Mitch Harris: > >Possibly someone knows of references to special cases? > > I think this is hard in general: take r(z) = 1/(1-2z), so r(n) = 2^n. I > don't know of a 'good' (rational function or even special function) > representation of the ogf > > \sum z^n [n=2^k] [log2(n)]-[log2(n-1)], n>1, would be cheating? ralf