Is the sequence , 0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... equidistributed?
Is the sequence ,
$$0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... $$ equidistributed ?
A sequence $\{\xi_n\}$ is equidistributed in $[0,1),$ that is if $$\lim_{N\rightarrow\infty}\frac{Card\{1\leq n\leq N|\xi_n\in(a,b)\}}{N}=b-a$$
Well ,the sequence write in a formula is :$$\xi_n=0,\;\; whenever\; n=k(k-1)/2+1$$$$\xi_n=\frac{k}{m},\;\;whenever\;n=\frac{(m-1)(m–2)}{2}+k$$$\forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so $$Card\{k|\frac{k(k-1)}{2}+1\leq N\}=O(\sqrt{N})$$Next we let $$\frac{(m-1)(m–2)}{2}+k\leq N$$and$$am\leq k\leq bm$$Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
I don't know how to do next
$\endgroup$ 21 Answer
$\begingroup$Well ,the sequence write in a formula is :$$\xi_n=0,\;\; whenever\; n=k(k-1)/2+1$$$$\xi_n=\frac{k}{m},\;\;whenever\;n=\frac{(m-1)(m–2)}{2}+k$$$\forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so $$Card\{k|\frac{k(k-1)}{2}+1\leq N\}=O(\sqrt{N})$$Next we let $$\frac{(m-1)(m–2)}{2}+k\leq N$$and$$am\leq k\leq bm$$Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
$\endgroup$
