13. From empty to empty. In one vessel, there are $2a$ liters of water, and the other is empty. From the first vessel, half of the water in it is poured into the second, then from the second, $\frac{1}{3}$ of the water in it is poured back into the first, then from the first, $\frac{1}{4}$ of the water in it is poured into the second, and so on. How many liters of water will be in the first vessel after the $1997$-th pouring?
13. We will record the amount of water in the vessels after each transfer in the table.

| steps | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $2 a$ | $a$ | $\frac{4}{3} a$ | $a$ | $\frac{6}{5} a$ | $a$ | $\frac{8}{7} a$ | $a$ |
| 2 | 0 | $a$ | $\frac{2}{3} a$ | $a$ | $\frac{4}{5} a$ | $a$ | $\frac{6}{7} a$ | $a$ |

We will prove that if at some odd step the amount of water in the vessels is the same, then at the next odd step the amount of water in the vessels will also be the same. Indeed, by dividing the water in one vessel into $n$ parts and adding one of these parts to the other vessel, we will have $n+1$ such parts in it (since the amount of water in the vessels was equal before). Therefore, when on the next step we return one part from the second vessel back, we simply restore the previous state.