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• ... Aren t they related to Bernoulli numbers? Décio [Non-text portions of this message have been removed]
Message 1 of 10 , Feb 11, 2005
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]
• ... 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
> > 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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• ... 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
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@...>;
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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