Unfortunately, it can’t be done.
A sphere (made of quads except for tris at the poles) in hexagon always has an even number of longitudinal sections (a sphere of “x” points has (x-1)*2 longitudinal sections).
Because this number is alway even, it means that any such sphere is ABLE to be checkerboarded the way you want, but you can only do it manually. The “problem” with hexagon’s “1 over n” algorithm is that it doesn’t have any kind of “offsetting” capability for adjacent layers.
Using the “1 over n” on a sphere means that n=0 always yields no faces selected, n=1 always yields all faces selected, n=2 always yields vertical stripes, and n=3 always yields diagonal stripes. Any n>3 yields some oddball pattern dependant on the ratio of n:x
And as intriguing as these “oddball patterns” may appear, you can’t even use any n>3 as a starting point in hopes of “filling in” the blank spaces, since checkerboarding it requires an even number for “n”, but offsetting adjacent rows of the sphere requires an odd numbered “n”.
I can post a tedious logical proof of that if necessary…