Pop will eat itself

I checked out a layman's description of the Classic mediant, and I found
this to be interesting.

I find it hard to get my head around the term COMPLEXITY, due to the fact
that an annoying idea keeps coming up into my head, and this being the
number of digits in the numerator, and denominator.

I supposse that complexity would be related to the numerator and denominator
in the closness that each sahres to 2^n, and how big n is, thus something
like 17/16 is more complex than 5/8, and something like 367/112 is more
complex than 17/16.

I know that with respect to subtracting logarithmic intervals, effectively
we are divideing, and that:

a^b/a^c=a^(b-c).

and I have been tinkering around, with mathematical recreations with expanding:

(a+b)^n, and (a+b)^^n, or superpowers, and taking their mirror images.

On a wee bit of a tangent here, I will state that it's difficule to find
rules for:

(a+b)^^4 as:

(a+b)^^2= (a+b)^a * (a+b)^b, and

(a+b)^^3= (a+b)^[(a+b)^a * (a+b)^b]= [ (a+b)^[(a+b)^a] ] ^ [(a+b)^b]=
=(a+b)^ [ (a+b)^ [a+b] ]

but...

(a+b)^^4 = Is a lt more tricky.

I would also like to know if there is a set of rules for:

(a^(a^(a^x))) * (a^(a^(a^x)) = a^[a^(a^x)) +
a^(a^x)]=???===>a^[a^{(a^x)[k](a^x)}]

Applying this mediant formulae we get:

a/b <===> c/d = a[1]c/b[1]d, where the number in parenthesis is a number
representing order of operations.

This proceudre of taking mediants it similar to a sucessive sereis of ratios
that converge towards the square root of two, which has us start with:

1 3 5 12 29
-, -, ---, ---, ----
1 2 7 17 41...

The procedure for generateing these ratios is as follows, and isn't
immediately obvious, but we start with 1/1, and 3/2, and add all of these
terms to get 7, and place this in the denominator.

The sucessive ratios numerator and denominator gives 7.

So it is like adding:

Numerator Denominator
--- --------
| | | | |
| | | | |
| | --------
| | | | |
| | | | |
-- --------

Given that:

(5+9):(4+7) === (a+b):(c+d)

(9+14):(7+11) === (b+a+b):(c+d+d) = (a+2b):(c+2d)

(14+23):(11+18) === (a+2b+a+b):(c+2d+c+d)= (2a+3b):(2c+3d)

Looking at what you'd get from this convergeing series that gives the square
root of two, you will se an identical pattern!!!!

Anyhow, I will go onto this further.

What interests me is the concept of using other n-bonacci ratios to weight
these intervals:

a:b, and c:d

and to compare:

xa + yx
-------/
xb + yd

effectively we are having weighted mediants by what I call the RELATIVE
GOLDEN PROPORTION, which are a whole lot of "false constants" that we get
when leding up to the golden proportion.

I woundered if medianst could be made out of THREE ratios???

a:b, c:d, e:f

=

a+c+e xa+yc+ze
-----/, and from this: ----------------
b+d+f bx+yd+zf

and perhas x, y, and z could be 1, golden proportion, and the tribonacci
ratio 1.839...

I also found information about a very funny little sequence which involved a
definition in mathematics called "recursions", where by what previous terms
we added was defined by the terms THEMSELVES.

It is thus a case of double subscripts, and to illustrate:

Q-Sereis:

T_n=T_(n-T_(n-1)) + T_(n-T_(n-2)), with T_1=1=T_2 which also = 1.

This series is similar to the definition used for the Fibonacci sereis but
with the Fibonacci series, we sum the IMMEDIATE two values, and those values
IMMEDIATELY PREVIOUS to T_n.

With this Q-Series, the previous two terms tell us HOW FAR TO COUNT BACK.

Thus:

1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 8, 8, 10, | 9, 10 |...........
| | -------
5+6=11 |
| --------->How far left to move

To obtaint eh next term, move leftwards from the three dots 10 and 9 terms,
take the sum, and we'd get 11.

This is T_18.

Continiueing this sequence:

11, 9, 11, 11, 12, 12, 12, 12, 16...etc..

I have woundered now if any recursive sequences might generate music, many
prime numbers (...with a high density of primes), proportions of n-bonaccis,
and I suggest:

T_n=T_(n-T_(n-T_(n-1)) + T_(n-T_(n-T_(n-3))), with T_1=T_2=T_3=1

How far can we take theis recursion?

---Sarn.