Re: Towards a Formula for Primes

On May 6, 12:52 pm, "charlesweh...@xxxxxxxxxxx"
<charlesweh...@xxxxxxxxxxx> wrote:
My pages on the Pythagorean Perimeters Theorem, which I discovered,
are here:

http://www.wehner.org/pythag/
For the starting concept

http://www.wehner.org/pythag/ratios.htm
For further examples

The whole thing is based on DIOPHANTINE arithmetic.

In each case, there is a rectangle with a Diophantine ratio.

Using the procedure, one gets a Pythagorean triple.

When the rectangle is constructed on the smallest side of the
triangle, its perimeter matches that of the triangle.

Here is an example:

The ratio 3:2 delivers the 8:15:17 Pythagorean BASE triple.

I say that it is a BASE triple, which could be called BASIC triple,
because the 6:8:10 triangle is not unique - it is the 3:4:5 scaled by
2. However, 3:4:5 is a BASE triple.

For reasons that take too many words, the numbers 3:4:5, or whatever,
of a BASE triple will always be COPRIME.

So we take the coprime pair 3:2 and get the coprime triple 8:15:17

Unfortunately, the pair and the triple are not coprime to each other.
There are not - nor can there be - five coprime numbers. This is
because the 3:2 rectangle is constructed (in this version) upon the
smallest side of the triple, and will therefore be a scaled edition.

The 8:15:17 perimeter is 40.

A 3:2 rectangle constructed on side 8 is a 12:8 rectangle. Perimeter
40.

However, there is more to be found.

Because we have a coprime triple 8:15:17, we can now create six ratios
from it.

8:15
8:17
15:17
15:8
17:8
17:15

ALL are coprime.

Each of these will expand out into a BASE triple.

So prime numbers are being brought into the results in a manner that
they, due to the coprimality, are ISOLATED from one another.

Recursing around the algorithm should bring the extensio ad infinitum.
Perhaps within this concept lurks the narrow gateway - the key - to
entering into the solving of the problem of finding a test of, or
formula for, primes.

Charles Douglas Wehner

A fraction as ratio was the c^2=a^2+b^2 equation and the next was the
question, how to
not just test, but count a list of primes.

So the test is commonly taken to the limit of conception and given an
algebratic ring representation. And then called a counting. I am not
a mathematician, but just warn of the level of trial you seek.

Step one is to stay geometric. Forsake mathematics. Always call the
number a point reference. A second class of reader is then required.
A geomtery expert.

Step two. Make a therom:

A point to predict any triangle appears the triangle's center.

A

*a
* *
*E
* *
b*****************c
B C

ABC enclosing the first triangle in the next must have this distance
formula.

E a point on line (ba) appears a two equal triangle angle vertex.
Spliting triangles (abc) into (bEc) and (Eac)

Given this center what becomes the line distance formula?

(EBC) = Eand I get lost here. An idea but the formula for the line
was forgotten.

Except the line formula was a mathematical theory not geomtry, I need
to stay geometric.
I will try angles.


Angle BEC equals angle bEc, likewise for angle AEC and aEc.

For all changes of the segments, Ea to EA a constant angle was to be
found.

So for c^2 as line AB, sorry for the bad notation
or c^2 is now equilateral triangel functionalized. Splitting or
expanding equilateral triangles allow all side to represent. So define
E as the center.

For ca>Ec as line segment magnitudes segment Ec defines the unit
segment to count with. A restrict to cs>EC is a type of right
triangle. A change to equilateral triganle side.


NOW just NOW select the sides as a certain equilateral with segment Ec
as one.

Therom two:

For all factor of one changes to Ec, i,e. 1 to two, two to three etc,
a distance aA, bB and cC exist. If aA changes by LESS THAN Ec, unit
one, you are home free.

You can now count all cases where aA equals One unit change. And if aA
is one then Ec appears the fractional change.

A modified Sieve of E. Ec the line extends to define EC where point
C is not in line with the triangle edges!!! Calculate the necessary cC
for the aA of unit one.

A distance formula is not required just c. Set the triangel c, of
c^2=a^2+b^2 equal to c.

Infer the c! Always AC as a segment.

(Ec+cC)^2+(Ea+aA)^2=AC^2

Where aA equal one and Ec equals one.

(1+cC)^2+(Ea+1)^2=AC^2

Take Ea and test for factor existence.

Ea cC =AC^2
_______

2 1 sqr(9+4)

and sqr(13) is not a segment!

so incease Ea to 2!!!!!!!!! and try again

Ea cC
_______

3 2 sqr(16+9)=5


Ea must always be greater then cC for this to work!!!!!!!

A unit change of one allow all angles.

Ea cC
______

4 3 sqr(25+16)= 41 seven!!!! Seven Passes


So the valid integer Square Root counts.


A basic method was derived independent of the Sieve of E. And to
count was to increment only.
And to prime was to square root only!

It is not complicated. And speed is the issue only. How fast can
this count. And a valid square root takes time.

A relative method was this design. Absolute counting was found the
slowest.

A speed goes the square root is testing factors itself:(

Making the Sieve of E and this derivative Slow.

A factor as algebra was designed to allow all relative set to
increment. I forget the method name though.
Mathematical solution was found to allow the fastest speed.

Geometry is speed limited by the ratio factor solution for it has no
commutivity property.


.

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