A15 (9-6, Hungary) The sports meet lasted for $n$ days, and a total of $m$ medals were awarded. On the first day, 1 medal was awarded, as well as $\frac{1}{7}$ of the remaining $(m-1)$ medals. On the second day, 2 medals were awarded, as well as $\frac{1}{7}$ of the remaining medals after that, and this pattern continued each day. On the last day, the $n$-th day, the remaining $n$ medals were awarded. How many days did the sports meet last? How many medals were awarded in total?
Let $u_{k}$ be the number of medals remaining at the start of the $k$-th day of the competition. Then on that day, $k + \frac{1}{7}(u_{k} - k)$ medals were issued, leaving $u_{k} - \left[k + \frac{1}{7}(u_{k} - k)\right] = \frac{6}{7}(u_{k} - k)$ medals.
Thus, we have the recurrence relation $u_{k+1} = \frac{6}{7}(u_{k} - k)$.
Given $u_{n} = \dot{n}$ and $u_{1} = m$. Therefore, we have
$$
\begin{array}{c}
m - \frac{7}{6} u_{2} = 1, \\
u_{2} - \frac{7}{6} u_{3} = 2, \\
u_{3} - \frac{7}{6} u_{4} = 3, \\
\ldots \ldots \\
u_{n-2} - \frac{7}{6} u_{n-1} = n-2, \\
u_{n-1} - \frac{7}{6} n = n-1 .
\end{array}
$$
$$
\begin{array}{c}
u_{n-2} - \frac{7}{6} u_{n-1} = n-2, \\
u_{n-1} - \frac{7}{6} n = n-1 .
\end{array}
$$

Multiplying the above equations by the factors $1, \frac{7}{6}, \left(\frac{7}{6}\right)^{2}, \cdots, \left(\frac{7}{6}\right)^{n-2}$, and then summing them, we get
$$
m = 1 + 2\left(\frac{7}{6}\right) + 3\left(\frac{7}{6}\right)^{2} + \cdots + (n-1)\left(\frac{7}{6}\right)^{n-2} + n\left(\frac{7}{6}\right)^{n-1} .
$$
(1) $\times \frac{7}{6}$, we get
$$
\frac{7}{6} m = \frac{7}{6} + 2\left(\frac{7}{6}\right)^{2} + 3\left(\frac{7}{6}\right)^{3} + \cdots + n\left(\frac{7}{6}\right)^{n} .
$$
(1) - (2), we get
$$
\begin{aligned}
-\frac{1}{6} m & = 1 + \frac{7}{6} + \left(\frac{7}{6}\right)^{2} + \cdots + \left(\frac{7}{6}\right)^{n-1} - n\left(\frac{7}{6}\right)^{n} \\
& = \frac{\left(\frac{7}{6}\right)^{n} - 1}{\frac{7}{6} - 1} - n\left(\frac{7}{6}\right)^{n} \\
& = \left(\frac{7}{6}\right)^{n}(6 - n) - 6
\end{aligned}
$$

Rearranging, we get
$$
m = \frac{7^{n}(n - 6)}{6^{n-1}} + 36
$$

When $n > 1$, $|n - 6| < 6^{n-1}$, and $7^{n}$ is coprime with $6^{n-1}$, so $n - 6 = 0$. Solving this, we get
$$
n = 6, m = 36 .
$$

Note: This problem actually provides a method for finding the general term formula for a simple recurrence relation.