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• Bonjour Simon, ... If you remove p1 and p2 from the right, then you should remove all the multiples of p1 and all the multiples of p2 from the left and not
Message 1 of 6 , Jun 2, 2002
Bonjour Simon,

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

If you remove p1 and p2 from the right, then you should remove all the
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*...
Message 2 of 6 , Jun 8, 2002
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}
Message 3 of 6 , Jun 9, 2002
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)...

> 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] ... So for s 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
Message 4 of 6 , Jun 9, 2002
Correcting a typo:

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

> f(p)=-1 for prime p
> f(a*b)=f(a)*f(b)
> chi(2*n)=0, chi(4*n+/1)=+/-1

So for s>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
• ... looking for ... SNIP ... more easily, ... Idea, instead of the classic trial division of n, is prime if it has no divisors =
Message 5 of 6 , Oct 14 10:17 PM
Simon Plouffe wrote:
> 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,
looking for
> the <whole>.

SNIP

> I believe that if we could do that then we could factor numbers
more easily,
> something like
> 5000 digits in a snap, ;-).
>
> Simon Plouffe

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