You are wrong. Let $a$ is even. If $u_0=2^km,$ were $m$ is odd, then $u_k=m$ and $u_{k+n}=na+m$.

WakeUp wrote: ...where $a$ is a fixed odd positive integer... Rust wrote: You are wrong. Let $a$ is even.. If $u_{n+1}=u_n+a$ then $u_{n+2} = \frac{u_{n+1}+a}{2}$, so if $u_m>a$ for some $m$ then there is a strictly decreasing subsequence of $u$ that eventually reaches $u_{m+k} <a$. At this point $u_n < a$ for all $n \ge m+k$, so two terms are equal etc etc. the sequence is periodic.