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July 11th, 2018, 07:24 AM   #1
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Prime Number Sequence

Has someone study previously the following structure/pattern in prime numbers:


a) genesis

1 (1) - > the numbers in the parenthesis indicates the pattern in the consecutive differences between the elements of the sequence, in this case the infinite set of integer numbers, from 1 to infinite.

Starting in 1 it continues by adding (1) as 2 3 4 5 6 7 8 9 …. up to infinite

This patter in the prime numbers will be seen up to 2² → 2 3

the total distance of the pattern is 1.

b) prime number patterns

b.2)

using two repetitive patterns of the previous sequence, then we add to it the influence of 2.

(1,1) -the first two components are 2 and 3-, but as 2 is the middle and it is removed then the previous expression combines to (2):

1 (2) → 3 5 7 9 11 …….. up to infinite

This pattern in the prime numbers will be seen up to 3² → 3 5 7

The total pattern of the distance is 2.

b.3)

Now we use three repetitive patterns of the past sequence:

1 (2,2,2)

As the 3 enters into action is removed then the first two patterns are combined and the final sequence remains as:

1 (4,2) → 1 5 7 11 13 17 19 21 … up to infinite, duplicating the same pattern. On the prime number list it is only visible up to 5², the square of the next prime number.

The total distance of the pattern is 6

b.5)

We replicate the previous pattern 5 times, having:

1 (4,2,4,2,4,2,4,2,4,2)

In this case we remove 5, that is between the first 4 and 2, and 5*5, that it is placed just before the last 4. The pattern recombines to:

1 (6,4,2,4,2,4,6,2)

The 5 multiples that fixed this sequence are spaced exactly as the previous sequence (4,2)

5*1
+4*5
5*5
+2*5
5*7
+4*5
5*11
…..

The output of this sequence shows the sequence of prime numbers up to 7²

1 (6,4,2,4,2,4,6,2) → 1 7 11 13 17 19 23 29 31 37 41 43 ….. up to infinitum repeating the same pattern.

The total distance of the pattern is 30

b.7 ) With seven it can be done exactly as in the previous cases.

1.- A replication of the previous pattern is done 7 times,

1 ( 6,4,2,4,2,4,6,2 ,6,4,2,4,2,4,6,2 ,6,4,2,4,2,4,6,2 , 6,4,2,4,2,4,6,2 , 6,4,2,4,2,4,6,2 6,4,2,4,2,4,6,2)

The length of the pattern is 210

Over this pattern are introduced the modifications induced by 7, those are indicated in the previous pattern:

7*1
7*7
7*11
7*13
7*17
7*19
7* 23
7* 29
7* 31 = 217 < 210

The same sequence follows afterwards, repeating the pattern again.

etc

The last factor modifying is the one below the patter

Once the pattern is modified taking into account the effect of 7 the sequence will provide up to 11² the list of prime numbers. The last part of the sequence will continue up to 210 but it will be modified by the 11 prime number.

Once the pattern for 7 has been calculated it is possible to continue with 11.



So basically each consecutive prime number modifies the previous sequence, originating a new pattern that replicates infinite times.

Also this sequences fixes how the next sequence will be modified by the next prime number.

Moreover it shows that the distribution of prime number follows a pattern, very simple in the very first prime numbers, and very complicated as an increasing number of prime numbers starts to modify the simpler patterns.

But, well, although complicated it is a pattern
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July 11th, 2018, 08:02 AM   #2
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If I'm understanding your post, I think you've discovered the sieve of Erotosthenes.

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

The problem is that it doesn't give you a pattern. Nobody knows if there's a pattern to the primes. The question's connected to a lot of deep number theory.
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July 11th, 2018, 08:20 AM   #3
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Yeah, it is a sieve, but not as the Erastosthenes one.

Erastosthenes sieve uses brute force. In this one the previous structures “tells” what are the numbers to be removed in the next round.

For instance in 5:

In the Erastosthenes sieve, the removed ones will be: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65,70, 75, 80, 85 ….

In the one described the removed ones will be: 5, 25, 35, 55, 65, 85, ….
So one every 4 and 2 of the previous list

And this patter is derived from the previous pattern generated by the 3 number 1(4,2)
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July 11th, 2018, 09:09 AM   #4
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Quote:
Originally Posted by ma1975 View Post
Yeah, it is a sieve, but not as the Erastosthenes one.

Erastosthenes sieve uses brute force. In this one the previous structures “tells” what are the numbers to be removed in the next round.

For instance in 5:

In the Erastosthenes sieve, the removed ones will be: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65,70, 75, 80, 85 ….
No, because 10 was already removed with the 2's. 15 with the 3's. 20 with the 2's. 25 is the next one to be removed. I think that's your idea.



Quote:
Originally Posted by ma1975 View Post
In the one described the removed ones will be: 5, 25, 35, 55, 65, 85, ….
Right. The ones that weren't already removed with the 2's and 3's. That's Erotosthenes.

