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AP
Posts:
134
Registered:
3/4/09
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gcd Fibonacci Lucas
Posted:
Dec 12, 2012 2:54 PM
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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 ...
one can read :
Using basic identities Lucas proved this
this result is more subtle than the first
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Author
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12/12/12
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AP
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12/14/12
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AP
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