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New prime series?

  1. Nov 15, 2003 #1
    Consider all primes

    2, 3, 5, 7, 11, 13...

    and their products such that

    2x3=6, 2x3x5=30, 2x3x5x7=210, 2x3x5x7x11=2310, 2x3x5x7x11x13=30030...

    Is this latter series used in number theory?


    Likewise, can one determine

    lim (2+3+5+7+11+13...pn-1)/(2+3+5+7+11+13...pn)
    n-->[oo]

    in analogy to phi of Fibonacci numbers?
     
  2. jcsd
  3. Nov 16, 2003 #2

    Hurkyl

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    This function is called the primorial function.
     
  4. Nov 16, 2003 #3
    Thanks for the helpful hyperlink, Hurkyl.

    Do you or anyone else have a hint about the second series I mentioned, the prime Fibonacci analog, and its limit:

     
  5. Nov 16, 2003 #4

    Hurkyl

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    I'm not sure why you call it a Fibonacci analog...

    Anyways, your fraction can be rewritten as:

    [tex]
    \frac{2 + 3 + \dots + p_{n-1}}{2 + 3 + \dots + p_n}
    = 1 - \frac{p_n}{2 + 3 + \dots + p_n}
    = 1 - \frac{1}{1 + \frac{2 + 3 + \dots + p_{n-1}}{p_n} }
    [/tex]

    So solving your limit reduces to finding

    [tex]
    \lim_{n\rightarrow\infty}\frac{2 + 3 + \dots + p_{n-1}}{p_n}
    [/tex]

    No proof of any value for this leaps to mind, however.
     
    Last edited: Nov 16, 2003
  6. Nov 16, 2003 #5

    Hurkyl

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    I don't know how I managed not to read the big thread on TeX. :frown:

    (PS you got some groupings wrong)
     
    Last edited: Nov 16, 2003
  7. Nov 16, 2003 #6

    chroot

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    Whoops! Feel free to edit my posts to reflect the correct TeX, or delete them altogether if you wish.

    Also, please note that I don't intend to coerce people into using TeX if they are already comfy and happy with basic HTML.

    - Warren
     
  8. Nov 16, 2003 #7

    Hurkyl

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    I need to learn LaTeX eventually anyways, might as do it here where I can get some practical benefit out of it. :smile: The only LaTeX I've written thus far was for writing a tutorial to use some code I had written, so I haven't gotten to play with any of the math stuff!
     
  9. Nov 16, 2003 #8

    Hurkyl

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    All right, here goes.

    Define

    [tex]
    S(n) := \frac{2 + 3 + ... + q_n}{n}
    [/tex]

    where [tex]q_n[/tex] is the largest prime less than [tex]n[/tex]. I aim to prove:

    [tex]
    \lim_{n\rightarrow\infty} S(n) = \infty
    [/tex]

    From which we can deduce

    [tex]
    \lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{p_n}
    = \infty
    [/tex]

    and thus

    [tex]
    \lim_{n\rightarrow\infty} \frac{2 + 3 + \dots + p_{n-1}}{2 + 3 + \dots + p_n}} = 1
    [/tex]


    It's clear that

    [tex]
    \lim_{n\rightarrow\infty} S(n) = \lim_{\substack{ n\rightarrow\infty \\ n~{\it even}} } S(n)
    [/tex]

    So I will restrict my attention to the case where [tex]n[/tex] is even.


    The general approach is to estimate the numerator of [tex]S(2n)[/tex] by just looking at the primes in the range [tex][n, 2n)[/tex], and underestimating [tex]S(2n)[/tex] by [tex]n[/tex] times the number of primes in this range. To do this, I will use Chebyshev's bound on the prime counting function:

    [tex]
    \frac{7}{8} < \frac{ \pi(n) }{ \frac{n}{ln~n} } < \frac{9}{8}
    [/tex]

    So here goes:

    [tex]
    \begin{equation*}
    \begin{split}
    S(2n) &= \frac{2 + 3 + \dots + q_{2n}}{2n} > \frac{1}{2n} (\pi(2n) - \pi(n)) n \\
    &> \frac{1}{2} \left (\frac{7}{8} \frac{2n}{ln~2n} - \frac{9}{8} \frac{n}{ln~n} \right)
    \end{split}
    \end{equation*}
    [/tex]

    A bit of elementary calculus proves that

    [tex]
    \lim_{n\rightarrow\infty} \frac{1}{2} \left( \frac{7}{8} \frac{2n}{ln~2n} - \frac{9}{8} \frac{n}{ln~n} \right) = \infty
    [/tex]

    And we're done.
     
