Let the terms of the sequences $u_{n}, v_{n}$ be formed from the arbitrarily chosen numbers $u_{1}, u_{2}, v_{1}, v_{2}$ according to the relations

$$
u_{n}=2 u_{n-1}+u_{n-2}, \quad v_{n}=3 v_{n-1}-v_{n-2}
$$

Then, based on these sequences, we formed the terms of the sequence $s_{n}=u_{n}+v_{n}(n=1,2, \ldots)$. Can $s_{k}$ be determined if we know the numbers $s_{1}, s_{2}, \ldots, s_{k-1}$?
Given $u_{1}, u_{2}, v_{1}, v_{2}$, the recursion allows us to determine $u_{n}$ and $v_{n}$, and therefore $s_{n}$ for any natural number $n$. Therefore, we need to examine whether the knowledge of $s_{1}, s_{2}, \ldots, s_{k-1}$ is sufficient to determine $u_{1}, u_{2}, v_{1}, v_{2}$. Since $s_{1}$, $s_{2}, \ldots, s_{k-1}$ can each be expressed as a linear combination of $u_{1}, u_{2}, v_{1}, v_{2}$, the expressions for $s_{1}, s_{2}, s_{3}, s_{4}$ yield a system of linear equations in four unknowns $u_{1}, u_{2}, v_{1}, v_{2}$. Indeed,

\[
\begin{gathered}
s_{3}=u_{3}+v_{3}=\left(2 u_{2}+u_{1}\right)+\left(3 v_{2}-v_{1}\right) \\
s_{4}=u_{4}+v_{4}=\left(2 u_{3}+u_{2}\right)+\left(3 v_{3}-v_{2}\right)= \\
=2\left(2 u_{2}+u_{1}\right)+u_{2}+3\left(3 v_{2}-v_{1}\right)-v_{2}=5 u_{2}+2 u_{1}+8 v_{2}-3 v_{1}
\end{gathered}
\]

Thus, the system of equations is:

\[
\begin{aligned}
u_{1}+v_{1} & =s_{1} \\
u_{2}+v_{2} & =s_{2} \\
u_{1}+2 u_{2}-v_{1}+3 v_{2} & =s_{3} \\
2 u_{1}+5 u_{2}-3 v_{1}+8 v_{2} & =s_{4}
\end{aligned}
\]

This system can be easily solved:

\[
\begin{aligned}
& u_{1}=-s_{2}+3 s_{3}-s_{4}, \\
& u_{2}=-s_{1}+s_{2}+5 s_{3}-2 s_{4} \\
& v_{1}=s_{1}+s_{2}-3 s_{3}+s_{4} \\
& v_{2}=s_{1} \quad-5 s_{3}+2 s_{4} .
\end{aligned}
\]

Therefore, if $k \geq 5$, i.e., at least $s_{1}, s_{2}, s_{3}, s_{4}$ are given, then $s_{1}, \ldots, s_{k-1}$ not only determine $s_{k}$ but also every $s_{n}(n \geq k)$. However, if $k \leq 4$, then $s_{k}$ cannot be determined uniquely. For example, if $s_{1}, s_{2}, s_{3}$ are given but $s_{4}$ is not, substituting any arbitrary number $p$ for $s_{4}$ in (4) yields a solution system where $s_{4}$ is replaced by $p$ everywhere. Thus, the solution system will differ for different values of $p$.

Remarks. 1. We naturally interpreted the question to mean that $s_{1}, s_{2}, \ldots, s_{k-1}$ are terms of a well-formed sequence, not arbitrarily chosen numbers. We have seen that $s_{1}, s_{2}, s_{3}, s_{4}$ are sufficient to reconstruct the sequences $u_{n}, v_{n}, s_{n}$.

2. The recursive definition of the terms $u_{n}, v_{n}$, and $s_{n}$ implies that the system of equations and its solution remain valid if the same integer is added to each index. Consequently, for $k \geq 5$,

\[
\begin{array}{lr}
u_{k-4} & -s_{k-3}+3 s_{k-2}-s_{k-1} \\
u_{k-3} & =-s_{k-4}+s_{k-3}+5 s_{k-2}-2 s_{k-1}
\end{array}
\]

\[
\begin{aligned}
& v_{k-4}=s_{k-4}+s_{k-3}-3 s_{k-2}+s_{k-1} \\
& v_{k-3}=s_{k-4}-5 s_{k-2}+2 s_{k-1}
\end{aligned}
\]

and from (4),

\[
\begin{aligned}
s_{k} & =5 u_{k-2}+2 u_{k-3}+8 v_{k-2}-3 v_{k-3}= \\
& =5\left(2 u_{k-3}+u_{k-4}\right)+2 u_{k-3}+8\left(3 v_{k-3}-v_{k-4}\right)-3 v_{k-3}= \\
& =12 u_{k-3}+5 u_{k-4}+21 v_{k-3}-8 v_{k-4}
\end{aligned}
\]

From this and (5), it easily follows that if $k \geq 5$, then

\[
s_{k}=s_{k-4}-s_{k-3}-6 s_{k-2}+5 s_{k-1}
\]

Accordingly, if we only need $s_{k}$, its calculation is simpler if we use the four largest indexed known terms.