4A. Let the sequence $\left(a_{n}\right)$ be defined by $a_{0}=3$ and $a_{n}=2+a_{0} a_{1} \cdot \ldots \cdot a_{n-1}$ for every $n \geq 1$. Find the term $a_{2013}$.
Solution. We can write the condition as $a_{n}-2=a_{0} a_{1} \cdot \ldots \cdot a_{n-1}$. From this, $a_{n-1}-2=a_{0} a_{1} \ldots \cdot a_{n-2}$. Then, we have $a_{n-1}\left(a_{n-1}-2\right)=a_{n}-2$ from which $a_{n-1}^{2}-2 a_{n-1}=a_{n}-2$. The latter is equivalent to $\left(a_{n-1}-1\right)^{2}=a_{n}-1$. Then,

$a_{n}-1=\left(a_{n-1}-1\right)^{2}=\left(a_{n-2}-1\right)^{4}=\ldots=\left(a_{n-k}-1\right)^{2^{k}}=\left(a_{0}-1\right)^{2^{n}}=2^{2^{n}}$.

Finally, we get $a_{2013}=2^{2^{2013}}+1$.