Well, to see one way of doing it, make a numbered grid on a piece of paper, turn it over and fold each corner into the centre. You will see that you now have an unbrocken surface that with a little imagination, and stretching, could be inflated into a globe. Each edge wraps to itself thus
6x6 grid
along first edge, 1,1 wraps to 1,6 : 1,2 to 1,5 : 1,3 to 1,4
along second edge 1,1 to 6,1 : 2,1 to 5,1 : 3,1 to 4,1
along 3rd edge 6,1 to 6,6 : 6,2 to 6,5 : 6,3 to 6,4
along 4th edge 6,6 to 1,6 : 5,6 to 2,6 : 4,6 to 3,6
Probably a maths formula in there somewhere, and bound to have been worked out by someone so try a google for it.
If you stick with the standard mercator projection then each 'square' along the top and bottom rows should really wrap to any other square on that row.
You could go for a cube and have a uniform shape and size of 'square'. The pyramid solution is just half a cube anyway. Still need 3 axes then to cover the face of the cube, and the wrap-mapping would have to be a static table rather than a calculated function.
For a realistic globe then hexagons and 3 axes will give the best fit. I don't know what the formula is for the number on a sphere, bound to be a limited number of possible totals for a complete sphere, so google for it.