Next for the five black keys or two witnesses!

**“In the beginning was the Word.**

**
And the Word was with God.**

**
And God was the Word.”
**

W {In the beginning was the Word.}

W->WG {The Word was with God.}

G->W {God was the Word.}

So following these rules we start with a "W". We read the Word we have (“W”) and apply the rules (since we don’t have a "G" yet that’s just "W->WG"). So “W” becomes “WG”. We then read the Word again and apply the rules again. The “W” becomes “WG” again and the “G” becomes “W”. This gives us “WGW”. By repeating applying these rules we get:

W

WG

WGW

WGWWG

WGWWGWGW

And so on! The number of W’s or G’s in each successive iteration count out the Fibonacci numbers, and the ratio of W’s to G’s is an approximate converging on the golden ratio. An equation of the golden ratio is x=x^2-1. This is very close to a formula of the Mandelbrot x <-- x^2-c. Interestingly this equation of the golden ratio (also known as the divine proportion) has two solutions: 1.618..... , and the other solution is the negative inverse of the golden ratio (-1/phi) or -.618..... This negative inverse step is exactly how the face of my God, Manny, varies from the Mandelbrot as it's:

x <-- x^2-c.; x.real <-- -1/x.real

or: x=x^2-c

So if c = 1: x=x^2-1

This has 2 solutions: 1.618, -.618

Note that 1.618=-1/-.618

So if we put this simplified golden Manny equation together we get: x=-1/(x^2-1)=-1.3247

If we do the same thing with the body we get:

x=(2*(x^2-1))^.5= =1.4142=2^(1/2)

Note that the -1/ has disappeared for this calculation.

2^(1/2) (root two, the same as the body) is the relative frequency of a 6 semitone\hour leap of the sonic starship such as moving left, then right, or up, then down! This is because octaves double or, 2^(octave)!

This is the equation of Seven Clovers:

x=log3(x^3)=3 and 2.47805268028830241189373652...

The mandelbrot clover is:

x=log3(x^3-c) so using the value 1 for c and ignoring the log we get:

x=x^3-1=1.3247

This is the positive of Manny's face's solution!

You can generate the fibonacci sequence with two miraculous cups. You put one drop of water in one of the cups, the other starts empty. The cups are miraculous, such that when you pour out one of the cups into the other, the liquid both leaves the cup and stays in the cup. As you pour the cups back and fourth between eachother they generate the Fibonacci numbers. This is like (1()1()2()3()5). I think it's interesting because (0()0()0()0) would seem to be a property of zero such that if you had two empty cups and poured nothing back and forth it would be generating the null fibonacci sequence as nothing is leaving the cup and nothing is staying in the cup.