Actual timed Mathomatic output from the fibonacci script

Mathomatic version 16.0.1
Copyright © 1987-2012 George Gesslein II.
200 equation spaces available in RAM; 2 megabytes per equation space.
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Reading in files specified on the command line...
1−> 
1−> ; This Mathomatic input file contains the mathematical formula to
1−> ; directly calculate the "n"th Fibonacci number.
1−> ; The formula presented here is called Binet's formula, found at
1−> ; http://en.wikipedia.org/wiki/Fibonacci_number
1−> ;
1−> ; The Fibonacci sequence is the endless integer sequence:
1−> ; 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 ...
1−> ; Any Fibonacci number is always the sum of the previous two Fibonacci numbers.
1−> ;
1−> ; Easy to understand info on the golden ratio can be found here:
1−> ; http://www.mathsisfun.com/numbers/golden-ratio.html
1−> 
1−> -1/phi=1-phi ; Derive the golden ratio (phi) from this quadratic polynomial.

    -1
#1: ––– = 1 − phi
    phi

1−> 0 ; show it is quadratic
Solve successful:

#1: 0 = ((1 − phi)·phi) + 1

1−> unfactor

#1: 0 = phi − phi^2 + 1

1−> solve verifiable for phi ; The golden ratio will help us directly compute Fibonacci numbers.
Solving equation #1 for phi with required verification...
Equation is a degree 2 polynomial equation in phi.
Equation was solved with the quadratic formula.
Solve and "simplify quick" successful:

                   1
          (1 − ((5^–)·sign))
                   2
#1: phi = ––––––––––––––––––
                  2

All solutions verified.
1−> replace sign with -1 ; the golden ratio constant:

                  1
          (1 + (5^–))
                  2
#1: phi = –––––––––––
               2

1−> fibonacci = ((phi^n) - ((1 - phi)^n))/(phi - (1 - phi)) ; Binet's Fibonacci formula.

                (phi^n − ((1 − phi)^n))
#2: fibonacci = –––––––––––––––––––––––
                    (phi − 1 + phi)

2−> eliminate phi ; Completed direct Fibonacci formula:
Eliminating variable phi using solved equation #1...

                          1                      1
                  (1 + (5^–))            (1 + (5^–))
                          2                      2
                ((–––––––––––^n) − ((1 − –––––––––––)^n))
                       2                      2
#2: fibonacci = –––––––––––––––––––––––––––––––––––––––––
                                     1
                                  (5^–)
                                     2

2−> simplify ; Note that Mathomatic rationalizes the denominator here.

                   1            1                 1
                (5^–)·(((1 + (5^–))^n) − ((1 − (5^–))^n))
                   2            2                 2
#2: fibonacci = –––––––––––––––––––––––––––––––––––––––––
                                 (5·2^n)

2−> for n 1 20 ; Display the first 20 Fibonacci numbers by plugging in values 1-20:
n = 1: 1
n = 2: 1
n = 3: 2
n = 4: 3
n = 5: 5
n = 6: 8
n = 7: 13
n = 8: 21
n = 9: 34
n = 10: 55
n = 11: 89
n = 12: 144
n = 13: 233
n = 14: 377
n = 15: 610
n = 16: 987
n = 17: 1597
n = 18: 2584
n = 19: 4181
n = 20: 6765
2−> ; Note that this formula should work for any positive integer value of n.
Successfully finished reading script file "fibonacci.in".
2−> 
End of input.