Find the number of increasing sequences $a_{1}, a_{2}, \ldots$ containing only integers between 1 and 20 and such that $a_{i} \equiv i[2]$ for all $i$.
Let $u_{k}$ be the number of sequences with terms between 1 and $k$ satisfying the condition of the statement, and $U(X)$ the associated generating function. We set $a_{0}=0$ and $b_{i}=a_{i}-a_{i-1}$ for all $i \geqslant 1$. A sequence of length $n$ then corresponds to $n$ odd numbers $b_{i}$ whose sum is less than or equal to $k$. We can therefore add a last term $c \geqslant 0$ such that the total sum of the $b_{i}$ and $c$ is exactly $k$.

The series associated with the odd numbers $b_{i}$ is $X+X^{3}+X^{5}+\ldots=\frac{X}{1-X^{2}}$ and the series associated with the choice of $c$ is $1+X+X^{2}+\ldots=\frac{1}{1-X}$, so:

![](https://cdn.mathpix.com/cropped/2024_05_10_cbd6df20257d6519feabg-247.jpg?height=166&width=1046&top_left_y=294&top_left_x=505)

$$
\begin{aligned}
& =\frac{1}{1-x} \frac{1}{1-\frac{X}{1-X^{2}}}=\frac{1}{1-x} \frac{1-X^{2}}{1-X-X^{2}}=\frac{1+X}{1-X-X^{2}}
\end{aligned}
$$

The series $\frac{1}{1-X-X^{2}}$ is that of the Fibonacci numbers (with $F_{0}=F_{1}=1$), so the desired number is $F_{20}+F_{19}=F_{21}$.