Good work on the AB=B.AA problem! This might even lead to a formula for the number of tables of each size. My data for N=2,3,4,5 is 2: dist 2 tabtot 4 idemspec 0 1 1 isospec 1.2 3: dist 11 tabtot 48 idemspec 0 1 3 7 isospec 1.6 2.3 3 6 4: dist 234 tabtot 4964 idemspec 0 1 5 36 192 isospec 1.185 2.38 3.6 6.5 5: dist 85196 tabtot 10089780 idemspec 0 1 7 108 2725 82355 isospec 1.83035 2.2023 3.57 4.34 5.6 24.3 6.36 8 20 Dist is the number of distinct, non-isomorphic tables; Tabtot is the number of tables with isomorphic tables counted. Idemspec is the number of (non-isomorphic) tables with each number of idempotents -- the "all elements are idempotent" class seems dominant. Isospec reports on the number of tbales with each (size of) automorphism group: For the 85196 tables with five elements, 83035 have no automorphisms (group size =1, self), while three tables have a size-24 group, and there's one table each for group sizes 8 and 20. The observations that squares (AA) are idempotent, and that an element and its square have identical rows and columns, constrains the table structure quite a bit. --------- Gosper's packing puzzle is to fit as many 4x4x1 bricks as possible into a 7x7x7 cube. I'm assuming an "orthogonal, registered" fit, with each brick oriented parallel to a cube face, and placed at integral subcube boundaries. The answer is 18 bricks. A view DOWN on the bottom half of the packing is AAAABBB AAAABBB AAAABBB CCCCBBB CCCC--- CCCC--- CCCC--- where the As and Bs are sideways stacks of 3 bricks each, and the Cs are a vertical stack of 3 bricks. My proof is the same as Michael Reid's: Consider the (orthogonal) lines going through the center of the 7x7x7 cube. Take the union, remove the center cell. Resulting collection of subcubes has 18 cells. Any 4x4x1 brick must occupy at least one of these cells. Note that the simple volume constraint gives an upper bound of 7^3/4^2 = 343/16 = 21.4375. Rich