## Goldbach (again)

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• Hello you all Recently there was a discussion about this topic which ended up in obscure fashion, but shortly after, there was a discussion about twin primes
Message 1 of 6 , Nov 17, 2004
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Hello you all

fashion,
but shortly after, there was a discussion about twin primes which triggered
some thinking
that can be also developped about Goldbach, as I venture here.

As a starter I propose a method to evaluate the count of Goldbach
partitions, based on heuristic formulas

time being :
http://www.geocities.com/dhvanderstraten/gldbevtxt.html
http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

I will develop further as your feedbacks and feelings dictate.
Rgds

Didier van der Straten
• In a message dated 11/17/04 9:42:05 AM Eastern Standard Time, ... I didn t understand your table. 2n = 530 and its count is 14. What are you counting? What is
Message 2 of 6 , Nov 17, 2004
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In a message dated 11/17/04 9:42:05 AM Eastern Standard Time,
vdstrat@... writes:
> As a starter I propose a method to evaluate the count of Goldbach
> partitions, based on heuristic formulas
>
> I have created a page and a table about this, which is draft version for the
> time being :
> http://www.geocities.com/dhvanderstraten/gldbevtxt.html
> http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

2n = 530 and its count is 14.
What are you counting? What is "eval."
Where is the Excel spreadsheet formula?

A set of 2n elements has 2^2n subsets including the empty set and 2n itself.
Where r combinations are expressed as (2^2n/r) then
2^2n = (2^2n/1) + (2^2n/2) + (2^2n/3) + ... + (2^2n/2^2n)
Eg. 2^5 = 32 = 1 + 5 + 10 + 10 + 5 + 1
Alternatively, an unrestricted partition of 5 includes these 5 sums:
4 + 1,
3 + 2,
3 + 1 + 1,
2 + 2 + 1,
2 + 1 + 1 + 1,
1 + 1+ 1 + 1 + 1

Regardless of whatever it is you are counting, I don't see the
"clue" that links such counts to the Goldbach Conjecture.

See if my Stepladder approach offers any clue at:
http://gmazur.freeyellow.com/Goldbach.com

Greg

[Non-text portions of this message have been removed]
• In a message dated 11/17/04 2:39:36 PM Eastern Standard Time, ... Of course I meant http://gmazur.freeyellow.com/Goldbach.htm Sorry for the typo [Non-text
Message 3 of 6 , Nov 17, 2004
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In a message dated 11/17/04 2:39:36 PM Eastern Standard Time,
Greg@... writes:
> See if my Stepladder approach offers any clue at:
> http://gmazur.freeyellow.com/Goldbach.com
>
Of course I meant
http://gmazur.freeyellow.com/Goldbach.htm

Sorry for the typo

[Non-text portions of this message have been removed]
• Dear Greg Thank you for your prompt feedbacks. First I ll try to answer to this one made of 4 questions : A Goldbach partition of an even number 2n is a set
Message 4 of 6 , Nov 18, 2004
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Dear Greg

Thank you for your prompt feedbacks.

First I'll try to answer to this one made of 4 questions :

A "Goldbach partition" of an even number 2n is a set of two prime numbers
the sum of which equals 2n. If n is prime, the set (n, n) is also a Goldbach
partition.
I used this vocabulary after reading lots of good internet pages about the
Goldbach conjecture.

The "count" I show in my table is the number of distinct Goldbach partitions
for an even number 2n. One can find an applet on internet that gives the
list
of Goldbach partitions for "any" proposed even number.

The "evaluation" is an attempt to evaluate that said count by a predicting
formula. Observing how "precise" the result is may give a clue as to "solve"
the Goldbach conjecture.

For the moment I do not disclose that formula. I might do so in the future
depending on further feedbacks and feelings.

Feel free to continue fruitful interaction. Offline or online as you wish.

something

Rgds.
Didier

-----Message d'origine-----
De : Greg@... [mailto:Greg@...]
Envoye : mercredi 17 novembre 2004 20:39
Objet : Re: [PrimeNumbers] Goldbach (again)

In a message dated 11/17/04 9:42:05 AM Eastern Standard Time,
vdstrat@... writes:
> As a starter I propose a method to evaluate the count of Goldbach
> partitions, based on heuristic formulas
>
the
2n = 530 and its count is 14.
What are you counting? What is "eval."
Where is the Excel spreadsheet formula?

A set of 2n elements has 2^2n subsets including the empty set and 2n itself.
Where r combinations are expressed as (2^2n/r) then
2^2n = (2^2n/1) + (2^2n/2) + (2^2n/3) + ... + (2^2n/2^2n)
Eg. 2^5 = 32 = 1 + 5 + 10 + 10 + 5 + 1
Alternatively, an unrestricted partition of 5 includes these 5 sums:
4 + 1,
3 + 2,
3 + 1 + 1,
2 + 2 + 1,
2 + 1 + 1 + 1,
1 + 1+ 1 + 1 + 1

Regardless of whatever it is you are counting, I don't see the
"clue" that links such counts to the Goldbach Conjecture.

See if my Stepladder approach offers any clue at:
http://gmazur.freeyellow.com/Goldbach.com

Greg

[Non-text portions of this message have been removed]

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

• ... http://www.ieeta.pt/~tos/goldbach.html Best, Andrey :) [Non-text portions of this message have been removed]
Message 5 of 6 , Nov 18, 2004
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http://www.ieeta.pt/~tos/goldbach.html

Best,

Andrey :)

[Non-text portions of this message have been removed]
• Dear interested primers Have appreciated feedbacks about my initial draft. I start disclosing my Goldbach partition count evaluation method and formula.
Message 6 of 6 , Dec 2, 2004
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Dear interested primers

Have appreciated feedbacks about my initial draft.

I start disclosing "my" Goldbach partition count evaluation method and
formula.

Please go to my updated page :
http://www.geocities.com/dhvanderstraten/gldbevtxt.html

Thanks. I look forward to your continued interest.
Didier
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