Let the function $f$ be defined on the set $\mathbb{N}$ such that

$\text{(i)}\ \ \quad f(1)=1$
$\text{(ii)}\ \quad f(2n+1)=f(2n)+1$
$\text{(iii)}\quad f(2n)=3f(n)$

Determine the set of values taken $f$.
To determine the set of values taken by the function \( f \) defined on the set \(\mathbb{N}\) with the given properties:
1. \( f(1) = 1 \)
2. \( f(2n+1) = f(2n) + 1 \)
3. \( f(2n) = 3f(n) \)

we will analyze the function step by step.

1. **Base Case:**
   \[
   f(1) = 1
   \]

2. **Evaluate \( f(2) \):**
   Using property (iii):
   \[
   f(2) = 3f(1) = 3 \cdot 1 = 3
   \]

3. **Evaluate \( f(3) \):**
   Using property (ii):
   \[
   f(3) = f(2 \cdot 1 + 1) = f(2) + 1 = 3 + 1 = 4
   \]

4. **Evaluate \( f(4) \):**
   Using property (iii):
   \[
   f(4) = 3f(2) = 3 \cdot 3 = 9
   \]

5. **Evaluate \( f(5) \):**
   Using property (ii):
   \[
   f(5) = f(4 + 1) = f(4) + 1 = 9 + 1 = 10
   \]

6. **Evaluate \( f(6) \):**
   Using property (iii):
   \[
   f(6) = 3f(3) = 3 \cdot 4 = 12
   \]

7. **Evaluate \( f(7) \):**
   Using property (ii):
   \[
   f(7) = f(6 + 1) = f(6) + 1 = 12 + 1 = 13
   \]

8. **Evaluate \( f(8) \):**
   Using property (iii):
   \[
   f(8) = 3f(4) = 3 \cdot 9 = 27
   \]

From these calculations, we observe that the values of \( f \) are:
\[
1, 3, 4, 9, 10, 12, 13, 27, \ldots
\]

These values can be written in base 3 as:
\[
1_3, 10_3, 11_3, 100_3, 101_3, 110_3, 111_3, 1000_3, \ldots
\]

This suggests that the values of \( f \) are numbers that can be written with only digits 0 and 1 in base 3. Therefore, the set of values taken by \( f \) is:
\[
S = \{0, 1, 3, 4, 9, 10, 12, 13, 27, \ldots\}
\]

The final answer is \( \boxed{ \{0, 1, 3, 4, 9, 10, 12, 13, 27, \ldots\} } \).