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Old 2010-07-06, 15:44   #1
CRGreathouse
 
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Default The constant in Hardy-Littlewood's Conjecture F

The Hardy-Littlewood Conjecture F includes calculation of the infinite product
\prod_{\varpi\ge3,\varpi\not|a}\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)
where (I believe!) \varpi ranges over the primes and \left(\frac D\varpi\right) is the Jacobi symbol.

Is there a good way to calculate this? Or, how can one calculate a reasonable number of decimal places of
f(D)=\prod\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)
where \varpi ranges over the odd primes? (Recovering the original problem for any factorable a is easy.)
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Old 2010-07-06, 16:47   #2
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The problem can be solved if there are good ways to accelerate the calculation of sums over primes in congruence classes:
\prod_{\varpi\equiv m\pmod n}1-\frac{1}{\varpi-1}
and
\prod_{\varpi\equiv m\pmod n}1+\frac{1}{\varpi-1}
for \varpi prime and integer m,n. Can this be done?

Last fiddled with by CRGreathouse on 2010-07-06 at 16:56
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Old 2010-07-06, 17:33   #3
ccorn
 
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Quote:
Originally Posted by CRGreathouse View Post
The Hardy-Littlewood Conjecture F includes calculation of the infinite product
\prod_{\varpi\ge3,\varpi\not|a}\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)
where (I believe!) \varpi ranges over the primes and \left(\frac D\varpi\right) is the Jacobi symbol.

Is there a good way to calculate this? Or, how can one calculate a reasonable number of decimal places of
f(D)=\prod\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)
where \varpi ranges over the odd primes? (Recovering the original problem for any factorable a is easy.)
Have you checked Shanks' paper?
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Old 2010-07-06, 17:34   #4
ccorn
 
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Quote:
Originally Posted by CRGreathouse View Post
The problem can be solved if there are good ways to accelerate the calculation of sums over primes in congruence classes:
\prod_{\varpi\equiv m\pmod n}1-\frac{1}{\varpi-1}
and
\prod_{\varpi\equiv m\pmod n}1+\frac{1}{\varpi-1}
for \varpi prime and integer m,n. Can this be done?
(Logs of) These would diverge, I'm afraid.
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Old 2010-07-06, 17:47   #5
CRGreathouse
 
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Thanks! I'll look it over.

Quote:
Originally Posted by ccorn View Post
(Logs of) These would diverge, I'm afraid.
(facepalm)

Of course if I had a Mertens' Theorem-like estimate for
\prod_{p\equiv m\pmod n,p\le x}1-\frac{1}{p-1}
and
\prod_{p\equiv m\pmod n,p\le x}1+\frac{1}{p-1}
that would suffice.

Last fiddled with by CRGreathouse on 2010-07-06 at 17:59
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