Our objective is to show all natural numbers  $m = \sum_{i=0}^n (a_i\cdot 10^i),\ (0 \leq a_i \leq 9,\  a_i \in \mathbb{N},\ a_n \not= 0)$ goes to $1$ or $169$ by taking finitely many successive operations of $\Delta$ such as $\Delta(m) = m/3$ (if $3|m$) or  $\Delta(m) = \sum_{i=0}^n(a_i)^2$ (otherwise). 
It is easy to see that $\Delta(169) = 16^2 =256$ and $\Delta(256) = 13^2 = 169$. We prove that for  any $m \in \mathbb{N}$, either $\Delta^k(m) = 1$, or $\Delta^k(m) = 169$ for some $k \in \mathbb{N}$.