Define a game on a number line as follows: at the beginning, there are (possibly) some tokens at some negative integers on the number line. Each integer can hold any number of tokens. At each step, before you move a token to an adjacent integer, an adjacent token (not at the same integer) must be $\color{red}{\text{removed}}$ as a cost. If you can let a token reside at some integer after a finite number of moves, it is said that the integer is "reachable". Your goal is to obtain the maximum "reachable" integer.
For example, suppose there are two tokens at $-2$ and three tokens at $-1$. Via the following moves, you can reach the integer $2$: \begin{array}{|c|c|c|c|c|c|} \hline \text{integers} & -2 & -1 & 0 & 1 & \color{red}{\textrm{2}} \\ \hline \text{beginning} & 2 & 3 & 0 & 0 & 0 \\ \hline \text{after move } 1 & 1 & 2 & 1 & 0 & 0 \\ \hline \text{after move } 2 & 0 & 1 & 2 & 0 & 0 \\ \hline \text{after move } 3 & 0 & 0 & 1 & 1 & 0 \\ \hline \text{after move } 4 & 0 & 0 & 0 & 0 & 1 \\ \hline \end{array} In the table, the first row means the integers while each of the following row represents the tokens after a move. We can see that the maximum reachable integer is $2$.
I'd like to ask, given an initial distribution of tokens, how to calculate the maximum reachable integer? An analysis on simple cases (for example, there are tokens on integers $-1$ and $-2$ only) are also appreciated.
This is the one-dimensional version of the (cool and interesting!) two-dimensional "Conway's soldiers" problem. In particular, you're essentially replacing a pair of tokens on spaces $n-2$ and $n-1$ with a token on space $n$, which means that assigning weight $\phi^n$ to space $n$ (where $\phi$ is the golden ratio) is preserved by this token move. In your example, the initial weighted sum of tokens is $$ 2\cdot\phi^{-2}+3\cdot\phi^{-1} = \frac{3+\sqrt5}2 = \phi^2, $$ which is a proof that no integer greater than $2$ is reachable for this example.