Give it a try

This question should probably be posted in the Number Theory section since it deals mainly with property of divisibility.

Let's write out a few values of $f(n)$ . We will need them later for reference.

$f(0)=0, f(1)=0, f(2)=1, f(3)=2, f(4)=3, f(5)=2, f(6)=4, f(7)=2, f(8)=3$ .

If $n$ is an odd number, then $f(n)=2$ , so $g(n)=f(f(f(n)))=f(f(2))=f(1)=0$ .

Therefore, $\displaystyle \sum_{n=1}^{2016} g(n)= g(2)+g(4)+g(6)+ \cdots +g(2016)$ .

To organize my solution, I will form a sequence which comprise the numbers $2,4,6 \cdots, 2016$ and remove numbers which are not divisible by $3$ after computing the sum of $g(n)$ for such numbers. Then, rinse and repeat for $4,5 \cdots$

Note that there are $1008$ numbers in the sequence $2,4,6, \cdots, 2016$ . Out of these, $\frac{2016}{3}=336$ are divisible by $3$ , while $672$ are not. For these $672$ numbers, with the exception of $g(2)=f(f(f(2)))=f(f(1))=f(0)=0$ , we have $f(n)=3$ , which implies $g(n)=f(f(3))=f(2)=1$ . These $g(n)$ sum up to $\boxed{671}$ .

Out of the remaining $336$ numbers in the sequence $6,12,18 \cdots, 2016$ , we find that $\frac{336}{2}=168$ are divisible by $4$ while the other $168$ aren't. For these $168$ numbers which aren't divisible by $4$ , we have $f(n)=4$ , which implies $g(n)=f(f(4))=f(3)=2$ . These $g(n)$ sum up to $2.168=\boxed{336}$ .

Now, $168$ numbers $(12,24, \cdots, 2016)$ remain in our sequence. Out of these, only $33$ , namely $60,120, \cdots, 1980$ are divisible by $5$ while the other $135$ aren't. For these $135$ numbers, we have $f(n)=5$ , which implies $g(n)=f(f(5))=f(2)=1$ . These $g(n)$ sum up to $\boxed{135}$ .

All the $33$ numbers remaining in the sequence are divisible by $6$ since they are divisible by both $2$ and $3$ . Therefore, we focus our attention on divisibility by $7$ .

It is easy to see that only $4$ members in the sequence are divisible by $7$ , which are $420,840,1260$ and $1680$ . For the remaining $29$ which aren't divisible by $7$ , we have $f(n)=7$ , which implies $g(n)=f(f(7))=f(2)=1$ . These $g(n)$ sum up to $\boxed{29}$ .

It is easy to see that $f(420)=f(1260)=8$ , while $f(840)=f(1680)=9$ . Therefore, $g(420)=g(1260)=f(f(8))=f(3)=2$ while $g(840)=g(1680)=f(f(9))=f(2)=1$ . Summing these $g(n)$ up, we get $\boxed{6}$ .

The answer follows from $671+336+135+29+6=\boxed{1177}$ .