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 $a_{2013}$.
Let $q$ and $t$ be prime numbers. Substitute $k=q, p=t$:

$$
a_{q t+1}=t a_{q}-3 a_{t}+13
$$

Also substitute $k=t, p=q$:

$$
a_{q t+1}=q a_{t}-3 a_{q}+13
$$

Both expressions on the right are thus equal to each other, which implies

$$
t a_{q}-3 a_{t}=q a_{t}-3 a_{q}
$$

or equivalently

$$
(t+3) a_{q}=(q+3) a_{t}
$$

In particular, $5 a_{3}=6 a_{2}$ and $5 a_{7}=10 a_{2}$. Substitute $k=3, p=2$:

$$
a_{7}=2 a_{3}-3 a_{2}+13=2 \cdot \frac{6}{5} a_{2}-3 a_{2}+13
$$

Since $a_{7}=\frac{10}{5} a_{2}$, we now find

$$
\frac{13}{5} a_{2}=13
$$

so $a_{2}=5$. For any prime number $p$, it holds that $a_{p}=\frac{(p+3) a_{2}}{5}=p+3$.
Substitute $k=4, p=3$:

$$
a_{13}=3 a_{4}-3 a_{3}+13
$$

We know that $a_{13}=16$ and $a_{3}=6$, so this gives $3 a_{4}=21$, or $a_{4}=7$.
Finally, substitute $k=4$ and $p=503$:

$$
a_{2013}=a_{4 \cdot 503+1}=503 \cdot a_{4}-3 a_{503}+13=503 \cdot 7-3 \cdot(503+3)+13=503 \cdot 4-9+13=2016
$$

Thus, $a_{2013}=2016$ and this is the only possible value.