User blog comment:D57799/Ranging FGH before omega/@comment-5529393-20141008144817/@comment-5307762-20141009110119

fix : (2^{m-2})^n (2^{m-1} n) = 2^{m-1} (2n)\

Isn't that  f_m(n) = (f_{m-1})^n (n)<=(2^{m-2})^n (2n) ?