Let $(a_{n})_{n\geq 1}$ be a sequence defined by $a_{n}=2^{n}+49$. Find all values of $n$ such that $a_{n}=pg, a_{n+1}=rs$, where $p,q,r,s$ are prime numbers with $p<q, r<s$ and $q-p=s-r$.
1. Given the sequence \( (a_n)_{n \geq 1} \) defined by \( a_n = 2^n + 49 \), we need to find all values of \( n \) such that \( a_n = pq \) and \( a_{n+1} = rs \), where \( p, q, r, s \) are prime numbers with \( p < q \), \( r < s \), and \( q - p = s - r \).

2. From the given conditions, we have:
   \[
   a_n = 2^n + 49 = pq
   \]
   \[
   a_{n+1} = 2^{n+1} + 49 = rs
   \]

3. We can derive the following equation by subtracting the two expressions:
   \[
   2^{n+1} + 49 - (2^n + 49) = rs - pq
   \]
   Simplifying, we get:
   \[
   2^n = rs - pq
   \]

4. Let \( k = q + r = p + s \). Then we can express \( q \) and \( s \) in terms of \( k \):
   \[
   q = k - p
   \]
   \[
   s = k - r
   \]

5. Substituting these into the equations for \( a_n \) and \( a_{n+1} \), we get:
   \[
   2^n + 49 = p(k - p)
   \]
   \[
   2^{n+1} + 49 = r(k - r)
   \]

6. Using the derived equation \( 2^n = rs - pq \), we substitute \( q \) and \( s \):
   \[
   2^n = r(k - r) - p(k - p)
   \]

7. Simplifying, we get:
   \[
   2^n = rk - r^2 - pk + p^2
   \]
   \[
   2^n = (r - p)k - (r^2 - p^2)
   \]
   \[
   2^n = (r - p)k - (r - p)(r + p)
   \]
   \[
   2^n = (r - p)(k - (r + p))
   \]

8. Since \( k = q + r = p + s \), we have \( k = p + s \) and \( k = q + r \). Therefore:
   \[
   2^n = (r - p)(p + s - (r + p))
   \]
   \[
   2^n = (r - p)(s - r)
   \]

9. If \( n \) is odd, then \( 2^n + 49 \equiv 0 \pmod{3} \). This implies \( p = 3 \). Substituting \( p = 3 \) into the equation:
   \[
   2^n + 49 = 3q
   \]
   \[
   2^{n+1} + 49 = rs
   \]

10. Using the derived equation \( 2pq = rs + 49 \), we get:
    \[
    2 \cdot 3q = rs + 49
    \]
    \[
    6q = rs + 49
    \]

11. Let \( r = 5 \) (since \( r \) must be a prime number greater than 3 and less than 6). Then:
    \[
    6q = 5s + 49
    \]
    \[
    6q - 5s = 49
    \]

12. Solving for \( q \) and \( s \), we find:
    \[
    q = 11, \quad s = 7
    \]

13. Substituting back, we get:
    \[
    2^n + 49 = 3 \cdot 11 = 33
    \]
    \[
    2^{n+1} + 49 = 5 \cdot 7 = 35
    \]

14. Checking the values, we find:
    \[
    2^7 + 49 = 128 + 49 = 177
    \]
    \[
    2^8 + 49 = 256 + 49 = 305
    \]

15. Therefore, the only solution is \( n = 7 \).

The final answer is \( \boxed{ n = 7 } \)