each eigenstate is plotted in a color corresponding to the expectation value of the $z$ component of the spin for that eigenstate. The plot for $V_\mathrm{Q} = 1.25$ is similar to that for $V_\mathrm{Q} = 0.8$ (not shown) except that the former is scaled horizontally, because of the duality of the lattice between $V_\mathrm{Q} = x$ and $1/x$. $V_\mathrm{Q} = 1$ is the self-similar point, at which the modulation potential equals the unmodulated band width for $\Gamma = \alpha = 0$. []{data-label="fig:DoubleHofstadter"}](Full_Fig1_200){width="8.66cm"}

Double Hofstadter butterfly
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Let us first consider the case of $\hat{{\mbox{\boldmath $B$}}} = \hat{{\mbox{\boldmath $z$}}}$. For an infinitely long system having $\Delta=\epsilon_{l}=0$ we easily obtain the single particle energy as a function of the quasimomentum $k$, $$E^{\pm}(k) = t(1-\cos(k)) \pm \sqrt{\alpha^2\sin^2(k) + \Gamma^2}.
\label{eqn:RZbands}$$ We call them the upper and lower Rashba–Zeeman (RZ) bands. [@Tezuka2012] The mapping of the Hamiltonian to that of a spinless system is possible if $\mu$ lies in only one of the RZ bands. [@Lutchyn2010; @Oreg2010; @Stoudenmire2011] In such a case, by introducing the pairing $\Delta$ such that $|\Delta| \lesssim \Gamma$, we obtain the topological superconductor phase.

We consider a site potential which is given by $$\epsilon_{l} = V_\mathrm{Q} \cos( \kappa (l-l_\textrm{c}) + 