- 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:

Aren't they related to Bernoulli numbers?

>

> 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)

Décio

[Non-text portions of this message have been removed] > > There is no easy way to express these denominators (6, 90, 945, 9450,

Yes they are - as Michael Pharidon pointed out to me off list that the

> > 93555,...) as far as I know apart from finding pi^(2n)/zeta(2n)

>

> Aren't they related to Bernoulli numbers?

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.

__________________________________________________

Virus checked by MessageLabs Virus Control Centre.- 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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