Need description of Prime# distribution in Riemann hypothesis

[Trurl]

You are right there is a lot of Riemann stuff. I just thought the surface had to do with the graph of the distribution of zeros. I’m probably wrong but this stuff is complex.

 I’m am looking to find any application of Prime numbers I can. Other than cryptography. Why are Primes so popular in mathematics. Just because we can’t find their pattern? Why did Primes interest the ancient Greeks if you can’t built cities with them? Are the Riemann surfaces related to the Riemann Hypothesis?

I believe that Prime numbers relate to calculus in the analysis of the graph. If you could find the graphs that form the Riemann function you could find the graphs that when combined make up any function’s graph.

That is what I took from the book. But what other applications of Primes are there?

Edited June 12, 2023 by Trurl
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[Genady]

1 hour ago, Trurl said:

You are right there is a lot of Riemann stuff. I just thought the surface had to do with the graph of the distribution of zeros. I’m probably wrong but this stuff is complex.

 I’m am looking to find any application of Prime numbers I can. Other than cryptography. Why are Primes so popular in mathematics. Just because we can’t find their pattern? Why did Primes interest the ancient Greeks if you can’t built cities with them? Are the Riemann surfaces related to the Riemann Hypothesis?

I believe that Prime numbers relate to calculus in the analysis of the graph. If you could find the graphs that form the Riemann function you could find the graphs that when combined make up any function’s graph.

That is what I took from the book. But what other applications of Primes are there?

AFAIK, Riemann surfaces are not related to the Riemann hypothesis. Greeks were fascinated with numbers philosophically. As opposed to Romans, who built cities.

As for the other questions, I'd like to know the answers, too.

[noobinmath]

Primes pattern is simple, just look at the sieve:
 

First you have all numbers starting from 2 until infinity

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ...

now use first number you see (2) and use the number of that number to remove numbers (sieve repeats every 2 numbers)

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ...

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (1 out of every 2 numbers are removed)

now use first number you see (3) and use the number of that number to remove numbers (sieve repeats every 2*3 numbers)

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 6 numbers)

5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 ... (4 out of every 6 numbers are removed) 

now use first number you see (5) and use the number of that number to remove numbers (sieve repeats every 2*3*5 numbers)

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 5 )

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 3 )

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 30 numbers)

7 11 13 17 19 23 29 31 37 41 43 47 49 ... (22 out of every 30 numbers are removed) 

now use first number you see (7) and use the number of that number to remove numbers (sieve repeats every 2*3*5*7 numbers)

etc.

 

[Genady]

16 minutes ago, noobinmath said:

Primes pattern is simple, just look at the sieve:
 

First you have all numbers starting from 2 until infinity

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ...

now use first number you see (2) and use the number of that number to remove numbers (sieve repeats every 2 numbers)

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 ...

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (1 out of every 2 numbers are removed)

now use first number you see (3) and use the number of that number to remove numbers (sieve repeats every 2*3 numbers)

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 6 numbers)

5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 ... (4 out of every 6 numbers are removed) 

now use first number you see (5) and use the number of that number to remove numbers (sieve repeats every 2*3*5 numbers)

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 5 )

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (apply 3 )

3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 ... (all combinations, repeats every 30 numbers)

7 11 13 17 19 23 29 31 37 41 43 47 49 ... (22 out of every 30 numbers are removed) 

now use first number you see (7) and use the number of that number to remove numbers (sieve repeats every 2*3*5*7 numbers)

etc.

 

Yes, this is a procedure. What is a resulting pattern?

[Trurl]

@noobinmath

 

I wish I could create a pattern from a sieve. Here is a proposal about how to go about it. If you graph the 2 equations separately as pnp equals all integers zero to infinity, as the equation is less than one; pnp is a semi-Prime and the Prime number is conformed to be Prime.

 

This may or may not work. Also do not confuse x on the graph equaling pnp as x in the equation equaling the Prime number.

 

I will have more on this later. But since you want a pattern in the sieve I thought I would share my attempt.

 

Where x * y = pnp

 

When y = Sqrt[pnp^3/(pnp*x^2+x)]

 

So that, x* Sqrt[pnp^3/(pnp*x^2+x)] = pnp

 

pnp – pnp = 0

 

 

Let x equal any Prime number. 5 for instance.

 

Graph pnp = x ; at x on the graph at that instance = pnp

 

So where 5* Sqrt[pnp^3/(pnp*5^2+5)] – pnp = 0, then pnp is a semi-Prime and we can plug it into the equation and find y (the larger Prime number) 

 

And if we continue to graph over all real numbers we will find every Prime number in existence.

 

 

 

[(pnp / (Sqrt[(pnp^3 / (pnp*5^2+5)])) -5] == 0 == [5* Sqrt[pnp^3/(pnp*5^2+5)] – pnp]

 

 

 

 

[Trurl]

 

The shear difficulty in making a secure one-way-function that will produce a public and secure private key makes security based on intense computation impossible to test, verify, and rely on the system. It is one thing to create a one-way-function but how will that one-way-function define the cypher that encrypts the message.

 

My example which I don’t know if it has been tried before is to use the addition of large numbers, relates to the odometer problem I proposed earlier in this tread. We are always plugging numbers into the functions by using counting numbers. But what if we start counting by the pattern before numbers are placed into the function. Instead of 1.2.3.4.5…. to infinity, we modify the numbers by a function and then place it into the function machine?

 

I know I am talking very abstractly and this is difficult for me to explain. But what if you counted only those numbers in the odometer that had 2 or more instances of a digit. 101100. If you count by a function that solves those numbers than you only see the distance between the pattern. However, if you counted 1.2.3.4.5… it would not show a pattern. There would be arbitrary distances between reading of 2 or more digits. I hypothesize this is why we can’t see a pattern in Primes.

 

 

______________________-

 

I wrote this last week. It doesn’t make sense, but I post it so someone might see the abstract idea it represents. On the odometer you count linearly. So, a number 90909 may occur at a greater distance from the current pattern number minus the previous repeating of the similar digit. Simply put 90909 would occur after 50505. But the distance between such numbers is dependent on counting linearly. That is the same way we look for a pattern in Primes. Is that what is making it so difficult to factor semi-Primes?

 

Does this make any sense? Have you ever had an abstract idea that is hard to put to words?

 

90909 – 50505 = 40404 but does 50505 – 50500 greater occur before 90909 – 90900.

 

It depends on how I am classifying repeated digits but do larger digits occur after the smaller digits of reoccurring digits? For example, 111111 is linearly faster that 999999, but 099999 comes before 111111.

 

This was the original idea for the odometer example. I know it didn’t explain my idea. Can anyone put my thought into a workable idea?