Denote p(m,n) the number of paths from (1,1) to (m,n).

If the region was x>0, y>0, p(m,n) would copy the coefficients in the Pascal's triangle.

I don't know if an explicit formula already exists for the cone x>0, y>0 and 2x≥y≥x/2 (maybe you have one?).

Anyway, a recursive formula is possible:

1) p(0,0)=0, p(1,2)=p(2,1)=1

2) p(m,n)=p(m-1,n)+p(m,n-1) if m≤2n-2 and n≤2m-2

3) p(2n,n)=p(2n-1,n)=p(2n-2,n) and p(m,2m)=p(m,2m-1)=p(m,2m-2)

(see the attached figure)

Thanks to the symmetry of the region, 1),2),3) can be simplified.

A better simplification is obtained if we put p(m,n)=0 for (m,n) outside the cone:

3) may be omitted, 2) holds for any (m,n) in the cone.