More fun in 2015 (and beyond), Part 47

Aareyan and Alan have written fine solutions. Let me show a more basic approach for those members who are not familiar with Mobius and von Mangoldt.

By definition , we have the recursive formula

$f(n)=\frac{n}{\prod_{d} f(d)}$ where the product is taken over all proper divisors $d$ of $n$ .

We will show by induction on $n$ that $f(n)=p$ if $n=p^m$ is a prime power and $f(n)=1$ otherwise. The base case $f(1)=1$ is trivial.

If $n=p^m$ , then $f(p^m)=\frac{p^m}{\prod_{j=1}^{m-1}f(p^j)}=\frac{p^m}{p^{m-1}}=p$ .

If $n=\prod_{i=1}^{r}p_i^{m_i}=p_1^{m_1}\times ..\times p_i^{m_i}\times..\times p_r^{m_r}$ with $r>1$ , then it suffices to consider the divisors $d$ of $n$ that are prime powers, by induction: $f(n)=\frac{n}{\prod_{i=1}^{r}\prod_{j=1}^{m_i}f(p_i^j)}=\frac{n}{\prod_{i=1}^{r}\prod_{j=1}^{m_i}p_i}=\frac{n}{n}=1$

Since 2017 is prime while 2015 and 2016 have more than one prime factor each, we have $f(2015)\times f(2016)\times f(2017)=1\times 1 \times 2017=\boxed{2017}$

By Mobius Inversion,

$f(n) = \prod_{d | n}\left(\frac{n}{d}\right)^{\mu(d)}$

We calculate $\begin{aligned} f(2015) & = \prod_{d | 5 \cdot 13 \cdot 31}\left(\frac{2015}{d} \right)^{\mu(d)} = \frac{2015 \cdot 5 \cdot 13 \cdot 31}{65 \cdot 155 \cdot 403} = 1 \\ f(2016) & = \prod_{d | 2^5 \cdot 3^2 \cdot 7}\left(\frac{2016}{d} \right)^{\mu (d)} = \prod_{d | 2 \cdot 3 \cdot 7}\left(\frac{2016}{d} \right)^{\mu (d)} \\ & = (2016)\left(\frac{2016}{2}\right)^{-1}\left(\frac{2016}{3}\right)^{-1}\left(\frac{2016}{7}\right)^{-1}\left(\frac{2016}{2 \cdot 3}\right)\left(\frac{2016}{2 \cdot 7}\right)\left(\frac{2016}{3 \cdot 7}\right)\left(\frac{2016}{2 \cdot 3 \cdot 7}\right)^{-1} = 1 \\ f(2017) & = \prod_{d | 2017}\left(\frac{2017}{d}\right)^{\mu(d)} = 2017 \end{aligned}$ Thus, $f(2015) \times f(2016) \times f(2017) = \boxed{2017}$ .

take log $\ln(n)=\sum_{d|n}\ln(f(d))$ let $g(n)=\ln(f(n))$ .and $p(n)=\ln(n)$ .then in dirichlet convolution form: $p=g*u$ . convolute both sides with $\mu$ to get: $p*\mu=g*I=g$ . in other words: $\ln(f(d))=\sum_{d|n} \ln(d)\mu\left(\dfrac{n}{d}\right)$ it is well known $\ln(n)=\sum_{d|n} \Lambda(d)$ where $\Lambda(n)$ is the von mangoldt function . this is easy to prove, try it yourself.

in dirichlet convolution form this is: $p=\Lambda*u$ . convolute bothsides with $\mu$ to get: $p*\mu=\Lambda*I=\Lambda$ . in other words $\sum_{d|n} \ln(d)\mu\left(\dfrac{n}{d}\right)=\Lambda(n)$ . we get that $\ln(f(n))=\Lambda(n)$ or $f(n)=e^{\Lambda(n)}$ . the only power of prime among these are $2017=2017^1$ . we get $e^0\times e^0\times e^{\ln(2017)}=\boxed{2017}$ .