A new prime serie?

  • Thread starter Sariaht
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  • #1
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This is strange... I can sort of proove this.

( n(1/2 + 1/3 + ... + 1/pa) - (1/3 + 2/5 + ... + (a-1)/pa) minus all whole queries ) <= ½

--> n = p

If it's true and I was the first to find the serie; can I name it after me?

In that case i would like to name my equation the EO-equation.


But pa is the last existing prime between the two numbers 1 and n.

Thereby it's a bit difficult to find really big primes, but anyway.
 
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  • #2
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Originally posted by Sariaht
This is strange... I can sort of proove this.

mod1( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) <= ½
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?

By the way, are you aware that both the series 1/2 + 1/3 + 1/5 + ... + 1/pa and the series 1/2 + 2/3 + 3/5 + ... + (a-1)/pa diverge in the limit to infinity? Actually, the second series diverges considerably faster, so the total series ( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) goes to [tex]-\infty[/tex] in the limit [tex]a\rightarrow\infty[/tex]...
 
  • #3
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I ment modulus.

mod1(23.5) = .5


If the first prime is two, how ever you think, the risk is fifty fifty for the next two primes to have a factor two.

If the risk is higher than... Well I can't explain it without a new small numbertheory
 
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  • #4
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Originally posted by suyver
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?



modulus 12.34 = .34

(Am I not right in this?)

Anyway what i ment was everything after the dot in tu.vxyz....

So .342563448 in 23.342563448 is modulus 23.342563448.
 
  • #5
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Originally posted by suyver
I must say that I don't really understand what you are trying to say.

What do you mean with mod1( ... )?

By the way, are you aware that both the series 1/2 + 1/3 + 1/5 + ... + 1/pa and the series 1/2 + 2/3 + 3/5 + ... + (a-1)/pa diverge in the limit to infinity? Actually, the second series diverges considerably faster, so the total series ( n(1/2 + 1/3 + ... + 1/pa) - (1/2 + 2/3 + 3/5 + ... + (a-1)/pa) ) goes to [tex]-\infty[/tex] in the limit [tex]a\rightarrow\infty[/tex]...

The n don't cover the second serie, it only covers the first.

Change sign on [tex]-\infty[/tex]
 
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  • #6
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Originally posted by Sariaht
the n don't cover the second serie, it only covers the first.
But the first series diverges for n->oo! If you won't take the limit in the second series as well, then the result will just be +oo.

Originally posted by Sariaht
Change sign on [tex]-\infty[/tex]
I do not understand this.

Anyway: can't you just give an example with the first 10 (or so) primes?
 
  • #7
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Originally posted by Sariaht
modulus 12.34 = .34

(Am I not right in this?)

Edit: I have been informed that this is an accepted notation (see below).
 
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  • #8
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Originally posted by suyver
Sorry, I was wrong.

Good try anyway.

Good night Erik-Olof Wallman
 
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  • #9
Hurkyl
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Actually, I've seen "mod 1" used in this way somewhat frequently.
 
  • #10
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Originally posted by Hurkyl
Actually, I've seen "mod 1" used in this way somewhat frequently.
Thank you, I stand corrected. (I had never seen it.)
 
  • #11
HallsofIvy
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Suyver, note that in the original post it was "mod1()". That might be more often written "( ) mod(1)" but the notation is perfectly reasonable.
 

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