Fib(n) =  Phi^{n} – (–Phi)^{–n}  =  Phi^{n} – (–phi)^{n}  
5 
5 
Fib(n) = round  Phi^{n}  
5 

The rounded values are certainly the first numbers of the Fibonacci series  but it is a counterfeit:
The next row of the table is But the next Fibonacci number, being the sum of the last two in the list above, should be 89 + 144 or 233 and not 234! 
If you like, you can skip this section marked by the green line on the left. It explains why this is a good forgery.  
Why is it a close fit?The reason it almost works is that we can express Phi^{n}/5 as Phi^{n+c} for some number c.To do this we will need logs to base Phi. We met logs previously when finding the length of a number .
The LOG button on your calculator will do the opposite of
the x^{y} button, which raises one number (the base)
to the
power of another (the exponent): base^{exponent}.
So what about other bases?
For instance the log of 8 to base 2 is 3 since 2^{3} is 8.Logtobaseb of x means "What power of the base (b) will give the number x?". It is fairly easy to show that we can get away with just one log button on a calculator and with it we can find logs to any base.
First, we note that to write "log to base b" we will use the mathematical notation
log_{b}. So LOG(x) means log_{10}(x)
and LN(x) means lob_{e}(x).
Let's check that "The logtobase2 of 8 is 3.":
Here are some rules of logs that we will need:
log(1) = 0 (since b^{0} = 1 for all bases b
Since 5 = 5^{1/2} and
1/5 = 5^{1/2}
log( xy ) = log(x) + log(y) for all bases log( x^{z} ) = z log(x) then log_{Phi}(1/5) = log_{Phi}(5^{1/2}) = log_{Phi}(5)/2 Using the formula above for log_{Phi}(x), we have log_{Phi}(5) = LOG(5)/LOG(1.61803..) = 0.69897/0.20899 = 3.34455 so log_{Phi}(5)/2 is 3.34455/2=1.67227
round( Phi^{n+1/3} ) is for Fib(n+2)
and we know that
Fib(n+2) = round(Phi^{n+2}/ 5).
We have just seen that 5 is Phi^{1.67227}, so
So the true value of 0.32772 is very near 1/3 which is why
1/3 works so well.
If we calculate the values of Phi^{n+0.33}, this agrees with Fib(n+1) up to n=12 (so it is already more accurate than Phi^{n+1/3}) and
Phi^{n+0.3277} is better still, being correct up to n=21 whereas
0.327724 is correct up right up to n=34.
 
Stop skipping now! Here comes the next Fibonacci Forgery... 
n:  1  2  3  4  5  6  7  8 

G(n):  1  1  2  3  5  8  13  21 
P(x) = xbut that doesn't give P(4)=5, which is what we want for the real Fibonacci series, so this P(x) is a Fibonacci forgery.
In how many ways can n segments of equal length be connected in a plane if the beginning of one segment is to be connected to the end of the previous segment at a rightangle? Congruent configurations are to be counted as one.From the examples given, we can clear up a few questions left by this definition.
oooo oooo o o o o oooo oooo oooo o o o o o o = o = o o = o oooo = oooo = oooo = oooo and oooo oooo o o oooo o o o oThe sequences are not to overlap, that is, a later segment cannot lie on top of an earlier one, so that each diagram of n links has exactly n straight line segments in it. Links can cross over (at the ends where they join others) and we can have "squares" in our link chains for example:
ooooooo oooo ooooooo ooooooo o o or o o or o o o but not o o o oooo ooooooo ooooooo ooooooo o o o o o o o oooo oooo ooooooosince the last shape cannot be made from 12 links in a single chain (in other words, it cannot be drawn in one go, without taking your pen off the page and without going over any line twice).
number of number of segments configurations ways  1 ooo 1  2 ooo o 1 o  3 ooo o o o o o 2 ooo ooo  4 ooo ooo o ooo o o o o o 3 ooo ooo ooo o o  5 ooo o ooo o ooo ooo o o o o o o o o o ooo ooo o o ooo ooo ooo 5 o o o o o ooo ooo ooo  6 ooo ooo ooo ooo o ooo ooo o ooo ooo o o o o o o o o o o o o o o ooo ooo o ooo o ooo ooo ooo o ooooo ooo ooooo o o o o o o o o o ooo ooo ooo ooo ooo o 8 o o  7 ooo o ooo ooo o o o o o o o o o o o ooo ooo o ooo o o ooo o ooooo ooo ooo ooo ooo o o o o o o o o o o o o o o o ooo ooo o ooo ooo ooo ooo o ooo ooo o o o ooo ooo ooo ooo ooo o o o o o o o o o o 13 ooo ooo ooooo ooo ooo ooooo ooooo ooooo o o o o o o o o o o o o o ooooo ooo ooo o ooo ooooo [With thanks to Jeff Myers, Granville High School, OH, USA for part of this table and for pointing out this problem in The Mathematics Teacher.]
How many shapes are there with 7 links? Try it and you'll find the number of 7link shapes is 15. In her listing in the The Mathematics Teacher she only gave 13, and missed the following two shapes with 7 links:
ooo o o ooo ooo ooo ooo o o o o ooo ooo oooThese were generated by a computer program (in Prolog), so, if my programming is correct, there aren't any more shapes missing. The program also showed that The number of 8link shapes is 23 and this should be 21 if the Fibonacci numbers were the correct series. There are 43 9link shapes and again this should be 34 if the Fibonacci numbers were involved.
So  Deborah's conjecture looks interesting  that there are Fib(n) shapes that can be made from n links at rightangles with no overlapping and allowing for rotations and reflections, but it is another forgery!
[There is an online WWW page for The Mathematics Teacher.]
This fun paper also has several other Fibonacci Forgeries including ones on partitions of n, rooted trees with one label, the number of disconnected graphs on n+1 vertices and the number of connected graphs on n+2 vertices which have just one cycle.
There are many
other forgeries in the paper to do with primes, Catalan numbers, binomial
and trinomial numbers, mixing some genuine examples with the forgeries. His whole
point is that There are not enough small numbers to meet the many demands
made of them and so we are bound to be fooled with small examples of a problem
if we are not careful! Also see..
The Strong Law of Small Numbers Richard K Guy in The American Mathematical
Monthly, Vol 95, 1988, pages 697712.
Fibonacci  the man and His Times 
Fibonacci Home Page
The Fibonacci Numbers in formulae for Pi WHERE TO NOW??? 
The next topics... Fibonacci, Phi and Lucas numbers Formulae Links and References 
© 19962006 Dr Ron Knott