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

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AP

Posts: 137
Registered: 3/4/09
Re: Fibonacci sequence
Posted: Oct 9, 2012 2:56 AM
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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(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).

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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