Let f:N→N be strictly increasing function such that f(f(n))=3n then f(11)=?

We first note that we must have f(n) >n. To prove this first its easy to see that for no n can we have f(n) = n as f([f(n)] = 3n would then yield f(n) = 3n which is a contradiction. Suppose f(n)<n, then since f is strictly increasing we have f[(f(n)] < f(n) so that 3n<f(n)<n which is again a contradiction. Thus we have established that f(n)>n for all n

Now, if f(1) = k>1 (since f(1) = k>1 as proved above), we have f(k) = 3 which means 3>k so that k =1 or 2. Since k>1, this gives k = 2

Hence f(2) =3 and so f(3) = f(f(2)) = 6 and similarly f(6) = 9. Now, we will attempt to plug the gap by finding f(4) and f(5). This is pretty much immediate since f(4) and f(5) must lie between f(3) = 6 and f(6) = 9 and since f(5)>f(4) we must have f(4) = 7 and f(5) = 8

Now f(7) = 12 and f(8) = 15, f(9) = f(f(6)) = 18

f(12) = f(f(7)) = 21. From a similar argument as previous, we must have f(10) = 19 and so f(11) = 20