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Re: [PrimeNumbers] primes and pi

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  • Décio Luiz Gazzoni Filho
    ... Aren t they related to Bernoulli numbers? Décio [Non-text portions of this message have been removed]
    Message 1 of 10 , Feb 11, 2005
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      On Friday 11 February 2005 07:51, Paul Jobling wrote:
      > For even positive integer values of s (eg 2, 4, 6), this value involves pi:
      >
      > zeta(2) = pi^2 / 6
      >
      > zeta(4) = pi^4 / 90
      >
      > zeta(6) = pi^6 / 945
      >
      > There is no easy way to express these denominators (6, 90, 945, 9450,
      > 93555,...) as far as I know apart from finding pi^(2n)/zeta(2n)

      Aren't they related to Bernoulli numbers?

      Décio


      [Non-text portions of this message have been removed]
    • Paul Jobling
      ... Yes they are - as Michael Pharidon pointed out to me off list that the denominators in the zeta(n) values are equal to 1 / (2^(n-1)* Bn/n!), when Bn is the
      Message 2 of 10 , Feb 11, 2005
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        > > There is no easy way to express these denominators (6, 90, 945, 9450,
        > > 93555,...) as far as I know apart from finding pi^(2n)/zeta(2n)
        >
        > Aren't they related to Bernoulli numbers?

        Yes they are - as Michael Pharidon pointed out to me off list that the
        denominators in the zeta(n) values are equal to 1 / (2^(n-1)* Bn/n!), when
        Bn is the nth Bernouille number.

        Another off-list contribution was that

        pi = 2/product(prime p>2,1+(-1)^((p-1)/2)/p)
        = 2/(1-1/3)/(1+1/5)/(1-1/7)/(1-1/11)/(1+1/13)/(1+1/17)....

        which is Gregory's formula written in terms of a product over odd primes.

        Regards,

        Paul.


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      • Dr. Michael Paridon
        ... As you probably will have noticed, it should have run when Bn is the absolute value of the nth Bernoulli number
        Message 3 of 10 , Feb 11, 2005
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          Just a little completion:

          >> 1 / (2^(n-1)* Bn/n!), when Bn is the nth Bernoulli number <<

          As you probably will have noticed, it should have run >> when Bn is the
          absolute value of the nth Bernoulli number <<, as some Bn are negativ.

          Best regards

          Michael Paridon



          ----- Original Message -----
          From: "Paul Jobling" <Paul.Jobling@...>
          To: "Décio Luiz Gazzoni Filho" <decio@...>;
          <primenumbers@yahoogroups.com>
          Sent: Friday, February 11, 2005 5:14 PM
          Subject: RE: [PrimeNumbers] primes and pi


          > > > There is no easy way to express these denominators (6, 90, 945, 9450,
          > > > 93555,...) as far as I know apart from finding pi^(2n)/zeta(2n)
          > >
          > > Aren't they related to Bernoulli numbers?
          >
          > Yes they are - as Michael Pharidon pointed out to me off list that the
          > denominators in the zeta(n) values are equal to 1 / (2^(n-1)* Bn/n!), when
          > Bn is the nth Bernouille number.
          >
          > Another off-list contribution was that
          >
          > pi = 2/product(prime p>2,1+(-1)^((p-1)/2)/p)
          > = 2/(1-1/3)/(1+1/5)/(1-1/7)/(1-1/11)/(1+1/13)/(1+1/17)....
          >
          > which is Gregory's formula written in terms of a product over odd primes.
          >
          > Regards,
          >
          > Paul.
          >
          >
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