You are wrong, if $a_0=\frac 13$, then $a_n=\frac 13 \ \forall n$. If $a_{n-1}\le \frac 14$, then $a_n=2a_{n-1}$, if $a_{n-1>\frac 14}$, then $a_n=1-2a_{n-1}$. Therefore always exist infinetely many $n$, suth that $a_n\ge \frac 13$. But may be never $a_n<\frac 13$.

$a_0$ is irrational.

By induction $a_n=m_n\pm 2^na_0, m_n\in Z,$ were $m_n=\mp [2^na_0+\frac 12]$. If $a_0$ irrational and $a_0=\sum_{k=2}^{\infty} b_k*2^{-k}$ ($b_k=0,1$ -are digits in base 2), then $a_n=\sum_{k=2}^{\infty}b_{k+n}2^{-k}$ if $b_{n+1}=1$, else $a_n=\sum_{k=2}^{\infty}(1-b_{k+n})2^{-k}$. a) If exist digits $b_{n+1}=0, b_{n+2}=0, b_{n+3}=0$ (three - 0), then $a_n<\frac 18$. If never 3 consequitive 0, then if exist 3 consequitive 1: $b_{n+1}=b_{n+2}=b_{n+3}=1$, then $a_n<\frac 18$. If never tree consequitive 1 and never three consequitive 0 and always after 2 -0 we had 2 -1, after 2 -1 had 2-0, then $a_0$ - rational. For irrational we had infinetely many $b_{n+1}=0=b_{n+2}=b_{n+4},b_{n+1}=1$ or $b_{n+1}=b_{n+2}=1=b_{n+4},b_{n+3}=0$. For suth n we had $a_n<\frac{3}{16}$. b)In this case never 3 consequitive 1 and never 3 consequitive 0. $\frac{7}{40}=0,....001100110011...$ ($0011$ in period) Let $b_{n+1}=0=b_{n+2}=0,b_{n+3}=1.$ then $a_n>\frac{7}{49}$. If $b_{n+1}=0,b_{n+2}=1$, then $a_n>\frac 14$. If $b_{n+1}=1,b_{n+2}=0$, then $a_n>\frac 14$. We can consruct stohastic sequense by 2 types digits $00101001$ and $00100101$, then if $b_{n+1}=1$ we get $a_n>\frac 14$ (because never 2 consequitive 1). If $b_{n+1}=b_{n+2}=0$, then $a_n>\frac 18+\frac{1}{64}+\frac{1}{512}+\frac{1}{2^{11}}+\frac{1}{2^{14}}+...=\frac{256+32+4+1}{2^{11}-1}=\frac{293}{2047}>\frac{7}{40}.$