Longest Lucas seq. from start to "n"

Max Alekseyev maxale at gmail.com
Thu Oct 18 22:53:27 CEST 2007

On 10/18/07, Max Alekseyev <maxale at gmail.com> wrote:

> > ... v(1) = A019446(n) =ceil(n/tau) or A060143(n) = floor(n/tau)
> >  but "round" cannot be used to select the correct value.

> This is an explanation of what is the correct value of v(1).

Oh, I mess up the notation in the previous message.
Let restate my arguments:

Let v(1) = m. Then we have a sequence:

N, m, N-m, 2m-N, 2N - 3m, ..., F(2k)*m - F(2k-1)*N, F(2k)*N - F(2k+1)*m, ...

F(2k+1)*m - F(2k)*N >= 0 is equivalent to N/m =< F(2k+1)/F(2k)
F(2k-1)*N - F(2k)*m >= 0 is equivalent to N/m >= F(2k)/F(2k-1).

Trivially we have

N / ceil(N/phi) < phi < N / floor(N/phi)

As the sequence F(n+1)/F(n) approaches phi,
there is the smallest value k such that F(k+1)/F(k) falls into the
interval ( N/ceil(N/phi), N/floor(N/phi) ).

If k is even then phi < F(k+1)/F(k) < N/floor(N/phi), implying that we
better choose v(1)=ceil(N/phi).
If k is odd then N/ceil(N/phi) < F(k+1)/F(k) < phi, implying that we
better choose v(1)=floor(N/phi).


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