The sequence $ (a_n)$ is given by $ a_1\equal{}1,a_2\equal{}0$ and:

$ a_{2k\plus{}1}\equal{}a_k\plus{}a_{k\plus{}1}, a_{2k\plus{}2}\equal{}2a_{k\plus{}1}$ for $ k \in \mathbb{N}.$

Find $ a_m$ for $ m\equal{}2^{19}\plus{}91.$
To find \( a_m \) for \( m = 2^{19} + 91 \), we need to analyze the given sequence and verify the hypothesis that \( a_{2^n + k} = k \) or \( 2^n - k \), whichever is smaller, for \( 0 \leq k \leq 2^n \).

1. **Base Cases:**
   - \( a_1 = 1 \)
   - \( a_2 = 0 \)

2. **Recursive Definitions:**
   - \( a_{2k+1} = a_k + a_{k+1} \)
   - \( a_{2k+2} = 2a_{k+1} \)

3. **Hypothesis Verification:**
   - We hypothesize that \( a_{2^n + k} = k \) or \( 2^n - k \), whichever is smaller, for \( 0 \leq k \leq 2^n \).

4. **Induction Step:**
   - Assume the hypothesis holds for \( n \). We need to show it holds for \( n+1 \).

5. **Induction Basis:**
   - For \( n = 0 \):
     - \( a_1 = 1 \) (distance from \( 2^0 = 1 \) is 0)
     - \( a_2 = 0 \) (distance from \( 2^1 = 2 \) is 0)

6. **Induction Hypothesis:**
   - Assume \( a_{2^n + k} = k \) or \( 2^n - k \), whichever is smaller, for \( 0 \leq k \leq 2^n \).

7. **Induction Step:**
   - For \( m = 2^{n+1} + k \):
     - If \( k \leq 2^n \), then \( a_{2^{n+1} + k} = k \).
     - If \( k > 2^n \), then \( a_{2^{n+1} + k} = 2^{n+1} - k \).

8. **Application to Given Problem:**
   - For \( m = 2^{19} + 91 \):
     - Here, \( k = 91 \) and \( 0 \leq 91 \leq 2^{19} \).
     - According to the hypothesis, \( a_{2^{19} + 91} = 91 \).

The final answer is \( \boxed{91} \).