I'll confess to not following your post in detail but that's what I'm getting from what you wrote.
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July 11th, 2018, 09:56 AM   #5
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It is true that in Eratosthenes the 10,15,20 and similar have been already removed by 3 or 2 but still you should evaluate before if they were removed by two or three.

In the other system you know in "advance" that they have been removed and you go directly to the 25 and 35, so the 10,15 and 20 do not need to be evaluated. (Instead in Eratosthenes you need to perform the evaluation)
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July 11th, 2018, 11:43 AM   #6
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Quote:
Originally Posted by ma1975 View Post
It is true that in Eratosthenes the 10,15,20 and similar have been already removed by 3 or 2 but still you should evaluate before if they were removed by two or three.

In the other system you know in "advance" that they have been removed and you go directly to the 25 and 35, so the 10,15 and 20 do not need to be evaluated. (Instead in Eratosthenes you need to perform the evaluation)
I guess I'm not following your "patterns" but perhaps someone else will. I don't really see what you're doing. It really seems like you're crossing out the multiples (greater than 1) of successive primes.
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July 12th, 2018, 01:41 AM   #7
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Quote:
Originally Posted by ma1975 View Post
Yeah, it is a sieve, but not as the Erastosthenes one.

Erastosthenes sieve uses brute force. In this one the previous structures “tells” what are the numbers to be removed in the next round.

For instance in 5:

In the Erastosthenes sieve, the removed ones will be: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65,70, 75, 80, 85 ….

In the one described the removed ones will be: 5, 25, 35, 55, 65, 85, ….
So one every 4 and 2 of the previous list

And this patter is derived from the previous pattern generated by the 3 number 1(4,2)

I found something similar, the sieve of Euler

https://en.wikipedia.org/wiki/Proof_...roduct_formula

But still is not the same, after the square of prime number being sieved, the sieve of Euler starts to evaluate all the multiples of the prime number being sieved, independently if they have been sieved before by a lower prime number.

so we have:

1.- Erastosthenes' sieve: that evaluates all the multiples of the prime number being evaluated.
2.- Euler's Sieves: that evaluates all the multiples of the prime number being evaluated after the square of that number.
3.- X Sieve: The one presented, that sieves -or evaluates if you prefer- only the multiples of prime number that have not been sieved before by lower prime numbers.

Someone knows its name or who has studied it before?
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July 12th, 2018, 02:37 AM   #8
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I found a method that implements the Sieve explained in the entry: It is called Sieve of Giraldo Franco

https://arxiv.org/pdf/1101.3919.pdf

The only think it is that although he claims that there is a clear pattern between the different structures created, he says that the structure has to been obtained before its application, but the case is that the pattern can be "learnt" from previous patterns, as exposed previously.
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July 15th, 2018, 11:14 PM   #9
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Legendre

These are known patterns described by the Legendre formula $\phi(x,a)$ which count the number of integers less than $x$ which are not divisibles by the first $a$ primes.

A well-known form widely used to describe prime numbers is the $6n+1$/$6n+5$ form, where all 2 and 3 multiples are removed [$\phi(x,2)$].

If you remove all multiple of 5 [$\phi(x,3)$], remaining numbers are of the form
$30n+1$
$30n+7$
$30n+11$
$30n+13$
$30n+17$
$30n+19$
$30n+23$
$30n+29$

In this one, in fact you removed all multiple of 5 or more precisely:
$30n+5$
$30n+25$

which is equivalent to what you do: $5\cdot(6n+1)$ and $5\cdot(6n+5)$ and is reflected in the formula $\phi(x,a)=\phi(x,a-1)-\phi(\frac{x}{p_a},a-1)$

If you continue (remove 7 multiple), you end up with a form $210n+1, 210n+11, 210n+13,...$ where the 7 multiples removed are equivalent to what you do or what you find in your paper:
$7\cdot(30n+11)=7\cdot11+210n$
$7\cdot(30n+13)=7\cdot13+210n$
$7\cdot(30n+17)=7\cdot17+210n$
$7\cdot(30n+19)=7\cdot19+210n$
$7\cdot(30n+23)=7\cdot23+210n$
$7\cdot(30n+29)=7\cdot29+210n$
$7\cdot(30n+31)=7\cdot31+210n=7\cdot(30n+1)$

$7\cdot(30n+7)=7\cdot7+210n$ removed earlier in the paper.

Theoretically, this is not something new, and practically these kind of optimisations are already used (e.g. here http://sweet.ua.pt/tos/bib/5.4.pdf)

Last edited by skipjack; July 16th, 2018 at 03:44 AM.
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July 16th, 2018, 01:34 AM   #10
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Some time ago, I worked with this same idea.. , nice post.

Last edited by skipjack; July 16th, 2018 at 03:44 AM.
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