Re: um.....this is...!
From: mina_world (mina_world_at_hanmail.net)
Date: 10/06/04
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Date: Thu, 7 Oct 2004 00:47:32 +0900
"Ignacio Larrosa Ca?stro" <ilarrosaQUITARMAYUSCULAS@mundo-r.com> wrote in
message news:2si5bhF1lljqbU1@uni-berlin.de...
>
> "mina_world" <mina_world@hanmail.net> escribi?en el mensaje
> news:cjvfcl$gc0$1@news.hananet.net...
> >
> > "Ignacio Larrosa Ca?stro" <ilarrosaQUITARMAYUSCULAS@mundo-r.com> wrote
in
> > message news:2sfst4F1kkvtbU1@uni-berlin.de...
> >> En el mensaje:cjudam$qor$1@news.hananet.net,
> >> mina_world <mina_world@hanmail.net> escribi?
> >> > hello.....doctor~
> >> >
> >> > a_1 = 1
> >> >
> >> > 2*{a_(n+1)} = 3*a_n + sqrt{(5(a_n)^2) + 4} (n >= 1)
> >> >
> >> > show that there does not exist n such that 1989 | a_(2n).
> >> > (n is positive integer)
> >> >
> >> > ----------------------------------------------------
> >> >
> >> > um....it's so difficult to me.
> >> >
> >> > i need your advice.
> >> >
> >> > thank you very much for your advice.
> >>
> >> Hint: Prove that a(n) = Fibonacci(2n)
> >>
> >>
> >
> > um.....
> > a_1 = 0
> > a_2 = 1
> > a_3 = 3
> > a_4 = 8
> > a_5 = 21
> > a_6 = 55
> No,
>
> a(1) = 1 (your definition!)
> a(2) = 3
> a(3) = 8
> a(4) = 21
> ...
>
>
> > a_(n+1) - 3.a_n + a_(n-1) = 0 by rule.
>
> Ok, writing
>
> 2a(n+1) - 3a(n) = sqrt(5a(n)^2 + 4)
>
> 4a(n+1)^2 + 9a(n)^2 - 12a(n+1)a(n) = 5a(n)^2 + 4 ===>
>
> 4a(n+1)^2 + 4a(n)^2 - 12a(n+1)a(n) = 4 ===>
>
> a(n+1)^2 + a(n)^2 - 3a(n+1)a(n) = 1 ===>
>
> replacing n with n-1,
>
> a(n)^2 + a(n-1)^2 - 3a(n)a(n-1) = 1 ===>
>
> substrctinge that last equations,
>
> a(n+1)^2 - a(n-1)^2 - 3a(n)(a(n+1)- a(n-1)) = 0
>
> a(n+1) + a(n - 1) - 3a(n) = 0
>
> (because a(n+1) =/= a(n-1) !)
>
> Its says,
>
> a(n+1) = 3a(n) - a(n-1), a(1) = 1
>
> The Fibonacci sequence is
>
> F(1) = F(2) = 1, F(n+1) = F(n) + F(n-1)
>
> Then F(n+2) = F(n+1) + F(n) = 2F(n) + F(n-1)
>
> F(n+3) = F(n+2) + F(n+1) = 3F(n) + 2F(n-1)
>
> F(n+4) = F(n+3) + F(n+2) = 5F(n) + 3F(n-1) = 3(2F(n) + F(n-1)) - F(n) =
> 3F(n+2) - F(n)
>
> Then, the sequence b(n) = F(2n) verify
>
> b(n+2) = F(2n + 4) = 3F(2n + 2) - F(2n) = 3b(n+1) - b(n)
>
> b(1) = F(2) = 1
>
> b(2) = F(4) = 3
>
> Then b(n) and a(n) verify the same requrrence relation with the same
initial
> values, then are equal.
>
> Is says,
>
> a(n) = F(2n)
>
> It is easy to prove by induction that k | F(n) ===> k | F(mn)
>
> As 1989 = 3^2*13*17 and
>
> F(7) = 13 = 0 (mod 13)
>
> F(9) = 34 = 0 (mod 17)
>
> F(12) = 144 = 0 (mod 9)
>
> F(LCM(7, 9, 12)) = F(252) = 0 (mod 9*13*17)
>
> Then a(126) = 0 (mod 1989) and 1989 | a(2*63)
>
>
oh....thank you very much for your advice.
in fact, a_1 = 0. it's my big mistake. very sorry.
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