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Topic: gcd Fibonacci Lucas
Replies: 1   Last Post: Dec 14, 2012 4:52 PM

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AP

Posts: 137
Registered: 3/4/09
Re: gcd Fibonacci Lucas
Posted: Dec 14, 2012 4:52 PM
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On Wed, 12 Dec 2012 20:54:33 +0100, AP
<marc.pichereau@wanadoo.fr.invalid> wrote:

>Let a, b the zeos of x^2-x-1
>
>and for n>=0
>F_n=(1/sqrt(5))(a^n-b^n)
>
> F_0=0 F_1=1
>F_(n+1)=F_n+F_(n-1)
>
>L_n=a^n+b^n
>
> L_0=2 L_1=1
>L_(n+1)=L_n+L_(n-1)
>
>the following result (and his proof) is well known
>
>gcd(F_n, F_p)=F_gcd(n,p)
>(first, one can use F_(n+p)=F_(n+1)F_p+F_nF_(n-1)
>to show gcd(F_n,F_p)=gcd(F_n,F_(n+p))
>
>but if this other result is known
>
>gcd(L_n,L_p)
>
>=L_d if n/d and p/d are odd
>
>= 2 if n/d or p/d is even and 3|d
>
>=1 if n/d or p/d is even and 3 do not divide d
>
>(with d= gcd(n,p) )
>
>but I don't find a proof ...

I have find a proof ...I think
>one can read :
> Using basic identities Lucas proved this
> this result is more subtle than the first




Date Subject Author
12/12/12
Read gcd Fibonacci Lucas
AP
12/14/12
Read Re: gcd Fibonacci Lucas
AP

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