Need description of Prime# distribution in Riemann hypothesis

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?

  On 6/12/2023 at 6:15 PM, 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?

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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.

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.

  On 6/13/2023 at 6:23 PM, 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.

 

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

@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]