By induction $ a_{i - 1} < a_i < 2 a_{i - 1}$ $ i = 1$ $ a_0 < a_1 < 2 a_0$ $ \forall i < j$ $ a_{i - 1} < a_i < 2 a_{i - 1}$ Operation $ 1)$ $ a_{j - 1} < 2 a_{j - 2}$ $ a_{j - 2} < a_{j - 1}$ $ a_{j - 1} < a_j = 2a_{j - 2} < 2a_{j - 1}$ Operation $ 2)$ $ a_{j - 2} < a_{j - 1}$ $ a_{j - 1} < 2a_{j - 2}$ $ a_{j - 1} < a_j = 3a_{j - 1} - 2a_{j - 2} < 2a_{j - 1}$ It remains to prove that $ a_i$ is a sum of at most two power of two then there is no $ k$ such $ 1024 + 512 = 1536 < a_k < 2048$

A simple induction shows $ (a_n,a_{n+1})$ must have the form $ (2^a+2^b,2^{a+1}+2^b)$

Exactly $ a_n = 2^{b_n} + 2^n,b_0 = b_1 = 0$. If $ a_{n + 1} = 2a_n$, then $ b_{n + 1} = b(n) + 1$. If $ a_{n + 1} = 3a_n - 2a_{n - 1}$, then $ 2^{b_{n + 1}} = 2^{b_n + 1} + 2^{b_n} - 2^{b_{n - 1} + 1}$. If $ b_{n - 1} = b_n$,then $ b_{n + 1} = b_n$ else $ b_{n + 1} = b_n + 1$. Therefore $ a_n = 2^{n - c} + 2^n$, were $ c\ge 1$ is first case, when used $ a_{c + 1} = 2a_c$ (if it is not used c=n $ a_n = 1 + 2^n$).