# New prime series?

Originally posted by Sariaht
That makes no difference of course: 1*x = x for all x. So including your p1=1 doesn't change the fact that your equation is incorrect...

Originally posted by Sariaht
What is your definition of a prime?
The commonly accepted definition for a prime number is any number having no factor except itself and one. From this rule it follows that 1 is not a prime number, but 2 is.

2*3 = 6 | +/- 1 | 5, 7

2*3*5 = 30 | +/- 1 | 29, 31

2*2*3 = 12 | +/- 1 | 11, 13

2*3*3 = 18 | +/- 1 | 17, 19

2*3*5*5 = 150 | +/- 1 | 149, 151

2*3*5*7*7 = 1470 | +/- 1 | 1469, 1471

allright.

Only one primesquare is aloud.

3*5*7*7 = 735 | +/- 2 | 733, 737
3*5*5*7 = 525 | +/- 2 | 523, 527
3*3*5*7 = 315 | +/- 2 | 313, 317
5*7*11 = 385 | +/- 2*3 | 379, 391
7*7 = 49 | +/- 30 | 19, 79

etc?

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Originally posted by Sariaht
2*3 = 6 5 7

2*3*5 = 30 29 31

2*2*3 = 12 11 13

2*3*3 = 18 17 19

2*3*5*5 = 150 149 151

2*3*5*7*7 = 1470 1469 1471

allright.

Only one primesquare is aloud.
You should really learn to write your ideas (or whatever they are) more clearly. I have NO IDEA what you are trying to say....

Originally posted by suyver
You should really learn to write your ideas (or whatever they are) more clearly. I have NO IDEA what you are trying to say....

Were does my equation error?

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The prime numbers have a RANDOM distribution over the natural numbers. You will not succeed in finding such an (easy or not) algorithm to always generate a new prime number from a set of already known ones. However, the numbers tend to become large, making it difficult to see that they are not prime.

If you find a simple algorithm and PROVE that this algorithm works for at least the first 100 prime numbers, then I will look at it again. But like this it's becomming a waste of my time: you're just guessing new algorithms without any idea about why they should work in the first place...

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Originally posted by suyver
(uip = un-included prime in serie)

Yes. I'll check. Sorry... (only one square is aloud).

No prime is aloud to have a negative conjugate.

.................................uip.(pp)...prime or square
2........................= 2.....+-1............1 | 3
3........................= 3.....+-2............1 | 5
2*2......................= 4.....+-1............3 | 5
2*3......................= 6.....+-1............5 | 7
2*3*3....................= 18....+-1...........17 | 19
3*3......................= 9.....+-2............7 | 11
2*2*3....................= 12....+-1...........11 | 13
2*5......................= 10....+-3............7 | 13
3*5......................= 15....+-2...........13 | 17
3*5*7....................= 105...+-2..........103 | 107
2*3*7....................= 42....+-5...........37 | 47
5*7*7....................= 245...+-6..........239 | 251
3*7*7....................= 147...+-10.........137 | 157
2*7*7....................= 98....+-15..........83 | 113
2*5*7....................= 70....+-3...........67 | 73
5*5*7 ..................= 175...+-6..169(square) | 181
5*7......................= 35....+-6...........29 | 41
2*3*5*7*11...............= 2310..+-1.........2309 | 2311
The numbers in the serie must be a prime (3 5 7 11 are four numbers)
2*5*11...................= 110...+-21..........89 | 131
5*7*11...................= 385...+-6..........379 | 391

3*7*11...................= 231...+-10....17*13(?) | 241
(maybe this has got something to do with that diff (3,7) = diff (7,11)
is not a prime, and that diff(x,y) cannot be a non-prime twice in a row?)

3*5*11...................= 165...+-14.........151 | 179
11*5.....................= 55....+-42..........13 | 97

Damit... Who cares if it works anyway: the permutations becomes to many.

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Originally posted by suyver

In the middle of two conjugative primes is a third number oftenly divideable
with 6.

Can two such queries have the same factor-sum if the sum is a prime?

2*2 = 4 | 2 + 2 = 4

2*3 = 6 | 2 + 3 = 5

2*2*3 = 12 | 2 + 2 + 3 = 7

2*3*3 = 18 | 2 + 3 + 3 = 8

2*3*5 = 30 | 2 + 3 + 5 = 10

2*3*7 = 42 | 2 + 7 + 3 = 12

2*2*3*5 = 60 | 2 + 2 + 3 + 5 = 12

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Originally posted by suyver
Have you read this thread? You might find some of the contents interesting...

But if this was true, you could find alot higher primes alot easier.

No: this is indeed a function that generates all the primes. But it is a function with 26 parameters that can vary.... Computationally very intensive!

Originally posted by suyver

You mean that the answer of my question is: No?

How can you get all primes through this?

Or did you mean that his function generates all primes?

I'm comfused...

Good night!

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In that other thread that I linked to, there is (about halfway through) a short discussion about this monster. That is a set that yields every prime number, as well as that it only yields prime numbers. However, there is one catch: you have to restrict its domain to parameters that give positive values (i.e. ignore all results <0).

I suggest that you spend some time reading that other thread. There is also a rough proof of the fact that it is fundamentally impossible to construct a nonconstant polynomial in a single variable over the integers that will generate all primes...

The first n factors...

Originally posted by suyver
Does the x number of factors to form the n first numbers follow a maclaurin serie?

1 | 1f
2 | 2f
3 | 3f
4 | 5f
5 | 6f
6 | 8f etc.

You must agree in that it's a good question anyway...

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p*p

"A multiple between two primes is always right in the middle of two primes."

Has that been proven, that the product of two primes is always the average of two primes?

Are all numbers > 2 the average of two primes?