Incredible Function

We have, $f(f(n))=3n$ and hence, $f(3n)=f(f(f(n)))=3f(n)$

$f(3)=3f(1)$ .

Suppose $f(1)=1$

Hence, $3=3\cdot 1=3f(1)=f(f(1))=f(1)=1$ which is absurd.So $f(1)>1$

Now, $3=f(f(1))>f(1)>1$

Hence, $f(1)=2$ .So, $f(2)=f(f(1))=3f(1)=3$

Also, $f(3)=3f(1)=6$ .Again, $f(6)=f(3.2)=3f(2)=9$

Now, $6=f(3)<f(4)<f(5)<f(6)=9$ .So, $f(4)=7$ and $f(5)=8$

Hence, $f(7)=f(f(4))=3.4=12$ , $f(8)=f(f(5))=15$ , $f(9)=f(f(6))=18$ and $f(1)=f(f(7))=21$ .

Now, $f(9)=18<f(10)<f(11)<f(12)=21$

So, $f(10)=19,f(11)=20$

From the above computation of several initial values of $f(n)$ one may guess the pattern how f(n) are formed.But,if still one cannot find f(n),look below.

Let $3^{m}≤n<2.3^{m}$ .

Now, $f(3^{m})=3^{m}f(1)=2.3^{m}$ and $f(2.3^{m})=f(f(3^{m})=3^{m+1}$

Now, $2.3^{m}=f(3^{m})<f(3^{m}+1)<...<f(3^m+3^m-1)<f(2.3^{m})=3^{m+1}$ .

Hence, $f(3^m+j)=2.3^{m}+j$ for $0≤j≤3^{m}$

Thus, $f(n)=n+3^{m}$ for all n such that $3^{m}≤n≤2.3^{m}$

If $2.3^{m}≤n≤3^{m+1}$ then $n=2.3^{m}+j$ where $0≤j≤2.3^{m}$

Therefore, $f(n)=f(2.3^{m}+j)=f(f(3^{m}+j))=3^{m+1}+3j=3n-3^{m+1}$

Now,note that $2.3^{6}<2014<3^{7}$ .

So, $f(2014)=3.2014-3^{7}=3855$