Determine the number of ways to serve $n$ foods in the cafeteria, knowing that apples are taken in groups of 3, yogurts come in pairs, and one is allowed at most 2 pieces of bread and one bowl of cereal due to a change in provider.
Without generating function: Let $a$ be the number of apples + pieces of bread, and $b$ be the number of yogurts + bowl of cereals. For fixed $a$ and $b$, it is observed that there is a unique way to take the necessary food items, which is to take the number of apples as close as possible to $a$ and complete with bread, and to take the number of yogurts as close as possible and add a bowl if necessary. Therefore, there are as many ways to serve oneself as there are ways to write $n=a+b$ with $a, b$ natural numbers, which is $n+1$ ways.

With generating function: The generating function for apples is $1+X^{3}+X^{6}+\cdots=\frac{1}{1-X^{3}}$, that for yogurts is $1+X^{2}+X^{4}+\cdots=\frac{1}{1-X^{2}}$, that for pieces of bread is $1+X+X^{2}=\frac{1-X^{3}}{1-X}$, and that for the bowl is $1+X$. Therefore, the generating function for the breakfast is

$$
\frac{1}{1-X^{3}} \frac{1}{1-X^{2}} \frac{1-X^{3}}{1-X}(1+X)=\frac{1}{(1-X)^{2}}=1+2 X+3 X^{2}+\ldots
$$

Therefore, the answer is $n+1$.