    Last edited: Nov 16, 2003
  10. Nov 16, 2003 #9
    Hurkyl,

    What experience do you have in math? You seem the most competant of a talented bunch here at PF. I hope you have seen the "Booda Theorem" at my website, my greatest accomplishment in mathematics.

    LB
     
  11. Nov 16, 2003 #10

    Hurkyl

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    It's the LaTeX. It makes one look smarter. :smile:

    I'm relatively fresh out of school, actually. I got my BS's in math and computer science two years ago, and started work this January. I've been hired as a mathematician, but my work thus far has been leaning more towards the programming.

    However, math has been my hobby since I was little, so I have experienced a lot more than these credentials would suggest.
     
  12. Nov 25, 2003 #11
    Sorry.

    But is this right then?

    2p - 1 = p

    2 - 1
    4 - 1
    8 - 1
    32 - 1
    128 - 1
    2048 - 1


    Am I wrong?
     
    Last edited by a moderator: Dec 2, 2003
  13. Dec 8, 2003 #12
    p*p

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

    1, 2*1, 3, 2*2, 5, 2*3, 7, /, 3*3, 2*5, 11, /, 13 etc.

    3*3 is right in the middle of 7 and 11
    2*5 is right in the middle of 7 and 13
    etc.

    (11*3 is right in the middle of 37 and 29)


    Erik-Olof Wallman
     
    Last edited: Dec 8, 2003
  14. Dec 8, 2003 #13

    HallsofIvy

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    ?? Yes, "2<sup>p</sup>-1= p" is wrong. In fact, the example you give show that :
    4-1= 2<sup>2</sup>-1= 2 NOT 2.
    8-1= 2<sup>3</sup>-1= 7 NOT 3, etc.

    What exactly did you intend to say?
     
  15. Dec 8, 2003 #14
    I think he means to suggest that for every p, 2p-1 is a prime number? These are the so-called Mersenne-numbers and not all of them are prime. Although the largest prime numbers found to date are typically Mersenne-numbers, not every p generates a prime number. Simplest example: 211-1 is composite (23*89).
     
  16. Dec 8, 2003 #15
     
  17. Dec 8, 2003 #16
    I had never seen that before
     
  18. Dec 9, 2003 #17
    The Mersenne-numbers are very interesting because of the so-called Lucas-Lehmer Test, which is a (relatively) easy method of deciding if any Mersenne-number is prime or composite.
     
  19. Dec 9, 2003 #18
    p1 = 1
    p2 = 2
    p3 = 3
    p4 = 5
    p5 = 7
    p6 = 11

    All numbers within the serie:
    p1*p2*p3*p4*...*pn +/- 1 are primenumbers.

    In this serie, pn can be raised to all primenumbers between 1 and p; the serie will still be giving primes.

    All primes, that in the serie are raised to zero, multiplied with eachother becomes the degree of conjugative.

    I can prove that this equation is new:

    p1*p2*p3*p4*...*pn = ( can't we call it p?, when n! = 1*2*3*...*n ? ).

    It's to good to be old!
     
    Last edited: Dec 9, 2003
  20. Dec 9, 2003 #19
    Two observations:

    1) The smallest prime number is 2, not 1.

    2) p1*p2*p3*p4*p5*p6*p7 = 2*3*5*7*11*13*17 = 510510
    510510 - 1 = 510509 = 8369 * 61

    Sorry, but if things were so simple...
     
  21. Dec 9, 2003 #20
    I'm sorry

    I'm sorry. Thanks anyway. but... wait.. 510 510 is a double query.

    Maybe it don't work on them? THANX!!

    by the way, 510 is 11111110 binary.

    Maybe both sides don't have to be primes either? Or?
     
    Last edited: Dec 9, 2003
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