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Fibonacci sequence

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AP

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Oct 8, 2012, 2:46:06 PM10/8/12
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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

dilettante

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Oct 8, 2012, 3:19:40 PM10/8/12
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"AP" <marc.pi...@wanadoo.fr.invalid> wrote in message
news:jq76785qi8948cunp...@4ax.com...
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).

quasi

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Oct 8, 2012, 4:26:01 PM10/8/12
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This was previously discussed in the thread:

<https://groups.google.com/forum/?hl=en&fromgroups#!topic/sci.math/jWr8psHzoug>

quasi

AP

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Oct 9, 2012, 2:56:08 AM10/9/12
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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 ?


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