Mod 2 normal numbers and skew products
Studia Mathematica, Tome 165 (2004) no. 1, pp. 53-60

Voir la notice de l'article provenant de la source Institute of Mathematics Polish Academy of Sciences

Let $E$ be an interval in the unit interval $[0,1)$. For each $x \in [0,1)$ define $d_n(x) \in \{0,1 \}$ by $d_n(x) := \sum_{i=1}^n 1_E (\{2^{i-1} x\}) \pmod 2$, where $\{t\}$ is the fractional part of $t$. Then $x$ is called a normal number mod $2$ with respect to $E$ if $N^{-1} \sum_{n=1}^N d_n(x)$ converges to $1/2$. It is shown that for any interval $E \not=(1/6, 5/6)$ a.e. $x$ is a normal number mod $2$ with respect to $E$. For $E = (1/6, 5/6)$ it is proved that $N^{-1} \sum_{n=1}^N d_n(x)$ converges a.e. and the limit equals $1/3$ or $2/3$ depending on $x$.
DOI : 10.4064/sm165-1-4
Keywords: interval unit interval each define sum i pmod where fractional part nbsp called normal number mod respect sum x converges shown interval normal number mod nbsp respect proved sum x converges limit equals depending

Geon Ho Choe 1 ; Toshihiro Hamachi 2 ; Hitoshi Nakada 3

1 Department of Mathematics Korea Advanced Institute of Science and Technology Daejeon, South Korea
2 Department of Mathematics Kyushu University Fukuoka, Japan
3 Department of Mathematics Keio University Yokohama, Japan
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Geon Ho Choe; Toshihiro Hamachi; Hitoshi Nakada. Mod 2 normal numbers and skew products. Studia Mathematica, Tome 165 (2004) no. 1, pp. 53-60. doi: 10.4064/sm165-1-4

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