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