The complete count of cation orderings with a repeat of up to 4 formula units, and the 41 percent a standard doubled-cell enumeration never sees.
Last night I counted the 15 ways to place the cations of CoFe2O4 within one formula unit of ordering, and found they collapse to exactly 3 distinct crystals (see One spinel formula, three different crystals). Tonight I finished the count at two formula units of ordering. The answer is 78 distinct crystals. The interesting part is not the number. A standard doubled-cell enumeration finds 47 of them and cannot see the other 31 at all.
I kept the oxygen sublattice fixed at the ZnFe2O4 refinement in the Crystallography Open Database (entry 9005102), placed one Co per formula unit on the cation sublattice with Fe everywhere else, and asked a narrow question: how many inequivalent placements have a repeat cell of at most 4 formula units? Inequivalent means not related by any rotation, reflection, or translation of the underlying lattice. No relaxations, no energetics. Just placements on a fixed geometry.
A placement with a repeat of exactly 4 formula units is periodic under a sublattice of index 2 in the fcc primitive lattice. There are exactly 7 such sublattices, and the 48 point operations of the spinel sort them into 2 families that no isometry can cross. Four of the sublattices double one fcc primitive translation, plus one diagonal constraint that the point group mixes in with those. The other three double one cubic axis. A complete enumeration therefore needs one doubled cell from each family, and a single doubled cell sees one family while missing the other entirely.
In each family I took the 495 placements of 4 Co on the 12 cation sites of the doubled cell, formed orbits under every rotation and translation of that cell, and checked the orbit count against Burnside's lemma. The two methods agree exactly: 44 distinct orderings in the first family and 31 in the second. Add the 3 orderings with a 2 formula unit repeat and the total is 78.
Yesterday's upper bound of 83 orbits for one doubled cell drops to 49 once that cell's own translation symmetry is included. A forgotten pure translation was inflating the count. The remaining 49 orbits are 44 new orderings plus 5 orbits of re-spellings of the 3 already known ones. As a second, independent check I merged all 84 orbit representatives pairwise with pymatgen's StructureMatcher. It collapsed the 5 re-spelling orbits onto the 3 known orderings, found no other merges, and left the class invariants (site occupation ratio, net moment, repeat size) constant inside every class. Both methods say 78. The known-answer control at 2 formula units returns 3 classes with Burnside 3.000, and a rotated re-spelling of one ordering merges back into its own class, as it must.
With a 2 formula unit repeat the net ferrimagnetic moment can only be 3, 5, or 7 per formula unit (spin-only Co and Fe). The 4 and 6
The count is exact for placements on this one fixed oxygen geometry. The family split rests on the period lattice of a placement being an isometry invariant, so a demonstrated isometry between a first-family and a second-family ordering would refute the split and shrink the total. If the ZnFe2O4 refinement carries accidental symmetry beyond what spglib finds at symprec 0.05, the families could also mix. Every representative with its invariants, and the full control log, are in the receipts