We have $$a_k=\sum_{i=1}^{k-1}m_ka_i$$ $m_k\in R$.
If we know $a_1,\dots,a_{k-1}$, we know every $a_n$. My question is: If we have $a_1,\dots a_{i-1},a_{i+1},\dots,a_{k-1}$ and an additional $a_l$ with $l\ge k$, can $a_n$ still be uniquely determined?
In general, yes. Leaving $a_i$ unknown -- call it $x$, say, you can solve for $a_k, a_{k+1},...\;$ in terms of $x$. They will all be linear functions of $x$, so given an actual value for $a_l$, you can solve for $x$, and you then have the value of $a_i$. The only problem that might arise is if the linear function of $x$ that corresponds to $a_l$ is constant. Barring that, there's no problem.
In the case where the linear function of $x$ corresponding to $a_l$ is constant, equal to c, say, there are two possibilities. If $c \ne a_l$, then the given value of $a_l$ is inconsistent with the given initial values. In other words, that value of $a_l$ is not possible, regardless of the value of $a_i$. On the other hand, if $c = a_l$, then any value of $x$ works, so $a_i$ can take any value. In that case, the sequence is definitely not uniquely determined.