- Bonjour Simon,

> Conversely, if we remove a composite number from the left : here

If you remove p1 and p2 from the right, then you should remove all the

> we simply take Pi^2/6 - 1/C^2

> then what happens on the right, aren't there 2 wholes ?

> (we suppose C = p1*p2).

multiples of p1 and all the multiples of p2 from the left and not just

p1*p2. What happen to the right if C is removed from the left... I don't

think that there is a relation with p1 and p2 (?)

The method may work but not with Euler's product :o(

Yves - Has anyone thought about approaching Euler's other equation.

Ofcourse the classic zeta:

1/1+1/4+1/9+1/16+1/25+1/36+...=4/3*9/8*25/24*49/48*121/120*169/168*...

Then:

Pi =2/1 * 3/2 * 5/6 * 7/6 * 11/10 * 13/14*...

? = 1/1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 - 1/7 ...

Which depends on, p+-1 mod 4, or simply if the product above is

proper or inproper.

Does anybody know what this converges or diverges to?

Sum of the inverses.

p^-1 ? - First specify a character

chi(a*b)=chi(a)*chi(b).

Then the Dirichlet L series

L(s)=sum_{n>0} chi(n)/n^s

has the Euler product form

L(s)=prod_{prime p} 1/(1-chi(p)/p^s).

So you can now ask what happens at s=1.

Example: chi(2*n)=0, chi(4*n+/1)=+/-1

is a character mod 4 and at s=1 the

sum is Gregory's formula for

L(1)=pi/4=1-1/3+1/5-1/7...

whose product form is

L(1)=pi/4=(3/4)*(5/4)*(7/8)*(11/12)...

You asked about

> pi=2/1 * 3/2 * 5/6 * 7/6 * 11/10 * 13/14 *...

Take out the 2/1, which does not belong there!

You are interested in 2*K(1) where

K(s)=prod_{prime p} 1/(1+chi(p)/p^s) ... [1]

Multiply by the Dirichlet series L(s):

K(s)*L(s)=prod_{prime p>2} 1/(1-1/p^(2*s))

=(1-1/4^s)*zeta(2*s)

since chi(p)^2=1 for each odd prime.

Hence we get

2*K(1)=2*(1-1/4)*zeta(2)/L(1)

=(3/2)*(pi^2/6)/(pi/4)=pi

as you claimed.

Now you are seeking, I guess,

a summation formula for [1] at s=1?

For s>1 it is clear that

K(s)=1+sum_{n>1} chi(n)*(-1)^f(n)/n^s ... [2]

where, f(p)=-1 for prime p

and f(a*b)=f(a)*f(b).

The problem, as I see it, is that

we cannot easily group terms in [2] at s=1,

while this was trivial for

L(1) = pi/4 = 1 - sum_{k>0} 2/(16*k^2-1)

Rather, you are playing with the reciprocal

of this series

1/L(1) = K(1)/(p^2/8)

Hope this helps!

David - Correcting a typo:

K(s)=1+sum_{n>1} chi(n)*f(n)/n^s ... [2]

> f(p)=-1 for prime p

So for s>1

> f(a*b)=f(a)*f(b)

> chi(2*n)=0, chi(4*n+/1)=+/-1

K(s)=1+1/3^s-1/5^s+1/7^s+1/9^s+1/11^s-1/13^s-1/15^s-1/17^s...

with the sign structure

1,1,-1,1,1,1,-1,-1,-1,1,1,1,1,1,-1,1,1,-1,-1,-1...

[check: K(5)=31*pi^5/9450]

I would be intrigued to see a way of grouping these

signs to get a convergent sum at s=1. Naively

I would not expect to be able to do,

since we are effectively taking the reciprocal

a bona fide Dirichlet series:

K(s)=(1-1/4^s)*zeta(2*s)/L(s)

L(s)=sum_{n>0} chi(n)/n^s

K(1)=(3/4)*(pi^2/6)/(pi/4)=pi/2

David - Simon Plouffe wrote:
> MontrĂ©al, june 2, 2002

looking for

> Hello there,

>

> Searching for an anti-electron in normal matter is like searching

> for the inverse of a rock : a whole in the flat land.

>

> Apply this idea to factoring, instead of trying to factor directly

> a number maybe we should look for factors in a sea of numbers,

> the <whole>.

SNIP

> I believe that if we could do that then we could factor numbers

more easily,

> something like

Idea, instead of the classic trial division of n, is prime if it has

> 5000 digits in a snap, ;-).

>

> Simon Plouffe

no divisors =< the square root of n.

How about the use of Fibonacci(n)

For odd n, if F(n) has no divisors F(p), where p=< square root of n.

F(n)/F(p) are divided in golden base.

Where Fibonacci numbers are represented:

1:1

2:10

3:100

5:1000

8:10000

...

It's just an idea anyway!

Any techniques for dividing Fn/Fp in golden base?