[seqfan] Re: A000032 Lucas numbers - suggestion
Fred Lunnon
fred.lunnon at gmail.com
Mon Jun 3 14:15:22 CEST 2019
Certainly, the quotation makes no sense as it stands. WFL
On 6/3/19, Jeff Anderson <peawormsworth at gmail.com> wrote:
> Hi,
>
> re: https://oeis.org/A000032
>
> === quote ===
> Let f(n) = phi^n + phi^( - n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1)
> - 2*Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan
> numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide
> Colazingari, Jul 01 2016
> === end ===
>
> It is known:
> L(n) = phi^n + (-phi)^-(n)
>
> and so...
> L(2n) = phi^2n + phi^-(2n)
> L(2n+1) = phi^(2n+1) - phi^-(2n+1)
>
> Given:
> f(n) = phi^n + phi^-(n)
>
> calculate...
> f(2n+1) = phi^(2n+1) + phi^-(2n+1)
>
> Assuming this from quote:
> L(2n+1) = f(2n + 1) - 2×Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)
>
> substitute...
> phi^(2n+1) - phi^-(2n+1) = phi^(2n+1) + phi^-(2n+1) -2×Sum_{k=0..infinity}
> C(k)/f(2n + 1)^(2k + 1)
> -2×phi^-(2n+1) = -2×Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)
> phi^-(2n+1) = Sum_{k=0..infinity} C(k)/f(2n + 1)^(2k + 1)
>
> === conclude ===
>
> The relationship and formula shown is not about Lucas numbers.
> Quote may be more appropriate for inclusion on a different series.
>
> Jeff
>
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>
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