k k-1 a= Nd+or 即→M=(Mk-1)2+6x) →)0=M62-2-6 →=。6x(Mk-1) 得R==M-12M(k≠0)0N>) 6 个 R≈N 个f→个P6 R = = Nk −1  Nk，(k  0)    得 (N>>1) R k N →         → =  (Nk −1) 即 R  Nk ( ) 1   +  − = Nd Nk d k → Nk = (Nk −1)( + ) →0 = Nk − − 