广义黄金比例基下具有可数展开的数
数论
2015-04-08 v1 动力系统
摘要
Sidorov 和 Vershik 表明，在基 G = 5 + 1 2 G=\frac{\sqrt{5}+1}{2} G = 2 5 + 1 0 , 1 0,1 0 , 1 n ∈ Z n\in\mathbb{Z} n ∈ Z x = n G ( mod 1 ) x=nG ~(\text {mod} 1) x = n G ( mod 1 ) ℵ 0 \aleph_{0} ℵ 0 ( 0 , 1 G − 1 ) (0, \frac{1}{G-1}) ( 0 , G − 1 1 ) 2 ℵ 0 2^{\aleph_{0}} 2 ℵ 0 β = G ( m ) \beta=\mathcal{G}(m) β = G ( m ) { 0 , 1 , ⋯ , m } \{0,1,\cdots, m\} { 0 , 1 , ⋯ , m } m = 2 k + 1 m=2k+1 m = 2 k + 1 G ( m ) = k + 1 + k 2 + 6 k + 5 2 \mathcal{G}(m)=\frac{k+1+\sqrt{k^{2}+6k+5}}{2} G ( m ) = 2 k + 1 + k 2 + 6 k + 5 x = p β + q ( k + 1 ) n ∈ ( 0 , m β − 1 ) x=\frac{p\beta+q}{(k+1)^{n}}\in(0, \frac{m}{\beta-1}) x = ( k + 1 ) n pβ + q ∈ ( 0 , β − 1 m ) n , p , q ∈ Z n, p, q\in\mathbb{Z} n , p , q ∈ Z ℵ 0 \aleph_{0} ℵ 0 ( 0 , m β − 1 ) (0, \frac{m}{\beta-1}) ( 0 , β − 1 m ) 2 ℵ 0 2^{\aleph_{0}} 2 ℵ 0 m = 2 k m=2k m = 2 k G ( m ) = k + 1 \mathcal{G}(m)=k+1 G ( m ) = k + 1 p ( k + 1 ) n ∈ ( 0 , 2 ) ( n , p ∈ N ∪ { 0 } ) \frac{p}{(k+1)^{n}}\in(0, 2) ~(n, p\in\mathbb{N}\cup\{0\}) ( k + 1 ) n p ∈ ( 0 , 2 ) ( n , p ∈ N ∪ { 0 })
引用
@article{arxiv.1504.01704,
title = {Numbers with countable expansions in base of generalized golden ratios},
author = {Yuehua Ge and Bo Tan},
journal= {arXiv preprint arXiv:1504.01704},
year = {2015}
}