= Normality and equidistribution under integer multiplication
{title2=$x\text{ normal in base }K\iff(\{K^n x\})_{n\geq0}\text{ equidistributed}$}
A <real number> $x$ is a <normal number> in base $K$ exactly when its iterates under the <integer multiplication map on the circle>, $T_K(x)=\{Kx\}$, form an <equidistributed sequence>. Braces denote the <fractional part>. Multiplication by $K$ shifts the <positional notation> one digit to the left:
$$
T_K^n(x)=\{K^nx\}=0.d_{n+1}d_{n+2}\ldots\quad\text{in base }K.
$$
A length-$r$ <word> with base-$K$ <integer> value $a$ specifies the half-open <interval> $[a/K^r,(a+1)/K^r)$, whose length is $K^{-r}$. Visiting that <interval> is exactly encountering that <word> at the next digit position. These digit <intervals> form arbitrarily fine grids, so their correct limiting visiting frequencies imply <equidistribution> on every <interval>. This converts a question about digits into one about a <dynamical system>. Bailey and Crandall develop this connection using <equidistribution> of the <fractional parts> of $K^nx$. https://www.davidhbailey.com/dhbpapers/bcnormal.pdf[David H. Bailey and Richard E. Crandall, Random Generators and Normal Numbers, 2003 author manuscript, Section 2]{ref}
In base ten the <word> $12$ corresponds to $[0.12,0.13)$. For <Champernowne constant>, counting the visits $T_{10}^n(x)$ for $0\leq n\leq99998$ gives precisely the $2088$ occurrences of $12$ in its first $100000$ digits. For $x=1/3$, the base-two <integer multiplication map on the circle> instead alternates between $1/3$ and $2/3$; it never visits $[0,1/4)$, the <interval> for $00$. This two-point <orbit> is not an <equidistributed sequence>.
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