[FormerLurker]

Fibonacci numbers are related to "phi", but they aren't "derived" from it. Thus, "phi" is not foundational (i.e., part of the underlying axiom system) for Fibonacci numbers

Do you care to restate that Mr. Wizard?

From Phi and Mathematics

You can use phi to compute the nth number in the Fibonacci series (f_n):

f_n =  Øⁿ / 5^½

This method actually provides an estimate which always rounds to the correct Fibonacci number.

You can compute any number of the Fibonacci series (f_n) exactly with a little more work:

f_n = [ Øⁿ - (-Ø)^-n ] / (2Ø-1)

Note:  2Ø-1 = 5^½= The square root of 5 

---

In fact, have you ever heard of Binet's Formula for the nth Fibonacci number ?

---

Binet's Formula for the nth Fibonacci number

We have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first - quite a task, even with a calculator!
A natural question to ask therefore is: Can we find a formula for F(n) which involves
only n and does not need any other (earlier) Fibonacci values? Yes! It involves our golden section number Phi and its reciprocal phi:
Here it is:

Fib(n) =	Phin – (–Phi)–n	=	Phin – (–phi)n
	5		5

where Phi = 1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ... .
The next version uses just one of the golden section values: Phi, and all the powers are positive:

Fib(n) = 

Phin –	(–1)n
	Phin

5

Since phi is the name we use for 1/Phi on these pages, then we can remove the fraction in the numerator here and make it simpler, giving the second form of the formula at the start of this section.

We can also write this in terms of 5 since Phi =	1 + 5	and –phi =	1 – 5	:
	2		2

If you prefer values in your formulae, then here is another form:-

Fib(n) =	1.6180339..n – (–0.6180339..)n
	2.236067977..

This is a surprising formula since it involves square roots and powers of Phi (an irrational number) but it always gives an integer for all (integer) values of n!

Here's how it works:

 
 Let X=  Phin  =(1·618..)n
 and Y=(-Phi)-n=(-1·618..)-n=(-0·618..)n then we have:
   
    n:	X=Phin :     Y=(-Phi)-n:     X-Y:       (X-Y)/sqrt(5):
    0   1             1             0            0
    1   1·618033989  -0·61803399    2·23606798   1
    2   2·618033989   0·38196601    2·23606798   1
    3   4·236067977  -0·23606798    4·47213595   2
    4   6·854101966   0·14589803    6·70820393   3
    5   11·09016994  -0·09016994   11·18033989   5
    6   17·94427191   0·05572809   17·88854382   8
    7   29·03444185  -0·03444185   29·06888371  13
    8   46·97871376   0·02128624   46·95742753  21
    9   76·01315562  -0·01315562   76·02631123  34
   10   122·9918694   0·00813062  122·9837388   55
   ..   ....           ....            ..
   

You might want to look at two ways to prove this formula: the first way is very simple and the second is more advanced and is for those who are already familiar with matrices.

Since phi is less than one in size, its powers decrease rapidly. We can use this to derive the following simpler formula for the n-th Fibonacci number F(n):

F(n) = round( Phiⁿ / 5 ) where the round function gives the nearest integer to its argument.

   n:   Phin/sqrt(5) ..rounded
   0	0·447213595      0
   1	0·723606798      1
   2	1·170820393      1
   3	1·894427191      2
   4	3·065247584      3
   5	4·959674775      5
   6	8·024922359      8
   7	12·98459713     13
   8	21·00951949     21
   9	33·99411663     34
  10	55·00363612     55
  ..    ...              ..
  

Notice how, as n gets larger, the value of Phiⁿ/5 is almost an integer.

---

So longshadow, would you say that Binet derived the formula for the nth Fibonacci number from phi, or did he relate the formula TO phi?

Perhaps you should study up a little on some of the links I've posted...

[longshadow]

Fibonacci numbers are related to "phi", but they aren't "derived" from it. Thus, "phi" is not foundational (i.e., part of the underlying axiom system) for Fibonacci numbers

Do you care to restate that Mr. Wizard?

You really don't understand, do you....

from YOUR link ("Binet's Formula for the nth Fibonacci number"):

---

the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th.

---

That's the definition of the numbers in the Fibonacci sequence. Nowhere is "phi" or "Sacred Geometry" including its "vibrational resonances" needed to define it! Fibonacci didn't use "phi" to define his sequence; the fact that it can be used to compute the nth term of the sequence is all very nice, but doesn't contradict a thing I wrote.

The fact that the Fibonacci numbers can be (and were) defined without "phi" means, by definition, that "phi" is not foundational or fundamental to the Fibonacci sequence.

You seem to be struggling very hard not to get this very simple point.