12*. In how many ways can milk be transferred from a 12-liter barrel, filled with milk, to another empty barrel of the same volume using two empty cans of 1 liter and 2 liters? Transferring milk from one can to another is not allowed.

Note that the question in this problem is different from the previous problems.
Let's denote by $a_{n}$ the number of ways to transfer milk from a full container of $n$ liters (where $n \in \boldsymbol{N}$) using two empty cans of 1 liter and 2 liters.

If we first use the 1-liter can, there will be $n-1$ liters left in the first container, and it can be transferred in $a_{n-1}$ ways.

If we first use the 2-liter can, there will be $n-2$ liters left, and it can be transferred in $a_{n-2}$ ways. Therefore, we get:

$$
a_{n}=a_{n-1}+a_{n-2}
$$

This formula can be used only when $n>2$. (In mathematics, a formula that expresses a general term of a sequence in terms of one or several of its preceding terms is called a recurrence formula.)

If initially there were 1 or 2 liters of milk in the container, it can be transferred to another container in one or two ways, respectively:

$$
a_{1}=1, \quad a_{2}=2
$$

Let's introduce the zeroth term of the sequence, assuming $a_{0}=1$. Then the obtained formula can be used even when $n=2$:

$$
a_{2}=a_{1}+a_{0}=1+1=2
$$

Now let's calculate all terms from $a_{3}$ to $a_{12}$ in sequence:

$$
\begin{gathered}
a_{3}=a_{2}+a_{1}=2+1=3, \quad a_{4}=a_{3}+a_{2}=3+2=5, \quad a_{5}=a_{4}+a_{3}=5+3=8, \\
a_{6}=a_{5}+a_{4}=8+5=13, \quad a_{7}=21, a_{8}=34, \\
a_{9}=55, \quad a_{10}=89, \quad a_{11}=144, \quad a_{12}=233 .
\end{gathered}
$$

Answer: 233.

In solving this problem, we obtained the sequence

$$
1,1,2,3,5,8,13,21,34, \ldots
$$

where the first two terms are 1, and each subsequent term, starting from the third, is the sum of the two preceding terms. This sequence plays a significant role in number theory and is called the Fibonacci sequence, named after the Italian scholar Leonardo Fibonacci (1180-1240).