## Re: Almost amazing fact

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• ... But I ve just recalled a truly amazing littleomega fact: sum(n 0,2^omega(n)/n^(2*k)) is _rational_ for all integers k 0. (It s 90/6^2=5/2 at k=1 and in
Message 1 of 5 , Jul 1, 2002
I said to Jon Perry:

> If you want to generate exact old truths,
> instead of approximate new nonsense,
> use bigomega.

But I've just recalled a
truly amazing littleomega fact:

sum(n>0,2^omega(n)/n^(2*k))

is _rational_ for all integers k>0.

(It's 90/6^2=5/2 at k=1 and in general
zeta(2*k)^2/zeta(4*k))

David
• Equations 20 & 21 of Riemann s Zeta Function at Mathworld. What intrigues me about these equations is the apparent lack of any symmetry, which is why I was
Message 2 of 5 , Jul 1, 2002
Equations 20 & 21 of Riemann's Zeta Function at Mathworld.

What intrigues me about these equations is the apparent lack of any
symmetry, which is why I was fairly gobsmacked when sigma(n>0,omega(n)/n^2)
was zeta(zeta(2)*2)/zeta(zeta(2)).

P.S. I (or Proth to be more precise - using a NewPGen sieve), determined
that 4472*1001^567+1 is prime. Do I win a prize?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
Sent: 01 July 2002 10:29
Subject: [PrimeNumbers] Re: Almost amazing fact

I said to Jon Perry:

> If you want to generate exact old truths,
> instead of approximate new nonsense,
> use bigomega.

But I've just recalled a
truly amazing littleomega fact:

sum(n>0,2^omega(n)/n^(2*k))

is _rational_ for all integers k>0.

(It's 90/6^2=5/2 at k=1 and in general
zeta(2*k)^2/zeta(4*k))

David

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