G.f.: hypergeom([1/3,1/3],[1],-27*x*(2*x+1)^2/((3*x+1)*(6*x-1)^2))/((1-x)*(3*x+1)^(1/3)*(1-6*x)^(2/3)). - Mark van Hoeij, Apr 17 2013
EXAMPLE
a(0) = 1 because there is a unique null walk on no points.
a(1) = 1 because there are no closed paths of length 1 (which connects the origin with one of 6 other points before symmetry is considered).
a(2) = 7 because one adds the 6 closed paths of length 2 (which go from origin to one of 6 surrounding points on the lattice, and return in the opposite directions).