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A new prime serie?

  1. Dec 11, 2003 #1
    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.
     
    Last edited: Dec 11, 2003
  2. jcsd
  3. Dec 11, 2003 #2
    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]...
     
  4. Dec 11, 2003 #3
    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
     
    Last edited: Dec 11, 2003
  5. Dec 11, 2003 #4
    Re: Re: A new prime serie?



    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.
     
  6. Dec 11, 2003 #5
    Re: Re: A new prime serie?


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

    Change sign on [tex]-\infty[/tex]
     
    Last edited: Dec 11, 2003
  7. Dec 11, 2003 #6
    Re: Re: Re: A new prime serie?

    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.

    I do not understand this.

    Anyway: can't you just give an example with the first 10 (or so) primes?
     
  8. Dec 11, 2003 #7
    Re: Re: Re: A new prime serie?


    Edit: I have been informed that this is an accepted notation (see below).
     
    Last edited: Dec 12, 2003
  9. Dec 11, 2003 #8
    Re: Re: Re: Re: A new prime serie?

    Sorry, I was wrong.

    Good try anyway.

    Good night Erik-Olof Wallman
     
    Last edited: Dec 11, 2003
  10. Dec 11, 2003 #9

    Hurkyl

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    Actually, I've seen "mod 1" used in this way somewhat frequently.
     
  11. Dec 12, 2003 #10
    Thank you, I stand corrected. (I had never seen it.)
     
  12. Dec 12, 2003 #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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