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Re: Almost amazing fact

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  • djbroadhurst
    ... But I ve just recalled a truly amazing littleomega fact: sum(n 0,2^omega(n)/n^(2*k)) is _rational_ for all integers k 0. (It s 90/6^2=5/2 at k=1 and in
    Message 1 of 5 , Jul 1, 2002
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      I said to Jon Perry:

      > If you want to generate exact old truths,
      > instead of approximate new nonsense,
      > use bigomega.

      But I've just recalled a
      truly amazing littleomega fact:

      sum(n>0,2^omega(n)/n^(2*k))

      is _rational_ for all integers k>0.

      (It's 90/6^2=5/2 at k=1 and in general
      zeta(2*k)^2/zeta(4*k))

      David
    • Jon Perry
      Equations 20 & 21 of Riemann s Zeta Function at Mathworld. What intrigues me about these equations is the apparent lack of any symmetry, which is why I was
      Message 2 of 5 , Jul 1, 2002
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        Equations 20 & 21 of Riemann's Zeta Function at Mathworld.

        What intrigues me about these equations is the apparent lack of any
        symmetry, which is why I was fairly gobsmacked when sigma(n>0,omega(n)/n^2)
        was zeta(zeta(2)*2)/zeta(zeta(2)).

        P.S. I (or Proth to be more precise - using a NewPGen sieve), determined
        that 4472*1001^567+1 is prime. Do I win a prize?

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com


        -----Original Message-----
        From: djbroadhurst [mailto:d.broadhurst@...]
        Sent: 01 July 2002 10:29
        To: primenumbers@yahoogroups.com
        Subject: [PrimeNumbers] Re: Almost amazing fact


        I said to Jon Perry:

        > If you want to generate exact old truths,
        > instead of approximate new nonsense,
        > use bigomega.

        But I've just recalled a
        truly amazing littleomega fact:

        sum(n>0,2^omega(n)/n^(2*k))

        is _rational_ for all integers k>0.

        (It's 90/6^2=5/2 at k=1 and in general
        zeta(2*k)^2/zeta(4*k))

        David



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