Just want to put in a partial solution to provide an independent count to validate other solutions.

Note that I take both zero & dummy to be legal units.

A couple of general points:

- If a (p,q)rider is present, then both (p,q)leaper and (kp,kq)rider for k>0 would be redundant.

- If q>3, then a (p,q)rider can never do better than a (p,q)leaper, so stick with the latter.

Now consider the (0,1) direction.

With no riders, there are 2^7 = 128 possibilities.

If there is (0,1)rider, then no other (0,q)riders add anything. There are apparently 2^6 = 64 choices of leaper but if all 6 leaper powers are present then (0,1)rider can never be blocked. So 63.

(0,2)rider and no other riders. Can’t have both (0,4)leaper & (0,6)leaper else (0,2)rider can never be blocked. So there are 3*2^4 = 48 choices of leapers.

(0,3)rider only is even simpler. Can’t have (0,6)leaper else (0,3)rider can never be blocked. So 2^5 = 32 choice of leaper.

This leaves the corner case where (0,2)rider & (0,3)rider are combined. This offers unique feature of two routes from e.g. a1 to b7, so (0,6)leaper cannot be present. So 2^4 = 16 choices of leaper.

So I make it 128 + 63 + 48 + 32 + 16 = 287 possible combinations for the (0,1) direction. Please confirm.

The (1,1) direction will work similarly.