Let $ \ \mathbb{N} = \{ 0,1,2,3,4,...\} \, $, $ \ O = \{ n \in \mathbb{N} : n \text{ is odd} \} \ $ and $ \ T: O \to O \ $ be such that, for all $ \ n \in \mathbb{N} \, $,
\begin{align*}T(8n+1) & = 6n+1 \\T(8n+3) & = 12n+5 \\T(8n+5) & = 2n+1 \\T(8n+7) & = 12n+11\end{align*}
For each $ \ n \in O$, let $ \ S(n) = \big( n, T(n) , T(T(n)), T(T(T(n))), ... \big)$.
Prove that $S(n)$ is eventually $1$, for all $ \ n \in O \, $.
A sequence of real numbers $ \ (x_1,x_2,x_3,x_4, ...) \ $ is eventually $ \, c \, $ if, and only if, there exists $ \ N \in \mathbb{N} \ $ such that $ \ x_n = c \, $, for all $ \ n > N$.
My attempt: I tried to separate classes where this function were closed.
\begin{align*}T(8n+1) & = 6n+1 & = & \left\{ \begin{array}{ll}8 \cdot (3k) +1 \ , & n=4k \ ; \\8 \cdot (3k) +7 \ , & n=4k+1 \ ; \\8 \cdot (3k+1) +5 \ , & n=4k+2 \ ; \\8 \cdot (3k+2) +3 \ , & n=4k+3 \ . \end{array} \right. \\T(8n+3) & = 12n+5 & = & \left\{ \begin{array}{ll}8 \cdot (3k) +5 \ , & n=2k \ ; \\8 \cdot (3k+2) +1 \ , & n=2k+1 \ . \end{array} \right. \\T(8n+5) & = 2n+1 & = & \left\{ \begin{array}{ll}8k+1 \ , & n=4k \ ; \\8k+3 \ , & n=4k+1 \ ; \\8k+5 \ , & n=4k+2 \ ; \\8k+7 \ , & n=4k+3 \ . \end{array} \right. \\T(8n+7) & = 12n+11 & = & \left\{ \begin{array}{ll}8 \cdot (3k+1) +3 \ , & n=2k \ ; \\8 \cdot (3k+2) +7 \ , & n=2k+1 \ . \end{array} \right. \\\end{align*}
But I failed and I don't know how to proceed from here.