The sequence $ (x_n)_{n \in \mathbb{N}}$ is defined by $ x_1\equal{}2, x_2\equal{}3,$ and

$ x_{2m\plus{}1}\equal{}x_{2m}\plus{}x_{2m\minus{}1}$ for $ m \ge 1;$
$ x_{2m}\equal{}x_{2m\minus{}1}\plus{}2x_{2m\minus{}2}$ for $ m \ge 2.$

Determine $ x_n$ as a function of $ n$.
To determine \( x_n \) as a function of \( n \), we will first derive the recursive formula for \( x_n \) and then solve it.

1. **Initial Conditions:**
   \[
   x_1 = 2, \quad x_2 = 3
   \]

2. **Recursive Definitions:**
   \[
   x_{2m+1} = x_{2m} + x_{2m-1} \quad \text{for} \quad m \geq 1
   \]
   \[
   x_{2m} = x_{2m-1} + 2x_{2m-2} \quad \text{for} \quad m \geq 2
   \]

3. **Prove the Recursive Formula for Even \( n \):**
   \[
   x_{2m} = x_{2m-1} + 2x_{2m-2}
   \]
   Using the recursive definition for \( x_{2m-1} \):
   \[
   x_{2m-1} = x_{2m-2} + x_{2m-3}
   \]
   Substituting this into the equation for \( x_{2m} \):
   \[
   x_{2m} = (x_{2m-2} + x_{2m-3}) + 2x_{2m-2} = 3x_{2m-2} + x_{2m-3}
   \]
   Now, using the recursive definition for \( x_{2m-2} \):
   \[
   x_{2m-2} = x_{2m-3} + 2x_{2m-4}
   \]
   Substituting this into the equation for \( x_{2m} \):
   \[
   x_{2m} = 3(x_{2m-3} + 2x_{2m-4}) + x_{2m-3} = 3x_{2m-3} + 6x_{2m-4} + x_{2m-3} = 4x_{2m-3} + 6x_{2m-4}
   \]
   Simplifying further:
   \[
   x_{2m} = 4x_{2m-2} - 2x_{2m-4}
   \]
   Thus, we have:
   \[
   \boxed{x_{2m} = 4x_{2m-2} - 2x_{2m-4}}
   \]

4. **Prove the Recursive Formula for Odd \( n \):**
   \[
   x_{2m+1} = x_{2m} + x_{2m-1}
   \]
   Using the recursive definition for \( x_{2m} \):
   \[
   x_{2m} = 4x_{2m-2} - 2x_{2m-4}
   \]
   And for \( x_{2m-1} \):
   \[
   x_{2m-1} = x_{2m-2} + x_{2m-3}
   \]
   Substituting these into the equation for \( x_{2m+1} \):
   \[
   x_{2m+1} = (4x_{2m-2} - 2x_{2m-4}) + (x_{2m-2} + x_{2m-3}) = 5x_{2m-2} + x_{2m-3} - 2x_{2m-4}
   \]
   Simplifying further:
   \[
   x_{2m+1} = 4x_{2m-1} - 2x_{2m-3}
   \]
   Thus, we have:
   \[
   \boxed{x_{2m+1} = 4x_{2m-1} - 2x_{2m-3}}
   \]