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Topic:
Fibonacci sequence
Replies:
3
Last Post:
Oct 9, 2012 2:56 AM



AP
Posts:
137
Registered:
3/4/09


Re: Fibonacci sequence
Posted:
Oct 9, 2012 2:56 AM


On Mon, 8 Oct 2012 14:19:40 0500, "dilettante" <no@nonono.no> wrote:
> >"AP" <marc.pichereau@wanadoo.fr.invalid> wrote in message >news:jq76785qi8948cunpaqto5icqflbdmkjeq@4ax.com... >> Fibonacci sequence is >> F_0=0 F_1=1 >> and F_n=F(n1)+F(n2) >> >> 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 mF_T(m) ad also mF_kT(m) (because F_pF_(pq) ) >> >> But for many examples we can find >> k<T(m) such as kF_q >> >> >> if m=6, T(6)=24 but 6F_12=144 >> >> if m=7 T(7)=16 but 7F_8=21 >> >> question : one can find always q<T(m) such as mF_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).
then T(m)/g(m) is 1 or 2 or 4 cf thread of Quasi
and for m=4 , T(m)/g(m)=1
But now, it seems that , always, g(m)<=2m ?



