On Mon, 8 Oct 2012 14:19:40 -0500, "dilettante" <email@example.com> wrote:
> >"AP" <firstname.lastname@example.org> wrote in message >news:email@example.com... >> Fibonacci sequence is >> F_0=0 F_1=1 >> and F_n=F(n-1)+F(n-2) >> >> F'_n=F_ mod m with m>=2 >> >> F'_n is periodic with period T(m)<=m^2 >> >> because among the m^2+1 couples >> >> (F'_0,F'_1) ......(F'_m^2,F'_(m^2+1)) >> two are egals >> >> so F'_T(m)=0 and m|F_T(m) ad also m|F_kT(m) (because F_p|F_(pq) ) >> >> But for many examples we can find >> k<T(m) such as k|F_q >> >> >> if m=6, T(6)=24 but 6|F_12=144 >> >> if m=7 T(7)=16 but 7|F_8=21 >> >> question : one can find always q<T(m) such as m|F_q ? >> Thanks > > What do you mean by always? If you mean for every m, then clearly not: if m >= 4 the period is 6, and F_6 is the first F_q to be divisible by 4(excluding >the trivial case q = 0, of course). yes , it is because if g(m) = the least positive integer n such that F[n] = 0 (mod m).