prove by induction (or perhaps otherwise?) that if $ n$ is odd then $ x_{\frac {(n + 3)(n + 1)}{8}} = n$, and that it is the only time $ n$ appears in the sequence. After that it follows immediately that 1992 appears at position $ \frac {258\cdot 256}{8} - 3 = 66045$, and that it is the only time it appears. It also follows that every number appears in the sequence exactly once, so $ \eta(k) = f^k(1) = x_k$ is a bijection on $ \mathbb{N}\to \mathbb{N}$.

Observation 1: Say $ x_n = (2k+1)2^p$ where $ p$ is positive. Then, $ x_{n+1} = (2k+1)2^{p-1} + 2^p = (2k+1)2^{p-1} + 2(2^{p-1}) = (2k+3)2^{p-1}$ Observation 2: From repeated use of Observation 1, we can see that \[ x_n = 2^p \implies x_{n+k} = (2k+1)2^{p-k}\] given $ p \ge k$. Observation 3: It follows that if there exist $ n_1$ such that $ x_{n_1} = 2^p$, then there exist $ n_2 = n_1+p$ such that $ x_{n_2} = (2p+1)$. Since $ 2p+1$ is odd, $ x_{n_2+1} = 2^\frac{2p+1+1}{2} = 2^{p+1}$. Thus \[ x_{n} = 2^p \implies x_{n+p+1} = 2^{p+1}\] Observation 4: Since $ x_1 = 1 = 2^0$ is in the sequence, it follows $ 2^p$ is in the sequence for any $ p$, which in turn implies $ (2k+1)2^{p-k}$ is in the sequence for any given $ p \ge k$. In fact, $ 1992 = (2*124 + 1)2^{127-124}$, so $ x_1 = 2^0$ so, $ x_{1+(1)+(2)+...(127)}=2^{127}$ so, $ x_{1+(1)+...(127)+124} = (2*124 + 1)2^{127-124} = 1992$ It's not hard to see that $ n = 1 + 1... + 124 = 8253$ is the smallest and the only value such that $ x_n = 1992$