Find the smallest prime which is not the difference (in some order) of a power of $2$ and a power of $3$.
To find the smallest prime which is not the difference (in some order) of a power of $2$ and a power of $3$, we need to check each prime number and see if it can be expressed as $2^x - 3^y$ or $3^y - 2^x$ for non-negative integers $x$ and $y$.

1. **Check small primes:**
   - $2 = 3^1 - 2^0$
   - $3 = 2^2 - 3^0$
   - $5 = 2^3 - 3^1$
   - $7 = 2^3 - 3^0$
   - $11 = 3^3 - 2^4$
   - $13 = 2^4 - 3^1$
   - $17 = 3^4 - 2^6$
   - $19 = 3^3 - 2^3$
   - $23 = 3^3 - 2^2$
   - $29 = 2^5 - 3^1$
   - $31 = 2^5 - 3^0$
   - $37 = 2^6 - 3^3$

2. **Check if $41$ can be expressed as $2^x - 3^y$ or $3^y - 2^x$:**
   - Assume $41 = 2^x - 3^y$. Then $2^x = 41 + 3^y$.
     - Since $2^x \equiv 0 \pmod{2}$, $41 + 3^y \equiv 1 + 3^y \equiv 0 \pmod{2}$, so $3^y \equiv 1 \pmod{2}$, which is always true since $3$ is odd.
     - Consider modulo $8$: $2^x \equiv 41 + 3^y \pmod{8}$. Since $2^x \equiv 0, 2, 4 \pmod{8}$ and $41 \equiv 1 \pmod{8}$, we have $1 + 3^y \equiv 0, 2, 4 \pmod{8}$.
     - $3^y \equiv -1 \pmod{8}$, but $3^y \equiv 1, 3 \pmod{8}$, which is a contradiction.

   - Assume $41 = 3^y - 2^x$. Then $3^y = 41 + 2^x$.
     - Since $3^y \equiv 0 \pmod{3}$, $41 + 2^x \equiv 2^x + 2 \equiv 0 \pmod{3}$, so $2^x \equiv 1 \pmod{3}$, which implies $x$ is even.
     - Let $x = 2u$. Then $2^x = 2^{2u} = (2^u)^2$.
     - Consider modulo $4$: $3^y \equiv 41 + 2^x \pmod{4}$. Since $3^y \equiv 3 \pmod{4}$ and $41 \equiv 1 \pmod{4}$, we have $1 + 2^x \equiv 3 \pmod{4}$.
     - $2^x \equiv 2 \pmod{4}$, which implies $x$ is odd, a contradiction.

Since $41$ cannot be expressed as $2^x - 3^y$ or $3^y - 2^x$, it is the smallest prime that cannot be represented in this form.

The final answer is $\boxed{41}$.