Let $f: \mathbb{Z}_{>0} \rightarrow \mathbb{R}$ be a function for which it holds that for all $n>1$ there is a prime divisor $p$ of $n$ such that

$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$

Moreover, it is given that $f\left(2^{2014}\right)+f\left(3^{2015}\right)+f\left(5^{2016}\right)=2013$.
Calculate $f\left(2014^{2}\right)+f\left(2015^{3}\right)+f\left(2016^{5}\right)$.
When $n=q$ with $q$ prime, there is only one prime divisor of $n$, namely $q$, so it must hold that $f(q)=f(1)-f(q)$, thus $f(q)=\frac{1}{2} f(1)$. If $n=q^{2}$ with $q$ prime, then $n$ also has only one prime divisor, so $f\left(q^{2}\right)=f(q)-f(q)=0$. We now prove by induction on $k$ that $f\left(q^{k}\right)=\frac{2-k}{2} f(1)$ if $q$ is a prime number and $k$ is a positive integer. For $k=1$ and $k=2$, we have already shown this. Suppose now that $f\left(q^{k}\right)=\frac{2-k}{2} f(1)$ for some $k \geq 2$ and substitute $n=q^{k+1}$. It holds that

$$
f\left(q^{k+1}\right)=f\left(q^{k}\right)-f(q)=\frac{2-k}{2} f(1)-\frac{1}{2} f(1)=\frac{2-(k+1)}{2} f(1)
$$

This completes the induction.
Now we use our second given. It holds that

$$
\begin{aligned}
2013 & =f\left(2^{2014}\right)+f\left(3^{2015}\right)+f\left(5^{2016}\right) \\
& =\frac{2-2014}{2} f(1)+\frac{2-2015}{2} f(1)+\frac{2-2016}{2} f(1) \\
& =-\frac{6039}{2} f(1)
\end{aligned}
$$

so $f(1)=\frac{2013 \cdot 2}{-6039}=-\frac{2}{3}$. And then for any prime number $q$, it holds that $f(q)=\frac{1}{2} f(1)=-\frac{1}{3}$.
We then prove that if $n=p_{1} p_{2} \cdots p_{m}$ with $p_{1}, p_{2}, \ldots, p_{m}$ not necessarily distinct primes and $m \geq 0$, then $f(n)=\frac{m-2}{3}$. We do this by induction on $m$. For $m=0$, $n=1$ and $f(1)=-\frac{2}{3}=\frac{0-2}{3}$, so it holds for this case. Suppose now that we have proven it for some $m \geq 0$. Consider any $n$ of the form $n=p_{1} p_{2} \cdots p_{m+1}$. Then $n>1$, so there is a prime factor $p \mid n$ for which $f(n)=f\left(\frac{n}{p}\right)-f(p)$; without loss of generality, this is $p=p_{m+1}$. It follows that

$$
f(n)=f\left(p_{1} p_{2} \cdots p_{m}\right)-f\left(p_{m+1}\right)=\frac{m-2}{3}--\frac{1}{3}=\frac{(m+1)-2}{3} .
$$

This completes the induction.
We can now calculate the desired result. The prime factorizations of 2014, 2015, and 2016 are $2014=2 \cdot 19 \cdot 53, 2015=5 \cdot 13 \cdot 31$, and $2016=2^{5} \cdot 3^{2} \cdot 7$, so

$$
f\left(2014^{2}\right)+f\left(2015^{3}\right)+f\left(2016^{5}\right)=\frac{6-2}{3}+\frac{9-2}{3}+\frac{40-2}{3}=\frac{49}{3} .
$$