The Golden Key

Riemann's zeta function is known as contraction.

This is also contraction.

Then you can get the golden key.

Zeros of Riemann's zeta function is related to π(x).

Therefore, if you can know the golden key, you may understand zeros of Riemann's zeta function.

As long as I use PC to count prime numbers, it seems to be the same with J(x). This is assumption. I know what I don't know.

A part is everything

It is very difficult to see what Riemann's zeta function is. I don't write about 0 and I can't understand it. However, it is connected with prime numbers.

Now I got this formula.
logζ(s)=1/2^s+(1/2*1/2^2s)+(1/3*1/2^3s)+・・+1/3^s+(1/2*1/3^2s)+(1/3*1/3^3s)+・・1/5^s+(1/2*1/5^2s)+(1/3*1/5^3s)+・・・

Then, you pick up (1/2*1/3^2s) from it. I don't know the reason, but I follow Prime obsession.

I can't understand this process, but this leads to J function

You may see a form which is almost a rectangular (width="3^2→∞", height="1/2").

J(x) include ∞, and 1/2*s*∫[3^2→∞]x^(-s-1)dx is also ∞.
Is this like a black hole?

Riemann's zeta function

Riemann's zeta function is expressed like this. This depends on Sieve of Eratosthenes.

According to index principle,

log1/A=－logA,so

Isaac Newton found

x=1/p^s (p is prime)

∴

This pattern is really complicated and miraculous. I'm in the spiral road.
Then, you need infinitesimal calculus.

I referred to Prime obsession.

Power Law

I counted prime numbers up to 100000000 by using visual basic. The graph is like this.Prime numbers increase infinitely.

However, when you add more integral numbers, there are less prime numbers. This is known as power law.

Inflation

I programmed to count prime numbers by PC, so I can see the same pattern from "Prime Obsession". There is the function about prime numbers.

π means the function of prime numbers up to x.
For example, you can put 100 to x.
10 is the square root of 100, 4.6415・・is the cube root of 100, 3.1622・・is the fourth root of 100・・,2.1544・・is the sixth root of 100,and 1.9306・・is the seventh root of 100. 2 is the first prime number,so you can ignore the seventh root of 100.

Therefore
J(100)=π(100)+1/2π(10)+1/3π(4.64・・)+1/4π(3.16・・)+1/5π(2.51・・)+1/6π(2.15・・)+0+0・・・

You can count prime numbers by 100, so you see π(100)=25, π(10)=4, π(4.64・・)=2, π(3.16・・）=2, π(2.51・・)=1, π(2.15・・)=1.

J(100)=25+(1/2×4)+(1/3×2)+(1/4×2)+(1/5×1)+(1/6×1)

This is the graph.

X is a prime number.
When x is 2, J(2)=1. When x is 3, J(3) jump to 2. When x is 5, J(5) jump to 3.5.

This graph would go up infinitely.

The pattern of prime numbers

I found the pattern of prime numbers. It must be ridiculous but it seems to be almost perfect as long as I know. The realm includes infinity, so it is impossible for me to prove it. However, it keeps moving with the same pattern.
You may say that it must be a fractal which means that a part includes everything. If you are interested in a long spiral road, read this blog. I will extend a stupid story from prime numbers to the universe.

ρ must be optional odd numbers.
Ш is apparent prime numbers such as 2,3,5.
In this case, you can ignore 2 because only odd numbers must be the target to expand prime numbers.

If ρ is divided by Ш and you can get the integral answer, it is Шn^2. It must be Composite numbers before Шn^2 such as 15=5*3.

These are not prime numbers.

∴

For example, you can find prime numbers up to 100 by following this theory.
《３、５、７、９、１１、１３，１５、１７、１９、２１、２３、２５、２７、２９、３１、３３、３５、３７、３９、４１、４３、４５、４７、４９、５１、５３、５５、５７、５９、６１、６３、６５、６７、６９、７１、７３、７５、７７、７９、８１、８３、８５、８７、８９、９１、９３、９５、９７、９９》is the odd numbers.

3,5,7 must be the apparent prime numbers because of √100=10.

Ш1=3,Ш2=5,Ш3=7

9 is divided by Ш1=3, so ρ=9 and ９+2(д-1)Ш1 is working.

д＞1

Therefore
９＋２×３＝１５、９＋４×３＝２１、９＋６×３＝２７、９＋８×３＝３３、９＋１０×３＝３９、９＋１２×３＝４５、９＋１４×３＝５１、９＋１６×３＝５７、９＋１８×３＝６３、９＋２０×３＝６９、９＋２２×３＝７５、９＋２４×３＝８１、９＋２６×３＝８７、９＋２８×３＝９３、９＋３０×３＝９９

１５、２１、２７、３３、３９、４５、５１、５７、６３、６９、７５、８１、８７、９３、９９ are not prime numbers.

The next is Ш2=5 and 25 is divided by this number.
ρ=25 and ２５＋２（д－１）Ш2 is working.
Therefore
２５＋２×５＝３５、２５＋４×５＝４５、２５＋６×５＝５５、２５＋８×５＝６５、２５＋１０×５＝７５、２５＋１２×５＝８５、２５＋１４×５＝９５

３５、４５、５５、６５、７５、８５、９５ are not prime numbers.

The next is Ш3=7 and 49 is divided by this number.
ρ=49 and ４９＋２（д－１）Ш3 is working.
Therefore
４９＋２×７＝６３、４９＋４×７＝７７、４９＋６×７＝９１

６３、７７、９１ are not prime numbers.

Finally you can find prime numbers by 100.

３、５、７、１１、１３、１７、１９、２３、２９、３１、３７、４１、４３、４７、５３、５９、６１、６７、７１、７３、７９、８３、８９、９７

You can't break this pattern until the end, and you never know.
Шn+1 must be expansion of Шn^2+2(д-1)Шn.

C=2(д-1)Шn

∴

This would be fractal.
Moreover, this is expressed by squares, so you can say C=αi^2 (i is an imaginary number).
For example, Ш2=5=3^2+4i^2=9-4 and Ш3=7=5^2+18i^2=25-18.
There is no pattern in C, but it is based on prime numbers.



Moreover, Шn must be P.
Here is Riemann's zeta function.



I exclude 2, but ζ(s) is still fractal because of multiplication.