(15C) Fibonacci Numbers

[Werner]

(08-07-2023 10:09 PM)Albert Chan Wrote:  F(n) ≈ (2+√20)^n / (2^(2n-1)*√20)

Hello Albert!
This overflows already for n=123.
perhaps better

F(n) ≈ 2*(0.2+√0.2)^n / (0.4^n*√20)

which can also very easily be turned into a program, eg 41-style
(I don't have a 15C.. should arrive any minute now ;-)
can handle up till n=248, correct for n=0..44

Update: snipped older version, see two posts down for new

Cheers, Werner