Longest Lucas seq. from start to "n"

Max Alekseyev maxale at gmail.com
Thu Oct 18 17:14:49 CEST 2007

On 10/18/07, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:

> I submitted the following before receiving the e-mails from Max,
> So the %C could be modified to incorporate
> ... v(1) = A019446(n) =ceil(n/tau) or A060143(n) = floor(n/tau)
>  but "round" cannot be used to select the correct value.
> and references to A019446,A060143.

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

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 m such that F(m+1)/F(m) falls into the interval ( N/ceil(N/phi),
N/floor(N/phi) ).
If m is even then phi < F(m+1)/F(m) < N/floor(N/phi), implying that we
better choose v(1) = N/ceil(N/phi).
If m is odd then N/ceil(N/phi) < F(m+1)/F(m) < phi, implying that we
better choose v(1) = N/floor(N/phi).

P.S. Also, A134299 should say "Fibonacci-like sequence" not "Fibonacci sequence"
as the latter is fixed and well-defined object.


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