Consecutive prime powers $p^m<q^n$ such that $q^n-p^m=11$

Question

Define two prime powers $p^m<q^n$, ($m,n\in\mathbb N$) to be consecutive if it doesn't exist a prime power $r^k$ such that $p^m\!<r^k\!<q^n$. It seems that every even number is the difference between two consecutive prime powers, I haven't ran any large tests yet, but what about the odd numbers?

Are there two consecutive prime powers $p^m<q^n$ and $i\in\mathbb N^+$ such that $q^n-p^m=11^i$?

I have tested all consecutive prime powers less than $10,000,000$ without finding any solution to $q^n-p^m=11^i$.

(A pattern seems to suggest that no odd numbers $N\ge 31$ have a solution to $q^n-p^m=N$, consecutive prime powers. But this must be wrong since prime powers has similar asymptotic distribution as the primes).

Answer 1

See the OEIS: https://oeis.org/A013603 gives the difference between $2^n$ and the largest prime less than $2^n$. We can read off that:

• $2^n - 11$ is prime and there are no primes between it and $2^{n}$ for $n = 42, 78, 114, 190, 322, 546, 3894$.
• $2^n - 11^2$ is prime and there are no primes between it and $2^n$ for $n = 219, 303, 1443, 2333, 2589, 3315, 3693$.
• $2^n - 11^3$ is prime and there are no primes between it and $2^n$ for $n = 1390, 1670$.

These are the only examples where $n < 5000$ and $q = 2$. The "mirror image" sequence of the gap between $2^m$ and the smallest prime greater than $2^m$, which doesn't exist in OEIS, would be needed for the $p = 2$ case.