Is the Product of Primes Significant in Number Theory?

[Sariaht]

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.

 

[Sariaht]

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

 

[suyver]

Have you read this thread? You might find some of the contents interesting...

 

[Sariaht]

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

No, i have not read his thread. What does he mean?

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

 

[suyver]

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!

 

[Sariaht]

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!

 

[suyver]

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...

 

[Sariaht]

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...

 

[geraldmcgarvey]

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?