I can see what you mean now. Intreerstingly while I've been offline I've noticed a pattern that explains why even-SOLL moves can be expressed in more ways than odd-SOLL ones. Taking m as the diagonal and n as the orthogonal, I looked at how the moves can be expressed as two orthogonal moves - p and q steps, say, where p>=q - with a 60° turn in between. Where 3m>n>m, p=2m and q=n-m, which translates into m=p/2 and n=q+p/2. Where n>3m, p=n-m and q=2m, which translates into m=q/2 and n=p+q/2. Where m>n, p=m+n and q=m-n, which translates into m=(p+q)/2 and n=(p-q)/2.
In the first case m must be even, in the second n must be even, and in the third m+n must be even. With an odd SOLL exactly one of these is the case and so only one of the three pairs of equations holds true. With an even SOLL all three are true and so all three pairs work, and not necessarily for the same values of m and n.
In the special case of 3m=n, p and q are equal and the first two pairs of equations are identical. In the special case of m=n, all equations except the second hold and q is zero and m and n both p/2. The missing equation may hold for different values of m and n if p and q are even.
In the first case m must be even, in the second n must be even, and in the third m+n must be even. With an odd SOLL exactly one of these is the case and so only one of the three pairs of equations holds true. With an even SOLL all three are true and so all three pairs work, and not necessarily for the same values of m and n.
In the special case of 3m=n, p and q are equal and the first two pairs of equations are identical. In the special case of m=n, all equations except the second hold and q is zero and m and n both p/2. The missing equation may hold for different values of m and n if p and q are even.