Problem (Algebra):
Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows:

$$
\left\{\begin{array}{l}
a_{1}=1 \\
\text { for } n \geq 2, a_{n} \text { is the smallest possible positive value of } n-a_{k}^{2}, \text { for } 1 \leq k<n .
\end{array}\right.
$$

For example, $a_{2}=2-1^{2}=1$, and $a_{3}=3-1^{2}=2$. Compute $a_{1}+a_{2}+\cdots+a_{50}$.

Solution:
The requirement that $a_{n}$ be the smallest positive value of $n-a_{k}^{2}$ for $k<n$ is equivalent to determining the largest value of $a_{k}$ such that $a_{k}^{2}<n$. For $n=3$, use either $a_{1}=a_{2}=1$ to find $a_{3}=3-1^{2}=2$. For $n=4$, the strict inequality eliminates $a_{3}$, so $a_{4}=4-1^{2}=3$, but $a_{3}$ can be used to compute $a_{5}=5-2^{2}=1$. In fact, until $n=10$, the largest allowable prior value of $a_{k}$ is $a_{3}=2$, yielding the values $a_{6}=2, a_{7}=3, a_{8}=4, a_{9}=5$. In general, this pattern continues: from $n=m^{2}+1$ until $n=(m+1)^{2}$, the values of $a_{n}$ increase from 1 to $2 m+1$. Let $S_{m}=1+2+\cdots+(2 m+1)$. Then the problem reduces to computing $S_{0}+S_{1}+\cdots+S_{6}+1$, because $a_{49}=49-6^{2}$ while $a_{50}=50-7^{2}=1 . S_{m}=\frac{(2 m+1)(2 m+2)}{2}=2 m^{2}+3 m+1$, so

$$
\begin{aligned}
S_{0}+S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6} & =1+6+15+28+45+66+91 \\
& =252
\end{aligned}
$$

Therefore the desired sum is $252+1=\mathbf{2 5 3}$.

Final answer: 253