A Problem of Function

I think I got the solution. However, Will you explain it a bit more lucidly.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328608#

Solution :( )

Note that f(x)>x

( because if we let f(x)=x we dont get f(f(x))= 3x , and f is nothing but increasing function of natural number,so at the moment we realize f(1)≥2 we get f(x) > x )

What if f(1) >2 ?
let f(1) = p (p≥3)

f(f(1)) = 3 or f(p) = 3 which contradicts f(x)>x hence f(1) = 2

so f(f(1))=f(2) = 3

f(f(2))=f(3) = 6

f(f(6)) = f(9) = 18 using induction we conclude f(3ⁿ)= 2* 3ⁿ

f(f(3ⁿ)) =f(2*3ⁿ) = 3ⁿ⁺¹

there are exactly 3ⁿ-1 numbers in Domain and in the range as well (in between 3ⁿ
and2* 3ⁿ)

As f is increasing function obviously f(3ⁿ+k) = 2*3ⁿ+ k ;(0<k<3ⁿ)

Now f(f(3ⁿ+k)) = f(2*3ⁿ+k)= 3ⁿ⁺¹+ 3k

f(2001) = f(2* 3⁶+ 543)=3⁷+ 3* 543 = 3816

Mujhe kuch bhi samjh nai aa raha :( :(

the qstn is a given from advanced problems in algebra by titu andreescu.... quite difficult ...