Recursive sequences

[Max Alekseyev]

On 2/27/07, Tanya Khovanova <tanyakh at tanyakhovanova.com> wrote:
>
> >
> >Nice work! Do you consider more general second order recurrences:
> >a(n) = c_1 * a(n-1) + c_2 * a(n-2)
> >and the inhomogenous case:
> >a(n) = c_1 * a(n-1) + c_2 * a(n-2) + b(n)?
> >
>
> Thank you! I would like to expand my webpage. Unfortunately for me and fortunately for humanity OEIS has too many sequences. :-) For the page I did, I had to check and copy manually more than 1,000 sequences.

Actually, it is easy to detect Lucas sequences algorithmically:

For any 5 consecutive terms of such a sequence t1, t2, t3, t4, t5 the
determinant

| t1 t2 t3 |
| t2 t3 t4 | must be 0.
| t3 t4 t5 |

If at least one of the determinants is non-zero, implies that the
sequence is not Lucas one. So, a simple program can easily filter out
most of the non-Lucas sequences.

Max



I wrote in part:


>First, form the recursive sequence of rationals:
>
>b(1) = 1. b(n) = 1 + 1/b(n-1)^2.
>
>1, 2, 5/4, 41/25, 2306/1681,...
>

>....


>So, if x = the root of x^3 - x -1 = 0,* (I am assuming that b(n) -> x.)...
>
>*(What is this root? I have no access to a computer algebra program, and 
>I do not know how to solve the cubic off the top of my head. I am 
>embarrassed that I don't know the root to even this simple cubic.)
> 
>
>...then limit(n -> infinity} r(n)*r(n+1) = (x^3 /2) - 2 + (2 /x^3).
>
>...'


Woops...
Thanks to Hugo for getting the numerical real root, but....

I made a mistake. I REALLY want x to be a real root of:

x^3 - x^2 -1 = 0.

(I noticed that Hugo's numerical root was much different than what I 
estimated on the calculator.)

Thanks, sorry,
Leroy Quet