20121124, 01:02  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}·3·79 Posts 
Fibonacci Formula
Hi all,
I thought I would tell about a little known Fibonacci formula that I found, and then later found it in the literature. A deffinition of Fibonacci numbers is  F(n) = F(n1) + F(n2) with F(0) = F(1) = 1 This is a formula that lets you compute F(2n), knowing F(n) and F(n1). The most general form is  F(n) = F(k)*F(nk) + F(k1)*F(nk1) substituting n=2k or n=2k+1 gives interesting results. I will try to link to my derivation https://docs.google.com/viewer?a=v&p...AwMGIzOTU5NjRk Regards, Matt 
20121124, 01:18  #2 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1C35_{16} Posts 
Take a look at the bottom of this section, and this section. The fourth formula of the latter seems to contradict one of your special cases, and it also has a general formula for F_{kn+c}.
Though your formula doesn't appear, I'd guess that someone's probably thought of it before given it's rather simple derivation. 
20121124, 03:23  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
I believe that you "destroy" a lot of mathematical (and also biological ) things with this unintentional shift...
For example you can not say anymore that "the only fibo numbers that can be prime are those with a prime index".... and many other such affirmations... 
20121124, 10:17  #4 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}×3×79 Posts 
Hi,
Thanks for the replies. You are right, the standard definition of Fibonacci numbers includes F(1) = F(2) = 1. Mentioning that F(0) = 1 was an error. With this change, the formulas currently at the bottom of the Wikipedia article sections cited are exactly the same ones I mentioned. Matt 
20121124, 23:28  #5 
"Daniel Jackson"
May 2011
14285714285714285714
677 Posts 
Exact formula.
Here's a formula for the Fibonacci numbers. It even works for negative indices:
F_{n}=(phi^{n}(phi)^{n})/sqrt(5), where phi is the Golden Ratio: (sqrt(5)+1)/2. 
20130108, 20:05  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{2}·3·79 Posts 
I corrected my derivation. This formula is useful because a person can calculate large Fibonacci numbers F(n) with integer arithmatic and fewer steps than the definition equation. For example, if someone wants to calculate F(100), Start with F(1), F(2), and F(3) then find F(6), F(12), F(24), F(25), F(50), and finally F(100). This is shorter than iterating 100 times.

20130111, 13:12  #7 
Dec 2012
The Netherlands
11011011110_{2} Posts 
Hendrik Lenstra wrote a magazine article about profinite Fibonacci numbers:
http://www.math.leidenuniv.nl/~hwl/papers/fibo.pdf With profinite integers, you get 8 extra fixed points. 
20130114, 23:29  #8 
"Daniel Jackson"
May 2011
14285714285714285714
1245_{8} Posts 
There is also a formula that uses the Lucas numbers:
F_{n}*L_{n}=F_{2n} :) 
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