1412.1 ** The general term formula for the sequence $101,104,116, \cdots$ is
$$
a_{n}=100+n^{2}, n=1,2,3, \cdots
$$

For each $n$, let $d_{n}$ denote the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $d_{n}$ when $n$ takes all positive integers.
Parse: Since $d_{n}=\left(100+n^{2}, 100+(n+1)^{2}\right)=\left(100+n^{2}, 2 n+1\right)=$ $\left(2\left(100+n^{2}\right), 2 n+1\right)=(200-n, 2 n+1)=(200-n, 401)$, therefore, when $n=200$, $d_{n}$ is maximized, with the maximum value being $d_{200}=401$.

Comment: We have a general conclusion: If $a$ is a positive integer, and $d_{n}$ is the greatest common divisor of $a+n^{2}$ and $a+(n+1)^{2}$, then when $n=2 a$, $d_{n}$ reaches its maximum value of $4 a+1$.