quantum mechanics - Landau level degeneracy in symmetry gauge, finite system

As we know, Landau level degeneracy in a finite rectangular system is $\Phi/\Phi_0$, where $\Phi=BS$ is the total magnetic flux and $\Phi_0=h/q$ is the flux quanta. This can be easily derived using Landau gauge and assuming the periodic boundary condition.

However, if you choose the symmetric gauge, $l_z=x\hat p_y-y\hat p_x$ commutes with Hamiltonian, the corresponding quantum number $m$ is thus a good one. After some calculation, at last the energy level is written as

$$E=\left[n+(m+|m|)/2+1/2\right]\hbar\omega.$$ I want to derive the degeneracy of Landau level for a finite system with radius $R$.

The thought I got is that because in a finite system the angular momentum is bounded, the maximum value of $l_z$ is $qBR^2$ when you consider the particle doing the circular motion classically. Hence the maximum $m=l_z/\hbar=2\Phi/\Phi_0$, and the degeneracy is thus at least doubled.

What we have learned is that the gauge choice won't change the observable effect, the finite degeneracy can obviously be observed. So what's wrong with my derivation? Or because they are different just because the system we considered is simply not equivalent?