Question 78
Let $f:\mathbb Z_{\gt 0}\to\mathbb R$ satisfy the following: for every $n\gt 1$ there is a prime divisor $p$ of $n$ such that
$$f(n)=f\left(\frac np\right)-f(p).$$
It is also given that $f(2^{2014})+f(3^{2015})+f(5^{2016})=2013$.
Calculate $f(2014^2)+f(2015^3)+f(2016^5)$.
$$f(n)=f\left(\frac np\right)-f(p).$$
It is also given that $f(2^{2014})+f(3^{2015})+f(5^{2016})=2013$.
Calculate $f(2014^2)+f(2015^3)+f(2016^5)$.