Task 3. Given is an unknown sequence $a_{1}, a_{2}, a_{3}, \ldots$ of integers that satisfies the following property: for every prime number $p$ and every positive integer $k$ it holds that

$$
a_{p k+1}=p a_{k}-3 a_{p}+13
$$

Determine all possible values of $a_{2013}$.
Solution II. Just as in Solution I, deduce that $a_{2}=5$ and $a_{p}=p+3$ for all prime numbers $p$. Therefore, $a_{3}=6$. Substituting $k=1$ and $p=2$ gives

$$
a_{3}=2 a_{1}-3 a_{2}+13=2 a_{1}-2,
$$

so $2 a_{1}=a_{3}+2=8$, thus $a_{1}=4$. Now we have calculated $a_{1}$ and $a_{2}$. For $n \geq 3$, there is always a prime number $p$ such that $p \mid n-1$, so we can always find $p$ and $k$ such that
$n=k p+1$. Therefore, the value of $a_{n}$ is fixed if we already know all values $a_{m}$ with $1 \leq m<n$. Thus, there can only be one sequence that satisfies the condition. We now try to see if $a_{n}=n+3$ satisfies the condition. Then we have

$$
p a_{k}-3 a_{p}+13=p(k+3)-3(p+3)+13=p k+3 p-3 p-9+13=p k+4=a_{p k+1} \text {. }
$$

So this indeed satisfies the condition. The only sequence that satisfies the condition is the one with $a_{n}=n+3$ for all $n \geq 1$, from which it follows that the only possible value for $a_{2013}$ is 2016.