1}+x_n \to 0$ (If $p=2$ then the LHS is identically zero).

Similarly, if $2<p<3$, then use the expansion $$(1+u)^p=1+pu+{p \choose 2}u^2 + O(u^3)\; \mbox{ as } \; u\to 0 \,,$$ to infer that $$(n+3)^p-3(n+2)^p+3(n+1)^p-n^p= O(n^{p-3}) \to 0 \,.$$ (This sum is identically 0 if $p=3$). In general, if $k-1<p \le k$, then as $n \to \infty$, $$ \sum_{j=0}^k (-1)^j {k \choose j } (n+j)^p \to 0\,, \; \;\;\;(*)$$ so as $n \to \infty$, $$\sum_{j=0}^k (-1)^j {k \choose j }x_{n+j} \to 0\,.$$ The formula (*) can be deduced from the calculus of finite differences (see [1] or [2]). Alternatively, following the arguments above, use the general Binomial series $$(1+u)^p= \sum_{\ell=0}^\infty {p \choose \ell } u^\ell \mbox{ for } \; |u|<1\,,$$ (applied with $u=j/n$) together with the identity for integer $0 \le \ell<k$: $$\sum_{j=0}^k (-1)^j {k \choose j } j^\ell=0 \,$$ [1] L.M. Milne-Thomson, "The calculus of finite differences" , Macmillan (1933) Zbl 0008.01801; reprinted Dover (1981) Zbl 0477.39001 [2] Finite-difference calculus. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Finite-difference_calculus&oldid=44401