True muonium energy levels
The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed.  See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments.

The non-relativistic energy levels are given by :

$$E_n =  \frac { \mu c^2 \alpha^2}{2n^2}$$

where $alpha$ is the fine structure constant and $\mu$ is the reduced mass, in this case because the masses are equal, it's $\mu=\frac m 2$ where $m$ is the muon mass.

Note that the muon mass is about $206.8$ times that of the electron and the reduced mass is then $103.4$ times than of the electron making the energy levels for the non-relativistic state about $103.4$ times that of the Hydrogen energy levels.

As the Wikipedia page explain (in more detail that you asked for), a more detailed relativistic calculation using the Dirac equation includes corrections for the spin-orbit interaction.


  The core is lighter for about 100 times but the muon is 100 times heavier than the electron


This is wrong.

Both the muon and antimoun have the same mass, hence the effect of reduced mass being a factor of one-half.

The reduced mass of two particles is given by :

$$\mu = \frac {m_1m_2}{m_1+m_2}$$

Strictly speaking there is no "core" as both the muon and antimoun are of the same mass whereas in Hydrogen the electron is of much smaller mass than the proton.