The upper 4 states are gap states, which are mainly derived from $p_{z}$ orbitals of two $4c$-site atoms.

As $n$ increases by one, the unoccupied states are filled by the additional 3 valence electrons, almost like a rigid-band shift. Although it is by no means a rigid-band shift, the valence-electron counting explains this rigid-band-like behavior very well.[@Shirai12] In $\alpha$-T boron, each equatorial ($e$ site) B atom of the B$_{12}$ icosahedron has one [*inter*]{}icosahedral bond. On placing a B atom at a $4c$ site, four [*inter*]{}icosahedral $e-e$ bonds are replaced by four tetrahedral bonds of $4c$-site atoms. In total, the number of orbitals is not changed, and consequently three electrons of a B atom are used for filling the unoccupied valence states. This way proceeds until $n$=2, where the valence states are completely filled. Further adding of interstitial atoms creates additional gap states, and it is therefore undesirable. B$_{52}$ with two $4c$-site B atoms is the best configuration to meet the valence requirement in a B$_{50+n}$ series.

![(Upper panel) Valence filling for B$_{50+n}$ as the number $n$ of $4c$-site B atoms increases. Only B$_{52}$ can completely fill up the valence bands, but 12 in-gap states remain. This indicates that the conditions 