Define \begin{eqnarray} f(n) &=& (n-1)^2 \; \textrm{if} \; (n \bmod 4) = 1\\ f(n) &=& \lfloor n/4 \rfloor \; \textrm{otherwise} \end{eqnarray} and let $f^k(n) = f(f( \cdots (n) \cdots ) )$ be the result of applying $f(\;)$ $k$ times to $n$. So $f^3(5)=0$ because the iterates produce the sequence $(5,16,4,1,0)$, $f^8(13)=0$ because the sequence is $(13,144,36,9,64,16,4,1,0)$, and so on. Many numbers map to $0$. But some seem not to, i.e., to grow without bound. E.g., here is the start of the sequence for $n=53$, the smallest "problematic" number: $$ (53,2704,676,169,28224,7056,1764,441,193600,48400,12100,3025,9144576,2286144,571536,142884,35721,1275918400,318979600,79744900,19936225,\ldots) $$ My question is:
What is special about the numbers $53, 85, 77, 101, \ldots$ that seem to grow without bound? Can one prove that, for any particular number $n$, $\lim_{k \to \infty}f^k(n) = \infty$ ?
If f(n) become $a_1→a_k→a_n→$
$a_k=b^2$
$a_{k+1}=(b+1)^2(b-1)^2$
$a_{k+2}=\frac{(b+1)^2(b-1)^2}{16}\geq (\frac{b-1}2)^4\geq [\frac{a_k}2]^2$
So, $a_{n} $ tend to diverge except the numbers with exponent 4.