We have an a sequence such that $a_n = 2 \cdot 10^{n + 1} + 19$. Determine all the primes $p$, with $p \le 19$, for which there exists some $n \ge 1$ such that $p$ divides $a_n$.
1. **Identify the sequence and the primes to be checked:**
   The sequence is given by \( a_n = 2 \cdot 10^{n + 1} + 19 \). We need to determine all primes \( p \leq 19 \) for which there exists some \( n \geq 1 \) such that \( p \) divides \( a_n \).

2. **Check for \( p = 2, 5, 19 \):**
   - For \( p = 2 \):
     \[
     a_n = 2 \cdot 10^{n + 1} + 19 \equiv 0 + 19 \equiv 1 \pmod{2}
     \]
     Since \( 1 \not\equiv 0 \pmod{2} \), \( p = 2 \) does not divide \( a_n \).

   - For \( p = 5 \):
     \[
     a_n = 2 \cdot 10^{n + 1} + 19 \equiv 0 + 19 \equiv 4 \pmod{5}
     \]
     Since \( 4 \not\equiv 0 \pmod{5} \), \( p = 5 \) does not divide \( a_n \).

   - For \( p = 19 \):
     \[
     a_n = 2 \cdot 10^{n + 1} + 19 \equiv 0 + 19 \equiv 0 \pmod{19}
     \]
     Since \( 19 \equiv 0 \pmod{19} \), \( p = 19 \) does not divide \( a_n \).

   Therefore, \( p \in \{3, 7, 11, 13, 17\} \).

3. **Check for \( p = 11 \):**
   - We need \( 2 \cdot 10^{n+1} + 19 \equiv 0 \pmod{11} \):
     \[
     2 \cdot 10^{n+1} \equiv -19 \pmod{11} \implies 2 \cdot 10^{n+1} \equiv -8 \pmod{11} \implies 10^{n+1} \equiv -4 \pmod{11}
     \]
     Since \( 10 \equiv -1 \pmod{11} \), we have:
     \[
     10^{n+1} \equiv (-1)^{n+1} \equiv \{1, -1\} \pmod{11}
     \]
     This implies \( -4 \not\equiv 1 \) and \( -4 \not\equiv -1 \pmod{11} \). Hence, \( p = 11 \) does not divide \( a_n \).

   Therefore, \( p \in \{3, 7, 13, 17\} \).

4. **Check for \( p = 3 \):**
   - We need \( 2 \cdot 10^{n+1} + 19 \equiv 0 \pmod{3} \):
     \[
     2 \cdot 10^{n+1} \equiv -19 \pmod{3} \implies 2 \cdot 10^{n+1} \equiv -1 \pmod{3} \implies 10^{n+1} \equiv 1 \pmod{3}
     \]
     Since \( 10 \equiv 1 \pmod{3} \), we have:
     \[
     10^{n+1} \equiv 1^{n+1} \equiv 1 \pmod{3}
     \]
     This is always true for any \( n \geq 1 \). Therefore, \( p = 3 \) divides \( a_n \).

5. **Check for \( p = 7 \):**
   - We need \( 2 \cdot 10^{n+1} + 19 \equiv 0 \pmod{7} \):
     \[
     2 \cdot 10^{n+1} \equiv -19 \pmod{7} \implies 2 \cdot 10^{n+1} \equiv -5 \pmod{7} \implies 10^{n+1} \equiv -\frac{5}{2} \pmod{7}
     \]
     Since \( 10 \equiv 3 \pmod{7} \), we have:
     \[
     10^{n+1} \equiv 3^{n+1} \pmod{7}
     \]
     We need \( 3^{n+1} \equiv -\frac{5}{2} \equiv -5 \cdot 4 \equiv -20 \equiv -6 \equiv 1 \pmod{7} \). This is true for \( n = 5 \). Therefore, \( p = 7 \) divides \( a_n \).

6. **Check for \( p = 13 \):**
   - We need \( 2 \cdot 10^{n+1} + 19 \equiv 0 \pmod{13} \):
     \[
     2 \cdot 10^{n+1} \equiv -19 \pmod{13} \implies 2 \cdot 10^{n+1} \equiv -6 \pmod{13} \implies 10^{n+1} \equiv -3 \pmod{13}
     \]
     Since \( 10 \equiv -3 \pmod{13} \), we have:
     \[
     10^{n+1} \equiv (-3)^{n+1} \pmod{13}
     \]
     This is true for \( n = 12 \). Therefore, \( p = 13 \) divides \( a_n \).

7. **Check for \( p = 17 \):**
   - We need \( 2 \cdot 10^{n+1} + 19 \equiv 0 \pmod{17} \):
     \[
     2 \cdot 10^{n+1} \equiv -19 \pmod{17} \implies 2 \cdot 10^{n+1} \equiv -2 \pmod{17} \implies 10^{n+1} \equiv -1 \pmod{17}
     \]
     Since \( 10 \equiv -7 \pmod{17} \), we have:
     \[
     10^{n+1} \equiv (-7)^{n+1} \pmod{17}
     \]
     This is true for \( n = 7 \). Therefore, \( p = 17 \) divides \( a_n \).

Thus, our answer is: there exists some \( n \ge 1 \) such that \( p \) divides \( a_n \) if and only if \( p \in \{3, 7, 13, 17\} \).

The final answer is \( \boxed{ \{3, 7, 13, 17\} } \)