46 (1)证明:
首先有$2n(n+1)=\left \lfloor 2n(n+1)+\frac{1}{2} \right \rfloor=\left \lfloor 2(n^{2}+n+\frac{1}{4}) \right \rfloor=\left \lfloor 2(n+\frac{1}{2})^{2} \right \rfloor$
其次,令$n+\theta =(\sqrt{2}^{l}+\sqrt{2}^{l-1})m=(1+\frac{\sqrt{2}}{2})\sqrt{2}^{l}m$,$n^{‘}+\theta^{‘} =(\sqrt{2}^{l+1}+\sqrt{2}^{l})m=(1+\sqrt{2})\sqrt{2}^{l}m$
如果$l$为偶数,那么$n+\theta =(1+\frac{\sqrt{2}}{2})k,n^{‘}+\theta^{‘} =(1+\sqrt{2})k$.所以$\theta =\left \{ \frac{\sqrt{2}}{2}k \right \},\theta^{‘} =\left \{ \sqrt{2}k \right \}$,所以$\theta$和$\theta^{‘}$的关系是要么$\theta^{‘}=2\theta$($\left \lfloor \sqrt{2}k \right \rfloor$为偶数),要么$\theta^{‘}=2\theta-1$($\left \lfloor \sqrt{2}k \right \rfloor$为奇数);
如果$l$为奇数,那么$n+\theta =(1+\sqrt{2})k,n^{‘}+\theta^{‘} =(2+\sqrt{2})k$,这时候$\theta = \theta^{‘}$
最后,假设要证明的式子成立,那么$n^{‘}=\left \lfloor \sqrt{2n(n+1)} \right \rfloor$
$=\left \lfloor \sqrt{\left \lfloor 2(n+\frac{1}{2})^{2} \right \rfloor} \right \rfloor$
$=\left \lfloor \sqrt{2}(n+\frac{1}{2}) \right \rfloor$ (这一步参见公式$3.9$)
$=\left \lfloor \sqrt{2}\left ( (1+\frac{\sqrt{2}}{2})\sqrt{2}^{l}m -\theta +\frac{1}{2}\right ) \right \rfloor$
$=\left \lfloor n^{‘}+\theta^{‘}+\sqrt{2}(\frac{1}{2}-\theta) \right \rfloor$
所以只要证明$0\leq \theta^{‘}+\sqrt{2}(\frac{1}{2}-\theta)<1$
首先当$\theta=\theta^{‘}$时成立,
其次,如果$\theta^{‘}=2\theta-d$时($d=0$或者$d=1$),那么$0\leq \theta^{‘}+\sqrt{2}(\frac{1}{2}-\theta)<1$
$\Leftrightarrow 0\leq \theta^{‘}+\sqrt{2}(\frac{1}{2}-\frac{\theta^{‘} +d}{2})<1$
$\Leftrightarrow 0\leq \theta^{‘}(2-\sqrt{2})+\sqrt{2}(1-d)<2$
最后这个式子明显成立
(2)由于$Spec(1+\frac{\sqrt{2}}{2}),Spec(1+\sqrt{2})$是一个划分,所以对于任何一个$a$一定存在唯一的$(l,m)$使得$a=(\sqrt{2}^{l}+\sqrt{2}^{l-1})m$,这时候$L_{n}=\left \lfloor\left ( \sqrt{2}^{l+n} -\sqrt{2}^{l+n-1} \right )m\right \rfloor$
原文地址:https://www.cnblogs.com/jianglangcaijin/p/9472572.html