Funky Functional Equation

Algebra Level 5

Let f:R+R+f:\mathbb{R^+} \rightarrow \mathbb{R^+} be a function satisfying

f(1)=2f(1)=2

f(2)=1f(2)=1

f(3n)=3f(n)f(3n)=3f(n)

f(3n+1)=3f(n)+2f(3n+1)=3f(n)+2

f(3n+2)=3f(n)+1.f(3n+2)=3f(n)+1.

Find the number of integers n2006n \leq 2006 such that f(n)=2n.f(n)=2n.


See Part 2 if you enjoyed this problem! :D

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