Given $a_1$,$a_2$ we can find $a_3$.Suppose we have got $a_1,a_2,...a_n$.Note that we have not yet use the pair of lowest positive numbers and the highest negative number among $a_1,a_2,...a_n$ to get a term of this sequence. Let say this term be $P$ and $N$.So we have $|a_{n+1}|\le P+N$.Wlog, assume $P+N$ is positive. Now we claim that there is a number in the sequence which belongs to $[N+1,P-1]$. Proof: If $a_{n+1}\in [N+1,P-1]$ then we are done. Otherwise,$a_{n+1} \in [-(P+N),N+1]$.If all $a_i$ after $a_{n+1}$ are in$[-(P+N),N+1]$ then as these $a_i$'s are sum of 2 previous $a_s$ and $a_t$ and if they are of different signs then their absolute value will both at least $P$.Only finitely many such term exists.So these terms can be represent as sum of several negative previous $a_i$.So if $P+N < r|N + 1|$ for some r then at most r terms less than $N+1$ will be involved in formimg these terms.But again only definitely many such terms exists.So here must be a term in $[N+1,P-1]$. So the difference between max negative and minimum positive decreases at least by 1.So after sometime we must get an $a_i$ with $|a_i| < 1$.