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• Sorry for the unhelpful subject line, but anything else that I tried was just too long or too vague. I have linked to Henri Cohen s paper which, among other
Message 1 of 8 , Jan 29, 2005
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Sorry for the unhelpful subject line, but anything else that I tried was just
too long or too vague.

I have linked to Henri Cohen's paper which, among other things, shows how to
compute certain sums over primes:

http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi

It turns out that the sum over all primes x < p of log(x)/x diverges to log(p)
+ k, a constant, which I've computed to 4-digit accuracy (I believe) as
-1.332 by the rather brute force-ish method of computing the sum over all
primes up to a billion.

Cohen's technique can evaluate sums of similar forms, but I can't tell if it
works for this particular one. I have reason to believe so, because in page 5
it reads: `By taking derivatives or limits, in an analogous manner we easily
compute:' and the first example is the sum over all primes of log(p)/p^2. It
turns out that the derivative of log(p)/p is log(p)/p^2 + 1. However, Cohen's
comment about derivatives is completely over my head, but I sense there's a
connection.

So, can anyone help out to derive a fast formula for the sum of log(x)/x over
primes using Cohen's technique? (or any other technique, for that matter)

Décio

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• Decio ... to ... All the techniques in that paper are to compute the sums of _convergent_ series. Yours is divergent, which is an essential difference, so I
Message 1 of 8 , Jan 29, 2005
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Decio

>I have linked to Henri Cohen's paper which, among other things, shows how
to
>compute certain sums over primes:
>
>http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi

All the techniques in that paper are to compute the sums of _convergent_
series.

Yours is divergent, which is an essential difference, so I see no way to
apply his methods.

>It turns out that the sum over all primes x < p of log(x)/x diverges to
log(p)
>+ k, a constant, which I've computed to 4-digit accuracy (I believe) as
>-1.332 by the rather brute force-ish method of computing the sum over all
>primes up to a billion.

That seems to be a remarkable result and one I've not seen.
Are you sure k is a constant? :-)

-Mike Oakes

[Non-text portions of this message have been removed]
• ... Yes, but have a look at page 5, where he claims his methods can be used to compute B1 = sum(1/p) - log log x, by taking a limit. ... Hardy & Wright,
Message 1 of 8 , Jan 29, 2005
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On Saturday 29 January 2005 21:48, you wrote:
> Decio
>
> >I have linked to Henri Cohen's paper which, among other things, shows how
>
> to
>
> >compute certain sums over primes:
> >
> >http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi
>
> All the techniques in that paper are to compute the sums of _convergent_
> series.
>
> Yours is divergent, which is an essential difference, so I see no way to
> apply his methods.

Yes, but have a look at page 5, where he claims his methods can be used to
compute B1 = sum(1/p) - log log x, by taking a limit.

> >It turns out that the sum over all primes x < p of log(x)/x diverges to
>
> log(p)
>
> >+ k, a constant, which I've computed to 4-digit accuracy (I believe) as
> >-1.332 by the rather brute force-ish method of computing the sum over all
> >primes up to a billion.
>
> That seems to be a remarkable result and one I've not seen.
> Are you sure k is a constant? :-)

Hardy & Wright, Theorem 425: sum_{p <= x) log(p)/p = log(x) + O(1).

I can type up the proof if you don't have the book.

Décio

[Non-text portions of this message have been removed]
• ... Of course! I didn t think to refresh my acquaintance with that chapter of the Bible. I ve got bookmarks all around that page so must have read it many
Message 1 of 8 , Jan 30, 2005
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>Hardy & Wright, Theorem 425: sum_{p <= x) log(p)/p = log(x) + O(1).

Of course!
I didn't think to refresh my acquaintance with that chapter of the Bible.
I've got bookmarks all around that page so must have read it many times.
Memory loss fast encroaching...

Have you followed the link to B3 at
_http://mathworld.wolfram.com/MertensConstant.html_
(http://mathworld.wolfram.com/MertensConstant.html)
which David Broadhurst supplied (off-list)? And I wonder just how he managed
to compute your constant to 60 decimals?...

Mike

[Non-text portions of this message have been removed]
• ... Please refrain from referring to a lesser book as the Bible. There s only one Bible in the world, handed down to us by God itself, Donald Ervin Knuth,
Message 1 of 8 , Jan 30, 2005
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On Sunday 30 January 2005 06:56, you wrote:
> >Hardy & Wright, Theorem 425: sum_{p <= x) log(p)/p = log(x) + O(1).
>
> Of course!
> I didn't think to refresh my acquaintance with that chapter of the Bible.

Please refrain from referring to a lesser book as the Bible. There's only one
Bible in the world, handed down to us by God itself, Donald Ervin Knuth,
typeset in His holy software TeX, and its name is `The Art of Computer
Programming.'

> I've got bookmarks all around that page so must have read it many times.
> Memory loss fast encroaching...
>
> Have you followed the link to B3 at
> _http://mathworld.wolfram.com/MertensConstant.html_
> (http://mathworld.wolfram.com/MertensConstant.html)
> which David Broadhurst supplied (off-list)? And I wonder just how he
> managed to compute your constant to 60 decimals?...

Have a look at section 3 of the following page from Xavier Gourdon's site,
http://numbers.computation.free.fr/Constants/Miscellaneous/constantsNumTheory.html

Maybe we're getting closer?

The alternative formulation on that Mathworld page may also help, at least it
converges faster than the original formulation (not fast enough for 60
digits, mind you.)

And finally there's a paper by Phillipe Flajolet (also linked from that
Mathworld page) which I'm still reading.

Décio

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• In a message dated 30/01/2005 11:16:17 GMT Standard Time, decio@decpp.net ... it ... Decio, Yes, you re right: that formula for B3 with gamma in it is just the
Message 1 of 8 , Jan 31, 2005
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In a message dated 30/01/2005 11:16:17 GMT Standard Time, decio@...
writes:

>The alternative formulation on that Mathworld page may also help, at least
it
>converges faster than the original formulation (not fast enough for 60
>digits, mind you.)

Decio,
Yes, you're right: that formula for B3 with gamma in it is just the key to
David's stupendous calculation - see his pages (communicated off-list):-

-Mike Oakes

[Non-text portions of this message have been removed]
• ... Mike: you re giving me too much credit here, I was just throwing stuff in the air to see whether anyone brighter than me could figure it out. Truth be
Message 1 of 8 , Jan 31, 2005
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On Monday 31 January 2005 08:56, you wrote:
> Decio,
> Yes, you're right: that formula for B3 with gamma in it is just the key to
> David's stupendous calculation - see his pages (communicated off-list):-
>

Mike: you're giving me too much credit here, I was just throwing stuff in the
air to see whether anyone brighter than me could figure it out. Truth be
told, I'm still working out the Postscript file written by David -- as the
joke goes, don't drink and derive...

Décio

[Non-text portions of this message have been removed]
• Let q a positive integer and 1
Message 1 of 8 , Feb 4, 2005
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Let q a positive integer and 1<=r<=q. Is there a theorem that tells
us how many integers a, 1<=a<=q, verify a^r=1 mod(q)?

I have found that:
(1) if q is prime or a power of a prime p, i.e. q=p^x, then there
are GCD(r,EulerPhi(q)) integers a that satisfy a^r=1 mod(q).
(2) if q is a product of two powers of primes p' and p'', i.e.
q=p'^x p''^y, then there are GCD(r,Carmichael(q))*GCD(r,Phi
(q)/Carmichael(q)) integers a that satisfy a^r=1 mod(q).

Are these results correct and can we generalize to any integer q?